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	<id>https://wiki.ubc.ca/api.php?action=feedcontributions&amp;feedformat=atom&amp;user=UnaVuckovic</id>
	<title>UBC Wiki - User contributions [en]</title>
	<link rel="self" type="application/atom+xml" href="https://wiki.ubc.ca/api.php?action=feedcontributions&amp;feedformat=atom&amp;user=UnaVuckovic"/>
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	<updated>2026-07-25T02:59:53Z</updated>
	<subtitle>User contributions</subtitle>
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	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:UnaVuckovic&amp;diff=93050</id>
		<title>User:UnaVuckovic</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:UnaVuckovic&amp;diff=93050"/>
		<updated>2011-04-30T23:14:58Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: Blanked the page&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75149</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75149"/>
		<updated>2011-02-04T02:28:16Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&amp;lt;center&amp;gt;http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif&amp;lt;center/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basicity of an aqueous solution, furthermore it is an inverse logarithmic representation of Hydrogen proton concentration within solution (Covington et al 1985). From an applications standpoint, pH on a logarithmic scale will range from 0 to 14 where any value measured to be below 7 is considered acidic and any value above 7 is considered basic and 7 being neutral. Now it is important to understand that each increase or decrease in value does not mean that you are simply increasing or decreasing the pH by 1 value within the solution. Rather each individual pH unit is a factor of 10 relating to either it’s lower or higher neighbour. Consider pH of 3, if I was to increase the pH to 4, I would have increased the basicity of my solution by 10 (or ten times that of the previous pH value). Also, if I was to have a solution with pH 4 and I worked to decrease the pH to a value of 1 I would have seen a 1000 times drop in basicity (10^3). These observations can be fully supported by the equation:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Equation.JPG]]&lt;br /&gt;
&lt;br /&gt;
The reasoning as to why pH scales are run on a logarithmic function can be traced to hydrogen ion concentration. Ion concentrations, more specifically hydrogen ions are incredibly small but addition or removal of these ions in just a small manner can work to significantly impact the acid or base. This significant impact can be traced to Hydrogen’s ability to donate electrons or receive electrons from other molecules within the solution and the subsequent electron flow will directly impact the pH of the solution by a large quantity given the fact that electrons carry a negative charge.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Below is the representation of values and their logarithmic counterparts:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===&#039;&#039;&#039;Number Representation and Logarithms&#039;&#039;&#039;===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
[[Image:Table.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
http://bcn.boulder.co.us/basin/data/NEW/info/pH.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
==References==  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Covington, A. K.; Bates, R. G.; Durst, R. A. (1985). &amp;quot;Definitions of pH scales, standard reference values, measurement of pH, and related terminology&amp;quot;&lt;br /&gt;
&lt;br /&gt;
http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75147</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75147"/>
		<updated>2011-02-04T02:27:04Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&amp;lt;center&amp;gt;http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif&amp;lt;center/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basicity of an aqueous solution, furthermore it is an inverse logarithmic representation of Hydrogen proton concentration within solution (Covington et al 1985). From an applications standpoint, pH on a logarithmic scale will range from 0 to 14 where any value measured to be below 7 is considered acidic and any value above 7 is considered basic and 7 being neutral. Now it is important to understand that each increase or decrease in value does not mean that you are simply increasing or decreasing the pH by 1 value within the solution. Rather each individual pH unit is a factor of 10 relating to either it’s lower or higher neighbour. Consider pH of 3, if I was to increase the pH to 4, I would have increased the basicity of my solution by 10 (or ten times that of the previous pH value). Also, if I was to have a solution with pH 4 and I worked to decrease the pH to a value of 1 I would have seen a 1000 times drop in basicity (10^3). These observations can be fully supported by the equation:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Equation.JPG]]&lt;br /&gt;
&lt;br /&gt;
The reasoning as to why pH scales are run on a logarithmic function can be traced to hydrogen ion concentration. Ion concentrations, more specifically hydrogen ions are incredibly small but addition or removal of these ions in just a small manner can work to significantly impact the acid or base. This significant impact can be traced to Hydrogen’s ability to donate electrons or receive electrons from other molecules within the solution and the subsequent electron flow will directly impact the pH of the solution by a large quantity given the fact that electrons carry a negative charge.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Below is the representation of values and their logarithmic counterparts:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Number Representation and Logarithms&#039;&#039;&#039;  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Table.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://bcn.boulder.co.us/basin/data/NEW/info/pH.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
==References==  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Covington, A. K.; Bates, R. G.; Durst, R. A. (1985). &amp;quot;Definitions of pH scales, standard reference values, measurement of pH, and related terminology&amp;quot;&lt;br /&gt;
&lt;br /&gt;
http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:Table.JPG&amp;diff=75146</id>
		<title>File:Table.JPG</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:Table.JPG&amp;diff=75146"/>
		<updated>2011-02-04T02:26:39Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75143</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75143"/>
		<updated>2011-02-04T02:09:30Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&amp;lt;center&amp;gt;http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif&amp;lt;center/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basicity of an aqueous solution, furthermore it is an inverse logarithmic representation of Hydrogen proton concentration within solution (Covington et al 1985). From an applications standpoint, pH on a logarithmic scale will range from 0 to 14 where any value measured to be below 7 is considered acidic and any value above 7 is considered basic and 7 being neutral. Now it is important to understand that each increase or decrease in value does not mean that you are simply increasing or decreasing the pH by 1 value within the solution. Rather each individual pH unit is a factor of 10 relating to either it’s lower or higher neighbour. Consider pH of 3, if I was to increase the pH to 4, I would have increased the basicity of my solution by 10 (or ten times that of the previous pH value). Also, if I was to have a solution with pH 4 and I worked to decrease the pH to a value of 1 I would have seen a 1000 times drop in basicity (10^3). These observations can be fully supported by the equation:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Equation.JPG]]&lt;br /&gt;
&lt;br /&gt;
The reasoning as to why pH scales are run on a logarithmic function can be traced to hydrogen ion concentration. Ion concentrations, more specifically hydrogen ions are incredibly small but addition or removal of these ions in just a small manner can work to significantly impact the acid or base. This significant impact can be traced to Hydrogen’s ability to donate electrons or receive electrons from other molecules within the solution and the subsequent electron flow will directly impact the pH of the solution by a large quantity given the fact that electrons carry a negative charge.   &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Below is the representation of values and their logarithmic counterparts:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Number Representation and Logarithms&#039;&#039;&#039;  &lt;br /&gt;
‎&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://bcn.boulder.co.us/basin/data/NEW/info/pH.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
==References==    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Covington, A. K.; Bates, R. G.; Durst, R. A. (1985). &amp;quot;Definitions of pH scales, standard reference values, measurement of pH, and related terminology&amp;quot;&lt;br /&gt;
&lt;br /&gt;
http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75142</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75142"/>
		<updated>2011-02-04T02:09:12Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&amp;lt;center&amp;gt;http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif&amp;lt;center/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basicity of an aqueous solution, furthermore it is an inverse logarithmic representation of Hydrogen proton concentration within solution (Covington et al 1985). From an applications standpoint, pH on a logarithmic scale will range from 0 to 14 where any value measured to be below 7 is considered acidic and any value above 7 is considered basic and 7 being neutral. Now it is important to understand that each increase or decrease in value does not mean that you are simply increasing or decreasing the pH by 1 value within the solution. Rather each individual pH unit is a factor of 10 relating to either it’s lower or higher neighbour. Consider pH of 3, if I was to increase the pH to 4, I would have increased the basicity of my solution by 10 (or ten times that of the previous pH value). Also, if I was to have a solution with pH 4 and I worked to decrease the pH to a value of 1 I would have seen a 1000 times drop in basicity (10^3). These observations can be fully supported by the equation:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Equation.JPG]]&lt;br /&gt;
&lt;br /&gt;
The reasoning as to why pH scales are run on a logarithmic function can be traced to hydrogen ion concentration. Ion concentrations, more specifically hydrogen ions are incredibly small but addition or removal of these ions in just a small manner can work to significantly impact the acid or base. This significant impact can be traced to Hydrogen’s ability to donate electrons or receive electrons from other molecules within the solution and the subsequent electron flow will directly impact the pH of the solution by a large quantity given the fact that electrons carry a negative charge.   &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Below is the representation of values and their logarithmic counterparts:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Number Representation and Logarithms&#039;&#039;&#039;  &lt;br /&gt;
‎&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://bcn.boulder.co.us/basin/data/NEW/info/pH.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
==References==    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Covington, A. K.; Bates, R. G.; Durst, R. A. (1985). &amp;quot;Definitions of pH scales, standard reference values, measurement of pH, and related terminology&amp;quot;&lt;br /&gt;
&lt;br /&gt;
http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75140</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75140"/>
		<updated>2011-02-04T02:07:37Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: /* References */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&amp;lt;center&amp;gt;http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif&amp;lt;center/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basicity of an aqueous solution, furthermore it is an inverse logarithmic representation of Hydrogen proton concentration within solution (Covington et al 1985). From an applications standpoint, pH on a logarithmic scale will range from 0 to 14 where any value measured to be below 7 is considered acidic and any value above 7 is considered basic and 7 being neutral. Now it is important to understand that each increase or decrease in value does not mean that you are simply increasing or decreasing the pH by 1 value within the solution. Rather each individual pH unit is a factor of 10 relating to either it’s lower or higher neighbour. Consider pH of 3, if I was to increase the pH to 4, I would have increased the basicity of my solution by 10 (or ten times that of the previous pH value). Also, if I was to have a solution with pH 4 and I worked to decrease the pH to a value of 1 I would have seen a 1000 times drop in basicity (10^3). These observations can be fully supported by the equation:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Equation.JPG]]&lt;br /&gt;
&lt;br /&gt;
The reasoning as to why pH scales are run on a logarithmic function can be traced to hydrogen ion concentration. Ion concentrations, more specifically hydrogen ions are incredibly small but addition or removal of these ions in just a small manner can work to significantly impact the acid or base. This significant impact can be traced to Hydrogen’s ability to donate electrons or receive electrons from other molecules within the solution and the subsequent electron flow will directly impact the pH of the solution by a large quantity given the fact that electrons carry a negative charge.   &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Below is the representation of values and their logarithmic counterparts:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Number Representation and Logarithms&#039;&#039;&#039;  &lt;br /&gt;
‎&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://bcn.boulder.co.us/basin/data/NEW/info/pH.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
==References==    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Covington, A. K.; Bates, R. G.; Durst, R. A. (1985). &amp;quot;Definitions of pH scales, standard reference values, measurement of pH, and related terminology&amp;quot;&lt;br /&gt;
&lt;br /&gt;
http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75139</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75139"/>
		<updated>2011-02-04T02:07:24Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: /* References */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&amp;lt;center&amp;gt;http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif&amp;lt;center/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basicity of an aqueous solution, furthermore it is an inverse logarithmic representation of Hydrogen proton concentration within solution (Covington et al 1985). From an applications standpoint, pH on a logarithmic scale will range from 0 to 14 where any value measured to be below 7 is considered acidic and any value above 7 is considered basic and 7 being neutral. Now it is important to understand that each increase or decrease in value does not mean that you are simply increasing or decreasing the pH by 1 value within the solution. Rather each individual pH unit is a factor of 10 relating to either it’s lower or higher neighbour. Consider pH of 3, if I was to increase the pH to 4, I would have increased the basicity of my solution by 10 (or ten times that of the previous pH value). Also, if I was to have a solution with pH 4 and I worked to decrease the pH to a value of 1 I would have seen a 1000 times drop in basicity (10^3). These observations can be fully supported by the equation:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Equation.JPG]]&lt;br /&gt;
&lt;br /&gt;
The reasoning as to why pH scales are run on a logarithmic function can be traced to hydrogen ion concentration. Ion concentrations, more specifically hydrogen ions are incredibly small but addition or removal of these ions in just a small manner can work to significantly impact the acid or base. This significant impact can be traced to Hydrogen’s ability to donate electrons or receive electrons from other molecules within the solution and the subsequent electron flow will directly impact the pH of the solution by a large quantity given the fact that electrons carry a negative charge.   &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Below is the representation of values and their logarithmic counterparts:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Number Representation and Logarithms&#039;&#039;&#039;  &lt;br /&gt;
‎&lt;br /&gt;
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http://bcn.boulder.co.us/basin/data/NEW/info/pH.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
==References==    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Covington, A. K.; Bates, R. G.; Durst, R. A. (1985). &amp;quot;Definitions of pH scales, standard reference values, measurement of pH, and related terminology&amp;quot;&lt;br /&gt;
&lt;br /&gt;
http://www.elmhurst.edu/~chm/vchembook/184ph.html  &lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75137</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75137"/>
		<updated>2011-02-04T02:07:02Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: /* References */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&amp;lt;center&amp;gt;http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif&amp;lt;center/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basicity of an aqueous solution, furthermore it is an inverse logarithmic representation of Hydrogen proton concentration within solution (Covington et al 1985). From an applications standpoint, pH on a logarithmic scale will range from 0 to 14 where any value measured to be below 7 is considered acidic and any value above 7 is considered basic and 7 being neutral. Now it is important to understand that each increase or decrease in value does not mean that you are simply increasing or decreasing the pH by 1 value within the solution. Rather each individual pH unit is a factor of 10 relating to either it’s lower or higher neighbour. Consider pH of 3, if I was to increase the pH to 4, I would have increased the basicity of my solution by 10 (or ten times that of the previous pH value). Also, if I was to have a solution with pH 4 and I worked to decrease the pH to a value of 1 I would have seen a 1000 times drop in basicity (10^3). These observations can be fully supported by the equation:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Equation.JPG]]&lt;br /&gt;
