UBC Wiki - User contributions [en]

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/14 Part3

2011-02-25T04:57:54Z

JustinHsu:

==Homework==

====Midterm Question 1====

For each derivative, compute the actual answer and explain what you didn't do correctly in your midterm (if you did your computation correctly, that's great).
Compute the derivative of each of the following functions.

* <math>f(x)= \displaystyle{3^x}</math>
* <math>g(x)= \ln \left( \frac{3x+3}{x^2+1} \right)</math>
* <math>h(x)= \ln(x \cdot x^2 \cdot x^3 \cdot x^4 \cdot x^5 \cdot x^6) + e^{x \cdot x^2 \cdot x^3 \cdot x^4 \cdot x^5 \cdot x^6}</math>

====Midterm Question 2====

As a woman walks away from a streetlight, her shadow lengthens. Prove that it does so at a rate which depends on her speed but not on her position.
Explain in details how to solve this problem and what you have learned from this problem. What could you have done differently during the midterm to solve the problem? Can you use some of those new ideas to get a good start at the problem in the second part of the homework?
HINT: The question 5 of this week's webwork should help you get started if you're stuck.

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/14 Part3

2011-02-25T04:57:08Z

JustinHsu: Blanked the page

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/14 Part3

2011-02-25T00:46:36Z

JustinHsu:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/14 Part3

2011-02-25T00:44:24Z

JustinHsu:

==Homework==

For each derivative, compute the actual answer and explain what you didn't do correctly in your midterm (if you did your computation correctly, that's great).
Compute the derivative of each of the following functions.

* <math>f(x)= \displaystyle{3^x}</math>
* <math>g(x)= \ln \left( \frac{3x+3}{x^2+1} \right)</math>
* <math>h(x)= \ln(x \cdot x^2 \cdot x^3 \cdot x^4 \cdot x^5 \cdot x^6) + e^{x \cdot x^2 \cdot x^3 \cdot x^4 \cdot x^5 \cdot x^6}</math>

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/14 Part3

2011-02-25T00:43:58Z

JustinHsu:

==Homework==

For each derivative, compute the actual answer and explain what you didn't do correctly in your midterm (if you did your computation correctly, that's great).
Compute the derivative of each of the following functions.

<math>f(x)= \displaystyle{3^x}</math>
* <math>g(x)= \ln \left( \frac{3x+3}{x^2+1} \right)</math>
* <math>h(x)= \ln(x \cdot x^2 \cdot x^3 \cdot x^4 \cdot x^5 \cdot x^6) + e^{x \cdot x^2 \cdot x^3 \cdot x^4 \cdot x^5 \cdot x^6}</math>

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/14 Part3

2011-02-25T00:40:54Z

JustinHsu:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/14 Part3

2011-02-25T00:40:12Z

JustinHsu:

For each derivative, compute the actual answer and explain what you didn't do correctly in your midterm (if you did your computation correctly, that's great).
Compute the derivative of each of the following functions.

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/14 Part3

2011-02-24T22:52:04Z

JustinHsu: Created page with "[Stuff Goes Here]"

[Stuff Goes Here]

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-04T01:14:40Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

Firstly team Uri has decided to explain the pH scale an how the concepts work in terms of logarithmic scale.

'''What is the pH scale?'''

It is defined as:

"A scale from 0 to 14 reflecting the concentration of hydrogen ions in solution; the lower numbers denote acidic conditions and the upper numbers denote basic, or alkaline, conditions."

Or in mathematical terms, the negative logarithm value of the hydrogen ion concentration in the solution.

As shown below are examples on how the pH scale measures different solutions:

 
http://img810.imageshack.us/img810/4394/phscaleenvironmentca.gif


'''How Does it Work?'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+].To find pH in a solution the following logarithmic formula is used:A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH

= -log [H3O+]

= -log(1.0x10^-7)

= -(-7.00) = 7.00

In general at 25 degrees Celsius

pH < 7 is acidic

pH >7 is basic

pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration.

'''Example'''

A lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Another example is using the pH formula to determine whether how acidic or alkaline the solution is. Say you are giving the information that the solution has a '''H+''' concentration of 0.001 or ( 1*10^-4 Moles).

We would have to plug it into our pH equation:

pH= -log [H+]

pH= -log [0.001]

pH= -log [10^-3]

pH= 3

Its pH would be 3 and the solution would be considered an acidic solution.

Below is a table showing the different concentrations of Hydromium and pH levels:
 
http://img98.imageshack.us/img98/78/phscale1.png


'''Reference'''

Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Course:MATH110/Archive/2010-2011/003/Teams/Uri

2011-02-04T00:52:39Z

JustinHsu:

{{
Infobox MATH110 Teams
| team name = Uri
| member 1 = Allie Miller
| member 2 = Justin Hsu
| member 3 = Matt Vetter
| member 4 = Victoria Wall
}}

== Discussion In-Class==

Magical.

==Workshop==

In workshop M.

==Homework==

<dpl>
titlematch={{PAGENAME}}/%
replaceintitle=$MATH110/003/Teams/$
namespace={{NAMESPACE}}
shownamespace=false
</dpl>

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T07:18:11Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

Firstly team Uri has decided to explain the pH scale an how the concepts work in terms of logarithmic scale.

'''What is the pH scale?'''

It is defined as:

"A scale from 0 to 14 reflecting the concentration of hydrogen ions in solution; the lower numbers denote acidic conditions and the upper numbers denote basic, or alkaline, conditions."

Or in mathematical terms, the negative logarithm value of the hydrogen ion concentration in the solution.

As shown below are examples on how the pH scale measures different solutions:

 
http://img810.imageshack.us/img810/4394/phscaleenvironmentca.gif


'''How Does it Work?'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+].To find pH in a solution the following logarithmic formula is used:A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH

= -log [H3O+]

= -log(1.0x10^-7)

= -(-7.00) = 7.00

In general at 25 degrees Celsius

pH < 7 is acidic

pH >7 is basic

pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration.

