https://wiki.ubc.ca/api.php?action=feedcontributions&user=Simontse&feedformat=atomUBC Wiki - User contributions [en]2024-03-28T19:59:14ZUser contributionsMediaWiki 1.39.6https://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2010/Question_04/Solution_1&diff=230886Science:Math Exam Resources/Courses/MATH101/April 2010/Question 04/Solution 12013-04-16T06:23:30Z<p>Simontse: added a checking of (the confusing imperial) units and a conclusion statement</p>
<hr />
<div>As the hint suggests, we split the cone into cylinders of height say <math> \Delta x </math> which are a height of <math> 1-x_{i}^{*} </math> from the ground (where the <math> x_{i}^{*} </math> is an arbitrary sample point)[[File:MATH101_April_2010_Question_4_Cone.png|500px|thumbnail|right]]. We would like to know the radius of each of these cylinders depending upon how high up the cone they are, then we can calculate the volume of these cylinders and add them all up. In order to find the radius ''r'' of each cylinder with respect to its height up the cone, we use ratios and the fact that the radius of the base of the cone is 1/2. I.e.: Since the base of the cone is 1/2 and that's 1 ft away from the tip of the cone, we know that (via similar triangles in the cone - see the diagram)<br />
:<math><br />
\frac{1}{1/2} = \frac{x_{i}^{*}}{r}<br />
</math><br />
and so<br />
:<math><br />
r = \frac{x_{i}^{*}}{2}<br />
</math><br />
<br />
In our diagram below, we have set it up so that the origin is at the apex (another way to solve this problem is to set up the 0 value at the bottom of the anthill). We now have, since the volume of a cylinder is <math> \pi r^2 h </math> where <math> r </math> is the radius of the base and <math> h </math> is the height of the cylinder, that<br />
:<math><br />
V=\pi \frac{(x_{i}^{*})^2}{4} \Delta x<br />
</math><br />
<br />
Since <br />
<br />
work done = weight <math> \times </math> distance <br />
<br />
and<br />
<br />
weight = density <math> \times </math> volume <br />
<br />
we have<br />
<br />
work = density <math> \times </math> volume <math> \times </math> distance <br />
<br />
where the distance is <math> 1-x_{i}^{*} </math>. Hence we can get the total volume by adding the contributions from each of the cylinders. In the limit as <math>\Delta{x}</math> goes to zero this becomes and integral and we get<br />
<br />
:<math>\begin{align}<br />
\text{work} &= \int_0^1 150\, \pi \frac{x^2}4\,(1-x) dx\\<br />
&= \left[\frac{150\pi x^3}{12} - \frac{150\pi x^4}{16}\right]_0^1 \\<br />
&= \frac{150 \pi}{12} -\frac{150\pi }{16} \\<br />
&= \frac{25\pi}{8}<br />
\end{align}</math><br />
<br />
Let us do a quick sanity check on the units to see if we have missed out anything. <br />
<br />
The volume is in <math>ft^3</math>, and the density is given in <math> lb / ft^3 </math>, so the weight is in <math> lb </math>, finally work is force (weight) times distance (which is in <math> ft </math>, so we get our work to be in <math> ft\cdot lb </math>, which is right.<br />
<br />
In conclusion, the work done by the ants to build the anthill is <math> \frac{25\pi}{8} ft\cdot lb </math>.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_08_(b)/Solution_1&diff=228232Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (b)/Solution 12013-04-03T22:57:11Z<p>Simontse: </p>
<hr />
<div>[[File:Sketch for Q8 b).png|thumb|The straight line below y=sin(x) that is selected to construct the inequality]]<br />
<br />
Following the hint, we want to find a straight line that is below <math> y=sin(x)</math> with minimal error. <br />
<br />
To do this, we require the straight line to pass through <math>y=sin(x)</math> at both ''x=0'' and ''x=π/2'', which are the endpoints of the domain of integration.<br />
<br />
Since ''y=sin(0)=0'' at ''x=0'', we look for a straight line that passes through the origin with formula ''y=&fnof;(x)=cx''. Then we put ''x=π/2'', ''y=sin(π/2)=1'' to find ''1=cπ/2'', so ''c=2/π''. <br />
<br />
Note that the slope of the line ''c=2/π<1'', while the initial slope of ''y=sin(x)'' is given by ''cos(0)=1'', so the linear function ''y=2/π x'' is really below ''y=sin(x)'' in the interval <math> [0, \pi/2] </math><br />
<br />
Hence, we can deduce the following inequalities:<br />
<br />
:<math> <br />
\begin{align}<br />
\frac{2}{\pi} x &\leq \sin x \\ \Rightarrow <br />
-a \sin x &\leq -\frac{2a}{\pi} x \\ \Rightarrow <br />
e^{-a \sin x} &\leq e^{-\frac{2a}{\pi} x} \\<br />
\end{align}<br />
</math><br />
<br />
because ''a>0'' and <math> e^x</math> is a strictly increasing function.<br />
<br />
So by comparison, we obtain<br />
:<math><br />
\begin{align}<br />
\int_0^{\pi/2} e^{-a \sin x} dx &\leq \int_0^{\pi/2} e^{-\frac{2a}{\pi} x} dx \\<br />
&= -\frac{\pi}{2a} \left[ e^{-\frac{2a}{\pi} x} \right]_0^{\pi/2} \\<br />
&= -\frac{\pi}{2a} \left(e^{-a}-1\right) \\<br />
&= \frac{\pi}{2a} \left(1- e^{-a}\right) < \frac{\pi}{2a} <br />
\end{align}<br />
</math><br />
<br />
because <math> e^{-a} > 0 </math> for any ''a''.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_08_(b)/Solution_1&diff=228231Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (b)/Solution 12013-04-03T22:56:36Z<p>Simontse: missing minus sign</p>
<hr />
<div>[[File:Sketch for Q8 b).png|thumb|The straight line below y=sin(x) that is selected to construct the inequality]]<br />
<br />
Following the hint, we want to find a straight line that is below <math> y=sin(x)</math> with minimal error. <br />
<br />
To do this, we require the straight line to pass through both <math>y=sin(x)</math> at ''x=0'' and ''x=π/2'', which are the endpoints of the domain of integration.<br />
<br />
Since ''y=sin(0)=0'' at ''x=0'', we look for a straight line that passes through the origin with formula ''y=&fnof;(x)=cx''. Then we put ''x=π/2'', ''y=sin(π/2)=1'' to find ''1=cπ/2'', so ''c=2/π''. <br />
<br />
Note that the slope of the line ''c=2/π<1'', while the initial slope of ''y=sin(x)'' is given by ''cos(0)=1'', so the linear function ''y=2/π x'' is really below ''y=sin(x)'' in the interval <math> [0, \pi/2] </math><br />
<br />
Hence, we can deduce the following inequalities:<br />
<br />
:<math> <br />
\begin{align}<br />
\frac{2}{\pi} x &\leq \sin x \\ \Rightarrow <br />
-a \sin x &\leq -\frac{2a}{\pi} x \\ \Rightarrow <br />
e^{-a \sin x} &\leq e^{-\frac{2a}{\pi} x} \\<br />
\end{align}<br />
</math><br />
<br />
because ''a>0'' and <math> e^x</math> is a strictly increasing function.<br />
<br />
So by comparison, we obtain<br />
:<math><br />
\begin{align}<br />
\int_0^{\pi/2} e^{-a \sin x} dx &\leq \int_0^{\pi/2} e^{-\frac{2a}{\pi} x} dx \\<br />
&= -\frac{\pi}{2a} \left[ e^{-\frac{2a}{\pi} x} \right]_0^{\pi/2} \\<br />
&= -\frac{\pi}{2a} \left(e^{-a}-1\right) \\<br />
&= \frac{\pi}{2a} \left(1- e^{-a}\right) < \frac{\pi}{2a} <br />
\end{align}<br />
</math><br />
<br />
because <math> e^{-a} > 0 </math> for any ''a''.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_02_(b)/Solution_1&diff=228229Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (b)/Solution 12013-04-03T22:55:07Z<p>Simontse: Wording tweak</p>
<hr />
<div>We use the method of washers.<br />
<br />
For any ''x'' in ''(-1,2)'', the larger and smaller radii ''R'' and ''r'' of the washer is given by <br />
:<math>R(x) = 5-(x+1) = 4-x \,</math> <br />
:<math>r(x) = 5-(4-(x-1)^2) = 1+(x-1)^2 \,</math><br />
(The simplifications given above are not really necessary)<br />
<br />
Notice that the larger radius ''R'' is given by the distance from the axis of revolution ''y=5'' to the lower curve, because it is further away. <br />
<br />
Therefore, the area of the washer is<br />
<br />
:<math> A(x) = (R^2 - r^2)\pi = \left[(4-x)^2 - \left(1+(x-1)^2\right)^2\right]\pi </math><br />
<br />
Now we can write down the integral that represents the volume of the solid of revolution:<br />
<br />
:<math> <br />
\begin{align}<br />
V &= \int_{-1}^2 A(x) dx \\<br />
&= \pi \left[\int_{-1}^2 (4-x)^2 - \left(1+(x-1)^2\right)^2 dx \right]\\<br />
\end{align}<br />
</math><br />
<br />
'''According to the question statement, we may stop here.'''<br />
<br />
''If you are interested to find the value of the integral, read on.''<br />
<br />
To solve this integral, we split it in two parts and write<br />
<br />
:<math> V = \pi \left[ \int_{-1}^2 (4-x)^2 dx - \int_{-1}^2 \left(1+(x-1)^2\right)^2 dx \right]</math><br />
<br />
We refrain from expanding the brackets but make a change of variables ''u=4-x, du=-dx'' and ''v=x-1, dv=dx'' for the first and second integral respectively. Not forgetting to update the integration limits, we arrive at<br />
<br />
:<math><br />
\begin{align}<br />
V &= \pi \left[\int_{-1}^2 (4-x)^2 dx + \int_{-1}^2 \left(1+(x-1)^2\right)^2 dx \right]\\<br />
&= \pi \left[\int_5^2 u^2 (-du) + \int_{-2}^1 \left(1+v^2\right)^2 dv \right]\\<br />
&= \pi \left[\int_2^5 u^2 du + \int_{-2}^1 \left(1+2v^2+v^4\right) dv \right]\\<br />
&= \pi \left( \left[u^3/3\right]^5_2 +\left[v+2v^3/3+v^5/5\right]_{-2}^1 \right)\\<br />
&= \pi \left( (125-8)/3 +\left[3+2(1+8)/3+(1+32)/5\right] \right)\\<br />
&= \pi \left( 39 + 3+ 6 + 6+ 3/5 \right)\\<br />
&= \pi \left( 54 + 3/5 \right) = 273\pi / 5\\<br />
\end{align}<br />
</math></div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_02_(a)/Hint_1&diff=228228Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (a)/Hint 12013-04-03T22:51:54Z<p>Simontse: small tweak of wordings</p>
<hr />
<div>Where does the two curves intersect?<br />
<br />
If you find the values of ''x'', namely ''a'' and ''b'' (where ''a<b'') so that the curves intersect, then the area bounded by the two curves is given by<br />
:<math>A=\int_a^b(f(x)-g(x))dx</math><br />
where ''f''(x) denotes the curve above the region ''R'' and ''g(x)'' denotes the curve below respectively.<br />
<br />
You will need the sketch to determine which curve is above and which is below.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_08_(a)&diff=228035Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (a)2013-04-03T03:29:17Z<p>Simontse: </p>
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{{MER Question page}}</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_02_(b)&diff=228034Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (b)2013-04-03T03:29:07Z<p>Simontse: </p>
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{{MER Question page}}</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_08_(b)/Solution_1&diff=228032Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (b)/Solution 12013-04-03T03:27:52Z<p>Simontse: </p>
<hr />
<div>[[File:Sketch for Q8 b).png|thumb|The straight line below y=sin(x) that is selected to construct the inequality]]<br />
<br />
Following the hint, we want to find a straight line that is below <math> y=sin(x)</math> with minimal error. <br />
<br />
To do this, we require the straight line to pass through both <math>y=sin(x)</math> at ''x=0'' and ''x=π/2'', which are the endpoints of the domain of integration.<br />
<br />
Since ''y=sin(0)=0'' at ''x=0'', we look for a straight line that passes through the origin with formula ''y=cx''. Then we put ''x=π/2'', ''y=sin(π/2)=1'' to find ''1=cπ/2'', so ''c=2/π''. <br />
<br />
Note that the slope of the line ''c=2/π<1'', while the initial slope of ''y=sin(x)'' is given by ''cos(0)=1'', so the linear function ''y=2/π x'' is really below ''y=sin(x)'' in the interval <math> [0, \pi/2] </math><br />
<br />
Hence, we can deduce the following inequalities:<br />
<br />
:<math> <br />
\begin{align}<br />
\frac{2}{\pi} x &\leq \sin x \\ \Rightarrow <br />
-a \sin x &\leq -\frac{2a}{\pi} x \\ \Rightarrow <br />
e^{-a \sin x} &\leq -\frac{2a}{\pi} x \\<br />
\end{align}<br />
</math><br />
<br />
because ''a>0'' and <math> e^x</math> is a strictly increasing function.<br />
<br />
So by comparison, we obtain<br />
:<math><br />
\begin{align}<br />
\int_0^{\pi/2} e^{-a \sin x} dx &\leq \int_0^{\pi/2} e^{-\frac{2a}{\pi} x} dx \\<br />
&= -\frac{\pi}{2a} \left[ e^{-\frac{2a}{\pi} x} \right]_0^{\pi/2} \\<br />
&= -\frac{\pi}{2a} \left(e^{-a}-1\right) \\<br />
&= \frac{\pi}{2a} \left(1- e^{-a}\right) < \frac{\pi}{2a} <br />
\end{align}<br />
</math><br />
<br />
because <math> e^x > 0 </math> for any ''x''.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_08_(b)/Solution_1&diff=228031Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (b)/Solution 12013-04-03T03:26:49Z<p>Simontse: </p>
<hr />
<div>[[File:Sketch for Q8 b).png|thumb|To linear function, i.e. straight line, selected that is below y=sin(x)]]<br />
<br />
Following the hint, we want to find a straight line that is below <math> y=sin(x)</math> with minimal error. <br />
<br />
To do this, we require the straight line to pass through both <math>y=sin(x)</math> at ''x=0'' and ''x=π/2'', which are the endpoints of the domain of integration.<br />
<br />
Since ''y=sin(0)=0'' at ''x=0'', we look for a straight line that passes through the origin with formula ''y=cx''. Then we put ''x=π/2'', ''y=sin(π/2)=1'' to find ''1=cπ/2'', so ''c=2/π''. <br />
<br />
Note that the slope of the line ''c=2/π<1'', while the initial slope of ''y=sin(x)'' is given by ''cos(0)=1'', so the linear function ''y=2/π x'' is really below ''y=sin(x)'' in the interval <math> [0, \pi/2] </math><br />
<br />
Hence, we can deduce the following inequalities:<br />
<br />
:<math> <br />
\begin{align}<br />
\frac{2}{\pi} x &\leq \sin x \\ \Rightarrow <br />
-a \sin x &\leq -\frac{2a}{\pi} x \\ \Rightarrow <br />
e^{-a \sin x} &\leq -\frac{2a}{\pi} x \\<br />
\end{align}<br />
</math><br />
<br />
because ''a>0'' and <math> e^x</math> is a strictly increasing function.<br />
<br />
So by comparison, we obtain<br />
:<math><br />
\begin{align}<br />
\int_0^{\pi/2} e^{-a \sin x} dx &\leq \int_0^{\pi/2} e^{-\frac{2a}{\pi} x} dx \\<br />
&= -\frac{\pi}{2a} \left[ e^{-\frac{2a}{\pi} x} \right]_0^{\pi/2} \\<br />
&= -\frac{\pi}{2a} \left(e^{-a}-1\right) \\<br />
&= \frac{\pi}{2a} \left(1- e^{-a}\right) < \frac{\pi}{2a} <br />
\end{align}<br />
</math><br />
<br />
because <math> e^x > 0 </math> for any ''x''.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_08_(b)/Solution_1&diff=228030Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (b)/Solution 12013-04-03T03:25:48Z<p>Simontse: Created page with "To select a correct linear function, i.e. straight line, below y=sin(x) Following the hint, we want to find a straight line that is below ..."</p>
<hr />
<div>[[File:Sketch for Q8 b).png|thumb|To select a correct linear function, i.e. straight line, below y=sin(x)]]<br />
<br />
Following the hint, we want to find a straight line that is below <math> y=sin(x)</math> with minimal error. <br />
<br />
To do this, we require the straight line to pass through both <math>y=sin(x)</math> at ''x=0'' and ''x=π/2'', which are the endpoints of the domain of integration.<br />
<br />
Since ''y=sin(0)=0'' at ''x=0'', we look for a straight line that passes through the origin with formula ''y=cx''. Then we put ''x=π/2'', ''y=sin(π/2)=1'' to find ''1=cπ/2'', so ''c=2/π''. <br />
<br />
Note that the slope of the line ''c=2/π<1'', while the initial slope of ''y=sin(x)'' is given by ''cos(0)=1'', so the linear function ''y=2/π x'' is really below ''y=sin(x)'' in the interval <math> [0, \pi/2] </math><br />
<br />
Hence, we can deduce the following inequalities:<br />
<br />
:<math> <br />
\begin{align}<br />
\frac{2}{\pi} x &\leq \sin x \\ \Rightarrow <br />
-a \sin x &\leq -\frac{2a}{\pi} x \\ \Rightarrow <br />
e^{-a \sin x} &\leq -\frac{2a}{\pi} x \\<br />
\end{align}<br />
</math><br />
<br />
because ''a>0'' and <math> e^x</math> is a strictly increasing function.<br />
<br />
So by comparison, we obtain<br />
:<math><br />
\begin{align}<br />
\int_0^{\pi/2} e^{-a \sin x} dx &\leq \int_0^{\pi/2} e^{-\frac{2a}{\pi} x} dx \\<br />
&= -\frac{\pi}{2a} \left[ e^{-\frac{2a}{\pi} x} \right]_0^{\pi/2} \\<br />
&= -\frac{\pi}{2a} \left(e^{-a}-1\right) \\<br />
&= \frac{\pi}{2a} \left(1- e^{-a}\right) < \frac{\pi}{2a} <br />
\end{align}<br />
</math><br />
<br />
because <math> e^x > 0 </math> for any ''x''.</div>Simontsehttps://wiki.ubc.ca/index.php?title=File:Sketch_for_Q8_b).png&diff=228029File:Sketch for Q8 b).png2013-04-03T03:07:17Z<p>Simontse: </p>
<hr />
<div>=={{int:filedesc}}==<br />
{{Information<br />
|description={{en|1=A sketch showing how a straight line approximate sin(x) from below to derive an inequality for a certain integral.}}<br />
|date=2013-04-02<br />
|source={{own}}<br />
|author=[[User:Simontse|Simontse]]<br />
|permission=<br />
|other_versions=<br />
|other_fields=<br />
}}<br />
<br />
=={{int:license-header}}==<br />
{{self|cc-by-sa-2.5-ca}}</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_03_(d)&diff=228028Science:Math Exam Resources/Courses/MATH101/April 2012/Question 03 (d)2013-04-03T03:05:48Z<p>Simontse: revision done, waiting for review</p>
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{{MER Question page}}</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_03_(d)/Solution_1&diff=228027Science:Math Exam Resources/Courses/MATH101/April 2012/Question 03 (d)/Solution 12013-04-03T03:04:51Z<p>Simontse: </p>
<hr />
<div>The integral is equal to the sum of the two standard Type I integrals:<br />
<br />
:<math>\int_{-\infty}^0\frac{x}{x^2+1}dx+\int_0^\infty\frac{x}{x^2+1}dx </math>.<br />
<br />
By the subtitution <math>u=x^2+1, du = 2dx</math>, we see that <br />
:<math> \int \frac{x}{x^2+1} dx = \frac{1}{2} \int \frac{1}{u}du = \frac{1}{2}\ln|u|+C = \frac{1}{2}\ln|x^2+1|+C </math><br />
<br />
So, we calculate<br />
:<math>\int_0^\infty\frac{x}{x^2+1}dx=\lim_{a\to \infty}\frac{1}{2}\ln(x^2+1)\big|_0^a=\lim_{a\to \infty}(\frac{1}{2}\ln(a^2+1)-\ln(1))=\infty</math><br />
<br />
Since this part diverges the whole integral diverges also.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_03_(d)/Hint_1&diff=228026Science:Math Exam Resources/Courses/MATH101/April 2012/Question 03 (d)/Hint 12013-04-03T03:00:07Z<p>Simontse: </p>
<hr />
<div>Note that this improper integral has '''both''' upper and lower limits of integration being improper. <br />
<br />
To write it as a standard Type I improper integral, one first has to split it into two integrals. Then, if both converges, the original integral converges. But if at least one of them diverges, then the original integral diverges.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_03_(d)/Hint_1&diff=228025Science:Math Exam Resources/Courses/MATH101/April 2012/Question 03 (d)/Hint 12013-04-03T02:59:26Z<p>Simontse: </p>
<hr />
<div>This improper integral which has both upper and lower limits of integration improper. <br />
<br />
To write it as a standard Type I improper integral, one first has to split it into two integrals. Then, if both converges, the original integral converges. But if at least one of them diverges, then the original integral diverges.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_08_(b)/Hint_1&diff=228024Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (b)/Hint 12013-04-03T02:54:40Z<p>Simontse: </p>
<hr />
<div>Further to the hint given in the question, what function is simpler than a linear function, i.e. a straight line? <br />
<br />
To draw a striaght line below <math> \sin x</math> to approximate, from above, the area represented by the integral, we should not choose a straight line is below but does not miss ''too much'' of the area from the graph of <math> \sin x</math>.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_08_(b)/Hint_1&diff=228023Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (b)/Hint 12013-04-03T02:53:55Z<p>Simontse: Created page with "Further to the hint given in the question, what function is simpler than a linear function, i.e. a straight line? To draw a striaght line below <math> \sin x</math> to appro..."</p>
<hr />
<div>Further to the hint given in the question, what function is simpler than a linear function, i.e. a straight line? <br />
<br />
To draw a striaght line below <math> \sin x</math> to approximate the area represented by the integral from above, we should not choose a straight line is below but does not miss ''too much'' of the area from the graph of <math> \sin x</math>.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_08_(a)/Hint_1&diff=228022Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (a)/Hint 12013-04-03T02:49:52Z<p>Simontse: </p>
<hr />
<div>Both Alternating Series Test (AST) and Alternating Series Esimation Theorem (ASET) ''DO NOT'' tell you the exact value of an alternating series. The key word ''evaluate'' implies that there should be a way to find the exact value of the series.<br />
<br />
Does the series look like a familiar Maclaurin series of a function evaluated at a certain point?</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_08_(a)/Solution_1&diff=228021Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (a)/Solution 12013-04-03T02:48:22Z<p>Simontse: Created page with "Recall that the Maclaurin series of <math> \ln(1+x) </math> is :<math> \ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n</math> with radius of convergence ''R=1''. If w..."</p>
<hr />
<div>Recall that the Maclaurin series of <math> \ln(1+x) </math> is<br />
<br />
:<math> \ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n</math> <br />
<br />
with radius of convergence ''R=1''.<br />
<br />
If we put ''x=1/2'', then we obtain the given series and hence we find<br />
<br />
<math> \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n 2^n} = \ln(1+1/2)= \ln(3/2)</math></div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_08_(a)/Hint_1&diff=228020Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (a)/Hint 12013-04-03T02:43:03Z<p>Simontse: </p>
<hr />
<div>Both Alternating Series Test (AST) and Alternating Series Esimation Theorem (ASET) '''DO NOT'' tell you the exact value of an alternating series. The key word ''evaluate'' implies that there should be a way to find the exact value of the series.<br />
<br />
Does the series look like a familiar Maclaurin series of a function evaluated at a certain point?</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_06&diff=228019Science:Math Exam Resources/Courses/MATH101/April 2012/Question 062013-04-03T02:41:17Z<p>Simontse: </p>
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{{MER Question page}}</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_08_(a)/Hint_1&diff=228018Science:Math Exam Resources/Courses/MATH101/April 2012/Question 08 (a)/Hint 12013-04-03T02:40:45Z<p>Simontse: Created page with "Both Alternating Series Test (AST) and Alternating Series Esimation Theorem (ASET) '''DO NOT'' tell you the exact value of an alternating series. The key word ''evaluate'' imp..."</p>
<hr />
<div>Both Alternating Series Test (AST) and Alternating Series Esimation Theorem (ASET) '''DO NOT'' tell you the exact value of an alternating series. The key word ''evaluate'' implies that there should be a way to find the exact value of the series.<br />
<br />
Does the series look like a familiar power series evaluated at a certain point?</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_06/Solution_1&diff=228017Science:Math Exam Resources/Courses/MATH101/April 2012/Question 06/Solution 12013-04-03T02:38:37Z<p>Simontse: first version</p>
<hr />
<div>Denote ''y'' as the distance from the top of the bowl to the water level and ''r'' the radius of the water surface. <br />
<br />
Then, consider a right-angled triangle with a water surface radius ''r'' as base, then the height is ''y'', and the hypoteneuse is 3, which is the radius of the bowl. (The top of the triangle is exactly at the center of the the sphere, half of which becomes the bowl).<br />
<br />
Then <math> r^2 + y^2 = 3^2 = 9 </math> implies <math> r^2 = 9-y^2</math>, so the area of the water surface is given as a function of ''y'' by<br />
<br />
:<math> A(y) = r^2 \pi = \pi (9-y^2) </math><br />
<br />
A thin slab of water thus has volume <math>A(y)\triangle y</math> and mass <br />
<br />
:<math> M(y) = 1000 A(y)\triangle y = 1000\pi(9-y^2)\triangle y</math><br />
<br />
This slab of water travels a distance of ''y+2'' to be pumped out (''+2'' because of the 2m vertical outlet) and requires a force of ''Mg''<br />
<br />
Therefore, summing the work done for each slab, and letting <math>\triangle y\to 0</math>, we get<br />
<br />
:<math> <br />
\begin{align}<br />
W &= \int_0^3 1000\pi g (2+y)(9-y^2) dy \\<br />
&= 1000\pi g \left[2 \int_0^3 (18 + 9y -2y^2 -y^3) dy \right]\\<br />
&= 1000\pi g \left[ 18y +9y^2/2 - 2y^3/3 - y^4/4]_0^3 \right]\\<br />
&= 1000\pi g \left( 18\cdot 3 + 9(3)^2/2 - 2(3)^3/3 - 3^4/4\right)<br />
\end{align}<br />
</math></div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_06/Hint_1&diff=228016Science:Math Exam Resources/Courses/MATH101/April 2012/Question 06/Hint 12013-04-03T02:16:51Z<p>Simontse: </p>
<hr />
<div>What is the shape of the water surface and how does it change during the process?<br />
<br />
To describe this, you need the crucial variable that changes with the process of pumping. <br />
<br />
You can use either the depth of water or the distance from the top of the bowl to the water level, but the latter will be easier.<br />
<br />
Then find a formula for the water surface so that you can find the work done to lift up a small slab of water from the water surface up to the top of the bowl and out of the outlet.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_06/Hint_1&diff=228015Science:Math Exam Resources/Courses/MATH101/April 2012/Question 06/Hint 12013-04-03T02:14:22Z<p>Simontse: Created page with "What is the shape of the water surface and how does it change during the process? To describe this, you need the crucial variable that changes with the process of pumping. ..."</p>
<hr />
<div>What is the shape of the water surface and how does it change during the process?<br />
<br />
To describe this, you need the crucial variable that changes with the process of pumping. <br />
<br />
You can use either the depth of water or the distance from the top of the bowl to the water level, then find a formula for the water surface so that you can find the work done to lift up a small slab of water from the water surface up to the top of the bowl and out of the outlet.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_02_(a)&diff=228014Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (a)2013-04-03T02:06:29Z<p>Simontse: </p>
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{{MER Question page}}</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_02_(b)/Solution_1&diff=228013Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (b)/Solution 12013-04-03T02:05:08Z<p>Simontse: </p>
<hr />
<div>We use the method of washers.<br />
<br />
For any ''x'' in ''(-1,2)'', the larger and smaller radii ''R'' and ''r'' of the washer is given by <br />
:<math>R(x) = 5-(x+1) = 4-x</math> <br />
:<math>r(x) = 5-(4-(x-1)^2) = 1+(x-1)^2</math><br />
<br />
Notice that the larger radius ''R'' is given by the distance from the axis of revolution ''y=5'' to the lower curve. (The simplifications given above are not really necessary)<br />
<br />
Therefore, the area of the washer is<br />
<br />
:<math> A(x) = (R^2 - r^2)\pi = \left[(4-x)^2 - \left(1+(x-1)^2\right)^2\right]\pi </math><br />
<br />
Now we can write down the integral that represents the volume of the solid of revolution:<br />
<br />
:<math> <br />
\begin{align}<br />
V &= \int_{-1}^2 A(x) dx \\<br />
&= \pi \left[\int_{-1}^2 (4-x)^2 - \left(1+(x-1)^2\right)^2 dx \right]\\<br />
\end{align}<br />
</math><br />
<br />
According to the question statement, we may stop here. <br />
<br />
If you are interested to find the value of the integral, read on.<br />
<br />
Now, we split the integral to write<br />
<br />
:<math> V = \pi \left[ \int_{-1}^2 (4-x)^2 dx - \int_{-1}^2 \left(1+(x-1)^2\right)^2 dx \right]</math><br />
<br />
We refrain from expanding the brackets but make a change of variables ''u=4-x, du=-dx'' and ''v=x-1, dv=dx'' for the first and second integral respectively. Not forgetting to update the integration limits, we arrive at<br />
<br />
:<math><br />
\begin{align}<br />
V &= \pi \left[\int_{-1}^2 (4-x)^2 dx + \int_{-1}^2 \left(1+(x-1)^2\right)^2 dx \right]\\<br />
&= \pi \left[\int_5^2 u^2 (-du) + \int_{-2}^1 \left(1+v^2\right)^2 dv \right]\\<br />
&= \pi \left[\int_2^5 u^2 du + \int_{-2}^1 \left(1+2v^2+v^4\right) dv \right]\\<br />
&= \pi \left( \left[u^3/3\right]^5_2 +\left[v+2v^3/3+v^5/5\right]_{-2}^1 \right)\\<br />
&= \pi \left( (125-8)/3 +\left[3+2(1+8)/3+(1+32)/5\right] \right)\\<br />
&= \pi \left( 39 + 3+ 6 + 6+ 3/5 \right)\\<br />
&= \pi \left( 54 + 3/5 \right) = 273\pi / 5\\<br />
\end{align}<br />
</math></div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_02_(b)/Solution_1&diff=228012Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (b)/Solution 12013-04-03T02:02:30Z<p>Simontse: first version</p>
<hr />
<div>We use the method of washers.<br />
<br />
For any ''x'' in ''(-1,2)'', the larger and smaller radii ''R'' and ''r'' of the washer is given by <br />
:<math>R(x) = 5-(x+1) = 4-x</math> <br />
:<math>r(x) = 5-(4-(x-1)^2) = 1+(x-1)^2</math><br />
<br />
Notice that since the larger radius ''R'' is given by the distance from the axis of revolution ''y=5'' to the lower curve. (The simplification is not really necessary)<br />
<br />
Therefore, the area of the washer is<br />
<br />
:<math> A(x) = (R^2 - r^2)\pi = \left[(4-x)^2 - \left(1+(x-1)^2\right)^2\right]\pi </math><br />
<br />
Note that we refrain from expanding until it really is necessary. <br />
<br />
Now we can write down the integral that represents the volume of the solid of revolution:<br />
<br />
:<math> <br />
\begin{align}<br />
V &= \int_{-1}^2 A(x) dx \\<br />
&= \pi \left[\int_{-1}^2 (4-x)^2 - \left(1+(x-1)^2\right)^2 dx \right]\\<br />
\end{align}<br />
</math><br />
<br />
According to the question statement, we may stop here. <br />
<br />
If you are interested to find the value of the integral, read on.<br />
<br />
Now, we split the integral to write<br />
<br />
:<math> V = \pi \left[ \int_{-1}^2 (4-x)^2 dx - \int_{-1}^2 \left(1+(x-1)^2\right)^2 dx \right]</math><br />
<br />
We refrain from expanding the brackets but make a change of variables ''u=4-x, du=-dx'' and ''v=x-1, dv=dx'' for the first and second integral respectively. Not forgetting to update the integration limits, we arrive at<br />
<br />
:<math><br />
\begin{align}<br />
V &= \pi \left[\int_{-1}^2 (4-x)^2 dx + \int_{-1}^2 \left(1+(x-1)^2\right)^2 dx \right]\\<br />
&= \pi \left[\int_5^2 u^2 (-du) + \int_{-2}^1 \left(1+v^2\right)^2 dv \right]\\<br />
&= \pi \left[\int_2^5 u^2 du + \int_{-2}^1 \left(1+2v^2+v^4\right) dv \right]\\<br />
&= \pi \left( \left[u^3/3\right]^5_2 +\left[v+2v^3/3+v^5/5\right]_{-2}^1 \right)\\<br />
&= \pi \left( (125-8)/3 +\left[3+2(1+8)/3+(1+32)/5\right] \right)\\<br />
&= \pi \left( 39 + 3+ 6 + 6+ 3/5 \right)\\<br />
&= \pi \left( 54 + 3/5 \right) = 273\pi / 5\\<br />
\end{align}<br />
</math></div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_02_(b)/Hint_1&diff=228011Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (b)/Hint 12013-04-03T01:27:26Z<p>Simontse: Created page with "Use the method of washers (cross-sections). Look at the sketch at part a), the cross-sections are formed by intersecting the region ''R'' by a striaght line ''x=c'' where ''-..."</p>
<hr />
<div>Use the method of washers (cross-sections).<br />
<br />
Look at the sketch at part a), the cross-sections are formed by intersecting the region ''R'' by a striaght line ''x=c'' where ''-1<c<2'', and then rotating the line segment about the axis ''y=5''. The result should look like a washer.<br />
<br />
To find the volume of revolution, find the area of this washer as a function of ''x'' and then integrate from the left to right endpoint of the region.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_02_(a)&diff=228007Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (a)2013-04-03T01:21:13Z<p>Simontse: </p>
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{{MER Question page}}</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_02_(a)/Solution_1&diff=228006Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (a)/Solution 12013-04-03T01:20:37Z<p>Simontse: </p>
<hr />
<div>[[File:Plot of the region for Q2 a).png|thumb|Plot of the region]]<br />
<br />
Since the two curves <math> y_1 = f(x) = 4 - (x-1)^2 </math> and <math> y_2 = g(x)=x+1 </math> intersects when <br />
<br />
:<math> <br />
\begin{align}<br />
0 &= f(x)-g(x) \\<br />
&= 4-(x-1)^2-(x-1)\\<br />
&= -(x^2-x-2) \\<br />
&= -(x-2)(x+1) \\<br />
\end{align}<br />
</math><br />
<br />
Thus, the curves intersect at ''x=-1'' and ''x=2''. Moreover, from the sketch, we can see that ''f(x) > g(x)'' in the interval ''[-1, 2]''. <br />
<br />
Therefore, the area ''A'' bounded by the curves is given by:<br />
<br />
:<math> <br />
\begin{align}<br />
A&=\int_{-1}^2 f(x)-g(x) dx\\<br />
&=\int_{-1}^2 (2+x-x^2) dx \\<br />
&= \left[ 2x + x^2/2 - x^3/3 \right]_{-1}^{2} \\<br />
&= \left(4+2-8/3\right) - \left(-2+1/2+1/3\right) \\<br />
&= 9/2 <br />
\end{align}<br />
</math></div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_02_(a)/Hint_1&diff=228004Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (a)/Hint 12013-04-03T01:19:22Z<p>Simontse: </p>
<hr />
<div>Where does the two curves intersect?<br />
<br />
If you find the values of ''x'', namely ''a'' and ''b'' (where ''a<b'') so that the curves intersect, then the area bounded is given by<br />
:<math>A=\int_a^b(f(x)-g(x))dx</math><br />
where ''f''(x) denotes the curve above and ''g(x)'' denotes the curve on below respectively.<br />
<br />
You will need the sketch to determine which curve is above and which is below.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_02_(a)/Solution_1&diff=227998Science:Math Exam Resources/Courses/MATH101/April 2012/Question 02 (a)/Solution 12013-04-03T01:14:44Z<p>Simontse: first version</p>
<hr />
<div>[[File:Plot of the region for Q2 a).png|thumb|Plot of the region]]<br />
<br />
Since the two curves <math> y_1 = f(x) = 4 - (x-1)^2 </math> and <math> y_2 = g(x)=x+1 </math> intersects when <br />
<br />
:<math> <br />
\begin{align}<br />
0 &= f(x)-g(x) \\<br />
&= 4-(x-1)^2-(x-1)\\<br />
&= -(x^2-x-2) \\<br />
&= -(x-2)(x+1) \\<br />
\end{align}<br />
</math><br />
<br />
Thus, the points of intersection is at ''x=-1'' and ''x=2''. Moreover, from the sketch, we can see that ''f(x) > g(x)'' in the interval ''[-1, 2]''. <br />
<br />
Therefore, the area ''A'' bounded by the curves is given by:<br />
<br />
:<math> <br />
\begin{align}<br />
A&=\int_{-1}^2 f(x)-g(x) dx\\<br />
&=\int_{-1}^2 (2+x-x^2) dx \\<br />
&= \left[ 2x + x^2/2 - x^3/3 \right]_{-1}^{2} \\<br />
&= \left(4+2-8/3\right) - \left(-2+1/2+1/3\right) \\<br />
&= 9/2 <br />
\end{align}<br />
</math></div>Simontsehttps://wiki.ubc.ca/index.php?title=File:Plot_of_the_region_for_Q2_a).png&diff=227990File:Plot of the region for Q2 a).png2013-04-03T00:58:11Z<p>Simontse: </p>
<hr />
<div>=={{int:filedesc}}==<br />
{{Information<br />
|description={{en|1=The region between the curve y = 4 − (x − 1)^2 and the line y = x + 1}}<br />
|date=2013-04-02<br />
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{{self|cc-by-sa-2.5-ca}}</div>Simontsehttps://wiki.ubc.ca/index.php?title=Course:MATH101/2012-2013/210/WebWork_Solutions/Assignment_6&diff=218346Course:MATH101/2012-2013/210/WebWork Solutions/Assignment 62013-02-28T07:33:48Z<p>Simontse: Created page with "== Problem 10 == Match the following improper integral with the improper integral below in which you can compare using the Comparison Test. Then determine whether the integra..."</p>
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<div>== Problem 10 ==<br />
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Match the following improper integral with the improper integral below in which you can compare using the Comparison Test. Then determine whether the integral converge or diverge.<br />
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A. <math>\int_1^\infty \frac{dx}{x}</math>, B. <math>\int_1^\infty \frac{dx}{x^2}</math>, C. <math>\int_1^\infty \frac{dx}{x^3}</math>, and D. <math>\int_1^\infty \frac{dx}{\sqrt{x}}</math><br />
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1. <math>\int_1^\infty \frac{dx}{x^2+5}</math>, ''Answer:'' B, Converges.<br />
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Key step:<br />
<math> \frac{1}{x^2+5} \leq \frac{1}{x^2} </math><br />
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2. <math>\int_1^\infty \frac{2+e^{-x}}{\sqrt{x}}dx</math>, 'Answer:'' D, Diverges.<br />
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Key step:<br />
<math> e^{-x}>0 ~</math>, so <math>\frac{2+e^{-x}}{\sqrt{x}}> \frac{1}{\sqrt{x}}</math><br />
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3. <math>\int_1^\infty \frac{x}{x^2-0.5}dx</math>, 'Answer:'' A, Diverges.<br />
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Key step:<br />
When the denominator is decreased, the fraction increases. So <br />
<math> \frac{x}{x^2-0.5} > \frac{x}{x^2} = \frac{1}{x} </math><br />
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4. <math>\int_1^\infty \frac{2+\cos x}{x} dx </math>, 'Answer:'' A, Diverges.<br />
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Key step:<br />
Recall the range of <math>\cos x </math> is [-1,1], so<br />
<math> 2 + \cos x \geq 1 </math></div>Simontsehttps://wiki.ubc.ca/index.php?title=Course:MATH101/2012-2013/210&diff=218345Course:MATH101/2012-2013/2102013-02-28T07:12:31Z<p>Simontse: </p>
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<div>This is a wiki page for student-instructor collaboration to prepare review materials and solutions.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Course:MATH101/2012-2013/210&diff=218344Course:MATH101/2012-2013/2102013-02-28T07:01:59Z<p>Simontse: </p>
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|subject code=MATH<br />
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|course number=101<br />
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This is a wiki page for student-instructor collaboration to prepare review materials and solutions.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Course:MATH101/2012-2013/210&diff=218343Course:MATH101/2012-2013/2102013-02-28T07:01:10Z<p>Simontse: </p>
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|subject code=MATH<br />
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|section number=210<br />
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|instructor=Simon Tse<br />
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This is a wiki page for student-instructor collaboration to prepare review materials and solutions.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Course:MATH101/2012-2013/210&diff=218342Course:MATH101/2012-2013/2102013-02-28T07:00:17Z<p>Simontse: first edit</p>
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|title=Integral Calculus with Applications to Physical Sciences and Engineering<br />
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|picture=Image:wiki.png<br />
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|subject code=MATH<br />
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|course number=101<br />
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|section number=210<br />
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|instructor=Simon Tse<br />
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|instructor 2=<br />
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|office=LSK303C<br />
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|office hours=Mon 10-11, Tue 2-4<br />
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This is a wiki page for student-instructor collaboration to prepare review materials and solutions.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science_talk:Math_Exam_Resources/Courses/MATH101/April_2012/Question_6&diff=218161Science talk:Math Exam Resources/Courses/MATH101/April 2012/Question 62013-02-26T01:22:14Z<p>Simontse: Talk page autocreated when first thread was posted</p>
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<div></div>Simontsehttps://wiki.ubc.ca/index.php?title=Thread:Science_talk:Math_Exam_Resources/Courses/MATH101/April_2012/Question_6/Wow,_that_bowl_is_beautiful._How_did_you_draw_it%3F&diff=218160Thread:Science talk:Math Exam Resources/Courses/MATH101/April 2012/Question 6/Wow, that bowl is beautiful. How did you draw it?2013-02-26T01:22:14Z<p>Simontse: New thread: Wow, that bowl is beautiful. How did you draw it?</p>
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<div>''[{{fullurl:{{FULLPAGENAMEE}}|action=edit}} Please enter your text for this thread. You should delete this line then.]''<br/></div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_07_(c)/Statement&diff=218159Science:Math Exam Resources/Courses/MATH101/April 2012/Question 07 (c)/Statement2013-02-26T01:21:13Z<p>Simontse: </p>
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<div>Let <math> I = \int_{1}^{2} (1/x)\, dx </math>. Without computing ''I'', find an upper bound for <math> |I - S_{4}|</math>. You may use the fact that if <math>|f^{(4)}(x)|\leq K</math> on the interval <math> [a,b]</math>, then the error in using ''S''<sub>''n''</sub> to approximate <math>\int_{a}^{b}f(x)\,dx</math> has absolute value less than or equal to <math>K(b-a)^{5}/180n^{4}</math>.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_07_(c)/Hint_1&diff=218158Science:Math Exam Resources/Courses/MATH101/April 2012/Question 07 (c)/Hint 12013-02-26T01:19:52Z<p>Simontse: Small mistake in the formula for K.</p>
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<div>As it says in the question statement, given <math>n</math>, an even integer, and an upper bound <math> K \geq |f^{(4)}(x)|</math> for all <math>x</math> in the interval <math>[a,b]</math>, the error that comes from using Simpson's Rule is as follows: <br />
<br />
:<math> |I - S_n| \leq \frac{K (b-a)^5}{180\cdot n^4}</math><br />
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How would you find ''K''?</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:Math_Exam_Resources/Courses/MATH101/April_2012/Question_01_(i)/Solution_1&diff=218157Science:Math Exam Resources/Courses/MATH101/April 2012/Question 01 (i)/Solution 12013-02-26T01:16:33Z<p>Simontse: add one more term in the initial expansion, and change \pm back to just +</p>
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<div>Recall the Maclaurin (power) series of sin(''x'') is<br />
<br />
:<math> \sin (x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots </math><br />
<br />
Note that the first two terms cancel exactly with the polynomial in the numerator. Thus,<br />
<br />
:<math> \begin{align}<br />
\lim_{x\rightarrow0}\frac{\sin (x)-x+x^{3}/6}{\sin (x^5)} &=\lim_{x\rightarrow0}\frac{x^{5}/(5!)-x^{7}/(7!)+\cdots}{\sin (x^5)} \\<br />
&= \lim_{x \to 0} \left( \frac{1}{5!}\frac{x^5}{\sin (x^5)} - \frac{x^2}{7!}\frac{x^5}{\sin (x^5)} + \dots \right) <br />
\end{align}</math><br />
<br />
Now, as shown in Hint 2, <br />
<br />
:<math>\lim_{x \to 0} \frac{y}{\sin (y)} = 1</math> <br />
<br />
we can also say that <br />
<br />
:<math> \lim_{x\rightarrow0}\frac{x^5}{\sin (x^5)}=1 </math> <br />
<br />
So the first term of the limit becomes <math>\frac{1}{5!}</math>. The remaining terms all have an <math>x^5 / \sin(x^5)</math> term which evaluates to <math>1</math>, but also have an extra power of <math>x</math> which will make the entire term go to ''0''. So the solution to the limit is <math>\frac{1}{5!}</math>.<br />
<br />
This question demonstrates the strength of power series in more difficult calculus problems.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Thread:Science_talk:Math_Exam_Resources/Courses/MATH101/April_2012/Question_1_(i)/great_update_and_a_minor_stuff:_%22%5Cpm%22_%3F/reply_(2)&diff=218058Thread:Science talk:Math Exam Resources/Courses/MATH101/April 2012/Question 1 (i)/great update and a minor stuff: "\pm" ?/reply (2)2013-02-24T08:05:46Z<p>Simontse: Reply to great update and a minor stuff: "\pm" ?</p>
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<div>I simply didn't understand it, and frankly I haven't seen people using that for omitting terms for the sine series. Perhaps Stewart wrote it that way? I haven't previewed that section yet. If you think it's right, sure just let it stay. I was thinking if it was a typo and realized it couldn't be, so there was something intentional that I could not catch.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Thread:Science_talk:Math_Exam_Resources/Courses/MATH101/April_2012/Question_1_(i)/great_update_and_a_minor_stuff:_%22%5Cpm%22_%3F&diff=218055Thread:Science talk:Math Exam Resources/Courses/MATH101/April 2012/Question 1 (i)/great update and a minor stuff: "\pm" ?2013-02-24T00:18:41Z<p>Simontse: New thread: great update and a minor stuff: "\pm" ?</p>
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<div>Hi Christina, thanks for the great update. I love your version more than the previous one. I was about to change sometime but just paused to confirm.<br />
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Was it intentional to have a \pm before the omitted terms in the sine expansion?</div>Simontsehttps://wiki.ubc.ca/index.php?title=Thread:Science_talk:Math_Exam_Resources/Courses/MATH101/April_2012/Question_1_(i)/Key_hint_missing._Wording_in_solution_should_be_improved./reply_(2)&diff=217744Thread:Science talk:Math Exam Resources/Courses/MATH101/April 2012/Question 1 (i)/Key hint missing. Wording in solution should be improved./reply (2)2013-02-20T23:12:06Z<p>Simontse: Reply to Key hint missing. Wording in solution should be improved.</p>
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<div>Thanks, please feel free go ahead to make the changes you envisioned. I agree enough to let you take over. I don't have time to implement it myself now.<br />
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The L'Hospital rule does invoke chain rule and possibly "impossibly-many" complications, however, a simple call to an algebra system gives the denominator to be something like (smaller terms) + 120*x*cos(x^5). So it either requires an inhuman length of calculation or an understanding to ignore "higher order terms" which is not probably not available to most students. (The numerator is simple though, and 4 differentiations suffice to end the case).<br />
<br />
So I agree we should simply omit it and say it's "impossible", in some sense.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Thread:Science_talk:Math_Exam_Resources/Courses/MATH101/April_2012/Question_1_(i)/Key_hint_missing._Wording_in_solution_should_be_improved./reply&diff=217743Thread:Science talk:Math Exam Resources/Courses/MATH101/April 2012/Question 1 (i)/Key hint missing. Wording in solution should be improved./reply2013-02-20T23:12:04Z<p>Simontse: Reply to Key hint missing. Wording in solution should be improved.</p>
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<div>Thanks, please feel free go ahead to make the changes you envisioned. I agree enough to let you take over. I don't have time to implement it myself now.<br />
<br />
The L'Hospital rule does invoke chain rule and possibly "impossibly-many" complications, however, a simple call to an algebra system gives the denominator to be something like (smaller terms) + 120*x*cos(x^5). So it either requires an inhuman length of calculation or an understanding to ignore "higher order terms" which is not probably not available to most students. (The numerator is simple though, and 4 differentiations suffice to end the case).<br />
<br />
So I agree we should simply omit it and say it's "impossible", in some sense.</div>Simontsehttps://wiki.ubc.ca/index.php?title=Science:MER/Solutions&diff=214266Science:MER/Solutions2013-01-24T10:00:39Z<p>Simontse: /* How to add/edit a solution? */</p>
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<div>{{MER Infobox}}<br />
Solutions to exam questions constitute the main part of the content this wiki offers to students (for the moment at least). Since our mission is to support student learning, we want to strive for solutions which encourage students to learn and think deeply of their work and the concepts involved with calculus and avoid promoting rote learning and procedural thinking as much as possible.<br />
<br />
== Criteria for good solutions ==<br />
A good solution should be:<br />
* Simple given the concepts and tools that are available to a typical student.<br />
* Clearly highlight the reasoning that is involved in making decisions when solving a problem.<br />
* Spend some time discussing how ''word problems'' are modelled. This element is critical for students and constitute one of their biggest weaknesses. As such, extra care should be taken to explain and guide students through that step while explaining alternate and equivalent choices which would also work.<br />
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== How to add/edit a solution? ==<br />
Question pages are automatically generated with a blank solution box. Just click edit to start adding content to that solution.<br />
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If you want to add a second (or more) solution to the page, you have to create that new page. How? Simple: let's say you are working on question 7 of the December 2011 exam of MATH 102. The question page address is:<br />
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<nowiki>http://wiki.ubc.ca/ ... /MATH102/December_2011/Question_7</nowiki><br />
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And if you type in the address bar<br />
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<nowiki>http://wiki.ubc.ca/ ... /MATH102/December_2011/Question_7/Solution_1</nowiki><br />
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You should actually see the first solution. (That's the actual page containing the solution, the question page only mirrors that content). To create a second solution, go to the address bar and change it to:<br />
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<nowiki>http://wiki.ubc.ca/ ... /MATH102/December_2011/Question_7/Solution_2</nowiki><br />
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If this page doesn't exist already, a message in it will say: '''There is currently no text in this page''' and the edit button should actually say ''create'' instead of ''edit''. Go ahead, hit ''create'' and write your second solution on that page. Once you are done, save the page and now go back to your question page. A second solution box should have magically appeared now (the question page has code that detects the existence of that type of subpage, so it isn't really magic).<br />
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Question pages support up to 5 solutions.<br />
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See the [[Help:Contents|UBC wiki Help pages]] if you have any questions about formatting (such as adding italics, bold, and inputting latex code).<br />
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== I added a solution, how do I flag it for review? ==<br />
If you have added a hint and think it is ready to be reviewed, go in the Question page and change the CH flag into a RH flag. This will automatically list the page for a hint review and someone will do that soon. When there are more than one hint present, it is worth adding in the discussion page a mention of which hint should be reviewed.<br />
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== Quality control - how to review a solution? ==<br />
Once a solution has been flagged for review (with a RS flag), a contributor who has NOT worked on that solution can review its content. There are two possible outcomes of the review:<br />
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** make some modifications to the solution and leave the RS flag to invite another contributor to review the new work;<br />
** OR leave the work as is and change the RS flag into a QBS flag to signal the problem to other contributors. In this case, the reviewer is asked to write the reasons for the bad flag in the discussion page to justify and inform other contributors of his decision and the work that needs to be done.<br />
* '''Good''': for a solution that is correct, well formatted AND support student learning. In this case, the reviewer change the RS flag into a QGS flag.<br />
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== There is already some solution(s) but I think we should have more solutions, what should I do? ==<br />
In that case, make sure that the Question page has an active CS flag and explain what kind of hint you are looking for in the discussion page. Also, you can go ahead, add a new solution box and start carving out a solution!<br />
<br />
[[Category:MER]]</div>Simontse