&lt;br /&gt;
The reasoning as to why pH scales are run on a logarithmic function can be traced to hydrogen ion concentration. Ion concentrations, more specifically hydrogen ions are incredibly small but addition or removal of these ions in just a small manner can work to significantly impact the acid or base. This significant impact can be traced to Hydrogen’s ability to donate electrons or receive electrons from other molecules within the solution and the subsequent electron flow will directly impact the pH of the solution by a large quantity given the fact that electrons carry a negative charge.   &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Below is the representation of values and their logarithmic counterparts:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Number Representation and Logarithms&#039;&#039;&#039;  &lt;br /&gt;
‎&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://bcn.boulder.co.us/basin/data/NEW/info/pH.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
==References==   &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.elmhurst.edu/~chm/vchembook/184ph.html Covington, A. K.; Bates, R. G.; Durst, R. A. (1985). &amp;quot;Definitions of pH scales, standard reference values, measurement of pH, and related terminology&amp;quot;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.elmhurst.edu/~chm/vchembook/184ph.html&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75136</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75136"/>
		<updated>2011-02-04T02:06:37Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&amp;lt;center&amp;gt;http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif&amp;lt;center/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basicity of an aqueous solution, furthermore it is an inverse logarithmic representation of Hydrogen proton concentration within solution (Covington et al 1985). From an applications standpoint, pH on a logarithmic scale will range from 0 to 14 where any value measured to be below 7 is considered acidic and any value above 7 is considered basic and 7 being neutral. Now it is important to understand that each increase or decrease in value does not mean that you are simply increasing or decreasing the pH by 1 value within the solution. Rather each individual pH unit is a factor of 10 relating to either it’s lower or higher neighbour. Consider pH of 3, if I was to increase the pH to 4, I would have increased the basicity of my solution by 10 (or ten times that of the previous pH value). Also, if I was to have a solution with pH 4 and I worked to decrease the pH to a value of 1 I would have seen a 1000 times drop in basicity (10^3). These observations can be fully supported by the equation:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Equation.JPG]]&lt;br /&gt;
&lt;br /&gt;
The reasoning as to why pH scales are run on a logarithmic function can be traced to hydrogen ion concentration. Ion concentrations, more specifically hydrogen ions are incredibly small but addition or removal of these ions in just a small manner can work to significantly impact the acid or base. This significant impact can be traced to Hydrogen’s ability to donate electrons or receive electrons from other molecules within the solution and the subsequent electron flow will directly impact the pH of the solution by a large quantity given the fact that electrons carry a negative charge.   &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Below is the representation of values and their logarithmic counterparts:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Number Representation and Logarithms&#039;&#039;&#039;  &lt;br /&gt;
‎&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://bcn.boulder.co.us/basin/data/NEW/info/pH.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
==References==  &lt;br /&gt;
http://www.elmhurst.edu/~chm/vchembook/184ph.html Covington, A. K.; Bates, R. G.; Durst, R. A. (1985). &amp;quot;Definitions of pH scales, standard reference values, measurement of pH, and related terminology&amp;quot;&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75135</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75135"/>
		<updated>2011-02-04T02:04:56Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&amp;lt;center&amp;gt;http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif&amp;lt;center/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basicity of an aqueous solution, furthermore it is an inverse logarithmic representation of Hydrogen proton concentration within solution (Covington et al 1985). From an applications standpoint, pH on a logarithmic scale will range from 0 to 14 where any value measured to be below 7 is considered acidic and any value above 7 is considered basic and 7 being neutral. Now it is important to understand that each increase or decrease in value does not mean that you are simply increasing or decreasing the pH by 1 value within the solution. Rather each individual pH unit is a factor of 10 relating to either it’s lower or higher neighbour. Consider pH of 3, if I was to increase the pH to 4, I would have increased the basicity of my solution by 10 (or ten times that of the previous pH value). Also, if I was to have a solution with pH 4 and I worked to decrease the pH to a value of 1 I would have seen a 1000 times drop in basicity (10^3). These observations can be fully supported by the equation:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Equation.JPG]]&lt;br /&gt;
&lt;br /&gt;
The reasoning as to why pH scales are run on a logarithmic function can be traced to hydrogen ion concentration. Ion concentrations, more specifically hydrogen ions are incredibly small but addition or removal of these ions in just a small manner can work to significantly impact the acid or base. This significant impact can be traced to Hydrogen’s ability to donate electrons or receive electrons from other molecules within the solution and the subsequent electron flow will directly impact the pH of the solution by a large quantity given the fact that electrons carry a negative charge.  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Below is the representation of values and their logarithmic counterparts:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Number Representation and Logarithms&#039;&#039;&#039; &lt;br /&gt;
‎&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://bcn.boulder.co.us/basin/data/NEW/info/pH.gif&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75130</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75130"/>
		<updated>2011-02-04T01:57:02Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&amp;lt;center&amp;gt;http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif&amp;lt;center/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basicity of an aqueous solution, furthermore it is an inverse logarithmic representation of Hydrogen proton concentration within solution (Covington et al 1985). From an applications standpoint, pH on a logarithmic scale will range from 0 to 14 where any value measured to be below 7 is considered acidic and any value above 7 is considered basic and 7 being neutral. Now it is important to understand that each increase or decrease in value does not mean that you are simply increasing or decreasing the pH by 1 value within the solution. Rather each individual pH unit is a factor of 10 relating to either it’s lower or higher neighbour. Consider pH of 3, if I was to increase the pH to 4, I would have increased the basicity of my solution by 10 (or ten times that of the previous pH value). Also, if I was to have a solution with pH 4 and I worked to decrease the pH to a value of 1 I would have seen a 1000 times drop in basicity (10^3). These observations can be fully supported by the equation:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:Equation.JPG]]&lt;br /&gt;
&lt;br /&gt;
The reasoning as to why pH scales are run on a logarithmic function can be traced to hydrogen ion concentration. Ion concentrations, more specifically hydrogen ions are incredibly small but addition or removal of these ions in just a small manner can work to significantly impact the acid or base. This significant impact can be traced to Hydrogen’s ability to donate electrons or receive electrons from other molecules within the solution and the subsequent electron flow will directly impact the pH of the solution by a large quantity given the fact that electrons carry a negative charge. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Below is the representation of values and their logarithmic counterparts. &lt;br /&gt;
‎&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
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&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://bcn.boulder.co.us/basin/data/NEW/info/pH.gif&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:Equation.JPG&amp;diff=75128</id>
		<title>File:Equation.JPG</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:Equation.JPG&amp;diff=75128"/>
		<updated>2011-02-04T01:56:12Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75126</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75126"/>
		<updated>2011-02-04T01:53:12Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
&amp;lt;center&amp;gt;http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif&amp;lt;center/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basicity of an aqueous solution, furthermore it is an inverse logarithmic representation of Hydrogen proton concentration within solution (Covington et al 1985). From an applications standpoint, pH on a logarithmic scale will range from 0 to 14 where any value measured to be below 7 is considered acidic and any value above 7 is considered basic and 7 being neutral. Now it is important to understand that each increase or decrease in value does not mean that you are simply increasing or decreasing the pH by 1 value within the solution. Rather each individual pH unit is a factor of 10 relating to either it’s lower or higher neighbour. Consider pH of 3, if I was to increase the pH to 4, I would have increased the basicity of my solution by 10 (or ten times that of the previous pH value). Also, if I was to have a solution with pH 4 and I worked to decrease the pH to a value of 1 I would have seen a 1000 times drop in basicity (10^3). These observations can be fully supported by the equation:&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75125</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 13</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_13&amp;diff=75125"/>
		<updated>2011-02-04T01:52:15Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: Created page with &amp;quot;  http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif  In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basi...&amp;quot;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
&lt;br /&gt;
http://soil.gsfc.nasa.gov/soil_pH/anphscal.gif&lt;br /&gt;
&lt;br /&gt;
In measuring pH, we need to first understand what pH is exactly. pH is used in chemistry and is a measure of the acidity or basicity of an aqueous solution, furthermore it is an inverse logarithmic representation of Hydrogen proton concentration within solution (Covington et al 1985). From an applications standpoint, pH on a logarithmic scale will range from 0 to 14 where any value measured to be below 7 is considered acidic and any value above 7 is considered basic and 7 being neutral. Now it is important to understand that each increase or decrease in value does not mean that you are simply increasing or decreasing the pH by 1 value within the solution. Rather each individual pH unit is a factor of 10 relating to either it’s lower or higher neighbour. Consider pH of 3, if I was to increase the pH to 4, I would have increased the basicity of my solution by 10 (or ten times that of the previous pH value). Also, if I was to have a solution with pH 4 and I worked to decrease the pH to a value of 1 I would have seen a 1000 times drop in basicity (10^3). These observations can be fully supported by the equation:&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73658</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73658"/>
		<updated>2011-01-28T10:22:30Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Starting with the function &lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:MSP220819e4e417b6708ich00001768587g039a5c38.gif]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Change the height of the horizontal asymptote on the right, we&#039;ll denote it by K. Change the y intercept to any number between 0 and K&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
‎ &lt;br /&gt;
 &lt;br /&gt;
To change the height of the horizontal asymptote, we need to change the numerator and the denominator. The horizontal asymptote depends on the numerator, and the part of the denominator independent of &#039;&#039;e&#039;&#039;. In this case 1 and 1. If &#039;&#039;t&#039;&#039; goes to infinity, p(t) = 1/1 = 1, therefore, in this case, the horizontal asymptote will be 1. When the equation is changed to: 5/(2.5 + e^-t), the asymptote is 2.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y intercept in this equation is always 1/2 of &#039;&#039;K&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;For example:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
For http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png   the y intercept is 0.5, in the case of &lt;br /&gt;
5/(2.5 + e^-t), it is 1. &lt;br /&gt;
So the vertical asymptote is always 1/2 of the horizontal asymptote at the right.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:52.5e.gif]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The slope of the curved part depends on the constant multiplied to &#039;&#039;t&#039;&#039; in the power of &#039;&#039;e&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For example: the slope of the curved part of http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than the slope of the curved part of 1/(1 + e^(-3t)) &lt;br /&gt;
&lt;br /&gt;
and the slope of the curved part for &lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
will be steeper than that of P(t) = 1/(1 + e^-0.5t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
[[File:3 graph.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
The function may be used to model the economic growth for developing countries. It is often seen that underdeveloped countries have a higher rate of economic growth than developed countries. When a country reaches its capacity level of output, the rate of economic growth decreases. For convenience let us assume economic growth is measured by the &#039;&#039;&#039;GDP&#039;&#039;&#039; of the country. Therefore, the P(t) axis shows the GDP in millions of dollars, and the t axis denote time. If the capacity of the economy’s GDP is 40 million dollars, and at t=0 the GDP is 20, and the rate of increase is such that the constant multiplied with t is 0.05. The function should be&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
P(t) = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:40.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The capacity level of GDP is 40 million. 95% of that is, 38 million.&lt;br /&gt;
Therefore we can say that:&lt;br /&gt;
&lt;br /&gt;
38 = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
38 + 38 e^-0.05t = 40&lt;br /&gt;
&lt;br /&gt;
38 e^-0.05t =2&lt;br /&gt;
&lt;br /&gt;
e^-0.05t = 2/38&lt;br /&gt;
&lt;br /&gt;
e^0.05t = 19&lt;br /&gt;
&lt;br /&gt;
ln e^-0.05t = ln 1/19&lt;br /&gt;
&lt;br /&gt;
-0.05t = ln(1/19)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;t&#039;&#039; is approximately &#039;&#039;&#039;58.888&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  &lt;br /&gt;
  Therefore in 58.8 years the economy will reach 95% of its capacity level of GDP&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73657</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73657"/>
		<updated>2011-01-28T10:21:14Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Starting with the function &lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:MSP220819e4e417b6708ich00001768587g039a5c38.gif]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Change the height of the horizontal asymptote on the right, we&#039;ll denote it by K. Change the y intercept to any number between 0 and K&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
‎ &lt;br /&gt;
 &lt;br /&gt;
To change the height of the horizontal asymptote, we need to change the numerator and the denominator. The horizontal asymptote depends on the numerator, and the part of the denominator independent of &#039;&#039;e&#039;&#039;. In this case 1 and 1. If &#039;&#039;t&#039;&#039; goes to infinity, p(t) = 1/1 = 1, therefore, in this case, the horizontal asymptote will be 1. When the equation is changed to: 5/(2.5 + e^-t), the asymptote is 2.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y intercept in this equation is always 1/2 of &#039;&#039;K&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For example:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
For http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png   the y intercept is 0.5, in the case of &lt;br /&gt;
5/(2.5 + e^-t), it is 1. &lt;br /&gt;
So the vertical asymptote is always 1/2 of the horizontal asymptote at the right.&lt;br /&gt;
&lt;br /&gt;
[[File:52.5e.gif]]&lt;br /&gt;
&lt;br /&gt;
The slope of the curved part depends on the constant multiplied to &#039;&#039;t&#039;&#039; in the power of &#039;&#039;e&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
For example: the slope of the curved part of http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than the slope of the curved part of 1/(1 + e^(-3t)) &lt;br /&gt;
&lt;br /&gt;
and the slope of the curved part for &lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than that of P(t) = 1/(1 + e^-0.5t)&lt;br /&gt;
&lt;br /&gt;
[[File:3 graph.JPG]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