'''Example'''

lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Below is a table showing the different concentrations of Hydromium and pH levels:
 
http://img98.imageshack.us/img98/78/phscale1.png


'''Reference'''

Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T07:17:54Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

Firstly team Uri has decided to explain the pH scale an how the concepts work in terms of logarithmic scale.

'''What is the pH scale?'''

It is defined as:

"A scale from 0 to 14 reflecting the concentration of hydrogen ions in solution; the lower numbers denote acidic conditions and the upper numbers denote basic, or alkaline, conditions."

Or in mathematical terms, the negative logarithm value of the hydrogen ion concentration in the solution.

As shown below are examples on how the pH scale measures different solutions:

 
http://img810.imageshack.us/img810/4394/phscaleenvironmentca.gif


'''How Does it Work?'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+].To find pH in a solution the following logarithmic formula is used:A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH

= -log [H3O+]

= -log(1.0x10^-7)

= -(-7.00) = 7.00

In general at 25 degrees Celsius

pH < 7 is acidic

pH >7 is basic

pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration.

'''Example'''

lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Below is a table showing the different concentrations of Hydromium and pH levels:
 
http://img98.imageshack.us/img98/78/phscale1.png


'''Reference'''

Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T07:17:36Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

Firstly team Uri has decided to explain the pH scale an how the concepts work in terms of logarithmic scale.

'''What is the pH scale?'''

It is defined as:

"A scale from 0 to 14 reflecting the concentration of hydrogen ions in solution; the lower numbers denote acidic conditions and the upper numbers denote basic, or alkaline, conditions."

Or in mathematical terms, the negative logarithm value of the hydrogen ion concentration in the solution.

As shown below are examples on how the pH scale measures different solutions:

 
http://img810.imageshack.us/img810/4394/phscaleenvironmentca.gif


'''How Does it Work?'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+].To find pH in a solution the following logarithmic formula is used:A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH

= -log [H3O+]

= -log(1.0x10^-7)

= -(-7.00) = 7.00

In general at 25 degrees Celsius

pH < 7 is acidic

pH >7 is basic

pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration.

'''Example'''

lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Below is a table showing the different concentrations of Hydromium and pH levels:

 
http://img98.imageshack.us/img98/78/phscale1.png


'''Reference'''

Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T07:17:04Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

Firstly team Uri has decided to explain the pH scale an how the concepts work in terms of logarithmic scale.

'''What is the pH scale?'''

It is defined as:

"A scale from 0 to 14 reflecting the concentration of hydrogen ions in solution; the lower numbers denote acidic conditions and the upper numbers denote basic, or alkaline, conditions."

Or in mathematical terms, the negative logarithm value of the hydrogen ion concentration in the solution.

As shown below are examples on how the pH scale measures different solutions:

 
http://img810.imageshack.us/img810/4394/phscaleenvironmentca.gif


'''How Does it Work?'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+].To find pH in a solution the following logarithmic formula is used:A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH

= -log [H3O+]

= -log(1.0x10^-7)

= -(-7.00) = 7.00

In general at 25 degrees Celsius

pH < 7 is acidic

pH >7 is basic

pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration.

'''Example'''

lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

 
http://img98.imageshack.us/img98/78/phscale1.png


'''Reference'''

Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T07:15:31Z

JustinHsu:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T07:14:47Z

JustinHsu:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T07:11:15Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

Firstly team Uri has decided to explain the pH scale an how the concepts work in terms of logarithmic scale.

'''What is the pH scale?'''

It is defined as:

"A scale from 0 to 14 reflecting the concentration of hydrogen ions in solution; the lower numbers denote acidic conditions and the upper numbers denote basic, or alkaline, conditions."

Or in mathematical terms, the negative logarithm value of the hydrogen ion concentration in the solution.

As shown below are examples on how the pH scale measures different solutions:

 
http://img810.imageshack.us/img810/4394/phscaleenvironmentca.gif


'''How Does it Work?'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+].
To find pH in a solution the following logarithmic formula is used:
A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH = -log [H3O+]
= -log(1.0x10^-7)
= -(-7.00)
= 7.00

In general at 25 degrees Celsius
- pH < 7 is acidic
- pH >7 is basic
- pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration

Example: lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Below is a table showing the different concentrations of Hydromium and pH levels:

 
http://img98.imageshack.us/img98/78/phscale1.png


'''Reference'''

Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T07:10:44Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

Firstly team Uri has decided to explain the pH scale an how the concepts work in terms of logarithmic scale.

'''What is the pH scale?'''

It is defined as:

"A scale from 0 to 14 reflecting the concentration of hydrogen ions in solution; the lower numbers denote acidic conditions and the upper numbers denote basic, or alkaline, conditions."

Or in mathematical terms, the negative logarithm value of the hydrogen ion concentration in the solution.

As shown below are examples on how the pH scale measures different solutions:

 
http://img810.imageshack.us/img810/4394/phscaleenvironmentca.gif


'''How Does it Work?'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+].
To find pH in a solution the following logarithmic formula is used:
A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH = -log [H3O+]
= -log(1.0x10^-7)
= -(-7.00)
= 7.00

In general at 25 degrees Celsius
- pH < 7 is acidic
- pH >7 is basic
- pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration

Example: lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Below is a table showing the different concentrations of Hydromium and pH levels:

 
http://img98.imageshack.us/img98/78/phscale1.png


'''Reference'''

Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T07:08:02Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

Firstly team Uri has decided to explain the pH scale an how the concepts work in terms of logarithmic scale.

'''What is the pH scale?'''

