This is the second exam for first semester single variable calculus.
Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.
 Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.
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 Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
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 Work the entire exam before checking any solutions.
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 Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.
IMPORTANT 
Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.
Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.
Exam Details  

Time 
2.5 hours 
Questions 
5 
Total Points 
70 
Tools  

Calculator 
not allowed 
Formula Sheet(s) 
one page 8.5x11 or A4; both sides okay 
Other Tools 
ruler 
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 your answers in exact, simplified form.
[20 points] Water is pouring into an inverted cone (point up) at the rate of 3 cubic meters per minute. The height of the cone is 10 meters and the radius of its base is 5 meters. How fast is the water level rising when the water stands 4 meters above the base?
Problem Statement 

[20 points] Water is pouring into an inverted cone (point up) at the rate of 3 cubic meters per minute. The height of the cone is 10 meters and the radius of its base is 5 meters. How fast is the water level rising when the water stands 4 meters above the base?
Hint 

The equation for the volume of a cone with height h and radius r is \(V=\pi r^2h/3\).
Problem Statement 

[20 points] Water is pouring into an inverted cone (point up) at the rate of 3 cubic meters per minute. The height of the cone is 10 meters and the radius of its base is 5 meters. How fast is the water level rising when the water stands 4 meters above the base?
Final Answer 

\(\displaystyle{ \frac{1}{3\pi} }\) meters/min
Problem Statement 

[20 points] Water is pouring into an inverted cone (point up) at the rate of 3 cubic meters per minute. The height of the cone is 10 meters and the radius of its base is 5 meters. How fast is the water level rising when the water stands 4 meters above the base?
Hint 

The equation for the volume of a cone with height h and radius r is \(V=\pi r^2h/3\).
Solution 

First, let's assign some variables. Call w the height of the water and y the distance from the top of the water to the point of the cone. Since the cone is 10 meters tall, we can write \(y=10w\).
The radius of the top of the water we will call r. Since the water is the larger part of the cone, the volume of the water is given by taking the volume of the entire cone and subtracting the empty space above the water. So the volume of water is given by \(\displaystyle{ V=\frac{\pi(5)^2 (10)}{3}  \frac{\pi r^2 y}{3} }\). We want to try to keep the variable w in the equation since we are given the height of the water is \(w=4\) meters. So substitute our equation for y into the volume equation to get
\(\displaystyle{ V=\frac{\pi(5)^2 (10)}{3}  \frac{\pi r^2 (10w)}{3} }\).
Now we need to find an expression for r in terms of w. To get this, we use the idea of similar triangles. In this problem we have \(\displaystyle{ \frac{r}{5} = \frac{y}{10} }\). Solving for r, we get \(\displaystyle{ r = \frac{10w}{2} }\). Now we are ready to put this expression into the equation for volume.
\(\displaystyle{ V = \frac{205\pi}{3}  \frac{\pi}{3}\left[ \frac{10w}{2} \right]^2 (10w) }\) 
\(\displaystyle{ V = \frac{250\pi}{3}  \frac{\pi}{12}(10w)^3 }\) 
Take the derivative with respect to t. 
\(\displaystyle{ \frac{dV}{dt} = \frac{\pi}{12}(3)(10w)^2 (1)\frac{dw}{dt} }\) 
\(\displaystyle{ \frac{dV}{dt} = \frac{\pi}{4}(10w)^2 \frac{dw}{dt} }\) 
We were given that \(dV/dt = 3 \) and \(w=4\). 
\(\displaystyle{ 3 = \frac{\pi}{4}(104)^2 \frac{dw}{dt} }\) 
\(\displaystyle{ \frac{dw}{dt} = \frac{1}{3\pi} }\) meters/min 
Final Answer 

\(\displaystyle{ \frac{1}{3\pi} }\) meters/min 
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[15 points] Calculate the derivative of \(\displaystyle{ f(x) = \frac{e^{2x}}{x^2+1} }\)
(a) using the product rule
(b) using the quotient rule
(c) Compare your answers to parts (a) and (b). Are they the same? Why or why not?
Problem Statement 