The function may be used to model the economic growth for developing countries. It is often seen that underdeveloped countries have a higher rate of economic growth than developed countries. When a country reaches its capacity level of output, the rate of economic growth decreases. For convenience let us assume economic growth is measured by the &#039;&#039;&#039;GDP&#039;&#039;&#039; of the country. Therefore, the P(t) axis shows the GDP in millions of dollars, and the t axis denote time. If the capacity of the economy’s GDP is 40 million dollars, and at t=0 the GDP is 20, and the rate of increase is such that the constant multiplied with t is 0.05. The function should be&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
P(t) = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:40.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The capacity level of GDP is 40 million. 95% of that is, 38 million.&lt;br /&gt;
Therefore we can say that:&lt;br /&gt;
&lt;br /&gt;
38 = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
38 + 38 e^-0.05t = 40&lt;br /&gt;
&lt;br /&gt;
38 e^-0.05t =2&lt;br /&gt;
&lt;br /&gt;
e^-0.05t = 2/38&lt;br /&gt;
&lt;br /&gt;
e^0.05t = 19&lt;br /&gt;
&lt;br /&gt;
ln e^-0.05t = ln 1/19&lt;br /&gt;
&lt;br /&gt;
-0.05t = ln(1/19)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;t&#039;&#039; is approximately &#039;&#039;&#039;58.888&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  &lt;br /&gt;
  Therefore in 58.8 years the economy will reach 95% of its capacity level of GDP&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:52.5e.gif&amp;diff=73655</id>
		<title>File:52.5e.gif</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:52.5e.gif&amp;diff=73655"/>
		<updated>2011-01-28T10:20:25Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:3_graph.JPG&amp;diff=73652</id>
		<title>File:3 graph.JPG</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:3_graph.JPG&amp;diff=73652"/>
		<updated>2011-01-28T10:15:56Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73638</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73638"/>
		<updated>2011-01-28T09:56:58Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Starting with the function &lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
 &lt;br /&gt;
[[Image:MSP220819e4e417b6708ich00001768587g039a5c38.gif]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Change the height of the horizontal asymptote on the right, we&#039;ll denote it by K. Change the y intercept to any number between 0 and K&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
‎ &lt;br /&gt;
 &lt;br /&gt;
To change the height of the horizontal asymptote, we need to change the numerator and the denominator. The horizontal asymptote depends on the numerator, and the part of the denominator independent of &#039;&#039;e&#039;&#039;. In this case 1 and 1. If &#039;&#039;t&#039;&#039; goes to infinity, p(t) = 1/1 = 1, therefore, in this case, the horizontal asymptote will be 1. When the equation is changed to: 5/(2.5 + e^-t), the asymptote is 2.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y intercept in this equation is always 1/2 of &#039;&#039;K&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For example:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
For http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png   the y intercept is 0.5, in the case of &lt;br /&gt;
5/(2.5 + e^-t), it is 1. &lt;br /&gt;
So the vertical asymptote is always 1/2 of the horizontal asymptote at the right.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The slope of the curved part depends on the constant multiplied to &#039;&#039;t&#039;&#039; in the power of &#039;&#039;e&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
For example: the slope of the curved part of http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than the slope of the curved part of 1/(1 + e^-3t) &lt;br /&gt;
&lt;br /&gt;
and the slope of the curved part for &lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than that of 1/(1 + e^-0.5t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
The function may be used to model the economic growth for developing countries. It is often seen that underdeveloped countries have a higher rate of economic growth than developed countries. When a country reaches its capacity level of output, the rate of economic growth decreases. For convenience let us assume economic growth is measured by the &#039;&#039;&#039;GDP&#039;&#039;&#039; of the country. Therefore, the P(t) axis shows the GDP in millions of dollars, and the t axis denote time. If the capacity of the economy’s GDP is 40 million dollars, and at t=0 the GDP is 20, and the rate of increase is such that the constant multiplied with t is 0.05. The function should be&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
P(t) = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:40.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The capacity level of GDP is 40 million. 95% of that is, 38 million.&lt;br /&gt;
Therefore we can say that:&lt;br /&gt;
&lt;br /&gt;
38 = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
38 + 38 e^-0.05t = 40&lt;br /&gt;
&lt;br /&gt;
38 e^-0.05t =2&lt;br /&gt;
&lt;br /&gt;
e^-0.05t = 2/38&lt;br /&gt;
&lt;br /&gt;
e^0.05t = 19&lt;br /&gt;
&lt;br /&gt;
ln e^-0.05t = ln 1/19&lt;br /&gt;
&lt;br /&gt;
-0.05t = ln(1/19)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;t&#039;&#039; is approximately &#039;&#039;&#039;58.888&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  &lt;br /&gt;
  Therefore in 58.8 years the economy will reach 95% of its capacity level of GDP&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73636</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73636"/>
		<updated>2011-01-28T09:55:08Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Starting with the function &lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
 &lt;br /&gt;
[[Image:MSP220819e4e417b6708ich00001768587g039a5c38.gif]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Change the height of the horizontal asymptote on the right, we&#039;ll denote it by K. Change the y intercept to any number between 0 and K&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
‎ &lt;br /&gt;
 &lt;br /&gt;
To change the height of the horizontal asymptote, we need to change the numerator and the denominator. The horizontal asymptote depends on the numerator, and the part of the denominator independent of &#039;&#039;e&#039;&#039;. In this case 1 and 1. If &#039;&#039;t&#039;&#039; goes to infinity, p(t) = 1/1 = 1, therefore, in this case, the horizontal asymptote will be 1. When the equation is changed to: 5/(2.5 + e^-t), the asymptote is 2.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y intercept in this equation is always 1/2 of &#039;&#039;K&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For example:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
For http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png   the y intercept is 0.5, in the case of &lt;br /&gt;
5/(2.5 + e^-t), it is 1. &lt;br /&gt;
So the vertical asymptote is always 1/2 of the horizontal asymptote at the right.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The slope of the curved part depends on the constant multiplied to &#039;&#039;t&#039;&#039; in the power of &#039;&#039;e&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
For example: the slope of the curved part of http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than the slope of the curved part of 1/(1 + e^-3t) &lt;br /&gt;
&lt;br /&gt;
and the slope of the curved part for &lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than that of 1/(1 + e^-0.5t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
The function may be used to model the economic growth for developing countries. It is often seen that underdeveloped countries have a higher rate of economic growth than developed countries. When a country reaches its capacity level of output, the rate of economic growth decreases. For convenience let us assume economic growth is measured by the &#039;&#039;&#039;GDP&#039;&#039;&#039; of the country. Therefore, the P(t) axis shows the GDP in millions of dollars, and the t axis denote time. If the capacity of the economy’s GDP is 40 million dollars, and at t=0 the GDP is 20, and the rate of increase is such that the constant multiplied with t is 0.05. The function should be&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
P(t) = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:40.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The capacity level of GDP is 40 million. 95% of that is, 38 million.&lt;br /&gt;
Therefore we can say that:&lt;br /&gt;
&lt;br /&gt;
38 = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
38 + 38 e^-0.05t = 40&lt;br /&gt;
&lt;br /&gt;
38 e^-0.05t =2&lt;br /&gt;
&lt;br /&gt;
e^-0.05t = 2/38&lt;br /&gt;
&lt;br /&gt;
e^0.05t = 19&lt;br /&gt;
&lt;br /&gt;
ln e^-0.05t = ln 1/19&lt;br /&gt;
&lt;br /&gt;
-0.05t = ln(1/19)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;t&#039;&#039; is approximately &#039;&#039;&#039;58.888&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  &lt;br /&gt;
  Therefore in 58.8 years the economy will reach 95% of its capacity level of GDP&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73634</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73634"/>
		<updated>2011-01-28T09:52:25Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Starting with the function &lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
 &lt;br /&gt;
[[Image:MSP220819e4e417b6708ich00001768587g039a5c38.gif]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Change the height of the horizontal asymptote on the right, we&#039;ll denote it by K. Change the y intercept to any number between 0 and K&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
‎ &lt;br /&gt;
 &lt;br /&gt;
To change the height of the horizontal asymptote, we need to change the numerator and the denominator. The horizontal asymptote depends on the numerator, and the part of the denominator independent of &#039;&#039;e&#039;&#039;. In this case 1 and 1. If &#039;&#039;t&#039;&#039; goes to infinity, p(t) = 1/1 = 1, therefore, in this case, the horizontal asymptote will be 1. When the equation is changed to:&lt;br /&gt;
&lt;br /&gt;
5/(2.5 + e^-t), the asymptote is 2.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y intercept in this equation is always 1/2 of &#039;&#039;K&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;For example:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
For http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png   the y intercept is 0.5, in the case of &lt;br /&gt;
5/(2.5 + e^-t), it is 1. &lt;br /&gt;
So the vertical asymptote is always 1/2 of the horizontal asymptote at the right.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The slope of the curved part depends on the constant multiplied to &#039;&#039;t&#039;&#039; in the power of &#039;&#039;e&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
For example: the slope of the curved part of http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than the slope of the curved part of 1/(1 + e^-3t) &lt;br /&gt;
&lt;br /&gt;
and the slope of the curved part for &lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than that of 1/(1 + e^-0.5t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
The function may be used to model the economic growth for developing countries. It is often seen that underdeveloped countries have a higher rate of economic growth than developed countries. When a country reaches its capacity level of output, the rate of economic growth decreases. For convenience let us assume economic growth is measured by the &#039;&#039;&#039;GDP&#039;&#039;&#039; of the country. Therefore, the P(t) axis shows the GDP in millions of dollars, and the t axis denote time. If the capacity of the economy’s GDP is 40 million dollars, and at t=0 the GDP is 20, and the rate of increase is such that the constant multiplied with t is 0.05. The function should be&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
P(t) = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:40.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The capacity level of GDP is 40 million. 95% of that is, 38 million.&lt;br /&gt;
Therefore we can say that:&lt;br /&gt;
&lt;br /&gt;
38 = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
38 + 38 e^-0.05t = 40&lt;br /&gt;
&lt;br /&gt;
38 e^-0.05t =2&lt;br /&gt;
&lt;br /&gt;
e^-0.05t = 2/38&lt;br /&gt;
&lt;br /&gt;
e^0.05t = 19&lt;br /&gt;
&lt;br /&gt;
ln e^-0.05t = ln 1/19&lt;br /&gt;
&lt;br /&gt;
-0.05t = ln(1/19)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;t&#039;&#039; is approximately &#039;&#039;&#039;58.888&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  &lt;br /&gt;
  Therefore in 58.8 years the economy will reach 95% of its capacity level of GDP&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73628</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73628"/>
		<updated>2011-01-28T09:47:59Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Starting with the function &lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
 &lt;br /&gt;
[[Image:MSP220819e4e417b6708ich00001768587g039a5c38.gif]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Change the height of the horizontal asymptote on the right, we&#039;ll denote it by K. Change the y intercept to any number between 0 and K&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
‎ &lt;br /&gt;
 &lt;br /&gt;
To change the height of the horizontal asymptote, we need to change the numerator and the denominator. The horizontal asymptote depends on the numerator, and the part of the denominator independent of &#039;&#039;e&#039;&#039;. In this case 1 and 1. If &#039;&#039;t&#039;&#039; goes to infinity, p(t) = 1/1 = 1, therefore, in this case, the horizontal asymptote will be 1. When the equation is changed to:&lt;br /&gt;
&lt;br /&gt;
5/(2.5 + e^-t), &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
the asymptote is 2.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y intercept in this equation is always 1/2 of &#039;&#039;K&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
For example: &lt;br /&gt;
For http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png   the y intercept is 0.5, in the case of &lt;br /&gt;
5/(2.5 + e^-t), it is 1. &lt;br /&gt;
So the vertical asymptote is always 1/2 of the horizontal asymptote at the right.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The slope of the curved part depends on the constant multiplied to &#039;&#039;t&#039;&#039; in the power of &#039;&#039;e&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
For example: the slope of the curved part of http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than the slope of the curved part of 1/(1 + e^-3t) &lt;br /&gt;
&lt;br /&gt;
and the slope of the curved part for &lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than that of 1/(1 + e^-0.5t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
The function may be used to model the economic growth for developing countries. It is often seen that underdeveloped countries have a higher rate of economic growth than developed countries. When a country reaches its capacity level of output, the rate of economic growth decreases. For convenience let us assume economic growth is measured by the &#039;&#039;&#039;GDP&#039;&#039;&#039; of the country. Therefore, the P(t) axis shows the GDP in millions of dollars, and the t axis denote time. If the capacity of the economy’s GDP is 40 million dollars, and at t=0 the GDP is 20, and the rate of increase is such that the constant multiplied with t is 0.05. The function should be&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
P(t) = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:40.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The capacity level of GDP is 40 million. 95% of that is, 38 million.&lt;br /&gt;
Therefore we can say that:&lt;br /&gt;
&lt;br /&gt;
38 = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
38 + 38 e^-0.05t = 40&lt;br /&gt;
&lt;br /&gt;
38 e^-0.05t =2&lt;br /&gt;
&lt;br /&gt;
e^-0.05t = 2/38&lt;br /&gt;
&lt;br /&gt;
e^0.05t = 19&lt;br /&gt;
&lt;br /&gt;
ln e^-0.05t = ln 1/19&lt;br /&gt;
&lt;br /&gt;
-0.05t = ln(1/19)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;t&#039;&#039; is approximately &#039;&#039;&#039;58.888&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  &lt;br /&gt;
  Therefore in 58.8 years the economy will reach 95% of its capacity level of GDP&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73627</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73627"/>
		<updated>2011-01-28T09:45:59Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Starting with the function &lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Change the height of the horizontal asymptote on the right, we&#039;ll denote it by K. Change the y intercept to any number between 0 and K&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:MSP220819e4e417b6708ich00001768587g039a5c38.gif]]&lt;br /&gt;
‎ &lt;br /&gt;
 &lt;br /&gt;
To change the height of the horizontal asymptote, we need to change the numerator and the denominator. The horizontal asymptote depends on the numerator, and the part of the denominator independent of &#039;&#039;e&#039;&#039;. In this case 1 and 1. If &#039;&#039;t&#039;&#039; goes to infinity, p(t) = 1/1 = 1, therefore, in this case, the horizontal asymptote will be 1. When the equation is changed to:&lt;br /&gt;
&lt;br /&gt;
5/(2.5 + e^-t), &lt;br /&gt;
&lt;br /&gt;
the asymptote is 2.&lt;br /&gt;
&lt;br /&gt;
The y intercept in this equation is always 1/2 of &#039;&#039;K&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
For example: &lt;br /&gt;
For http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png   the y intercept is 0.5, in the case of &lt;br /&gt;
5/(2.5 + e^-t), it is 1. &lt;br /&gt;
So the vertical asymptote is always 1/2 of the horizontal asymptote at the right.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The slope of the curved part depends on the constant multiplied to &#039;&#039;t&#039;&#039; in the power of &#039;&#039;e&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
For example: the slope of the curved part of http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than the slope of the curved part of 1/(1 + e^-3t) &lt;br /&gt;
&lt;br /&gt;
and the slope of the curved part for &lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than that of 1/(1 + e^-0.5t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
The function may be used to model the economic growth for developing countries. It is often seen that underdeveloped countries have a higher rate of economic growth than developed countries. When a country reaches its capacity level of output, the rate of economic growth decreases. For convenience let us assume economic growth is measured by the &#039;&#039;&#039;GDP&#039;&#039;&#039; of the country. Therefore, the P(t) axis shows the GDP in millions of dollars, and the t axis denote time. If the capacity of the economy’s GDP is 40 million dollars, and at t=0 the GDP is 20, and the rate of increase is such that the constant multiplied with t is 0.05. The function should be&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