It is defined as:

"A scale from 0 to 14 reflecting the concentration of hydrogen ions in solution; the lower numbers denote acidic conditions and the upper numbers denote basic, or alkaline, conditions."

Or in mathematical terms, the negative logarithm value of the hydrogen ion concentration in the solution.

As shown below are examples on how the pH scale measures different solutions:

 
http://img810.imageshack.us/img810/4394/phscaleenvironmentca.gif


'''How Does it Work?'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+].
To find pH in a solution the following logarithmic formula is used:
A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH = -log [H3O+]
= -log(1.0x10^-7)
= -(-7.00)
= 7.00

In general at 25 degrees Celsius
- pH < 7 is acidic
- pH >7 is basic
- pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration

Example: lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Below is a table showing the different concentrations of Hydromium and pH levels:

 
http://img97.imageshack.us/img97/1865/phscale.jpg


'''Reference'''

Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T07:00:05Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

Firstly team Uri has decided to explain the pH scale an how the concepts work in terms of logarithmic scale.

'''What is the pH scale?'''

It is defined as:

"A scale from 0 to 14 reflecting the concentration of hydrogen ions in solution; the lower numbers denote acidic conditions and the upper numbers denote basic, or alkaline, conditions."

Or in mathematical terms, the negative logarithm value of the hydrogen ion concentration in the solution.

As shown below are examples on how the pH scale measures different solutions:

 
http://img810.imageshack.us/img810/4394/phscaleenvironmentca.gif


'''How Does it Work?'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+].
To find pH in a solution the following logarithmic formula is used:
A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH = -log [H3O+]
= -log(1.0x10^-7)
= -(-7.00)
= 7.00

In general at 25 degrees Celsius
- pH < 7 is acidic
- pH >7 is basic
- pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration

Example: lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

'''Reference'''

Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T06:55:23Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

Firstly team Uri has decided to explain the pH scale an how the concepts work in terms of logarithmic scale.

'''What is the pH scale?'''

It is defined as:

"A scale from 0 to 14 reflecting the concentration of hydrogen ions in solution; the lower numbers denote acidic conditions and the upper numbers denote basic, or alkaline, conditions."

Or in mathematical terms, the negative logarithm value of the hydrogen ion concentration in the solution.

As shown below are examples on how the pH scale measures different solutions:

 
http://img810.imageshack.us/img810/4394/phscaleenvironmentca.gif


'''How Does it Work?'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+]
To find pH in a solution the following logarithmic formula is used:
A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH = -log [H3O+]
= -log(1.0x10^-7)
= -(-7.00)
= 7.00

In general at 25 degrees Celsius
- pH < 7 is acidic
- pH >7 is basic
- pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration

Example: lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Reference: Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T06:54:56Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

Firstly team Uri has decided to explain the pH scale an how the concepts work in terms of logarithmic scale.

'''What is the pH scale?'''

It is defined as:

"A scale from 0 to 14 reflecting the concentration of hydrogen ions in solution; the lower numbers denote acidic conditions and the upper numbers denote basic, or alkaline, conditions."

Or in mathematical terms, the negative logarithm value of the hydrogen ion concentration in the solution.

As shown below are examples on how the pH scale measures different solutions:

 
http://img810.imageshack.us/img810/4394/phscaleenvironmentca.gif


'''How Does it Work?'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+]
To find pH in a solution the following logarithmic formula is used:
A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH = -log [H3O+]
= -log(1.0x10^-7)
= -(-7.00)
= 7.00

In general at 25 degrees Celsius
- pH < 7 is acidic
- pH >7 is basic
- pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration

Example: lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Reference: Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

http://www.elmhurst.edu/~chm/vchembook/184ph.html

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T06:48:07Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

Firstly team Uri has decided to explain the pH scale an how the concepts work in terms of logarithmic scale.

'''What is the pH scale?'''

It is defined as:

"A scale from 0 to 14 reflecting the concentration of hydrogen ions in solution; the lower numbers denote acidic conditions and the upper numbers denote basic, or alkaline, conditions."

Or in other words its a scale that measures how strong an acid/base the solution is given the hydronium ionic bonds in the solution.

As shown below are examples on how the pH scale measures different solutions:

 
http://img810.imageshack.us/img810/4394/phscaleenvironmentca.gif


'''How Does it Work?'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+]
To find pH in a solution the following logarithmic formula is used:
A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH = -log [H3O+]
= -log(1.0x10^-7)
= -(-7.00)
= 7.00

In general at 25 degrees Celsius
- pH < 7 is acidic
- pH >7 is basic
- pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration

Example: lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Reference: Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T06:43:55Z

JustinHsu:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T06:36:41Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale.'''

'''pH'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+]
To find pH in a solution the following logarithmic formula is used:
A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH = -log [H3O+]
= -log(1.0x10^-7)
= -(-7.00)
= 7.00

In general at 25 degrees Celsius
- pH < 7 is acidic
- pH >7 is basic
- pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration

Example: lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Reference: Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T06:36:27Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a logarithmic scale'''

'''pH'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+]
To find pH in a solution the following logarithmic formula is used:
A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH = -log [H3O+]
= -log(1.0x10^-7)
= -(-7.00)
= 7.00

In general at 25 degrees Celsius
- pH < 7 is acidic
- pH >7 is basic
- pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration

Example: lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Reference: Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-03T06:36:08Z

JustinHsu:

==Homework==

'''Explain in your own words what it means that these concepts work on a
logarithmic scale'''

'''pH'''

pH works on a logarithmic scale because in chemistry pH is defined as pH=-log[H3O+]
To find pH in a solution the following logarithmic formula is used:
A solution with [H3O+]= 1.0 x 10^-7 M (neutral) has pH of...

pH = -log [H3O+]
= -log(1.0x10^-7)
= -(-7.00)
= 7.00

In general at 25 degrees Celsius
- pH < 7 is acidic
- pH >7 is basic
- pH = 7 neutral

Since pH scale is a logarithmic scale a change of one pH unit corresponds to a 10 fold change in [H3O+]concentration

Example: lime has a pH of 2.0 this is 10 times more acidic than plums with a pH of 3.0

Reference: Principles of Chemistry: A Molecular Approach by Nivaldo J. Tro

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/13 Part3

2011-02-02T06:38:29Z

JustinHsu: Created page with "[Stuff Goes Here]"

[Stuff Goes Here]

Thread:Course talk:MATH110/003/Teams/Uri/Homework/12 Part3/Comments/reply

2011-02-02T06:34:35Z

JustinHsu: Reply to Comments

Sorry about that. I corrected the mistake by adding to our model.

