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Calculus 2  Exam 1 ( Semester C ) 
Exam Overview 
This is the first exam for calculus 2. The material covered in this exam includes area between curves, moment and center of mass of a planar lamina, work moving water out of a container, volume of revolution, arc length and linear motion. 

Exam Details 

Tools 
Time  2.5 hours  
Calculators  no 
Questions  6  
Formula Sheet(s) 
2 pages, 8.5x11 or A4 
Total Points  100  
Other Tools  ruler for drawing graphs 
Instructions:
 Show all your work.
 For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.
 Correct notation counts (i.e. points will be taken off for incorrect notation).
 Give exact, simplified answers.
Question 1 
[15 points] A stone is thrown vertically into the air off the edge of a bridge 160ft above a river. If the initial velocity of the rock is 48ft/sec (upward) and the acceleration is a constant 32ft/sec^{2} (downward), find the following.
a. The velocity v(t) after t seconds.
b. The position s(t) of the stone above the river.
c. The velocity at which the stone hits the water. 


a. \(v(t)=32t+48\) ft/sec
b. \(s(t)=16t^2+48t+160\) ft
c. \(v(5)=112\) ft/sec

a. We know the acceleration is constant at \(a=32\) ft/sec^{2}. So we can integrate to get \(v(t)\).
\(v(t)=\int{a~dt} = \int{32~dt} = 32t+C\)
To determine \(C\), we use the information about the initial velocity, \(48\) ft/sec as follows.
\(v(0)=32(0)+C = 48 \to C=48\)
b. Now that we have the equation for velocity, we can integrate to get the equation for position.
\(s(t)=\int{v(t)~dt} =\) \(\int{32t+38~dt} = 16t^2+48t+C\)
Like we did for velocity, we use the initial position given in the problem statement to determine the value of C as follows.
\(s(0)=16(0)^2+48(0)+C=160 \)
\(s(t)=16t^2+48t+160\) ft
c. To determine the velocity when the stone hits the water, we use the velocity equation \(v(t)\). However, the parameter for velocity is time t. So we need to determine WHEN the stone hits the water.
We know the position of the stone \(s(t)\) and the position when the stone hits the water is when \(s(t)=0\). So we will set the position function equal to zero, solve for t and use that value in the velocity function.
\(
\begin{array}{rcl}
16t^2+48t+160 & = & 0 \\
t^23t10 & = & 0 \\
(t5)(t+2) & = & 0
\end{array}
\)
So \(t=5\) or \(t=2\). The value of t that makes sense is \(t=5\).
\(v(5)=32(5)+48=160+48=112\) ft/sec
Question 1 Final Answer 
a. \(v(t)=32t+48\) ft/sec
b. \(s(t)=16t^2+48t+160\) ft
c. \(v(5)=112\) ft/sec

Question 2 
[10 points] Calculate the length of the arc \(y=2x^{3/2}\) from \(x=1/3\) to \(x=8/3\). 


\(26/3\)

\(s=\int_{a}^{b}{\sqrt{1+[f'(x)]^2}}\)
\(f(x)=y=2x^{3/2}\)
\(f'(x)=2(3/2)x^{1/2} = 3x^{1/2}\)
\(\displaystyle{s=\int_{1/3}^{8/3}{\sqrt{1+9x}}}\)
Use integration by substitution.
\(u=1+9x \to du=9dx\)
Change the limits of integration.
\(x=1/3 \to u=4\)
\(x=8/3 \to u=25\)
\(\displaystyle{s=\int_{4}^{25}{u^{1/2}~\frac{du}{9}} = \left. \frac{1}{9} \frac{u^{3/2}}{3/2} \right_{4}^{25} = }\)
\(\displaystyle{ \frac{2}{27}(25)^{3/2}\frac{2}{27}(4)^{3/2} = \frac{2}{27}[1258] = }\)
\(\displaystyle{ \frac{2}{27}(117) = \frac{26}{3}}\)
Question 2 Final Answer 
\(26/3\)

Question 3 
[10 points] Consider the region bounded by \(y=\ln(x)\) and the line with slope \(m=(3+\ln(5))/5\) and yintercept at \(x=3\) above the xaxis as shown in the plot. Set up and but do not evaluate one integral that calculates the shaded area.