P(t) = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:40.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The capacity level of GDP is 40 million. 95% of that is, 38 million.&lt;br /&gt;
Therefore we can say that:&lt;br /&gt;
&lt;br /&gt;
38 = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
38 + 38 e^-0.05t = 40&lt;br /&gt;
&lt;br /&gt;
38 e^-0.05t =2&lt;br /&gt;
&lt;br /&gt;
e^-0.05t = 2/38&lt;br /&gt;
&lt;br /&gt;
e^0.05t = 19&lt;br /&gt;
&lt;br /&gt;
ln e^-0.05t = ln 1/19&lt;br /&gt;
&lt;br /&gt;
-0.05t = ln(1/19)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;t&#039;&#039; is approximately &#039;&#039;&#039;58.888&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  &lt;br /&gt;
  Therefore in 58.8 years the economy will reach 95% of its capacity level of GDP&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73626</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73626"/>
		<updated>2011-01-28T09:43:52Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Starting with the function &lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Change the height of the horizontal asymptote on the right, we&#039;ll denote it by K. Change the y intercept to any number between 0 and K&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:MSP220819e4e417b6708ich00001768587g039a5c38.gif]]&lt;br /&gt;
‎ &lt;br /&gt;
 &lt;br /&gt;
To change the height of the horizontal asymptote, we need to change the numerator and the denominator. The horizontal asymptote depends on the numerator, and the part of the denominator independent of &#039;&#039;e&#039;&#039;. In this case 1 and 1. If &#039;&#039;t&#039;&#039; goes to infinity, p(t) = 1/1 = 1, therefore, in this case, the horizontal asymptote will be 1. When the equation is changed to:&lt;br /&gt;
&lt;br /&gt;
5/(2.5 + e^-t), &lt;br /&gt;
&lt;br /&gt;
the asymptote is 2.&lt;br /&gt;
&lt;br /&gt;
The y intercept in this equation is always 1/2 of &#039;&#039;K&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
For example: &lt;br /&gt;
For http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png   the y intercept is 0.5, in the case of &lt;br /&gt;
5/(2.5 + e^-t), it is 1. &lt;br /&gt;
So the vertical asymptote is always 1/2 of the horizontal asymptote at the right.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The slope of the curved part depends on the constant multiplied to &#039;&#039;t&#039;&#039; in the power of &#039;&#039;e&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
For example: the slope of the curved part of http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than the slope of the curved part of 1/(1 + e^-3t) &lt;br /&gt;
&lt;br /&gt;
and the slope of the curved part for &lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than that of 1/(1 + e^-0.5t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
The function may be used to model the economic growth for developing countries. It is often seen that underdeveloped countries have a higher rate of economic growth than developed countries. When a country reaches its capacity level of output, the rate of economic growth decreases. For convenience let us assume economic growth is measured by the &#039;&#039;&#039;GDP&#039;&#039;&#039; of the country. Therefore, the P(t) axis shows the GDP in millions of dollars, and the t axis denote time. If the capacity of the economy’s GDP is 40 million dollars, and at t=0 the GDP is 20, and the rate of increase is such that the constant multiplied with t is 0.05. The function should be&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
P(t) = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
[[Image:40.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The capacity level of GDP is 40 million. 95% of that is, 38 million.&lt;br /&gt;
Therefore we can say that:&lt;br /&gt;
&lt;br /&gt;
38 = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
38 + 38 e^-0.05t = 40&lt;br /&gt;
&lt;br /&gt;
38 e^-0.05t =2&lt;br /&gt;
&lt;br /&gt;
e^-0.05t = 2/38&lt;br /&gt;
&lt;br /&gt;
e^0.05t = 19&lt;br /&gt;
&lt;br /&gt;
ln e^-0.05t = ln 1/19&lt;br /&gt;
&lt;br /&gt;
-0.05t = ln(1/19)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;t&#039;&#039; is approximately &#039;&#039;&#039;58.888&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  &lt;br /&gt;
  Therefore in 58.8 years the economy will reach 95%% of its capacity level of GDP&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:40.gif&amp;diff=73623</id>
		<title>File:40.gif</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:40.gif&amp;diff=73623"/>
		<updated>2011-01-28T09:42:48Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73621</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73621"/>
		<updated>2011-01-28T09:40:23Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Starting with the function &lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Change the height of the horizontal asymptote on the right, we&#039;ll denote it by K. Change the y intercept to any number between 0 and K&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:MSP220819e4e417b6708ich00001768587g039a5c38.gif]]&lt;br /&gt;
‎ &lt;br /&gt;
 &lt;br /&gt;
To change the height of the horizontal asymptote, we need to change the numerator and the denominator. The horizontal asymptote depends on the numerator, and the part of the denominator independent of &#039;&#039;e&#039;&#039;. In this case 1 and 1. If &#039;&#039;t&#039;&#039; goes to infinity, p(t) = 1/1 = 1, therefore, in this case, the horizontal asymptote will be 1. When the equation is changed to:&lt;br /&gt;
&lt;br /&gt;
5/(2.5 + e^-t), &lt;br /&gt;
&lt;br /&gt;
the asymptote is 2.&lt;br /&gt;
&lt;br /&gt;
The y intercept in this equation is always 1/2 of &#039;&#039;K&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
For example: &lt;br /&gt;
For http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png   the y intercept is 0.5, in the case of &lt;br /&gt;
5/(2.5 + e^-t), it is 1. &lt;br /&gt;
So the vertical asymptote is always 1/2 of the horizontal asymptote at the right.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The slope of the curved part depends on the constant multiplied to &#039;&#039;t&#039;&#039; in the power of &#039;&#039;e&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
For example: the slope of the curved part of http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than the slope of the curved part of 1/(1 + e^-3t) &lt;br /&gt;
&lt;br /&gt;
and the slope of the curved part for &lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than that of 1/(1 + e^-0.5t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
-----&lt;br /&gt;
&lt;br /&gt;
The function may be used to model the economic growth for developing countries. It is often seen that underdeveloped countries have a higher rate of economic growth than developed countries. When a country reaches its capacity level of output, the rate of economic growth decreases. For convenience let us assume economic growth is measured by the &#039;&#039;&#039;GDP&#039;&#039;&#039; of the country. Therefore, the P(t) axis shows the GDP in millions of dollars, and the t axis denote time. If the capacity of the economy’s GDP is 40 million dollars, and at t=0 the GDP is 20, and the rate of increase is such that the constant multiplied with t is 0.05. The function should be&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
P(t) = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The capacity level of GDP is 40 million. 95% of that is, 38 million.&lt;br /&gt;
Therefore we can say that:&lt;br /&gt;
&lt;br /&gt;
38 = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
38 + 38 e^-0.05t = 40&lt;br /&gt;
&lt;br /&gt;
38 e^-0.05t =2&lt;br /&gt;
&lt;br /&gt;
e^-0.05t = 2/38&lt;br /&gt;
&lt;br /&gt;
e^0.05t = 19&lt;br /&gt;
&lt;br /&gt;
ln e^-0.05t = ln 1/19&lt;br /&gt;
&lt;br /&gt;
-0.05t = ln(1/19)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;t&#039;&#039; is approximately &#039;&#039;&#039;58.888&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  &lt;br /&gt;
  Therefore in 58.8 years the economy will reach 95%% of its capacity level of GDP&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73620</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73620"/>
		<updated>2011-01-28T09:36:29Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Starting with the function &lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Change the height of the horizontal asymptote on the right, we&#039;ll denote it by K. Change the y intercept to any number between 0 and K&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:MSP220819e4e417b6708ich00001768587g039a5c38.gif]]&lt;br /&gt;
‎ &lt;br /&gt;
 &lt;br /&gt;
To change the height of the horizontal asymptote, we need to change the numerator and the denominator. The horizontal asymptote depends on the numerator, and the part of the denominator independent of &#039;&#039;e&#039;&#039;. In this case 1 and 1. If &#039;&#039;t&#039;&#039; goes to infinity, p(t) = 1/1 = 1, therefore, in this case, the horizontal asymptote will be 1. When the equation is changed to:&lt;br /&gt;
&lt;br /&gt;
5/(2.5 + e^-t), &lt;br /&gt;
&lt;br /&gt;
the asymptote is 2.&lt;br /&gt;
&lt;br /&gt;
The y intercept in this equation is always 1/2 of &#039;&#039;K&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
For example: &lt;br /&gt;
For http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png   the y intercept is 0.5, in the case of &lt;br /&gt;
5/(2.5 + e^-t), it is 1. &lt;br /&gt;
So the vertical asymptote is always 1/2 of the horizontal asymptote at the right.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The slope of the curved part depends on the constant multiplied to &#039;&#039;t&#039;&#039; in the power of &#039;&#039;e&#039;&#039;. &lt;br /&gt;
&lt;br /&gt;
For example: the slope of the curved part of http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than the slope of the curved part of 1/(1 + e^-3t) &lt;br /&gt;
&lt;br /&gt;
and the slope of the curved part for &lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
will be steeper than that of 1/(1 + e^-0.5t)&lt;br /&gt;
&lt;br /&gt;
--------&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The function may be used to model the economic growth for developing countries. It is often seen that underdeveloped countries have a higher rate of economic growth than developed countries. When a country reaches its capacity level of output, the rate of economic growth decreases. For convenience let us assume economic growth is measured by the &#039;&#039;&#039;GDP&#039;&#039;&#039; of the country. Therefore, the P(t) axis shows the GDP in millions of dollars, and the t axis denote time. If the capacity of the economy’s GDP is 40 million dollars, and at t=0 the GDP is 20, and the rate of increase is such that the constant multiplied with t is 0.05. The function should be&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
P(t) = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The capacity level of GDP is 40 million. 95% of that is, 38 million.&lt;br /&gt;
Therefore we can say that:&lt;br /&gt;
&lt;br /&gt;
38 = 40/(1 + e^-0.05t)&lt;br /&gt;
&lt;br /&gt;
38 + 38 e^-0.05t = 40&lt;br /&gt;
&lt;br /&gt;
38 e^-0.05t =2&lt;br /&gt;
&lt;br /&gt;
e^-0.05t = 2/38&lt;br /&gt;
&lt;br /&gt;
e^0.05t = 19&lt;br /&gt;
&lt;br /&gt;
ln e^-0.05t = ln 1/19&lt;br /&gt;
&lt;br /&gt;
-0.05t = ln(1/19)&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;t&#039;&#039; is approximately &#039;&#039;&#039;58.888&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
  &lt;br /&gt;
  Therefore in 58.8 years the economy will reach 95%% of its capacity level of GDP&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73609</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73609"/>
		<updated>2011-01-28T09:12:53Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Starting with the function &lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things: &#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Change the height of the horizontal asymptote on the right, we&#039;ll denote it by K. Change the y intercept to any number between 0 and K&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Image:MSP220819e4e417b6708ich00001768587g039a5c38.gif]]&lt;br /&gt;
‎ &lt;br /&gt;
 &lt;br /&gt;
To change the height of the horizontal asymptote, we need to change the numerator and the denominator. The horizontal asymptote depends on the numerator, and the part of the denominator independent of &#039;&#039;e&#039;&#039;. In this case 1 and 1. If &#039;&#039;t&#039;&#039; goes to infinity, p(t) = 1/1 =1, therefore, in this case, the horizontal asymptote will be 1. When the equation is changed to:&lt;br /&gt;
&lt;br /&gt;
5/(2.5 + e^-t), &lt;br /&gt;
the asymptote is 2.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
BONUS (just the point below, not what comes after) &lt;br /&gt;
Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical. (If you graph it, it should be quite clear). &lt;br /&gt;
&lt;br /&gt;
Once you&#039;ve played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it&#039;s carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). &lt;br /&gt;
When doing this last part, explain well where you&#039;re taking your data from (real data or imagined data), what it is that you&#039;re modelling and how you are doing the math to answer a predictive question. &lt;br /&gt;
This should all be done on a dedicated page of the wiki whose address should look like&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:MSP220819e4e417b6708ich00001768587g039a5c38.gif&amp;diff=73608</id>
		<title>File:MSP220819e4e417b6708ich00001768587g039a5c38.gif</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:MSP220819e4e417b6708ich00001768587g039a5c38.gif&amp;diff=73608"/>
		<updated>2011-01-28T09:11:30Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73604</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_12&amp;diff=73604"/>
		<updated>2011-01-28T09:01:57Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: Created page with &amp;quot; Starting with the function   http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png   Your goal is to modify the function so that we can use it to model a rea...&amp;quot;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
Starting with the function &lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/images/math/b/5/3/b532d798ec926e169adf3c1a55f0cdf0.png&lt;br /&gt;
 &lt;br /&gt;
Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things: &lt;br /&gt;
&lt;br /&gt;
Change the height of the horizontal asymptote on the right, we&#039;ll denote it by K. Change the y-intercept to any number between 0 and K &lt;br /&gt;
&lt;br /&gt;
BONUS (just the point below, not what comes after) &lt;br /&gt;
Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical. (If you graph it, it should be quite clear). &lt;br /&gt;
&lt;br /&gt;
Once you&#039;ve played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it&#039;s carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). &lt;br /&gt;
When doing this last part, explain well where you&#039;re taking your data from (real data or imagined data), what it is that you&#039;re modelling and how you are doing the math to answer a predictive question. &lt;br /&gt;
This should all be done on a dedicated page of the wiki whose address should look like&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/Homework_12&amp;diff=73533</id>
		<title>User:UnaVuckovic/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/Homework_12&amp;diff=73533"/>
		<updated>2011-01-28T07:01:30Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Generally people only think of calculus as a subject in school. In reality, they are just oblivious to the fact that calculus is common in daily life. When you pay your monthly bills, buy a car, and even follow a recipe, you are using the basic principles of calculus. For example how would you play a friendly game of basketball when you don’t know how to keep score? Most people wouldn’t associate sports and math, but they are actually very closely related.      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
In the Olympics, statistics are used in order to determine which country is in the lead. If you take a look at the table below, you will see the overall Olympic medals table for the 2010 Winter Olympics in Vancouver. As you may notice, Canada is in the third position even though it has received the most gold medals. This is because the medals are valued by quantity not by quality, resulting in the value of the gold, silver and bronze medals even. Hence the United States is in first position with an overall total of 37 medals, following Germany with 30 medals, then Canada in third place with 26 medals. &lt;br /&gt;
&lt;br /&gt;
       &lt;br /&gt;
&lt;br /&gt;
[[File:Medal_Count.JPG]]&lt;br /&gt;
‎ &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Without the knowledge of math, these statistics would be impossible to create. For instance in baseball you have to use statistics to determine a player’s batting average. A batting average represents the percentage of bats that result in hits for particular baseball players. The formula is:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
  Hits/ Bats = Batting average      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
For example:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
Bob and Steve are baseball players. Last month, Bob got 195 hits out of 535 bats while Steve got 97 hits in 113 bats. Determine the batting average for both players. Which player has a better average?      &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Bob:&#039;&#039;&#039;      &lt;br /&gt;
 &lt;br /&gt;
= 195 hits/ 535 bats&lt;br /&gt;
    &lt;br /&gt;
= 195/535      &lt;br /&gt;
 &lt;br /&gt;
= &#039;&#039;&#039;0.364&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steve:&#039;&#039;&#039;      &lt;br /&gt;
 &lt;br /&gt;
= 97 hits/ 113 bats     &lt;br /&gt;
 &lt;br /&gt;
= 97/113      &lt;br /&gt;
 &lt;br /&gt;
= &#039;&#039;&#039;0.858&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;NOTE:&#039;&#039;&#039; Notice that batting averages are NOT represented as percentages. They are represented as decimal numbers to three decimal places.      &lt;br /&gt;