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-02-02T06:33:22Z

JustinHsu:

==Homework==

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:

* Change the height of the horizontal asymptote on the right, we'll denote it by K.

* Change the y-intercept to any number between 0 and K

* 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).

Original function:
 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


Which is shown as:

 
http://img820.imageshack.us/img820/6396/msp631419e4ca1c0iaea3b2.gif


'''Change the height of the horizontal asymptote on the right, we'll denote it by
K.'''

P(t)= k /[1+e^-x]

P(t)= 5 /[1+e^-x]

P(t)= 10 /[1+e^-x]

As we can see the horizontal asymptote has been shifted higher.

Which is shown as:

The larger the number for K the horizontal asymptote increases.

''Lower Horizontal asymptote: P(t)= 5 /1+e^-x]''
 
http://img838.imageshack.us/img838/1862/92613195.gif


''High Horizontal asymptote: P(t)= 10 /[1+e^-x]''

 
http://img254.imageshack.us/img254/3043/msp1123419e4c80fdb40h5c.gif


'''Change the y-intercept to any number between 0 and K'''

P(t)= 1 /[(1+ e^(-x+y))]

P(t)= 1 /[(1+(e^(-t+5))]

P(t)= 1/(1+(e^(-t+1))]

Which is shown as:

By changing the value of Y into a larger number it can be observed that the Y-intercept becomes less and less.

''This one has a lower Y-Intercept: P(t)= 1 /[(1+(e^(-t+5))]''
 
http://img406.imageshack.us/img406/3913/13941006.gif


''This one has a higher Y-Interecept: P(t)= 1/(1+(e^(-t+1))]''
 
http://img146.imageshack.us/img146/5422/70293261.gif


'''Applying the Model'''

An eco-system which is supporting a population of birds can theoretically can hold up to 300 birds, which is its carrying capacity. Initially the population of birds is 50 with a growth rate of .6 each year. Using the model:

 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


We can input the data given to form:

P(t)= C/1+Ie^(-rt)

C = Carrying capacity of the eco-system

I = Initial starting population

R = Rate of growth

P(t)= 300/1+50e^(-.6t)

For the predictions how long does it take for the population of bird to reach 80% of its carrying capacity?

To find this we use the function: P(t)= 300/1+50e^(-.6t)

P= Population

t= Time it takes for the population to grow

300*.80= 540

So we are looking for how long its take for the population to grow to 540 Birds. To find that we plug 540 into P:

540= 300/1+50e^(-.6t)

Then Solve for T

 
http://img831.imageshack.us/img831/4021/16539198.gif

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-02-02T06:32:59Z

JustinHsu:

User:JustinHsu

2011-02-02T02:56:01Z

JustinHsu:

==Homework 12==

Calculus in Economics

Hello again! As we all know in to days world calculus is used in almost every field of work. Most people believe that calculus is only limited to areas in science and mathematics but calculus can be found in almost any professional field of work. In this short essay I will be discussing the use of calculus in the field of economics and how to apply it.The basic goals of micro economics include economists analyzing, creating models, and predicting elasticities of goods and services. Another key area of micro economics is firms trying to maximise profits for its company which means producing the most amount of goods at the lowest price. Calculus is practically required if anyone wants to pursue a career in economics.

One important accept of economics is that it uses functions as a tool for ways to examining relationships between dependent or independent variables. One Example in the applications of these functions in economics functions is determining the relationship between income and education and using it to predict standard to living. If the average standard of living rises as income and education increases it can be said that there is a positive relationship between these variables and standard of living in a function of income and education . Economists with the use of derivatives are able to measure the average change in the standard of living relative to increases in income and education.

 
http://img607.imageshack.us/img607/1779/oligopoly.jpg


When economists are examining profit maximization for firms economists would have to calculate the how much the firm should produce in order to maximize profits. When firms are maximizing profits the firms will have to equate marginal revenue to marginal cost . The word “marginal” means the use of derivatives in the calculations. If the firms wants to maximize profits it would have to take its deviate to make marginal profit equal to marginal revenue minus marginal cost. Profits will be maximized when marginal revenue equals marginal costs.

 
http://img407.imageshack.us/img407/677/typesofpriceelasticity.jpg


Another way Calculus is used in microeconomics is calculating the elasticity of goods or services. In economics elasticity is defined as the measurement of the responsiveness of supply or demand to changes in price. When calculating elasticity ecnoomists have to use some calculus because economists are dealing with marginal changes in a good.Using this formula we can calculate any elasticity: (Percentage change in X) / (Percentage change in Y) = (dX / dY) * (X/Y). dZ/dY is a derivative of Z with respect to Y. With this basic formula we can calculate different types of elasticity such as price elasticity of demand, income elasticity of demand, cross price elasticity of demand, and price elasticity of supply. These examples are just some of the many applications of calculus in economics.

By writing this essay I hope I have made more clear a way in which calculus is used and the various applications that require some form of calculus to compare and analyze large amounts of data in order to make accurate and clear predictions in various scenarios.