Area = \(\displaystyle{ \int_{0}^{\ln(5)}{\frac{5(y+3)}{3+\ln(5)}  e^y ~ dy}}\)

The key here is to notice that the question asks for ONE integral. The only way to do that is to evaluate in the ydirection.
The equation of the straight line is \(y=[(3+\ln(5))/5]x3\). All of this information is given in the problem statement. In order to find the point of intersection of the two curves, you can pull right off the graph that the point of intersection occurs at \(x=5\). So, \(y=[(2+\ln(5))/5](5)3 =
\ln(5) \to (5,\ln(5))\).
The limits of integration are from \(y=0\) to \(y=\ln(5)\).
Our general equation for area between curves is \(\int_{c}^{d}{f(y)g(y)~dy}\). In our case, \(f(y)\) is the straight line and \(g(y)\) is \(y=\ln(x)\).
The integral is in terms of y, so we need to solve both equations for y.
\(y=\ln(x) \to e^y = e^{\ln(x)} \to e^y = x\)
\(y=[(3+\ln(5))/5]x3 \to y+3=[(3+\ln(5))/5]x \to x=(y+3)[5/(3+\ln(5))]\)
So the integral is
\(\displaystyle{ \int_{0}^{\ln(5)}{\frac{5(y+3)}{3+\ln(5)}  e^y ~ dy}}\)
Question 3 Final Answer 
Area = \(\displaystyle{ \int_{0}^{\ln(5)}{\frac{5(y+3)}{3+\ln(5)}  e^y ~ dy}}\)

Question 4 
[20 points] Consider the region R bounded by \(y=x^3\) and \(y=\sqrt{x}\). a. Calculate the area of the region R. Sketch a plot and shade the region. b. Calculate the moments M_{x} and M_{y}. c. Determine the center of mass. 


a. Area = \(5/12\)
b. \(M_x = 5\rho/28\), \(M_y = \rho/5\)
c. \((\bar{x}, \bar{y} ) = (12/25, 3/7)\)

a. The plot is shown to the right. The green curve is \(y=x^3\) and the red curve is \(y=\sqrt{x}\). We choose to evaluate in the xdirection. The point of intersection is \((1,1)\) so our integral is
Area = \(\displaystyle{ \int_{0}^{1}{\sqrt{x}x^3~dx} = }\)
\(\displaystyle{ \left[ \frac{x^{3/2}}{3/2}  \frac{x^4}{4} \right]_{0}^{1} = }\)
\(\displaystyle{ \frac{2}{3}  \frac{1}{4} = \frac{5}{12} }\)
b. The general equations for moments of a planar lamina are
\(\displaystyle{ M_x = \rho \int_{a}^{b}{(1/2)[f(x)+g(x)][f(x)g(x)]~dx} }\)
\(\displaystyle{ M_y = \rho \int_{a}^{b}{x[f(x)g(x)]~dx} }\)
where \(f(x) \geq g(x)\)
Since the density is not given, we can assume a constant density of \(\rho\).
\(
\begin{array}{rcl}
M_x & = & \rho/2 \int_{a}^{b}{(\sqrt{x}+x^3)(\sqrt{x}x^3)~dx} \\
& = & \rho/2 \int_{0}^{1}{xx^6~dx} \\
& = & \displaystyle{ \frac{\rho}{2}\left[ \frac{x^2}{2}\frac{x^7}{7} \right]_{0}^{1} } \\
& = & \displaystyle{ \frac{\rho}{2} \left[ \frac{1}{2}  \frac{1}{7} \right] } \\
& = & \displaystyle{ \frac{\rho}{2} \left[ \frac{7}{14}  \frac{2}{14} \right] } \\
& = & \displaystyle{ \frac{5\rho}{28} }
\end{array}
\)
\(
\begin{array}{rcl}
M_y & = & \rho\int_{0}^{1}{x(\sqrt{x}x^3)~dx} \\
& = & \rho \int_{0}^{1}{x^{3/2}x^4~dx} \\
& = & \displaystyle{\rho \left[\frac{x^{5/2}}{5/2}  \frac{x^5}{5} \right]_{0}^{1} } \\
& = & \displaystyle{\rho \left[ \frac{2}{5}\frac{1}{5} \right] } \\
& = & \displaystyle{ \frac{\rho}{5} }
\end{array}
\)
c. To calculate the center of mass we use the moments \(M_x\) and \(M_y\) and we need the total mass. Since the density is constant, the mass is just the density times area, i.e. \(m=5\rho/12\).
\(\displaystyle{ \bar{x} = \frac{M_y}{m} = \frac{\rho}{5} \cdot \frac{12}{5\rho} = \frac{12}{25}}\)
\(\displaystyle{ \bar{y} = \frac{M_x}{m} = \frac{5\rho}{28} \cdot \frac{12}{5\rho} = \frac{3}{7}}\)
\((\bar{x},\bar{y}) = (12/25, 3/7)\)
Question 4 Final Answer 
a. Area = \(5/12\)
b. \(M_x = 5\rho/28\), \(M_y = \rho/5\)
c. \((\bar{x}, \bar{y} ) = (12/25, 3/7)\)