To clarify, a batting average of 1.000 means that the player gets a hit at every bat, while a batting average of 0.000 means that the player has no hits.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
         &lt;br /&gt;
  With this information, it is clear that Steve has the better batting average.         &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Although it would be impossible to determine batting averages without math, there are plenty of other uses for calculus in sports besides statistics.&lt;br /&gt;
&lt;br /&gt;
The use of angles is vital in every single sport. In basketball, a player must know his/her spot where they most frequently get the ball in the hoop, and remember where that spot is on the court. In soccer, goaltenders use angles to reduce the amount of net that a player has to shoot at, therefore increasing his/her chances of saving the shot. Angles are used in various other sports such as golf, bobsled, water polo, hockey etc. &lt;br /&gt;
&lt;br /&gt;
For instance, when a punter kicks a football, he can control three factors:       &lt;br /&gt;
- The velocity and/or speed of the kick     &lt;br /&gt;
&lt;br /&gt;
- The angle of the kick     &lt;br /&gt;
&lt;br /&gt;
- The rotation of the football     &lt;br /&gt;
&lt;br /&gt;
These three factors are crucial because they will determine how high and far the ball will go, the direction of the football and whether the kick will reach its desired location.&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
http://static.howstuffworks.com/gif/physics-of-football-kick.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
It is no coincidence that this action represents a parabolic function. This path is always curved because the force of gravity affects the movement of the ball in an upward direction.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Since the language of math is numbers,it can help us make important decisions and perform everyday tasks. In sports it determines the country in first place at the Olympics, teaches athletes how to use angles and has hundreds of other essential uses.&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/Homework_12&amp;diff=73532</id>
		<title>User:UnaVuckovic/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/Homework_12&amp;diff=73532"/>
		<updated>2011-01-28T06:59:07Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Generally people only think of calculus as a subject in school. In reality, they are just oblivious to the fact that calculus is common in daily life. When you pay your monthly bills, buy a car, and even follow a recipe, you are using the basic principles of calculus. For example how would you play a friendly game of basketball when you don’t know how to keep score? Most people wouldn’t associate sports and math, but they are actually very closely related.      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
In the Olympics, statistics are used in order to determine which country is in the lead. If you take a look at the table below, you will see the overall Olympic medals table for the 2010 Winter Olympics in Vancouver. As you may notice, Canada is in the third position even though it has received the most gold medals. This is because the medals are valued by quantity not by quality, resulting in the value of the gold, silver and bronze medals even. Hence the United States is in first position with an overall total of 37 medals, following Germany with 30 medals, then Canada in third place with 26 medals. &lt;br /&gt;
&lt;br /&gt;
       &lt;br /&gt;
&lt;br /&gt;
[[File:Medal_Count.JPG]]&lt;br /&gt;
‎ &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Without the knowledge of math, these statistics would be impossible to create. For instance in baseball you have to use statistics to determine a player’s batting average. A batting average represents the percentage of bats that result in hits for particular baseball players. The formula is:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
  Hits/ Bats = Batting average      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
For example:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
Bob and Steve are baseball players. Last month, Bob got 195 hits out of 535 bats while Steve got 97 hits in 113 bats. Determine the batting average for both players. Which player has a better average?      &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Bob:&#039;&#039;&#039;      &lt;br /&gt;
 &lt;br /&gt;
= 195 hits/ 535 bats&lt;br /&gt;
    &lt;br /&gt;
= 195/535      &lt;br /&gt;
 &lt;br /&gt;
= &#039;&#039;&#039;0.364&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steve:&#039;&#039;&#039;      &lt;br /&gt;
 &lt;br /&gt;
= 97 hits/ 113 bats     &lt;br /&gt;
 &lt;br /&gt;
= 97/113      &lt;br /&gt;
 &lt;br /&gt;
= &#039;&#039;&#039;0.858&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;NOTE:&#039;&#039;&#039; Notice that batting averages are NOT represented as percentages. They are represented as decimal numbers to three decimal places.      &lt;br /&gt;
To clarify, a batting average of 1.000 means that the player gets a hit at every bat, while a batting average of 0.000 means that the player has no hits.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
         &lt;br /&gt;
  With this information, it is clear that Steve has the better batting average.         &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Although it would be impossible to determine batting averages without math, there are plenty of other uses for calculus in sports besides statistics.&lt;br /&gt;
&lt;br /&gt;
The use of angles is vital in every single sport. In basketball, a player must know his/her spot where they most frequently get the ball in the hoop, and remember where that spot is on the court. In soccer, goaltenders use angles to reduce the amount of net that a player has to shoot at, therefore increasing his/her chances of saving the shot. Angles are used in various other sports such as golf, bobsled, water polo, hockey etc. &lt;br /&gt;
&lt;br /&gt;
For instance, when a punter kicks a football, he can control three factors:       &lt;br /&gt;
- The velocity and/or speed of the kick     &lt;br /&gt;
&lt;br /&gt;
- The angle of the kick     &lt;br /&gt;
&lt;br /&gt;
- The rotation of the football     &lt;br /&gt;
&lt;br /&gt;
These three factors are crucial because they will determine how high and far the ball will go, the direction of the football and whether the kick will reach its desired location.&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
http://static.howstuffworks.com/gif/physics-of-football-kick.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
It is no coincidence that this action represents a parabolic function. This path is always curved because the force of gravity affects the movement of the ball in an upward direction.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Since the language of math is numbers and it can help us make important decisions and perform everyday tasks. In sports it determines the country with the most gold medals, teaches athletes how to use angles and has hundreds of other essential uses.&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:Medal_Count.JPG&amp;diff=73530</id>
		<title>File:Medal Count.JPG</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:Medal_Count.JPG&amp;diff=73530"/>
		<updated>2011-01-28T06:54:00Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/Homework_12&amp;diff=73527</id>
		<title>User:UnaVuckovic/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/Homework_12&amp;diff=73527"/>
		<updated>2011-01-28T06:45:53Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Generally people only think of calculus as a subject in school. In reality, they are just oblivious to the fact that calculus is common in daily life. When you pay your monthly bills, buy a car, and even follow a recipe, you are using the basic principles of calculus. For example how would you play a friendly game of basketball when you don’t know how to keep score? Most people wouldn’t associate sports and math, but they are actually very closely related.      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
In the Olympics, statistics are used in order to determine which country is in the lead. If you take a look at the table below, you will see the overall Olympics medals table for the 2010 Winter Olympics in Vancouver. As you may notice, Canada is in the third position even though it has received the most gold medals. This is because the medals are valued by quantity not by quality, resulting in the value of the gold, silver and bronze medals even. Hence the United States is in first position with an overall total of 37 medals, following Germany with 30 medals, then Canada in thrid place with 27 medals. &lt;br /&gt;
&lt;br /&gt;
       &lt;br /&gt;
&lt;br /&gt;
http://sports.yahoo.com/olympics/vancouver/medals&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Without the knowledge of math, these statistics would be impossible to create. For instance in baseball you have to use statistics to determine a player’s batting average. A batting average represents the percentage of bats that result in hits for particular baseball players. The formula is:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
  Hits/ Bats = Batting average      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
For example:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
Bob and Steve are baseball players. Last month, Bob got 195 hits out of 535 bats while Steve got 97 hits in 113 bats. Determine the batting average for both players. Which player has a better average?      &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Bob:&#039;&#039;&#039;      &lt;br /&gt;
 &lt;br /&gt;
= 195 hits/ 535 bats&lt;br /&gt;
    &lt;br /&gt;
= 195/535      &lt;br /&gt;
 &lt;br /&gt;
= &#039;&#039;&#039;0.363&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steve:&#039;&#039;&#039;      &lt;br /&gt;
 &lt;br /&gt;
= 97 hits/ 113 bats     &lt;br /&gt;
 &lt;br /&gt;
= 97/113      &lt;br /&gt;
 &lt;br /&gt;
= &#039;&#039;&#039;0.858&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;NOTE:&#039;&#039;&#039; Notice that batting averages are NOT represented as percentages. They are represented as decimal numbers to three decimal places.      &lt;br /&gt;
To clarify, a batting average of 1.000 means that the player gets a hit at every bat, while a batting average of 0.000 means that the player has no hits.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
         &lt;br /&gt;
  With this information, it is clear that Steve has the better batting average.         &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Although it would be impossible to determine batting averages without math, there are plenty of other uses for calculus in sports besides statistics.&lt;br /&gt;
&lt;br /&gt;
The use of angles is vital in every single sport. In basketball, a player must know his/her spot where they most frequently get the ball in the hoop, and remember where that spot is on the court. In soccer, goaltenders use angles to reduce the amount of net that a player has to shoot at, therefore increasing his/her chances of saving the shot. Angles are used in various other sports such as golf, bobsled, water polo, hockey etc. For example, when a punter kicks a football, he can control three factors:       &lt;br /&gt;
# The velocity and/or speed of the kick     &lt;br /&gt;
&lt;br /&gt;
# The angle of the kick     &lt;br /&gt;
&lt;br /&gt;
# The rotation of the football     &lt;br /&gt;
&lt;br /&gt;
These three factors are crucial because they will determine how high and far the ball will go, the direction of the football and whether the kick will reach its desired location.&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
http://static.howstuffworks.com/gif/physics-of-football-kick.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
It is no coincidence that this action represents a parabolic function. This path is always curved because the force of gravity affects the movement of the ball in an upward direction.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Since the language of math is numbers and it can help us make important decisions and perform everyday tasks. In sports it determines the country with the most gold medals, teaches athletes how to use angles and has hundreds of other essential uses.&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/Homework_12&amp;diff=73525</id>
		<title>User:UnaVuckovic/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/Homework_12&amp;diff=73525"/>
		<updated>2011-01-28T06:43:29Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Generally people only think of calculus as a subject in school. In reality, they are just oblivious to the fact that calculus is common in daily life. When you pay your monthly bills, buy a car, and even follow a recipe, you are using the basic principles of calculus. For example how would you play a friendly game of basketball when you don’t know how to keep score? Most people wouldn’t associate sports and math, but they are actually very closely related.      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
In the Olympics, statistics are used in order to determine which country is in the lead. If you take a look at the table below, you will see the overall Olympics medals table for the 2010 Winter Olympics in Vancouver. As you may notice, Canada is in the third position even though it has received the most gold medals. This is because the medals are valued by quantity not by quality, resulting in the value of the gold, silver and bronze medals even. Hence the United States is in first position with an overall total of 37 medals, following Germany with 30 medals, then Canada in thrid place with 27 medals. &lt;br /&gt;
&lt;br /&gt;
       &lt;br /&gt;
&lt;br /&gt;
http://sports.yahoo.com/olympics/vancouver/medals&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Without the knowledge of math, these statistics would be impossible to create. For instance in baseball you have to use statistics to determine a player’s batting average. A batting average represents the percentage of bats that result in hits for particular baseball players. The formula is:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
  Hits/ Bats = Batting average      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
For example:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
Bob and Steve are baseball players. Last month, Bob got 195 hits out of 535 bats while Steve got 97 hits in 113 bats. Determine the batting average for both players. Which player has a better average?      &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Bob:&#039;&#039;&#039;      &lt;br /&gt;
 &lt;br /&gt;
= 195 hits/ 535 bats&lt;br /&gt;
    &lt;br /&gt;
= 195/535      &lt;br /&gt;
 &lt;br /&gt;
= &#039;&#039;&#039;0.363&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steve:&#039;&#039;&#039;      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
= 97 hits/ 113 bats      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
= 97/113      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
= &#039;&#039;&#039;0.858&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;NOTE:&#039;&#039;&#039; Notice that batting averages are NOT represented as percentages. They are represented as decimal numbers to three decimal places.      &lt;br /&gt;
To clarify, a batting average of 1.000 means that the player gets a hit at every bat, while a batting average of 0.000 means that the player has no hits.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
         &lt;br /&gt;
  With this information, it is clear that Steve has the better batting average.         &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Although it would be impossible to determine batting averages without math, there are plenty of other uses for calculus in sports besides statistics.&lt;br /&gt;
&lt;br /&gt;
The use of angles is vital in every single sport. In basketball, a player must know his/her spot where they most frequently get the ball in the hoop, and remember where that spot is on the court. In soccer, goaltenders use angles to reduce the amount of net that a player has to shoot at, therefore increasing his/her chances of saving the shot. Angles are used in various other sports such as golf, bobsled, water polo, hockey etc. For example, when a punter kicks a football, he can control three factors:       &lt;br /&gt;
# The velocity and/or speed of the kick     &lt;br /&gt;
&lt;br /&gt;
## The angle of the kick     &lt;br /&gt;
&lt;br /&gt;
### The rotation of the football     &lt;br /&gt;
&lt;br /&gt;
These three factors are crucial because they will determine how high and far the ball will go, the direction of the football and whether the kick will reach its desired location.&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
http://static.howstuffworks.com/gif/physics-of-football-kick.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
It is no coincidence that this action represents a parabolic function. This path is always curved because the force of gravity affects the movement of the ball in an upward direction.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Since the language of math is numbers and it can help us make important decisions and perform everyday tasks. In sports it determines the country with the most gold medals, teaches athletes how to use angles and has hundreds of other essential uses.&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/Homework_12&amp;diff=73521</id>
		<title>User:UnaVuckovic/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/Homework_12&amp;diff=73521"/>
		<updated>2011-01-28T06:38:23Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Generally people only think of calculus as a subject in school. In reality, they are just oblivious to the fact that calculus is common in daily life. When you pay your monthly bills, buy a car, and even follow a recipe, you are using the basic principles of calculus. For example how would you play a friendly game of basketball when you don’t know how to keep score? Most people wouldn’t associate sports and math, but they are actually very closely related.      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
In the Olympics, statistics are used in order to determine which country is in the lead. If you take a look at the table below, you will see the overall Olympics medals table for the 2010 Winter Olympics in Vancouver. As you may notice, Canada is in the number one position even though it has received a fewer total of medals than its competitors Germany and the United States. This is because Canada has received the most gold medals, which are valued more than the silver, and bronze.      &lt;br /&gt;