References

http://economics.about.com/cs/micfrohelp/a/calculus_elast.htm

http://en.wikipedia.org/wiki/Calculus#Applications

http://www.answerbag.com/q_view/1964478

http://www.ehow.com/about_6662178_calculus-used-economics_.html

==Homework 1==

Hello,

My name is Justin Hsu. I am a first year student here at UBC and hoping to major in economics.

The Pythagorean Theorem:

To start off the Pythagorean Theorem is in a right triangle or able to take the 2 sides of the triangle and find the hypotenuse. The equation is a^2 + b^2 = c^2 C represents the hypotenuse and A and B represent the other sides of the right triangle. This equation will be very useful when your trying to solve basic algebraic or geometric problems. For example, you are give the side 5 and 12 on a right triangle and are told to find the hypotenuse. So how we solve is we plug 5 into A and 12 into B this should look like 5^2 + 12^2 = C^2. Then the next step is too simplify so it should look more like 25 + 144 = C^2. The next step is too add 25 and 144 together so it should look like 169 = C^2. The last step is to square root both sides and you should get C = 13. Some real life application is finding the length of a ladder if your trying to climb to the top of a building. First you must know the height of the building and how far away the ladder is from the building so you can find the hypotenuse which is the ladder. There are many different real life applications for the Pythagorean Theorem but this is mostly used for solving algebraic or geometric problems.

User:JustinHsu

2011-02-02T02:55:19Z

JustinHsu:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T06:44:03Z

JustinHsu:

==Homework==

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:

* Change the height of the horizontal asymptote on the right, we'll denote it by K.

* Change the y-intercept to any number between 0 and K

* 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).

Original function:
 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


Which is shown as:

 
http://img820.imageshack.us/img820/6396/msp631419e4ca1c0iaea3b2.gif


'''Change the height of the horizontal asymptote on the right, we'll denote it by
K.'''

P(t)= k /[1+e^-x]

P(t)= 5 /[1+e^-x]

P(t)= 10 /[1+e^-x]

As we can see the horizontal asymptote has been shifted higher.

Which is shown as:

The larger the number for K the horizontal asymptote increases.

''Lower Horizontal asymptote: P(t)= 5 /1+e^-x]''
 
http://img838.imageshack.us/img838/1862/92613195.gif


''High Horizontal asymptote: P(t)= 10 /[1+e^-x]''

 
http://img254.imageshack.us/img254/3043/msp1123419e4c80fdb40h5c.gif


'''Change the y-intercept to any number between 0 and K'''

P(t)= 1 /[(1+ e^(-x+y))]

P(t)= 1 /[(1+(e^(-t+5))]

P(t)= 1/(1+(e^(-t+1))]

Which is shown as:

By changing the value of Y into a larger number it can be observed that the Y-intercept becomes less and less.

''This one has a lower Y-Intercept: P(t)= 1 /[(1+(e^(-t+5))]''
 
http://img406.imageshack.us/img406/3913/13941006.gif


''This one has a higher Y-Interecept: P(t)= 1/(1+(e^(-t+1))]''
 
http://img146.imageshack.us/img146/5422/70293261.gif


'''Applying the Model'''

An eco-system which is supporting a population of birds can theoretically can hold up to 300 birds, which is its carrying capacity. Initially the population of birds is 50 with a growth rate of .6 each year. Using the model:

 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


We can input the data given to form:

P(t)= C/1+Ie^(-rt)

C = Carrying capacity of the eco-system

I = Initial starting population

R = Rate of growth

P(t)= 300/1+50e^(-.6t)

 
http://img831.imageshack.us/img831/4021/16539198.gif


At year four we can see that the population has grow almost to its carrying capacity.

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T06:39:43Z

JustinHsu:

==Homework==

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:

* Change the height of the horizontal asymptote on the right, we'll denote it by K.

* Change the y-intercept to any number between 0 and K

* 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).

Original function:
 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


Which is shown as:

 
http://img820.imageshack.us/img820/6396/msp631419e4ca1c0iaea3b2.gif


'''Change the height of the horizontal asymptote on the right, we'll denote it by
K.'''

P(t)= k /[1+e^-x]

P(t)= 5 /[1+e^-x]

P(t)= 10 /[1+e^-x]

As we can see the horizontal asymptote has been shifted higher.

Which is shown as:

The larger the number for K the horizontal asymptote increases.

''Lower Horizontal asymptote: P(t)= 5 /1+e^-x]''
 
http://img838.imageshack.us/img838/1862/92613195.gif


''High Horizontal asymptote: P(t)= 10 /[1+e^-x]''

 
http://img254.imageshack.us/img254/3043/msp1123419e4c80fdb40h5c.gif


'''Change the y-intercept to any number between 0 and K'''

P(t)= 1 /[(1+ e^(-x+y))]

P(t)= 1 /[(1+(e^(-t+5))]

P(t)= 1/(1+(e^(-t+1))]

Which is shown as:

By changing the value of Y into a larger number it can be observed that the Y-intercept becomes less and less.

''This one has a lower Y-Intercept: P(t)= 1 /[(1+(e^(-t+5))]''
 
http://img406.imageshack.us/img406/3913/13941006.gif


''This one has a higher Y-Interecept: P(t)= 1/(1+(e^(-t+1))]''
 
http://img146.imageshack.us/img146/5422/70293261.gif


'''Applying the Model'''

An eco-system which is supporting a population of birds can theoretically can hold up to 300 birds, which is its carrying capacity. Initially the population of birds is 50. Using the model:

 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


We can input the data given to form:

P(t)= C/1+Ie^(-rt)

C = Carrying capacity of the eco-system

I = Initial starting population

R = Rate of growth

P(t)= 300/1+50e^(-.6t)

 
http://img831.imageshack.us/img831/4021/16539198.gif


At year four we can see that the population has grow almost to its carrying capacity.