[15 points] Calculate the derivative of \(\displaystyle{ f(x) = \frac{e^{2x}}{x^2+1} }\)
(a) using the product rule
(b) using the quotient rule
(c) Compare your answers to parts (a) and (b). Are they the same? Why or why not?
Final Answer 

\(\displaystyle{ f'(x) = \frac{2e^{2x}(x^2x+1)}{(x^2+1)^2} }\)
Problem Statement 

[15 points] Calculate the derivative of \(\displaystyle{ f(x) = \frac{e^{2x}}{x^2+1} }\)
(a) using the product rule
(b) using the quotient rule
(c) Compare your answers to parts (a) and (b). Are they the same? Why or why not?
Solution 

(a) product rule
Rewrite \(f(x)\) as \( f(x) = e^{2x} (x^2+1)^{1} \) 
\(\displaystyle{ f'(x) = e^{2x}\frac{d}{dx}[(x^2+1)^{1}] + (x^2+1)^{1}\frac{d}{dx}[e^{2x}] }\) 
\(\displaystyle{ f'(x) = e^{2x}(1)(x^2+1)^{2}(2x) + (2e^{2x})(x^2+1)^{1} }\) 
\(\displaystyle{ f'(x) = \frac{2xe^{2x}}{(x^2+1)^2} + \frac{2e^{2x}}{x^2+1} }\) 
Multiply the numerator and denominator of the second term on the right by \(x^2+1\) to get a common denominator. 
\(\displaystyle{ f'(x) = \frac{2xe^{2x}+2(x^2+1)e^{2x}}{(x^2+1)^2} }\) 
\(\displaystyle{ f'(x) = \frac{2e^{2x}(x^2x+1)}{(x^2+1)^2} }\) 
(b) quotient rule
\(\displaystyle{ f'(x) = \frac{(x^2+1)2e^{2x}  e^{2x}(2x)}{(x^2+1)^2} }\) 
\(\displaystyle{ f'(x) = \frac{2e^{2x}(x^2x+1)}{(x^2+1)^2} }\) 
(c) Yes, they are the same as they should always be. I have simplied both expressions for my answers in parts a and b to the same form to show that they are equal.
Final Answer 

\(\displaystyle{ f'(x) = \frac{2e^{2x}(x^2x+1)}{(x^2+1)^2} }\) 
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[10 points] Calculate \(dy/dx\) for \( y = \sqrt{x}e^{x^2}(x^2+1)^{10} \) using logarithmic differentiation.
Problem Statement 

[10 points] Calculate \(dy/dx\) for \( y = \sqrt{x}e^{x^2}(x^2+1)^{10} \) using logarithmic differentiation.
Final Answer 

\(\displaystyle{ \frac{dy}{dx} = \left[ \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right] \sqrt{x}e^{x^2}(x^2+1)^{10} }\)
Problem Statement 

[10 points] Calculate \(dy/dx\) for \( y = \sqrt{x}e^{x^2}(x^2+1)^{10} \) using logarithmic differentiation.
Solution 

\( y = \sqrt{x}e^{x^2}(x^2+1)^{10} \) 
\(\displaystyle{ \ln y = \frac{1}{2}\ln x + x^2 + 10\ln(x^2+1) }\) 
\(\displaystyle{ \frac{1}{y}\frac{dy}{dx} = \frac{1}{2x} + 2x + \frac{10}{x^2+1}(2x) }\) 
\(\displaystyle{ \frac{dy}{dx} = \left[ \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right] \sqrt{x}e^{x^2}(x^2+1)^{10} }\) 
Final Answer 

\(\displaystyle{ \frac{dy}{dx} = \left[ \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right] \sqrt{x}e^{x^2}(x^2+1)^{10} }\) 
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[10 points] Calculate the equation of the tangent line, in slopeintercept form, to the graph of the equation \(y+\sqrt{xy} = 6\) at \(y=4\). Accurately plot your tangent line on the graph. Note: Your line on the graph must match your calculations.
Problem Statement 

[10 points] Calculate the equation of the tangent line, in slopeintercept form, to the graph of the equation \(y+\sqrt{xy} = 6\) at \(y=4\). Accurately plot your tangent line on the graph. Note: Your line on the graph must match your calculations.
Final Answer 