&lt;br /&gt;
       &lt;br /&gt;
&lt;br /&gt;
http://sports.yahoo.com/olympics/vancouver/medals&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Without the knowledge of math, these statistics would be impossible to create. For instance in baseball you have to use statistics to determine a player’s batting average. A batting average represents the percentage of bats that result in hits for particular baseball players. The formula is:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
  Hits/ Bats = Batting average      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
For example:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
Bob and Steve are baseball players. Last month, Bob got 195 hits out of 535 bats while Steve got 97 hits in 113 bats. Determine the batting average for both players. Which player has a better average?      &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Bob:&#039;&#039;&#039;      &lt;br /&gt;
 &lt;br /&gt;
= 195 hits/ 535 bats&lt;br /&gt;
    &lt;br /&gt;
= 195/535      &lt;br /&gt;
 &lt;br /&gt;
= &#039;&#039;&#039;0.363&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steve:&#039;&#039;&#039;      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
= 97 hits/ 113 bats      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
= 97/113      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
= &#039;&#039;&#039;0.858&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;NOTE:&#039;&#039;&#039; Notice that batting averages are NOT represented as percentages. They are represented as decimal numbers to three decimal places.      &lt;br /&gt;
To clarify, a batting average of 1.000 means that the player gets a hit at every bat, while a batting average of 0.000 means that the player has no hits.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
         &lt;br /&gt;
  With this information, it is clear that Steve has the better batting average.         &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Although it would be impossible to determine batting averages without math, there are plenty of other uses for calculus in sports besides statistics.&lt;br /&gt;
&lt;br /&gt;
The use of angles is vital in every single sport. In basketball, a player must know his/her spot where they most frequently get the ball in the hoop, and remember where that spot is on the court. In soccer, goaltenders use angles to reduce the amount of net that a player has to shoot at, therefore increasing his/her chances of saving the shot. Angles are used in various other sports such as golf, bobsled, water polo, hockey etc. For example, when a punter kicks a football, he can control three factors:       &lt;br /&gt;
# The velocity and/or speed of the kick     &lt;br /&gt;
&lt;br /&gt;
## The angle of the kick     &lt;br /&gt;
&lt;br /&gt;
### The rotation of the football     &lt;br /&gt;
&lt;br /&gt;
These three factors are crucial because they will determine how high and far the ball will go, the direction of the football and whether the kick will reach its desired location.&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
http://static.howstuffworks.com/gif/physics-of-football-kick.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
It is no coincidence that this action represents a parabolic function. This path is always curved because the force of gravity affects the movement of the ball in an upward direction.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Since the language of math is numbers and it can help us make important decisions and perform everyday tasks. In sports it determines the country with the most gold medals, teaches athletes how to use angles and has hundreds of other essential uses.&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/Homework_12&amp;diff=73517</id>
		<title>User:UnaVuckovic/Homework 12</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/Homework_12&amp;diff=73517"/>
		<updated>2011-01-28T06:32:58Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: Created page with &amp;quot;Generally people only think of calculus as a subject in school. In reality, they are just oblivious to the fact that calculus is common in daily life. When you pay your monthly b...&amp;quot;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Generally people only think of calculus as a subject in school. In reality, they are just oblivious to the fact that calculus is common in daily life. When you pay your monthly bills, buy a car, and even follow a recipe, you are using the basic principles of calculus. For example how would you play a friendly game of basketball when you don’t know how to keep score? Most people wouldn’t associate sports and math, but they are actually very closely related.      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
In the Olympics, statistics are used in order to determine which country is in the lead. If you take a look at the table below, you will see the overall Olympics medals table for the 2010 Winter Olympics in Vancouver. As you may notice, Canada is in the number one position even though it has received a fewer total of medals than its competitors Germany and the United States. This is because Canada has received the most gold medals, which are valued more than the silver, and bronze.      &lt;br /&gt;
&lt;br /&gt;
       &lt;br /&gt;
&lt;br /&gt;
http://sports.yahoo.com/olympics/vancouver/medals&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Without the knowledge of math, these statistics would be impossible to create. For instance in baseball you have to use statistics to determine a player’s batting average. A batting average represents the percentage of bats that result in hits for particular baseball players. The formula is:&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
  Hits/ Bats = Batting average      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
For example:&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
Bob and Steve are baseball players. Last month, Bob got 195 hits out of 535 bats while Steve got 97 hits in 113 bats. Determine the batting average for both players. Which player has a better average?      &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
&#039;&#039;&#039;Bob:&#039;&#039;&#039;      &lt;br /&gt;
 &lt;br /&gt;
= 195 hits/ 535 bats&lt;br /&gt;
    &lt;br /&gt;
= 195/535      &lt;br /&gt;
 &lt;br /&gt;
= &#039;&#039;&#039;0.363&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Steve:&#039;&#039;&#039;      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
= 97 hits/ 113 bats      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
= 97/113      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
= &#039;&#039;&#039;0.858&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;NOTE:&#039;&#039;&#039; Notice that batting averages are NOT represented as percentages. They are represented as decimal numbers to three decimal places.      &lt;br /&gt;
To clarify, a batting average of 1.000 means that the player gets a hit at every bat, while a batting average of 0.000 means that the player has no hits.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
         &lt;br /&gt;
  With this information, it is clear that Steve has the better batting average.         &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Although it would be impossible to determine batting averages without math, there are plenty of other uses for calculus in sports besides statistics.&lt;br /&gt;
&lt;br /&gt;
The use of angles is vital in every single sport. In basketball, a player must know his/her spot where they most frequently get the ball in the hoop, and remember where that spot is on the court. In soccer, goaltenders use angles to reduce the amount of net that a player has to shoot at, therefore increasing his/her chances of saving the shot. Angles are used in various other sports such as golf, bobsled, water polo, hockey etc. For example, when a punter kicks a football, he can control three factors:       &lt;br /&gt;
# The velocity and/or speed of the kick     &lt;br /&gt;
&lt;br /&gt;
# The angel of the kick     &lt;br /&gt;
&lt;br /&gt;
# The rotation of the football     &lt;br /&gt;
&lt;br /&gt;
These three factors are crucial because they will determine how high and far the ball will go, the direction of the football and whether the kick will reach its desired location.&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
http://static.howstuffworks.com/gif/physics-of-football-kick.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
It is no coincidence that this action represents a parabolic function. This path is always curved because the force of gravity affects the movement of the ball in an upward direction.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Since the language of math is numbers and it can help us make important decisions and perform everyday tasks. In sports it determines the country with the most gold medals, teaches athletes how to use angles and has hundreds of other essential uses.&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:UnaVuckovic&amp;diff=73503</id>
		<title>User:UnaVuckovic</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:UnaVuckovic&amp;diff=73503"/>
		<updated>2011-01-28T06:15:40Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Hi I&#039;m Una Vuckovic. I&#039;m in the Faculty of Commerce and I&#039;m a first year student. Below is my essay, I hope you enjoy it!   &lt;br /&gt;
&lt;br /&gt;
email: v_una@hotmail.com     &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
== Homework 12 ==  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/User:UnaVuckovic/Homework_12&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== The PythagoreanTheorem ==  &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/User:UnaVuckovic/ThePythagoreanTheorem&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:UnaVuckovic&amp;diff=73474</id>
		<title>User:UnaVuckovic</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:UnaVuckovic&amp;diff=73474"/>
		<updated>2011-01-28T05:58:23Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: Replaced content with &amp;quot;Hi I&amp;#039;m Una Vuckovic. I&amp;#039;m in the Faculty of Commerce and I&amp;#039;m a first year student. Below is my essay, I hope you enjoy it!  

email:  v_una@hotmail.com  

 
== The Pythagorean...&amp;quot;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Hi I&#039;m Una Vuckovic. I&#039;m in the Faculty of Commerce and I&#039;m a first year student. Below is my essay, I hope you enjoy it!  &lt;br /&gt;
&lt;br /&gt;
email:  v_una@hotmail.com  &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
== The PythagoreanTheorem == &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://wiki.ubc.ca/User:UnaVuckovic/ThePythagoreanTheorem&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/ThePythagoreanTheorem&amp;diff=73473</id>
		<title>User:UnaVuckovic/ThePythagoreanTheorem</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=User:UnaVuckovic/ThePythagoreanTheorem&amp;diff=73473"/>
		<updated>2011-01-28T05:57:19Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: Created page with &amp;quot;== The Pythagorean Theorem ==    In mathematics, the Pythagorean theorem or Pythagoras&amp;#039; theorem relates to the three sides of any right angle triangle. This theorem states:   In ...&amp;quot;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== The Pythagorean Theorem ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
In mathematics, the Pythagorean theorem or Pythagoras&#039; theorem relates to the three sides of any right angle triangle. This theorem states: &lt;br /&gt;
&lt;br /&gt;
In any right-angled triangle (triangle with one angle of 90°) the area the hypotenuse (side opposite from right angle) is equal to the sum of the areas of the other two sides.[1] This theorem can also be written as an equation using the variables a, b and c, to represent the sides of the triangle. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://www.mathsisfun.com/geometry/images/triangle-abc.gif http://www.coolmath.com/reference/images/dictionary-pythagorean-theorem-1.gif&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;a^2&amp;lt;/math&amp;gt; + &amp;lt;math&amp;gt;b^2&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;c^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
In this equation c represents the length of the hypotenuse while a and b represent the lengths of the other two sides&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Try It! ==&lt;br /&gt;
Lets say you are given a triangle like this: &lt;br /&gt;
&lt;br /&gt;
http://image.wistatutor.com/content/feed/tvcs/x20on20a20triangle.PNG&lt;br /&gt;
&lt;br /&gt;
You need to find the value of &#039;&#039;x&#039;&#039;. So you plug in the Pythagorean Theorem... &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&amp;lt;math&amp;gt;a^2&amp;lt;/math&amp;gt;&#039;&#039; + &#039;&#039;&amp;lt;math&amp;gt;b^2&amp;lt;/math&amp;gt;&#039;&#039; = &#039;&#039;&amp;lt;math&amp;gt;c^2&amp;lt;/math&amp;gt;&#039;&#039; &#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;a&#039;&#039; = 3 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;b&#039;&#039; = 4 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;c&#039;&#039; = &#039;&#039;x&#039;&#039; &lt;br /&gt;
  &#039;&#039;&#039;NOTE&#039;&#039;&#039;: &#039;&#039;a&#039;&#039; and &#039;&#039;b&#039;&#039; and interchangable&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;3^2&amp;lt;/math&amp;gt; + &amp;lt;math&amp;gt;4^2&amp;lt;/math&amp;gt; = &#039;&#039;&amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
9 + 16 = &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
25 = &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \sqrt{25} &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; \sqrt{x} &amp;lt;/math&amp;gt; &lt;br /&gt;
  &#039;&#039;&#039;NOTE&#039;&#039;&#039;: &amp;lt;math&amp;gt; \sqrt{x} &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;x^2&amp;lt;/math&amp;gt; cancel eachother out&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
x = 5&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Pythagoras ==&lt;br /&gt;
&lt;br /&gt;
http://www-history.mcs.st-and.ac.uk/BigPictures/Pythagoras_10.jpeg&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
This geometric theorem is named after the famous and controversial philosopher Pythagoras who was born in 569 BC in Samos, an island off the coast of Greece. Many would consider him a mathematician as he constantly studied even and odd numbers, triangular numbers, perfect numbers etc. Although the theorem was well known to the Babylonians 1000 years prior, Pythagoras was the first person to prove it, hence the theorem being named after him.[2] The whereabouts about his death are unclear but it is known that Pythagoras died in 475 BC.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== References ==&lt;br /&gt;
&lt;br /&gt;
[1] Judith D. Sally, Paul Sally (2007). &amp;quot;Chapter 3: Pythagorean triples&amp;quot;. Roots to research: a vertical development of mathematical problems. American Mathematical Society Bookstore. p. 63. ISBN 0821844032. http://books.google.com/books?id=nHxBw-WlECUC&amp;amp;pg=PA63.&lt;br /&gt;
&lt;br /&gt;
[2] J J O&#039;Connor and E F Robertson (1999). &amp;quot;Pythagoras of Samos&amp;quot;. University of St Andrews, Scotland. &lt;br /&gt;
http://www-history.mcs.st-and.ac.uk/Biographies/Pythagoras.html&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug&amp;diff=73425</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug&amp;diff=73425"/>
		<updated>2011-01-28T04:46:37Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{&lt;br /&gt;
Infobox MATH110 Teams&lt;br /&gt;
| team name = Zug&lt;br /&gt;
| member 1 = Christopher Wong&lt;br /&gt;
| member 2 = Ryan Dalen&lt;br /&gt;
| member 3 = Sifat Hasan&lt;br /&gt;
| member 4 = Una Vuckovic&lt;br /&gt;
}}&lt;br /&gt;
In workshop L.&lt;br /&gt;
&lt;br /&gt;
Our collective goal in this course is to excel.&lt;br /&gt;
&lt;br /&gt;
Key Words!&lt;br /&gt;
&lt;br /&gt;
| Ryan: Success&lt;br /&gt;
&lt;br /&gt;
| Chris: Diligence&lt;br /&gt;
&lt;br /&gt;
| Sifat: Teamwork&lt;br /&gt;
&lt;br /&gt;
| Una: Pass!&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
= Homework 11 =  &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
     http://wiki.ubc.ca/Course:MATH110/003/Teams/Zug/Homework11&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
  &lt;br /&gt;
&lt;br /&gt;
= Homework 12 =      &lt;br /&gt;
    http://wiki.ubc.ca/Course:MATH110/003/Teams/Zug/Homework_12&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_11&amp;diff=71194</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 11</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_11&amp;diff=71194"/>
		<updated>2011-01-20T00:23:06Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: /* - You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost. */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=Homework 11=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://upload.wikimedia.org/wikipedia/commons/thumb/3/36/Zoug-ville-blason.jpg/60px-Zoug-ville-blason.jpg&lt;br /&gt;
&lt;br /&gt;
====&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your team&#039;s Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.&#039;&#039;&#039;====                    &lt;br /&gt;
&lt;br /&gt;
==== * Describe your model====                      &lt;br /&gt;
&lt;br /&gt;
The model we came up with is:&lt;br /&gt;
&lt;br /&gt;
                        &lt;br /&gt;
 C = 7x – 40&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of flags                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;M&#039;&#039; = marginal cost                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The marginal cost is 7, and at the production level of 20 (i.e. x = 20) the cost is 100 (C=100)                      &lt;br /&gt;
&lt;br /&gt;
So, here 100 = 7(20) + m (where m is the constant in the linear equation)                      &lt;br /&gt;
&lt;br /&gt;
Therefore m = -40, thus the equation is                      &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = 7x - 40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
----     &lt;br /&gt;
       &lt;br /&gt;
==== * What does your model predict for a production of 150 items? ====&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
C = 7x - 40    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
plug in 150 for &#039;&#039;x&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Therefore:    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = 7(150) - 40    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = $1010&#039;&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