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T06:38:40Z

JustinHsu:

==Homework==

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:

* Change the height of the horizontal asymptote on the right, we'll denote it by K.

* Change the y-intercept to any number between 0 and K

* 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).

Original function:
 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


Which is shown as:

 
http://img820.imageshack.us/img820/6396/msp631419e4ca1c0iaea3b2.gif


'''Change the height of the horizontal asymptote on the right, we'll denote it by
K.'''

P(t)= k /[1+e^-x]

P(t)= 5 /[1+e^-x]

P(t)= 10 /[1+e^-x]

As we can see the horizontal asymptote has been shifted higher.

Which is shown as:

The larger the number for K the horizontal asymptote increases.

''Lower Horizontal asymptote: P(t)= 5 /1+e^-x]''
 
http://img838.imageshack.us/img838/1862/92613195.gif


''High Horizontal asymptote: P(t)= 10 /[1+e^-x]''

 
http://img254.imageshack.us/img254/3043/msp1123419e4c80fdb40h5c.gif


'''Change the y-intercept to any number between 0 and K'''

P(t)= 1 /[(1+ e^(-x+y))]

P(t)= 1 /[(1+(e^(-t+5))]

P(t)= 1/(1+(e^(-t+1))]

Which is shown as:

By changing the value of Y into a larger number it can be observed that the Y-intercept becomes less and less.

''This one has a lower Y-Intercept: P(t)= 1 /[(1+(e^(-t+5))]''
 
http://img406.imageshack.us/img406/3913/13941006.gif


''This one has a higher Y-Interecept: P(t)= 1/(1+(e^(-t+1))]''
 
http://img146.imageshack.us/img146/5422/70293261.gif


'''Applying the Model'''

An eco-system which is supporting a population of birds can theoretically can hold up to 300 birds, which is its carrying capacity. Initially the population of birds is 50. Using the model:

 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


We can input the data given to form:

P(t)= C/1+Ie^(-rt)

C = Carrying capacity of the eco-system

I = Initial starting population

R = Rate of growth

P(t)= 300/1+50e^(-.6t)

 
http://img831.imageshack.us/img831/4021/16539198.gif

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T06:38:07Z

JustinHsu:

==Homework==

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:

* Change the height of the horizontal asymptote on the right, we'll denote it by K.

* Change the y-intercept to any number between 0 and K

* 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).

Original function:
 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


Which is shown as:

 
http://img820.imageshack.us/img820/6396/msp631419e4ca1c0iaea3b2.gif


'''Change the height of the horizontal asymptote on the right, we'll denote it by
K.'''

P(t)= k /1+e^-x]

P(t)= 5 /[1+e^-x]

P(t)= 10 /[1+e^-x]

As we can see the horizontal asymptote has been shifted higher.

Which is shown as:

The larger the number for K the horizontal asymptote increases.

''Lower Horizontal asymptote: P(t)= 5 /1+e^-x]''
 
http://img838.imageshack.us/img838/1862/92613195.gif


''High Horizontal asymptote: P(t)= 10 /[1+e^-x]''

 
http://img254.imageshack.us/img254/3043/msp1123419e4c80fdb40h5c.gif


'''Change the y-intercept to any number between 0 and K'''

P(t)= 1 /[(1+ e^(-x+y))]

P(t)= 1 /[(1+(e^(-t+5))]

P(t)= 1/(1+(e^(-t+1))]

Which is shown as:

By changing the value of Y into a larger number it can be observed that the Y-intercept becomes less and less.

''This one has a lower Y-Intercept: P(t)= 1 /[(1+(e^(-t+5))]''
 
http://img406.imageshack.us/img406/3913/13941006.gif


''This one has a higher Y-Interecept: P(t)= 1/(1+(e^(-t+1))]''
 
http://img146.imageshack.us/img146/5422/70293261.gif


'''Applying the Model'''

An eco-system which is supporting a population of birds can theoretically can hold up to 300 birds, which is its carrying capacity. Initially the population of birds is 50. Using the model:

 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


We can input the data given to form:

P(t)= C/1+Ie^(-rt)

C = Carrying capacity of the eco-system

I = Initial starting population

R = Rate of growth

P(t)= 300/1+50e^(-.6t)

 
http://img831.imageshack.us/img831/4021/16539198.gif

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T06:36:14Z

JustinHsu:

==Homework==

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:

* Change the height of the horizontal asymptote on the right, we'll denote it by K.

* Change the y-intercept to any number between 0 and K

* 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).

Original function:
 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


Which is shown as:

 
http://img820.imageshack.us/img820/6396/msp631419e4ca1c0iaea3b2.gif


'''Change the height of the horizontal asymptote on the right, we'll denote it by
K.'''

P(t)= 1 /[(1/k)+e^-x]

P(t)= 1 /[(1/5)+e^-x]

P(t)= 1 /[(1/2)+e^-x]

As we can see the horizontal asymptote has been shifted higher.

Which is shown as:

The smaller the number for K the horizontal asymptote increases.

''Lower Horizontal asymptote: P(t)= 5 /1+e^-x]''
 
http://img838.imageshack.us/img838/1862/92613195.gif


''High Horizontal asymptote: P(t)= 10 /[1+e^-x]''

 
http://img254.imageshack.us/img254/3043/msp1123419e4c80fdb40h5c.gif


'''Change the y-intercept to any number between 0 and K'''

P(t)= 1 /[(1+ e^(-x+y))]

P(t)= 1 /[(1+(e^(-t+5))]

P(t)= 1/(1+(e^(-t+1))]

Which is shown as:

By changing the value of Y into a larger number it can be observed that the Y-intercept becomes less and less.