\(\displaystyle{ y = \frac{4x}{5}+ \frac{24}{5} }\)
Problem Statement 

[10 points] Calculate the equation of the tangent line, in slopeintercept form, to the graph of the equation \(y+\sqrt{xy} = 6\) at \(y=4\). Accurately plot your tangent line on the graph. Note: Your line on the graph must match your calculations.
Solution 

\( y+\sqrt{xy} = 6 \) 
\(\displaystyle{ \frac{dy}{dx} + \frac{d}{dx}[(xy)^{1/2}] = \frac{d}{dx}[6] }\) 
\(\displaystyle{ \frac{dy}{dx} + \frac{1}{2}(xy)^{1/2}\frac{d}{dx}[xy] = 0 }\) 
\(\displaystyle{ \frac{dy}{dx} + \frac{1}{2\sqrt{xy}}\left[ x\frac{dy}{dx}+ y(1) \right] = 0 }\) 
\(\displaystyle{ \frac{dy}{dx} + \frac{x}{2\sqrt{xy}} \frac{dy}{dx} + \frac{y}{2\sqrt{xy}} = 0 }\) 
\(\displaystyle{ \frac{dy}{dx} \left[ 1 + \frac{x}{2\sqrt{xy}} \right] = \frac{y}{2\sqrt{xy}} }\) 
Now substitute \(y=4\). 
\(\displaystyle{ \frac{dy}{dx}\left[ 1 + \frac{1}{4} \right] = \frac{4}{2(2)} }\) 
\(\displaystyle{ \frac{dy}{dx} (5/4) = 1 }\) 
\(\displaystyle{ \frac{dy}{dx} = 4/5 }\) 
So the slope is \(m=4/5\). To find the point, we plug in \(y=4\) in the original equation to get \(x=1\).
\(yy_1 = m(xx_1) \)
\(y4 = (4/5)(x1) \)
Final Answer 
\(\displaystyle{ y = \frac{4x}{5}+ \frac{24}{5} }\) 
Final Answer 

\(\displaystyle{ y = \frac{4x}{5}+ \frac{24}{5} }\) 
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[15 points] Calculate the derivative of \(f(t) = \tan(\arccos(3t)) \) two ways. First, directly and secondly, by converting \(f(t)\) to algebraic form then taking the derivative. You do not need to show that your answers are equal.
Problem Statement 

[15 points] Calculate the derivative of \(f(t) = \tan(\arccos(3t)) \) two ways. First, directly and secondly, by converting \(f(t)\) to algebraic form then taking the derivative. You do not need to show that your answers are equal.
Solution 

First, take the derivative directly.
\(f(t) = \tan(\arccos(3t)) \) 
\(\displaystyle{ f'(t) = \sec^2(\arccos(3t))\frac{d}{dt}[\arccos(3t)] }\) 
\(\displaystyle{ f'(t) = \sec^2(\arccos(3t))\frac{3}{\sqrt{19t^2}} }\) 
Final Answer  First Part 
\(\displaystyle{ f'(t) = \frac{3\sec^2(\arccos(3t))}{\sqrt{19t^2}} }\) 
Secondly, converting to algebraic terms then taking the derivative.
Starting with the inside function, \(\theta = \arccos(3t) \to \cos\theta = 3t\).
Now draw a triangle matching \( \cos\theta = 3t \). From this triangle, the expression for \(\tan\theta\) is
\(\displaystyle{ f(t) = \tan\theta = \frac{\sqrt{19t^2}}{3t} }\). Now take the derivative.
\(\displaystyle{ f(t) = \frac{\sqrt{19t^2}}{3t} }\) 
Use the quotient rule. 
\(\displaystyle{ f'(t) = \frac{3t/2(19t^2)^{1/2}(18t)  (19t^2)^{1/2}(3)}{9t^2} }\) 
Simplifying, we get the final answer.
Final Answer  Second Part 
\(\displaystyle{ f'(t) = \frac{1}{3t^2(19t^2)^{1/2}} }\) 
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You CAN Ace Calculus
related rates 
other exams from calculus 1 

The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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Single Variable Calculus 

MultiVariable Calculus 

Differential Equations 

Precalculus 

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