  Our model predicts that with the production of 150 items, the cost is $1010&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----        &lt;br /&gt;
&lt;br /&gt;
==== * According to your model, what happens to the average cost per item as production levels increase? ====&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
As the number of items rises from 150 to 200 for example, at 150 items&lt;br /&gt;
&lt;br /&gt;
C=1010&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Therefore the average cost 1010/150 = &#039;&#039;&#039;6.733&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
At x =200,&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C=7(200)-40 = 1360&lt;br /&gt;
&lt;br /&gt;
Therefore, the average cost is 1360/200 = &#039;&#039;&#039;6.8&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
   Thus, as the output level increases, the average cost increases&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== * Finally, find some other models (not necessarily linear) for which you get other behaviours such as: ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost remains constant as production increases.&#039;&#039; ====             &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
If the average cost is to remain constant the marginal cost must equal the average cost.                      &lt;br /&gt;
One model that would illustrate this would be:                      &lt;br /&gt;
&lt;br /&gt;
  C(m)= mx                 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost           &lt;br /&gt;
&lt;br /&gt;
                   &lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of flags          &lt;br /&gt;
&lt;br /&gt;
                    &lt;br /&gt;
&#039;&#039;m&#039;&#039; = marginal cost                      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
As the marginal cost of producing a flag is $7 and we are producing 20 flags, therefore                      &lt;br /&gt;
&lt;br /&gt;
C(20) =(20)(7)          &lt;br /&gt;
                    &lt;br /&gt;
&#039;&#039;&#039;C = $140&#039;&#039;&#039;                &lt;br /&gt;
              &lt;br /&gt;
&lt;br /&gt;
And the average cost is calculated by dividing the total cost by the output quantity (m)                      &lt;br /&gt;
&lt;br /&gt;
140/20 = &#039;&#039;&#039;$7&#039;&#039;&#039;                      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we were to increase the output quantity, the average cost should still stay the same                      &lt;br /&gt;
&lt;br /&gt;
C(30) =(30)(7)          &lt;br /&gt;
                &lt;br /&gt;
            &lt;br /&gt;
C = &#039;&#039;&#039;$210&#039;&#039;&#039;           &lt;br /&gt;
                &lt;br /&gt;
           &lt;br /&gt;
210/30 = &#039;&#039;&#039;$7&#039;&#039;&#039;                      &lt;br /&gt;
&lt;br /&gt;
 From this we can see that the average cost is indeed constant.                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----    &lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost diminishes as production increases.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
                  &lt;br /&gt;
&lt;br /&gt;
When the average cost is diminishing as the production increases, if we have a graph of with the y axis being cost and the x axis being number of items produced, the slope of the curve (denoting cost per unit) should be decreasing as output increases.  &lt;br /&gt;
&lt;br /&gt;
So when,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of units&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A good way to show that kind of a curve is by using the function:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = ln(x+m)&#039;&#039;&#039; [&#039;&#039;m&#039;&#039; being any positive constant!!]     &lt;br /&gt;
&lt;br /&gt;
The graph would probably look like this.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Decreasing.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y-intercept being the fixed cost.&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost increases as production increases.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Just like the previous answer, when the average cost is increasing, the slope of the cost/number of items graph should have an increasing slope.&lt;br /&gt;
&lt;br /&gt;
So when,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total Cost&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of units&lt;br /&gt;
&lt;br /&gt;
A good example of a curve like such would be&lt;br /&gt;
&#039;&#039;&#039;C = e^x + m &#039;&#039;&#039; [m being any positive constant]     &lt;br /&gt;
&lt;br /&gt;
The graph would look like this.&lt;br /&gt;
   &lt;br /&gt;
&lt;br /&gt;
[[File:Increasing.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The y-intercept being the fixed cost again.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Marginal cost is the cost of producing the last unit.&lt;br /&gt;
If we assume that the marginal cost is constant through all levels of items produced, then the Cost/number of items would be a linear equation like:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = 20x + 50&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
(20 being the marginal cost here, and 50 being the fixed cost)&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
[[File:Marginal_cost.gif]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Now for a company that is Economies of Scale, any curve with a constant or a decreasing slope to the right of the marginal cost graph, would fulfill the conditions!&lt;br /&gt;
&lt;br /&gt;
----&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:Marginal_cost.gif&amp;diff=71193</id>
		<title>File:Marginal cost.gif</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:Marginal_cost.gif&amp;diff=71193"/>
		<updated>2011-01-20T00:22:51Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_11&amp;diff=71192</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 11</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_11&amp;diff=71192"/>
		<updated>2011-01-20T00:22:29Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: /* - The average cost increases as production increases. */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=Homework 11=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://upload.wikimedia.org/wikipedia/commons/thumb/3/36/Zoug-ville-blason.jpg/60px-Zoug-ville-blason.jpg&lt;br /&gt;
&lt;br /&gt;
====&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your team&#039;s Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.&#039;&#039;&#039;====                    &lt;br /&gt;
&lt;br /&gt;
==== * Describe your model====                      &lt;br /&gt;
&lt;br /&gt;
The model we came up with is:&lt;br /&gt;
&lt;br /&gt;
                        &lt;br /&gt;
 C = 7x – 40&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of flags                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;M&#039;&#039; = marginal cost                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The marginal cost is 7, and at the production level of 20 (i.e. x = 20) the cost is 100 (C=100)                      &lt;br /&gt;
&lt;br /&gt;
So, here 100 = 7(20) + m (where m is the constant in the linear equation)                      &lt;br /&gt;
&lt;br /&gt;
Therefore m = -40, thus the equation is                      &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = 7x - 40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
----     &lt;br /&gt;
       &lt;br /&gt;
==== * What does your model predict for a production of 150 items? ====&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
C = 7x - 40    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
plug in 150 for &#039;&#039;x&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Therefore:    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = 7(150) - 40    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = $1010&#039;&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
  Our model predicts that with the production of 150 items, the cost is $1010&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----        &lt;br /&gt;
&lt;br /&gt;
==== * According to your model, what happens to the average cost per item as production levels increase? ====&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
As the number of items rises from 150 to 200 for example, at 150 items&lt;br /&gt;
&lt;br /&gt;
C=1010&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Therefore the average cost 1010/150 = &#039;&#039;&#039;6.733&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
At x =200,&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C=7(200)-40 = 1360&lt;br /&gt;
&lt;br /&gt;
Therefore, the average cost is 1360/200 = &#039;&#039;&#039;6.8&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
   Thus, as the output level increases, the average cost increases&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== * Finally, find some other models (not necessarily linear) for which you get other behaviours such as: ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost remains constant as production increases.&#039;&#039; ====             &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
If the average cost is to remain constant the marginal cost must equal the average cost.                      &lt;br /&gt;
One model that would illustrate this would be:                      &lt;br /&gt;
&lt;br /&gt;
  C(m)= mx                 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost           &lt;br /&gt;
&lt;br /&gt;
                   &lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of flags          &lt;br /&gt;
&lt;br /&gt;
                    &lt;br /&gt;
&#039;&#039;m&#039;&#039; = marginal cost                      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
As the marginal cost of producing a flag is $7 and we are producing 20 flags, therefore                      &lt;br /&gt;
&lt;br /&gt;
C(20) =(20)(7)          &lt;br /&gt;
                    &lt;br /&gt;
&#039;&#039;&#039;C = $140&#039;&#039;&#039;                &lt;br /&gt;
              &lt;br /&gt;
&lt;br /&gt;
And the average cost is calculated by dividing the total cost by the output quantity (m)                      &lt;br /&gt;
&lt;br /&gt;
140/20 = &#039;&#039;&#039;$7&#039;&#039;&#039;                      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we were to increase the output quantity, the average cost should still stay the same                      &lt;br /&gt;
&lt;br /&gt;
C(30) =(30)(7)          &lt;br /&gt;
                &lt;br /&gt;
            &lt;br /&gt;
C = &#039;&#039;&#039;$210&#039;&#039;&#039;           &lt;br /&gt;
                &lt;br /&gt;
           &lt;br /&gt;
210/30 = &#039;&#039;&#039;$7&#039;&#039;&#039;                      &lt;br /&gt;
&lt;br /&gt;
 From this we can see that the average cost is indeed constant.                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----    &lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost diminishes as production increases.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
                  &lt;br /&gt;
&lt;br /&gt;
When the average cost is diminishing as the production increases, if we have a graph of with the y axis being cost and the x axis being number of items produced, the slope of the curve (denoting cost per unit) should be decreasing as output increases.  &lt;br /&gt;
&lt;br /&gt;
So when,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of units&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A good way to show that kind of a curve is by using the function:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = ln(x+m)&#039;&#039;&#039; [&#039;&#039;m&#039;&#039; being any positive constant!!]     &lt;br /&gt;
&lt;br /&gt;
The graph would probably look like this.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Decreasing.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y-intercept being the fixed cost.&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost increases as production increases.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Just like the previous answer, when the average cost is increasing, the slope of the cost/number of items graph should have an increasing slope.&lt;br /&gt;
&lt;br /&gt;
So when,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total Cost&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of units&lt;br /&gt;
&lt;br /&gt;
A good example of a curve like such would be&lt;br /&gt;
&#039;&#039;&#039;C = e^x + m &#039;&#039;&#039; [m being any positive constant]     &lt;br /&gt;
&lt;br /&gt;
The graph would look like this.&lt;br /&gt;
   &lt;br /&gt;
&lt;br /&gt;
[[File:Increasing.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The y-intercept being the fixed cost again.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Marginal cost is the cost of producing the last unit.&lt;br /&gt;
If we assume that the marginal cost is constant through all levels of items produced, then the Cost/number of items would be a linear equation like:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = 20x + 50&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
(20 being the marginal cost here, and 50 being the fixed cost)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Now for a company that is Economies of Scale, any curve with a constant or a decreasing slope to the right of the marginal cost graph, would fulfill the conditions!&lt;br /&gt;
&lt;br /&gt;
----&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:Increasing.gif&amp;diff=71191</id>
		<title>File:Increasing.gif</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:Increasing.gif&amp;diff=71191"/>
		<updated>2011-01-20T00:21:57Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_11&amp;diff=71190</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 11</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_11&amp;diff=71190"/>
		<updated>2011-01-20T00:21:21Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: /* - The average cost diminishes as production increases. */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=Homework 11=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://upload.wikimedia.org/wikipedia/commons/thumb/3/36/Zoug-ville-blason.jpg/60px-Zoug-ville-blason.jpg&lt;br /&gt;
&lt;br /&gt;
====&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your team&#039;s Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.&#039;&#039;&#039;====                    &lt;br /&gt;
&lt;br /&gt;
==== * Describe your model====                      &lt;br /&gt;
&lt;br /&gt;
The model we came up with is:&lt;br /&gt;
&lt;br /&gt;
                        &lt;br /&gt;
 C = 7x – 40&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of flags                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;M&#039;&#039; = marginal cost                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The marginal cost is 7, and at the production level of 20 (i.e. x = 20) the cost is 100 (C=100)                      &lt;br /&gt;
&lt;br /&gt;
So, here 100 = 7(20) + m (where m is the constant in the linear equation)                      &lt;br /&gt;
&lt;br /&gt;
Therefore m = -40, thus the equation is                      &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = 7x - 40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
----     &lt;br /&gt;
       &lt;br /&gt;
==== * What does your model predict for a production of 150 items? ====&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
C = 7x - 40    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
plug in 150 for &#039;&#039;x&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Therefore:    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = 7(150) - 40    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = $1010&#039;&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
  Our model predicts that with the production of 150 items, the cost is $1010&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----        &lt;br /&gt;
&lt;br /&gt;
==== * According to your model, what happens to the average cost per item as production levels increase? ====&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
As the number of items rises from 150 to 200 for example, at 150 items&lt;br /&gt;
&lt;br /&gt;
C=1010&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Therefore the average cost 1010/150 = &#039;&#039;&#039;6.733&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
At x =200,&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C=7(200)-40 = 1360&lt;br /&gt;
&lt;br /&gt;
Therefore, the average cost is 1360/200 = &#039;&#039;&#039;6.8&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
   Thus, as the output level increases, the average cost increases&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== * Finally, find some other models (not necessarily linear) for which you get other behaviours such as: ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost remains constant as production increases.&#039;&#039; ====             &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
If the average cost is to remain constant the marginal cost must equal the average cost.                      &lt;br /&gt;
One model that would illustrate this would be:                      &lt;br /&gt;
&lt;br /&gt;
  C(m)= mx                 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost           &lt;br /&gt;
&lt;br /&gt;
                   &lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of flags          &lt;br /&gt;
&lt;br /&gt;
                    &lt;br /&gt;
&#039;&#039;m&#039;&#039; = marginal cost                      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
As the marginal cost of producing a flag is $7 and we are producing 20 flags, therefore                      &lt;br /&gt;
&lt;br /&gt;
C(20) =(20)(7)          &lt;br /&gt;
                    &lt;br /&gt;
&#039;&#039;&#039;C = $140&#039;&#039;&#039;                &lt;br /&gt;
              &lt;br /&gt;
&lt;br /&gt;
And the average cost is calculated by dividing the total cost by the output quantity (m)                      &lt;br /&gt;
&lt;br /&gt;
140/20 = &#039;&#039;&#039;$7&#039;&#039;&#039;                      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we were to increase the output quantity, the average cost should still stay the same                      &lt;br /&gt;
&lt;br /&gt;
C(30) =(30)(7)          &lt;br /&gt;
                &lt;br /&gt;
            &lt;br /&gt;
C = &#039;&#039;&#039;$210&#039;&#039;&#039;           &lt;br /&gt;
                &lt;br /&gt;
           &lt;br /&gt;
210/30 = &#039;&#039;&#039;$7&#039;&#039;&#039;                      &lt;br /&gt;
&lt;br /&gt;
 From this we can see that the average cost is indeed constant.                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----    &lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost diminishes as production increases.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
                  &lt;br /&gt;
&lt;br /&gt;
When the average cost is diminishing as the production increases, if we have a graph of with the y axis being cost and the x axis being number of items produced, the slope of the curve (denoting cost per unit) should be decreasing as output increases.  &lt;br /&gt;
&lt;br /&gt;
So when,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of units&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A good way to show that kind of a curve is by using the function:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = ln(x+m)&#039;&#039;&#039; [&#039;&#039;m&#039;&#039; being any positive constant!!]     &lt;br /&gt;
&lt;br /&gt;
The graph would probably look like this.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[File:Decreasing.gif]]&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y-intercept being the fixed cost.&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost increases as production increases.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Just like the previous answer, when the average cost is increasing, the slope of the cost/number of items graph should have an increasing slope.&lt;br /&gt;
&lt;br /&gt;
So when,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total Cost&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of units&lt;br /&gt;
&lt;br /&gt;
A good example of a curve like such would be&lt;br /&gt;
&#039;&#039;&#039;C = e^x + m &#039;&#039;&#039;     [m being any positive constant]&lt;br /&gt;
&lt;br /&gt;
The graph would look like this.&lt;br /&gt;
  &lt;br /&gt;
&lt;br /&gt;
IMAGE HERE!!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y-intercept being the fixed cost again.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Marginal cost is the cost of producing the last unit.&lt;br /&gt;
If we assume that the marginal cost is constant through all levels of items produced, then the Cost/number of items would be a linear equation like:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = 20x + 50&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
(20 being the marginal cost here, and 50 being the fixed cost)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Now for a company that is Economies of Scale, any curve with a constant or a decreasing slope to the right of the marginal cost graph, would fulfill the conditions!&lt;br /&gt;
&lt;br /&gt;
----&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=File:Decreasing.gif&amp;diff=71188</id>
		<title>File:Decreasing.gif</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=File:Decreasing.gif&amp;diff=71188"/>
		<updated>2011-01-20T00:18:46Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_11&amp;diff=71186</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 11</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_11&amp;diff=71186"/>
		<updated>2011-01-20T00:11:44Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: /* - The average cost increases as production increases. */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=Homework 11=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://upload.wikimedia.org/wikipedia/commons/thumb/3/36/Zoug-ville-blason.jpg/60px-Zoug-ville-blason.jpg&lt;br /&gt;
&lt;br /&gt;
====&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your team&#039;s Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.&#039;&#039;&#039;====                    &lt;br /&gt;
&lt;br /&gt;
==== * Describe your model====                      &lt;br /&gt;
&lt;br /&gt;
The model we came up with is:&lt;br /&gt;
&lt;br /&gt;
                        &lt;br /&gt;
 C = 7x – 40&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of flags                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;M&#039;&#039; = marginal cost                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The marginal cost is 7, and at the production level of 20 (i.e. x = 20) the cost is 100 (C=100)                      &lt;br /&gt;
&lt;br /&gt;
So, here 100 = 7(20) + m (where m is the constant in the linear equation)                      &lt;br /&gt;
&lt;br /&gt;
Therefore m = -40, thus the equation is                      &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = 7x - 40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
----     &lt;br /&gt;
       &lt;br /&gt;
==== * What does your model predict for a production of 150 items? ====&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
C = 7x - 40    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
plug in 150 for &#039;&#039;x&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Therefore:    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = 7(150) - 40    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = $1010&#039;&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
  Our model predicts that with the production of 150 items, the cost is $1010&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----        &lt;br /&gt;
&lt;br /&gt;
==== * According to your model, what happens to the average cost per item as production levels increase? ====&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
As the number of items rises from 150 to 200 for example, at 150 items&lt;br /&gt;
&lt;br /&gt;
C=1010&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Therefore the average cost 1010/150 = &#039;&#039;&#039;6.733&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
At x =200,&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C=7(200)-40 = 1360&lt;br /&gt;
&lt;br /&gt;
Therefore, the average cost is 1360/200 = &#039;&#039;&#039;6.8&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
   Thus, as the output level increases, the average cost increases&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== * Finally, find some other models (not necessarily linear) for which you get other behaviours such as: ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost remains constant as production increases.&#039;&#039; ====             &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
If the average cost is to remain constant the marginal cost must equal the average cost.                      &lt;br /&gt;
One model that would illustrate this would be:                      &lt;br /&gt;
&lt;br /&gt;
  C(m)= mx                 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost           &lt;br /&gt;
&lt;br /&gt;
                   &lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of flags          &lt;br /&gt;
&lt;br /&gt;
                    &lt;br /&gt;
&#039;&#039;m&#039;&#039; = marginal cost                      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
As the marginal cost of producing a flag is $7 and we are producing 20 flags, therefore                      &lt;br /&gt;
&lt;br /&gt;
C(20) =(20)(7)          &lt;br /&gt;
                    &lt;br /&gt;
&#039;&#039;&#039;C = $140&#039;&#039;&#039;                &lt;br /&gt;
              &lt;br /&gt;
&lt;br /&gt;
And the average cost is calculated by dividing the total cost by the output quantity (m)                      &lt;br /&gt;
&lt;br /&gt;
140/20 = &#039;&#039;&#039;$7&#039;&#039;&#039;                      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we were to increase the output quantity, the average cost should still stay the same                      &lt;br /&gt;
&lt;br /&gt;
C(30) =(30)(7)          &lt;br /&gt;
                &lt;br /&gt;
            &lt;br /&gt;
C = &#039;&#039;&#039;$210&#039;&#039;&#039;           &lt;br /&gt;
                &lt;br /&gt;
           &lt;br /&gt;
210/30 = &#039;&#039;&#039;$7&#039;&#039;&#039;                      &lt;br /&gt;
&lt;br /&gt;
 From this we can see that the average cost is indeed constant.                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----    &lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost diminishes as production increases.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
                 &lt;br /&gt;
&lt;br /&gt;
When the average cost is diminishing as the production increases, if we have a graph of with the y axis being cost and the x axis being number of items produced, the slope of the curve (denoting cost per unit) should be decreasing as output increases. &lt;br /&gt;
&lt;br /&gt;
So when,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of units&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A good way to show that kind of a curve is by using the function:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = ln(x+m)&#039;&#039;&#039;     [&#039;&#039;m&#039;&#039; being any positive constant!!]&lt;br /&gt;
&lt;br /&gt;
The graph would probably look like this.&lt;br /&gt;
&lt;br /&gt;
Image HERE!!&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y-intercept being the fixed cost.&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost increases as production increases.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Just like the previous answer, when the average cost is increasing, the slope of the cost/number of items graph should have an increasing slope.&lt;br /&gt;
&lt;br /&gt;
So when,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total Cost&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of units&lt;br /&gt;
&lt;br /&gt;
A good example of a curve like such would be&lt;br /&gt;
&#039;&#039;&#039;C = e^x + m &#039;&#039;&#039;     [m being any positive constant]&lt;br /&gt;
&lt;br /&gt;
The graph would look like this.&lt;br /&gt;
  &lt;br /&gt;
&lt;br /&gt;
IMAGE HERE!!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y-intercept being the fixed cost again.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Marginal cost is the cost of producing the last unit.&lt;br /&gt;
If we assume that the marginal cost is constant through all levels of items produced, then the Cost/number of items would be a linear equation like:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = 20x + 50&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
(20 being the marginal cost here, and 50 being the fixed cost)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Now for a company that is Economies of Scale, any curve with a constant or a decreasing slope to the right of the marginal cost graph, would fulfill the conditions!&lt;br /&gt;
&lt;br /&gt;
----&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
	<entry>
		<id>https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_11&amp;diff=71185</id>
		<title>Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework 11</title>
		<link rel="alternate" type="text/html" href="https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/Zug/Homework_11&amp;diff=71185"/>
		<updated>2011-01-20T00:10:12Z</updated>

		<summary type="html">&lt;p&gt;UnaVuckovic: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;=Homework 11=&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
http://upload.wikimedia.org/wikipedia/commons/thumb/3/36/Zoug-ville-blason.jpg/60px-Zoug-ville-blason.jpg&lt;br /&gt;
&lt;br /&gt;
====&#039;&#039;&#039;Write a linear model to predict the cost of producing flags of your team&#039;s Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.&#039;&#039;&#039;====                    &lt;br /&gt;
&lt;br /&gt;
==== * Describe your model====                      &lt;br /&gt;
&lt;br /&gt;
The model we came up with is:&lt;br /&gt;
&lt;br /&gt;
                        &lt;br /&gt;
 C = 7x – 40&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of flags                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;M&#039;&#039; = marginal cost                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
The marginal cost is 7, and at the production level of 20 (i.e. x = 20) the cost is 100 (C=100)                      &lt;br /&gt;
&lt;br /&gt;
So, here 100 = 7(20) + m (where m is the constant in the linear equation)                      &lt;br /&gt;
&lt;br /&gt;
Therefore m = -40, thus the equation is                      &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = 7x - 40&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
----     &lt;br /&gt;
       &lt;br /&gt;
==== * What does your model predict for a production of 150 items? ====&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
C = 7x - 40    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
plug in 150 for &#039;&#039;x&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Therefore:    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C = 7(150) - 40    &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = $1010&#039;&#039;&#039;    &lt;br /&gt;
&lt;br /&gt;
  Our model predicts that with the production of 150 items, the cost is $1010&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----        &lt;br /&gt;
&lt;br /&gt;
==== * According to your model, what happens to the average cost per item as production levels increase? ====&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
As the number of items rises from 150 to 200 for example, at 150 items&lt;br /&gt;
&lt;br /&gt;
C=1010&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
Therefore the average cost 1010/150 = &#039;&#039;&#039;6.733&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
At x =200,&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
C=7(200)-40 = 1360&lt;br /&gt;
&lt;br /&gt;
Therefore, the average cost is 1360/200 = &#039;&#039;&#039;6.8&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
   Thus, as the output level increases, the average cost increases&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== * Finally, find some other models (not necessarily linear) for which you get other behaviours such as: ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost remains constant as production increases.&#039;&#039; ====             &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
If the average cost is to remain constant the marginal cost must equal the average cost.                      &lt;br /&gt;
One model that would illustrate this would be:                      &lt;br /&gt;
&lt;br /&gt;
  C(m)= mx                 &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost           &lt;br /&gt;
&lt;br /&gt;
                   &lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of flags          &lt;br /&gt;
&lt;br /&gt;
                    &lt;br /&gt;
&#039;&#039;m&#039;&#039; = marginal cost                      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
As the marginal cost of producing a flag is $7 and we are producing 20 flags, therefore                      &lt;br /&gt;
&lt;br /&gt;
C(20) =(20)(7)          &lt;br /&gt;
                    &lt;br /&gt;
&#039;&#039;&#039;C = $140&#039;&#039;&#039;                &lt;br /&gt;
              &lt;br /&gt;
&lt;br /&gt;
And the average cost is calculated by dividing the total cost by the output quantity (m)                      &lt;br /&gt;
&lt;br /&gt;
140/20 = &#039;&#039;&#039;$7&#039;&#039;&#039;                      &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
If we were to increase the output quantity, the average cost should still stay the same                      &lt;br /&gt;
&lt;br /&gt;
C(30) =(30)(7)          &lt;br /&gt;
                &lt;br /&gt;
            &lt;br /&gt;
C = &#039;&#039;&#039;$210&#039;&#039;&#039;           &lt;br /&gt;
                &lt;br /&gt;
           &lt;br /&gt;
210/30 = &#039;&#039;&#039;$7&#039;&#039;&#039;                      &lt;br /&gt;
&lt;br /&gt;
 From this we can see that the average cost is indeed constant.                      &lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
----    &lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost diminishes as production increases.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
                 &lt;br /&gt;
&lt;br /&gt;
When the average cost is diminishing as the production increases, if we have a graph of with the y axis being cost and the x axis being number of items produced, the slope of the curve (denoting cost per unit) should be decreasing as output increases. &lt;br /&gt;
&lt;br /&gt;
So when,&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;C&#039;&#039; = total cost&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;x&#039;&#039; = number of units&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
A good way to show that kind of a curve is by using the function:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = ln(x+m)&#039;&#039;&#039;     [&#039;&#039;m&#039;&#039; being any positive constant!!]&lt;br /&gt;
&lt;br /&gt;
The graph would probably look like this.&lt;br /&gt;
&lt;br /&gt;
Image HERE!!&lt;br /&gt;
&lt;br /&gt;
 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The y-intercept being the fixed cost.&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;The average cost increases as production increases.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Just like the previous answer, when the average cost is increasing, the slope of the Cost/number of items graph should have an increasing slope.&lt;br /&gt;
So when,&lt;br /&gt;
C= Total Cost&lt;br /&gt;
X= number of units&lt;br /&gt;
A good example of a curve like such would be&lt;br /&gt;
C=e^x + m    [m being any positive constant]&lt;br /&gt;
The graph would look like this.&lt;br /&gt;
  &lt;br /&gt;
IMAGE HERE!!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Y intercept being the fixed cost again.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
----&lt;br /&gt;
&lt;br /&gt;
==== - &#039;&#039;You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.&#039;&#039; ====&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Marginal cost is the cost of producing the last unit.&lt;br /&gt;
If we assume that the marginal cost is constant through all levels of items produced, then the Cost/number of items would be a linear equation like:&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;C = 20x + 50&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
(20 being the marginal cost here, and 50 being the fixed cost)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Now for a company that is Economies of Scale, any curve with a constant or a decreasing slope to the right of the marginal cost graph, would fulfill the conditions!&lt;br /&gt;
&lt;br /&gt;
----&lt;/div&gt;</summary>
		<author><name>UnaVuckovic</name></author>
	</entry>
</feed>