''This one has a lower Y-Intercept: P(t)= 1 /[(1+(e^(-t+5))]''
 
http://img406.imageshack.us/img406/3913/13941006.gif


''This one has a higher Y-Interecept: P(t)= 1/(1+(e^(-t+1))]''
 
http://img146.imageshack.us/img146/5422/70293261.gif


'''Applying the Model'''

An eco-system which is supporting a population of birds can theoretically can hold up to 300 birds, which is its carrying capacity. Initially the population of birds is 50. Using the model:

 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


We can input the data given to form:

P(t)= C/1+Ie^(-rt)

C = Carrying capacity of the eco-system

I = Initial starting population

R = Rate of growth

P(t)= 300/1+50e^(-.6t)

 
http://img831.imageshack.us/img831/4021/16539198.gif

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T06:25:29Z

JustinHsu:

==Homework==

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:

* Change the height of the horizontal asymptote on the right, we'll denote it by K.

* Change the y-intercept to any number between 0 and K

* 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).

Original function:
 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


Which is shown as:

 
http://img820.imageshack.us/img820/6396/msp631419e4ca1c0iaea3b2.gif


'''Change the height of the horizontal asymptote on the right, we'll denote it by
K.'''

P(t)= 1 /[(1/k)+e^-x]

P(t)= 1 /[(1/5)+e^-x]

P(t)= 1 /[(1/2)+e^-x]

As we can see the horizontal asymptote has been shifted higher.

Which is shown as:

The smaller the number for K the horizontal asymptote increases.

''Lower Horizontal asymptote: P(t)= 1 /[(1/5)+e^-x]''
 
http://img248.imageshack.us/img248/9728/msp428519e4d4e1432d9h8i.gif


''High Horizontal asymptote: P(t)= 1 /[(1/2)+e^-x]''

 
http://img696.imageshack.us/img696/985/50880118.gif


'''Change the y-intercept to any number between 0 and K'''

P(t)= 1 /[(1+ e^(-x+y))]

P(t)= 1 /[(1+(e^(-t+5))]

P(t)= 1/(1+(e^(-t+1))]

Which is shown as:

By changing the value of Y into a larger number it can be observed that the Y-intercept becomes less and less.

''This one has a lower Y-Intercept: P(t)= 1 /[(1+(e^(-t+5))]''
 
http://img406.imageshack.us/img406/3913/13941006.gif


''This one has a higher Y-Interecept: P(t)= 1/(1+(e^(-t+1))]''
 
http://img146.imageshack.us/img146/5422/70293261.gif


'''Applying the Model'''

An eco-system which is supporting a population of birds can theoretically can hold up to 300 birds, which is its carrying capacity. Initially the population of birds is 50. Using the model:

 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


We can input the data given to form:

P(t)= C/1+Ie^(-rt)

C = Carrying capacity of the eco-system

I = Initial starting population

R = Rate of growth

P(t)= 300/1+50e^(-.6t)

 
http://img831.imageshack.us/img831/4021/16539198.gif

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T06:07:32Z

JustinHsu:

==Homework==

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:

* Change the height of the horizontal asymptote on the right, we'll denote it by K.

* Change the y-intercept to any number between 0 and K

* 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).

Original function:
 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


Which is shown as:

 
http://img820.imageshack.us/img820/6396/msp631419e4ca1c0iaea3b2.gif


'''Change the height of the horizontal asymptote on the right, we'll denote it by
K.'''

P(t)= 1 /[(1/k)+e^-x]

P(t)= 1 /[(1/5)+e^-x]

P(t)= 1 /[(1/2)+e^-x]

As we can see the horizontal asymptote has been shifted higher.

Which is shown as:

The smaller the number for K the horizontal asymptote increases.

''Lower Horizontal asymptote: P(t)= 1 /[(1/5)+e^-x]''
 
http://img248.imageshack.us/img248/9728/msp428519e4d4e1432d9h8i.gif


''High Horizontal asymptote: P(t)= 1 /[(1/2)+e^-x]''

 
http://img696.imageshack.us/img696/985/50880118.gif


'''Change the y-intercept to any number between 0 and K'''

P(t)= 1 /[(1+ e^(-x+y))]

P(t)= 1 /[(1+(e^(-t+5))]

P(t)= 1/(1+(e^(-t+1))]

Which is shown as:

By changing the value of Y into a larger number it can be observed that the Y-intercept becomes less and less.

''This one has a lower Y-Intercept: P(t)= 1 /[(1+(e^(-t+5))]''
 
http://img406.imageshack.us/img406/3913/13941006.gif


''This one has a higher Y-Interecept: P(t)= 1/(1+(e^(-t+1))]''
 
http://img146.imageshack.us/img146/5422/70293261.gif


'''Applying the Model'''

An eco-system which is supporting a population of birds can theoretically can hold up to 300 birds, which is its carrying capacity. Initially the population of birds is 50. Using the model:

 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


We can input the data given to form:

P(t)= C/1+Ie^(-rt)

C = Carrying capacity of the eco-system

I = Initial starting population

R = Rate of growth

P(t)= 300/1+50e^(-.6t)

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T06:05:51Z

JustinHsu:

==Homework==

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:

* Change the height of the horizontal asymptote on the right, we'll denote it by K.

* Change the y-intercept to any number between 0 and K

* 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).

Original function:
 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


Which is shown as:

 
http://img820.imageshack.us/img820/6396/msp631419e4ca1c0iaea3b2.gif


'''Change the height of the horizontal asymptote on the right, we'll denote it by
K.'''

P(t)= 1 /[(1/k)+e^-x]

P(t)= 1 /[(1/5)+e^-x]

P(t)= 1 /[(1/2)+e^-x]

As we can see the horizontal asymptote has been shifted higher.

Which is shown as:

The smaller the number for K the horizontal asymptote increases.