Question 5 [ exam link ] 
[20 points] For the region R bounded by the graphs of \(y=2x\), the xaxis and \(x=1\), consider the solid generated by revolving the region about the line \(y=3\). Sketch a different plot of region R for each method, clearly showing the region by shading, the axes of revolution, the representative rectangle, R, r, p and h. a. Set up the integral to calculate the volume using the discwasher method. b. Set up the integral to calculate the volume using the shellcylinder method. c. Calculate the volume of the solid by evaluating one of the integrals you set up in parts a and b. 


a. \(V=\pi\int_{0}^{1}{3^2(32x)^2~dx}\)
b. \(V=2\pi\int_{0}^{2}{(3y)(1y/2)~dy}\)
c. \(V = 14\pi/3\)

a. The general equation is \(V=\pi\int_{a}^{b}{R^2r^2~dx}\).
\(y+r=3 \to r=3y \to r=32x \) and \(R=3\)
\(V=\pi\int_{0}^{1}{3^2(32x)^2~dx}\)
b. The general equation is \(V=2\pi\int_{c}^{d}{ph~dy}\)
\( y+p=3 \to p=3y \)
\(x+h=1 \to h=1x = 1y/2 \)
\(V=2\pi\int_{0}^{2}{(3y)(1y/2)~dy}\)
c. The integral from part a is
\(
\begin{array}{rcl}
V & = & \pi\int_{0}^{1}{3^2(32x)^2~dx} \\
& = & \pi \int_{0}^{1}{9  (912x+4x^2)~dx} \\
& = & \pi \int_{0}^{1}{12x4x^2 ~dx} \\
& = & \displaystyle{ \pi \left[ \frac{12x^2}{2}  \frac{4x^3}{3} \right]_{0}^{1} } \\
& = & \displaystyle{ \pi \left[ 6  \frac{4}{3} \right] } \\
& = & \displaystyle{ \frac{14\pi}{3} }
\end{array}
\)
The integral from part b is
\(
\begin{array}{rcl}
V & = & 2\pi\int_{0}^{2}{(3y)(1y/2)~dy} \\
& = & \displaystyle{ 2\pi \int_{0}^{2}{3\frac{3y}{2}y+\frac{y^2}{2}~dy} } \\
& = & \displaystyle{ 2\pi \left[ 3y  \frac{5}{2}\frac{y^2}{2} + \frac{y^3}{6} \right]_{0}^{2}} \\
& = & \displaystyle{ 2\pi \left[ 6  \frac{5}{4}4 +\frac{8}{6} \right] } \\
& = & \displaystyle{ 2\pi\left[ 1 +\frac{4}{3}\right] } \\
& = & \displaystyle{ 2\pi\frac{7}{3} = \frac{14\pi}{3} }
\end{array}
\)
Question 5 Final Answer 
a. \(V=\pi\int_{0}^{1}{3^2(32x)^2~dx}\)
b. \(V=2\pi\int_{0}^{2}{(3y)(1y/2)~dy}\)
c. \(V = 14\pi/3\)

Question 6 
[25 points] A swimming pool is 20 m long and 10 m wide, with a bottom that slopes uniformly from a depth of 1 m at one end to a depth of 2 m at the other end. Assuming the pool is full, how much work is required to pump the water to a level 1 m above the top of the pool?



Total Work \(W = 1600\rho g/3\)

For this problem, we need to set up a plot of the side of the pool and establish our coordinate system. We have chosen to flip the side, so that the origin is at bottom of the deep end of the pool, as shown here. The sides of the pool are outlined in blue and the red line is at \(y=3\), the level to where the water is to be lifted.
The general equation for work that we need is \(W=\int_{a}^{b}{\rho g A(y)D(y)~dy}\).
From this plot, you should see that we need two integrals to calculate work, one from 0 to 1, the other from 1 to 2. This is true because the profile, and thus the area, is different in those two sections. But first let's discuss the equation for the distance moved, \(D(y)\).
If we think about a slice of water, \(dy\), at some point within the pool, the slice will be at position y on our axes. We would then need to move that slice of water up to 3. This would mean the slice is moved \(3y\). This is our equation for \(D(y)=3y\).
Now we need to describe the area of one slice of water. The area is different depending on the position y. Let's start with a slice of water when \(1 \leq y \leq 2\). Since the depth of the pool width of the pool is 10m and the length is always 20m in this range of y, the area is just a rectangle and \(A(y)=10(20) = 200\).
So we have one of our integrals, i.e.
\(\displaystyle{\int_{1}^{2}{\rho g (200)(3y)~dy}}\)
Let's go ahead and evaluate this integral.
\(
\begin{array}{rcl}
W_{12} & = & \displaystyle{\int_{1}^{2}{\rho g (200)(3y)~dy}} \\
& = & \displaystyle{200\rho g\int_{1}^{2}{(3y)~dy}} \\
& = & \displaystyle{ 200\rho g \left[ 3y  \frac{y^2}{2} \right]_{1}^{2}} \\
& = & 200\rho g \left[ (62)  (31/2) \right] \\
& = & 300\rho g
\end{array}
\)
Okay, so that is the work required to move the top 1m of water out of the pool, up to one meter above the pool. Now let's determine the integral for moving the bottom 1m of water.
We already know the equation for \(D(y)=3y\). We just need the equation for \(A(y)\). Notice that at each slice, the area is rectangular with width of 10m. But the length is different depending on the height y.
The equation of the line from \((0,0)\) to \((20,1)\) is \(y=x/20\) so \(x=20y\). This x is the length. So our area is \(A(y)=(10)(20y)=200y\). Now we are ready to set up our integral and evaluate.
\(
\begin{array}{rcl}
W_{01} & = & \displaystyle{\int_{0}^{1}{\rho g (200y)(3y)~dy}} \\
& = & \displaystyle{200\rho g\int_{0}^{1}{3yy^2~dy}} \\
& = & \displaystyle{ 200\rho g \left[ \frac{3y^2}{2]}  \frac{y^3}{3} \right]_{0}^{1}} \\
& = & \displaystyle{ 200\rho g \left[ \frac{3}{2}  \frac{1}{3} \right] } \\
& = & \displaystyle{ 200 \rho g \left[ \frac{9}{6}  \frac{2}{6} \right] } \\
& = & \displaystyle{ \frac{700\rho g}{3} }
\end{array}
\)
Now we combine the results of the two integrals to get the total work.
\(\displaystyle{ W = W_{01} + W_{12} = \frac{700\rho g}{3} + 300\rho g = \frac{1600\rho g}{3} }\)
Total Work \(W = 1600\rho g/3\)
Question 6 Final Answer 
Total Work \(W = 1600\rho g/3\)