''Lower Horizontal asymptote: P(t)= 1 /[(1/5)+e^-x]''
 
http://img248.imageshack.us/img248/9728/msp428519e4d4e1432d9h8i.gif


''High Horizontal asymptote: P(t)= 1 /[(1/2)+e^-x]''

 
http://img696.imageshack.us/img696/985/50880118.gif


'''Change the y-intercept to any number between 0 and K'''

P(t)= 1 /[(1+ e^(-x+y))]

P(t)= 1 /[(1+(e^(-t+5))]

P(t)= 1/(1+(e^(-t+1))]

Which is shown as:

By changing the value of Y into a larger number it can be observed that the Y-intercept becomes less and less.

''This one has a lower Y-Intercept: P(t)= 1 /[(1+(e^(-t+5))]''
 
http://img406.imageshack.us/img406/3913/13941006.gif


''This one has a higher Y-Interecept: P(t)= 1/(1+(e^(-t+1))]''
 
http://img146.imageshack.us/img146/5422/70293261.gif


'''Applying the Model'''

An eco-system which is supporting a population of birds can theoretically can hold up to 300 birds, which is its carrying capacity. Initially the population of birds is 50. Using the model:

 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


We can input the data given to form:

P(t)= C/1+Ie^(-rt)

C = Carrying capacity of the eco-system
I = Initial starting population
R = Rate of growth

P(t)= 300/1+50e^(-.6t)

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T05:55:48Z

JustinHsu:

==Homework==

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:

* Change the height of the horizontal asymptote on the right, we'll denote it by K.

* Change the y-intercept to any number between 0 and K

* 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).

Original function:
 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


Which is shown as:

 
http://img820.imageshack.us/img820/6396/msp631419e4ca1c0iaea3b2.gif


'''Change the height of the horizontal asymptote on the right, we'll denote it by
K.'''

P(t)= 1 /[(1/k)+e^-x]

P(t)= 1 /[(1/5)+e^-x]

P(t)= 1 /[(1/2)+e^-x]

As we can see the horizontal asymptote has been shifted higher.

Which is shown as:

The smaller the number for K the horizontal asymptote increases.

''Lower Horizontal asymptote: P(t)= 1 /[(1/5)+e^-x]''
 
http://img248.imageshack.us/img248/9728/msp428519e4d4e1432d9h8i.gif


''High Horizontal asymptote: P(t)= 1 /[(1/2)+e^-x]''

 
http://img696.imageshack.us/img696/985/50880118.gif


'''Change the y-intercept to any number between 0 and K'''

P(t)= 1 /[(1+ e^(-x+y))]

P(t)= 1 /[(1+(e^(-t+5))]

P(t)= 1/(1+(e^(-t+1))]

Which is shown as:

By changing the value of Y into a larger number it can be observed that the Y-intercept becomes less and less.

''This one has a lower Y-Intercept: P(t)= 1 /[(1+(e^(-t+5))]''
 
http://img406.imageshack.us/img406/3913/13941006.gif


''This one has a higher Y-Interecept: P(t)= 1/(1+(e^(-t+1))]''
 
http://img146.imageshack.us/img146/5422/70293261.gif


'''Applying the Model'''

An eco-system which is supporting a population of birds can theoretically can hold up to 300 birds, which is its carrying capacity. Initially the population of birds is 50. Using the model:

 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


We can input the data given to form:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T05:26:18Z

JustinHsu:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T05:25:33Z

JustinHsu:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T05:03:23Z

JustinHsu:

==Homework==

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:

* Change the height of the horizontal asymptote on the right, we'll denote it by K.

* Change the y-intercept to any number between 0 and K

* 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).

Original function:
 
http://img695.imageshack.us/img695/9875/b532d798ec926e169adf3c1.png


Which is shown as:

 
http://img820.imageshack.us/img820/6396/msp631419e4ca1c0iaea3b2.gif


'''Change the height of the horizontal asymptote on the right, we'll denote it by
K.'''

P(t)= 1 /[(1/k)+e^-x]

P(t)= 1 /[(1/5)+e^-x]

P(t)= 1 /[(1/2)+e^-x]

As we can see the horizontal asymptote has been shifted higher.

Which is shown as:

The smaller the number for K the horizontal asymptote increases.

Lower Horizontal asymptote: P(t)= 1 /[(1/5)+e^-x]
 
http://img248.imageshack.us/img248/9728/msp428519e4d4e1432d9h8i.gif


High Horizontal asymptote: P(t)= 1 /[(1/2)+e^-x]

 
http://img696.imageshack.us/img696/985/50880118.gif


'''Change the y-intercept to any number between 0 and K'''

P(t)= 1 /[(1+ e^(-x+y))]

P(t)= 1 /[(1+(e^(-t+5))]

P(t)= 1/(1+(e^(-t+1))]

Which is shown as:

By changing the value of Y into a larger number it can be observed that the Y-intercept becomes less and less.

This one has a lower Y-Intercept: P(t)= 1 /[(1+(e^(-t+5))]
 
http://img406.imageshack.us/img406/3913/13941006.gif


This one has a higher Y-Interecept: P(t)= 1/(1+(e^(-t+1))]
 
http://img146.imageshack.us/img146/5422/70293261.gif


'''Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.'''

P(t)= k(1/(1+e^(-t))

P(t)= 5(1/(1+e^(-t))

P(t)= 10(1/1+e^(-t))

Which is as shown:

 
http://img191.imageshack.us/img191/237/41043529.gif


 
http://img191.imageshack.us/img191/237/41043529.gif

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T04:50:24Z

JustinHsu:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T04:48:43Z

JustinHsu:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T04:44:54Z

JustinHsu:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T04:42:57Z

JustinHsu:

Course:MATH110/Archive/2010-2011/003/Teams/Uri/Homework/12 Part3

2011-01-28T04:35:52Z

JustinHsu: