This is the first exam for first semester single variable calculus. Topics include limits, continuity, basic derivatives, implicit differentiation, tangent lines and applications of the derivative, related rates, critical points, maximums and minimums, concavity and graphing.
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.
 Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.
 Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).
 Use your calculator as little as possible except for graphing and checking your calculations.
 Work the entire exam before checking any solutions.
 After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.
 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  1.5 hours 
Questions  17 
Total Points  100 
Tools  

Calculator  see instructions 
Formula Sheet(s)  none 
Other Tools  none 
Instructions:
 This exam is in two main parts, labeled parts A and B, with different instructions for each part.
 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.
Part A  Questions 111
Instructions for Part A   You have 30 minutes to complete this part of the exam. No calculators are allowed. All questions in this section are worth 3 points, except for question 10, which is worth 4 points.
Section 1 

Evaluate the indicated limit or determine that it doesn't exist.
\(\displaystyle{ \lim_{\theta \to 0}{ \left[ \frac{\sin(\theta)}{\tan(3\theta)} \right] } }\)
Problem Statement 

\(\displaystyle{ \lim_{\theta \to 0}{ \left[ \frac{\sin(\theta)}{\tan(3\theta)} \right] } }\)
Final Answer 

\(\displaystyle{ \lim_{\theta \to 0}{ \left[ \frac{\sin(\theta)}{\tan(3\theta)} \right] } = 1/3 }\)
Problem Statement 

\(\displaystyle{ \lim_{\theta \to 0}{ \left[ \frac{\sin(\theta)}{\tan(3\theta)} \right] } }\)
Solution 

Direct substitution yields \( 0/0 \) which is indeterminate. So we need to do some algebra to be able to use one of the trig limits.
\(\displaystyle{ \lim_{\theta \to 0}{\left[ \frac{\sin(\theta)}{\tan(3\theta)} \right]} }\) 
\(\displaystyle{\lim_{\theta \to 0}{\left[ \sin(\theta) \frac{\cos(3\theta)}{\sin(3\theta)} \right]}}\) 
\(\displaystyle{\lim_{\theta \to 0}{\left[\sin(\theta) \frac{\theta}{\theta}\frac{3}{3}\frac{\cos(3\theta)}{\sin(3\theta)} \right]}} \) 
\(\displaystyle{\lim_{\theta \to 0}{\left[\frac{\sin(\theta)}{\theta}\frac{3\theta}{\sin(3\theta)}\frac{\cos(3\theta)}{3} \right]}} \) 
\(\displaystyle{\lim_{\theta \to 0}{\left[\frac{\sin(\theta)}{\theta}\right]}\cdot \lim_{\theta \to 0}{\left[ \frac{3\theta}{\sin(3\theta)}\right]}\cdot \lim_{\theta \to 0}{\left[\frac{\cos(3\theta)}{3} \right]}} \) 
\(\displaystyle{(1)(1)\frac{1}{3} = 1/3} \) 
Final Answer 

\(\displaystyle{ \lim_{\theta \to 0}{ \left[ \frac{\sin(\theta)}{\tan(3\theta)} \right] } = 1/3 }\) 
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\(\displaystyle{ \lim_{x\to2}{ \left[ \frac{x^2+x6}{x2} \right] } }\)
Problem Statement 

\(\displaystyle{ \lim_{x\to2}{ \left[ \frac{x^2+x6}{x2} \right] } }\)
Final Answer 

\(\displaystyle{ \lim_{x\to2}{ \left[ \frac{x^2+x6}{x2} \right] } = 5 }\)
Problem Statement 

\(\displaystyle{ \lim_{x\to2}{ \left[ \frac{x^2+x6}{x2} \right] } }\)
Solution 

Direct substitution yields \(0/0\) which is indeterminate. Let's see if we can factor the numerator and, perhaps, cancel with the \((x2)\) factor in the denominator.
\(\displaystyle{\lim_{x\to 2}{\left[ \frac{x^2+x6}{x2} \right]} = }\) \(\displaystyle{\lim_{x \to 2}{\left[ \frac{(x+3)(x2)}{(x2)}\right]} = }\) \(\displaystyle{\lim_{x \to 2}{(x+3)} = 2+3 = 5}\)
Final Answer 

\(\displaystyle{ \lim_{x\to2}{ \left[ \frac{x^2+x6}{x2} \right] } = 5 }\) 
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\(\displaystyle{ \lim_{x\to3}{ \left[ \frac{x^24x+6}{x^29} \right] } }\)
Problem Statement 

\(\displaystyle{ \lim_{x\to3}{ \left[ \frac{x^24x+6}{x^29} \right] } }\)
Final Answer 

\(\displaystyle{ \lim_{x\to 3}{ \left[ \frac{x^24x+6}{x^29} \right] } }\) does not exist
Problem Statement 

\(\displaystyle{ \lim_{x\to3}{ \left[ \frac{x^24x+6}{x^29} \right] } }\)
Solution 

Final Answer 

\(\displaystyle{ \lim_{x\to 3}{ \left[ \frac{x^24x+6}{x^29} \right] } }\) does not exist 
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\(\displaystyle{ \lim_{x\to1}{ \left[ \frac{x1}{xx^3} \right] } }\)
Problem Statement 

\(\displaystyle{ \lim_{x\to1}{ \left[ \frac{x1}{xx^3} \right] } }\)
Final Answer 

\(\displaystyle{ \lim_{x\to1}{ \left[ \frac{x1}{xx^3} \right] } = 1/2 }\)
Problem Statement 

\(\displaystyle{ \lim_{x\to1}{ \left[ \frac{x1}{xx^3} \right] } }\)
Solution 

Direct substitution yields \(0/0\) which is indeterminate. So let's factor and see what we get.
\( \displaystyle{\lim_{x\to 1}{\left[\frac{x1}{xx^3}\right]}} \) 
\( \displaystyle{\lim_{x\to1}{\left[\frac{x1}{x(1x^2)}\right]}} \) 
\( \displaystyle{\lim_{x\to1}{\left[\frac{x1}{x(x^21)}\right]}} \) 
\( \displaystyle{\lim_{x\to1}{\left[\frac{x1}{x(x1)(x+1)}\right]}} \) 
\( \displaystyle{\lim_{x\to1}{\left[\frac{1}{x(x+1)}\right]}} \) 
\( \displaystyle{\frac{1}{1(1+1)}=1/2} \) 
Final Answer 

\(\displaystyle{ \lim_{x\to1}{ \left[ \frac{x1}{xx^3} \right] } = 1/2 }\) 
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\(\displaystyle{ \lim_{x\to\infty}{ \left[ \frac{5x1}{3x^2+1/4} \right] } }\)
Problem Statement 

\(\displaystyle{ \lim_{x\to\infty}{ \left[ \frac{5x1}{3x^2+1/4} \right] } }\)
Final Answer 

\(\displaystyle{ \lim_{x\to\infty}{ \left[ \frac{5x1}{3x^2+1/4} \right] } = 0 }\)
Problem Statement 

\(\displaystyle{ \lim_{x\to\infty}{ \left[ \frac{5x1}{3x^2+1/4} \right] } }\)
Solution 

Direct substitution yields \(\infty / \infty\) which is indeterminate. Since we can't factor, let's try dividing the numerator and denominator by the highest factor, in this case \( x^2 \), and then use the Infinite Limits Theorem.
\(\displaystyle{ \lim_{x\to\infty}{\left[\frac{5x1}{3x^2+1/4}\frac{1/x^2}{1/x^2}\right]}= }\) \(\displaystyle{ \lim_{x\to\infty}{\left[\frac{5/x1/x^2}{3+1/(4x^2)}\right]}= }\) \(\displaystyle{ \frac{0}{3}=0 }\)
Final Answer 

\(\displaystyle{ \lim_{x\to\infty}{ \left[ \frac{5x1}{3x^2+1/4} \right] } = 0 }\) 
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Section 2 

Evaluate the following derivatives and simplify.
\(\displaystyle{ \frac{d}{dx} \left[ (3x^2+x+1)^{2010} \right] }\)
Problem Statement 

\(\displaystyle{ \frac{d}{dx} \left[ (3x^2+x+1)^{2010} \right] }\)
Final Answer 

\( 2010 (6x+1) (3x^2+x+1)^{2009} \)
Problem Statement 

\(\displaystyle{ \frac{d}{dx} \left[ (3x^2+x+1)^{2010} \right] }\)
Solution 

Use the chain rule.
\(\displaystyle{ \frac{d}{dx}\left[(3x^2+x+1)^{2010}\right]= }\) \(\displaystyle{ 2010(3x^2+x+1)^{2009}\cdot }\) \(\displaystyle{ \frac{d}{dx}\left[3x^2+x+1\right]= }\) \(\displaystyle{ 2010(3x^2+x+1)^{2009}(6x+1) }\)
Final Answer 

\( 2010 (6x+1) (3x^2+x+1)^{2009} \) 
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\(\displaystyle{ \frac{d}{dx} \left[ \frac{\sin^2(x)}{\cos(x)} \right] }\)
Problem Statement 

\(\displaystyle{ \frac{d}{dx} \left[ \frac{\sin^2(x)}{\cos(x)} \right] }\)
Final Answer 

\(\displaystyle{ \frac{\sin(x)[ 2\cos^2(x) + \sin^2(x) ]}{\cos^2(x)} }\)
Problem Statement 

\(\displaystyle{ \frac{d}{dx} \left[ \frac{\sin^2(x)}{\cos(x)} \right] }\)
Solution 

Use the quotient rule.
\(\displaystyle{ \frac{d}{dx}\left[\frac{\sin^2(x)}{\cos(x)}\right] }\) 
\(\displaystyle{ \frac{\cos(x)(2\sin(x))\cos(x)\sin^2(x)(\sin(x))}{\cos^2(x)} }\) 
\(\displaystyle{ \frac{\sin(x)[2\cos^2(x)+\sin^2 (x)]}{\cos^2 (x)} }\) 
Final Answer 

\(\displaystyle{ \frac{\sin(x)[ 2\cos^2(x) + \sin^2(x) ]}{\cos^2(x)} }\) 
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Find \(D_x y\) for \(y=\sec^2(4x)\)
Problem Statement 

Find \(D_x y\) for \(y=\sec^2(4x)\)
Final Answer 

\( 8 \sec^2(4x) \tan(4x) \)
Problem Statement 

Find \(D_x y\) for \(y=\sec^2(4x)\)
Solution 

Use the chain rule.
\(\displaystyle{ D_x[\sec^2 (4x)] }\) 
\(\displaystyle{ 2\sec(4x)\frac{d}{dx}[\sec(4x)] }\) 
\( 2\sec(4x)[ \sec(4x)\tan(4x) ](4) \) 
\( 8\sec^2(4x)\tan(4x) \) 
Final Answer 

\( 8 \sec^2(4x) \tan(4x) \) 
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Find \(f'(3)\) for \(f(x)=(2+x^2)^{4/3}\)
Problem Statement 

Find \(f'(3)\) for \(f(x)=(2+x^2)^{4/3}\)
Final Answer 

\( f'(3) = 8(11)^{1/3} \)
Problem Statement 

Find \(f'(3)\) for \(f(x)=(2+x^2)^{4/3}\)
Solution 

Use the chain rule.
\(\displaystyle{ f'(x) = \frac{4}{3}(2+x^2)^{1/3}(2x) }\)
\(\displaystyle{ f'(3) = \frac{4}{3}(2+3^2)^{1/3}(6) = 8(11)^{1/3} }\)
Final Answer 

\( f'(3) = 8(11)^{1/3} \) 
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Find \(dy/dx\) using implicit differentiation for the curve \( 2x  y^2 = \cos(xy) + 5 \)
Problem Statement 

Find \(dy/dx\) using implicit differentiation for the curve \( 2x  y^2 = \cos(xy) + 5 \)
Final Answer 

\(\displaystyle{ \frac{dy}{dx} = \frac{2+y\sin(xy)}{2yx\sin(xy)} }\)
Problem Statement 

Find \(dy/dx\) using implicit differentiation for the curve \( 2x  y^2 = \cos(xy) + 5 \)
Solution 

\( \displaystyle{\frac{d}{dx}\left[2xy^2\right] = \frac{d}{dx}\left[\cos(xy)+5\right]} \) 
\( \displaystyle{22y \frac{dy}{dx} = \frac{d}{dx}\left[\cos(xy)\right]+0} \) 
\( \displaystyle{22y \frac{dy}{dx} = \sin(xy)\frac{d}{dx}[xy]} \) 
\( \displaystyle{22y \frac{dy}{dx} = \sin(xy)\left[x\frac{dy}{dx}+y(1)\right]} \) 
\( \displaystyle{22y \frac{dy}{dx} = x\sin(xy)\frac{dy}{dx}y\sin(xy)} \) 
\( \displaystyle{2y\frac{dy}{dx}+x\sin(xy)\frac{dy}{dx} = 2y\sin(xy) }\) 
\( \displaystyle{(2y+x\sin(xy))\frac{dy}{dx}} = 2y\sin(xy) \) 
\( \displaystyle{\frac{dy}{dx} = \frac{2y\sin(xy)}{2y+x\sin(xy)} }\) 
Final Answer 

\(\displaystyle{ \frac{dy}{dx} = \frac{2+y\sin(xy)}{2yx\sin(xy)} }\) 
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Find \( D_x^{(101)}[\cos(x)] \)
Problem Statement 

Find \( D_x^{(101)}[\cos(x)] \)
Final Answer 

\( D_x^{(101)}[\cos(x)]=\sin(x) \)
Problem Statement 

Find \( D_x^{(101)}[\cos(x)] \)
Solution 

In order to work this problem, we need to find a pattern. So let's build a table.
\(\begin{array}{lrcl} 0 & y & = & \cos(x) \\ 1 & y' & = & \sin(x) \\ 2 & y'' & = & \cos(x) \\ 3 & y^{(3)} & = & \sin(x) \\ 4 & y^{(4)} & = & \cos(x) \\ 5 & y^{(5)} & = & \sin(x) \end{array}\)
So this repeats every 4th derivative.
\(D_x^{(101)}[\cos(x)]=D_x^1[\cos(x)]=\sin(x)\)
Final Answer 

\( D_x^{(101)}[\cos(x)]=\sin(x) \) 
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Part B  Questions 1217
Instructions for Part B   You may use your calculator. You have one hour to complete this part of the exam. Show all your work. Correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions. Give exact, simplified answers.
Section 3 

Solve the following problems.
[12 points] Find the equation of the tangent line to \(\displaystyle{ y = \frac{1}{x^2+4} }\) at \( x = 2 \). Give your answer in slopeintercept form.
Problem Statement 

[12 points] Find the equation of the tangent line to \(\displaystyle{ y = \frac{1}{x^2+4} }\) at \( x = 2 \). Give your answer in slopeintercept form.
Final Answer 

\(\displaystyle{ y = \frac{x}{16} + \frac{1}{4} }\)
Problem Statement 

[12 points] Find the equation of the tangent line to \(\displaystyle{ y = \frac{1}{x^2+4} }\) at \( x = 2 \). Give your answer in slopeintercept form.
Solution 

The equation of a line requires a point and a slope. To find the slope we need the derivative. We could use the quotient rule, however, since the numerator is a constant, it is easier to rewrite the function and use the chain rule.
\(y=(x^2+4)^{1}\)
\(\displaystyle{y'=(1)(x^2+4)^{2}(2x)=\frac{2x}{(x^2+4)^2}}\)
Now we need the slope at \(x=2\).
\(\displaystyle{y'(2)=\frac{2(2)}{(2^2+4)^2}=\frac{4}{8^2}=\frac{4}{64}=\frac{1}{16}}\)
So the slope at \(x=2\) is \(m=1/16\). Now we need the point. To find the yvalue of the function at \(x=2\), we calculate \(y(2)\).
\(\displaystyle{y(2)=\frac{1}{2^2+4}=\frac{1}{8}}\)
So we have the point \((2,1/8)\) and slope \(m=1/16\). We can find the equation of the tangent line.
\(\begin{array}{rcl} \displaystyle{y\frac{1}{8}} & = & \displaystyle{\frac{1}{16}(x2)} \\ y & = & \displaystyle{\frac{x}{16}+\frac{1}{8}+\frac{1}{8}} \\ y & = & \displaystyle{\frac{x}{16}+\frac{1}{4}} \end{array}\)
Although we were not asked to graph the function and it's tangent line in the problem statement, if you have a graphing calculator with you, it's always a good idea to plot them quickly to make sure your answer makes sense.
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Final Answer 

\(\displaystyle{ y = \frac{x}{16} + \frac{1}{4} }\) 
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[12 points] Sketch the graph of a function \(g(x)\) that satisfies the following conditions. Be sure to label your graph and axes.
a. \(g(x)\) is continuous at \(x=1\); b. \(g'(1)\) does not exist; c. \(g(1)=g(2)=3\); d. \(\displaystyle{\lim_{x\to2^+}{g(x)}=\infty}\); e. \(\displaystyle{\lim_{x\to\infty}{g(x)}=1}\)
Problem Statement 

[12 points] Sketch the graph of a function \(g(x)\) that satisfies the following conditions. Be sure to label your graph and axes.
a. \(g(x)\) is continuous at \(x=1\); b. \(g'(1)\) does not exist; c. \(g(1)=g(2)=3\); d. \(\displaystyle{\lim_{x\to2^+}{g(x)}=\infty}\); e. \(\displaystyle{\lim_{x\to\infty}{g(x)}=1}\)
Solution 

one possible solution 

There are many possible answers. The key points are as follows.
a. \(g(x)\) is continuous at \(x=1\)
This means that the limit from the right and left of \(x=1\) must be equal to the value of \(g(1)\).
b. \(g'(1)\) does not exist
Together with part (a), this situation occurs when there is a sharp corner at that point.
c. \(g(1)=g(2)=3\)
This gives the actual values we need to have at \(x=1\) and \(x=2\). To satisfy this requirement, we need to have two closed circles on the graph indicating these points. We also need to connect these points so as to satisfy requirement (a).
d. \(\displaystyle{\lim_{x\to2^+}{g(x)}=\infty}\)
This requires that the righthanded limit at \(x=2\) go to negative infinity. So \(x=2\) is a vertical asymptote.
e. \(\displaystyle{\lim_{x\to\infty}{g(x)}=1}\)
Finally, this requires the limit of the function to go to \(1\) as \(x\) goes to \(+\infty\), i.e. we need a horizontal asymptote at \(y=1\). It doesn't matter if the limit is above or below the line \(y=1\).
As long as your graph follows these requirements and it is a function (passes the vertical line test), then it will be correct.
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[12 points] Identify all critical points and find the maximum and minimum value of the function \( f(x) = x^3  3x^2 + 1 \) on \( S = [1,4] \).
Problem Statement 

[12 points] Identify all critical points and find the maximum and minimum value of the function \( f(x) = x^3  3x^2 + 1 \) on \( S = [1,4] \).
Final Answer 

The critical points are \((0,1)\) and \((2,3)\). The minimums occurs at the points \((1,3)\) and \((2,3)\) and the maximum occurs at the point \((4,17)\).
Problem Statement 

[12 points] Identify all critical points and find the maximum and minimum value of the function \( f(x) = x^3  3x^2 + 1 \) on \( S = [1,4] \).
Solution 

Critical Points
Since \(f(x)\) is continuous for all real numbers, the only critical points will be where \(f'(x)=0\).
\(f'(x) = 3x^2  6x = 0\)
\(3x(x2)=0 \)
\(3x = 0 ~~ \to ~~ x=0\)
\(x2 = 0 ~~ \to ~~ x=2\)
So, \(x=0\) and \(x=2\) are the critical numbers. Since the question used the term critical 'points', we need to find each yvalue for each corresponding critical number.
\(f(0) = 0^33(0)^2+1 = 1\)
\(f(2) = 2^3  3(2)^2+1 = 8 3(4)+1 = 812+1 = 3\)
So the critical points are \((0,1)\) and \((2,3)\).
Maximum and Minimum
Now we need to find the maximum and minimum values of the function. Since we have a closed interval, we need to check the endpoints as well.
Previously calculated, we have
\(f(0) = 1\)
\(f(2) = 3\)
\(f(1) = (1)^33(1)^2+1 = 13+1 = 3\)
\(f(4) = 4^3 3(4)^2_1 = 6448+1 = 17\)
Comparing the numbers \(\{1, 3, 3, 17\}\), \(3\) is the smallest, so \((1,3)\) and \((2,3)\) are the minimums; \(17\) is the largest, so \((4,17)\) is the maximum. Since this is a closed interval, these points are the absolute minimums and absolute maximum.
Although the problem statement did not ask for a graph, if you have a graphing calculator and some extra time after you have worked all the problems on the exam, it's a good idea to quickly plot a graph to make sure your answers make sense.
Final Answer 

The critical points are \((0,1)\) and \((2,3)\). The minimums occurs at the points \((1,3)\) and \((2,3)\) and the maximum occurs at the point \((4,17)\). 
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[15 points] A ladder 3 meters long, with one end on the ground, is raised so that the angle \(\theta\) of the ladder with the ground increases by 2.5 radians per minute. Consider the right triangle formed by the ladder, the ground, and a vertical line through the top of the ladder. How fast is the area of this triangle increasing at the time when the angle \(\theta\) equals \(\pi/6\)?
Problem Statement 

[15 points] A ladder 3 meters long, with one end on the ground, is raised so that the angle \(\theta\) of the ladder with the ground increases by 2.5 radians per minute. Consider the right triangle formed by the ladder, the ground, and a vertical line through the top of the ladder. How fast is the area of this triangle increasing at the time when the angle \(\theta\) equals \(\pi/6\)?
Final Answer 

\(\displaystyle{ \frac{dA}{dt} = \frac{45}{8} }\) m^{2}/min
Problem Statement 

[15 points] A ladder 3 meters long, with one end on the ground, is raised so that the angle \(\theta\) of the ladder with the ground increases by 2.5 radians per minute. Consider the right triangle formed by the ladder, the ground, and a vertical line through the top of the ladder. How fast is the area of this triangle increasing at the time when the angle \(\theta\) equals \(\pi/6\)?
Solution 

First, let's draw the triangle. We have labeled the height \(h\) and base \(b\).
Given:
\(\displaystyle{\frac{d\theta}{dt}=2.5 rad/min}\) and ladder length \(3\) meters.
When \(\theta =\pi/6\), we are asked to calculate \(dA/dt\) where \(A\) is the area of the triangle.
Equations 

\( A = bh/2 \) 
\( \cos(\theta) = b/3 \to b = 3\cos(\theta) \) 
\( \sin(\theta) = h/3 \to h = 3\sin(\theta) \) 
Solving:
\(\displaystyle{A=\frac{1}{2}(3\cos(\theta))(3\sin(\theta))=\frac{9}{2}\cos(\theta)\sin(\theta)}\)
\(\displaystyle{\frac{dA}{dt}=}\) \(\displaystyle{\frac{9}{2}\left[\cos(\theta)(\cos(\theta))\frac{d\theta}{dt}+\sin(\theta)(\sin(\theta))\frac{d\theta}{dt}\right]=}\) \(\displaystyle{\frac{9}{2}\left[\cos^2(\theta)\sin^2(\theta)\right]\frac{d\theta}{dt}}\)
at \(\theta=\pi/6 ~~ d\theta/dt=2.5 rad/min\)
\( \displaystyle{\frac{dA}{dt}} \) 
\( \displaystyle{\frac{9}{2}\left[\cos^2(\pi/6)\sin^2(\pi/6)\right] (m^2)(2.5 rad/min)} \) 
\( \displaystyle{\frac{9}{2}\left[\frac{3}{4}\frac{1}{4}\right] 2.5 m^2/min} \) 
\( \displaystyle{\left[\frac{9}{4}\cdot\frac{5}{2}\right] m^2/min} \) 
\( \displaystyle{\frac{45}{8} m^2/min} \) 
Final Answer 

\(\displaystyle{ \frac{dA}{dt} = \frac{45}{8} }\) m^{2}/min 
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Section 4 

For the questions in this section, consider the function \( f(x) = x^44x^310 \).
[ click here to show/hide the plot of f(x) ]
[8 points] Find the intervals on which \(f(x)\) is decreasing and the intervals where \(f(x)\) is increasing.
Problem Statement 

[8 points] Find the intervals on which \(f(x)\) is decreasing and the intervals where \(f(x)\) is increasing.
Final Answer 

The function is decreasing on the intervals \((\infty,0)\) and \((0,3)\) and increasing on the interval \((3,\infty)\).
Problem Statement 

[8 points] Find the intervals on which \(f(x)\) is decreasing and the intervals where \(f(x)\) is increasing.
Solution 

Increasing and decreasing intervals are determined by the critical points. The critical points are points where the derivative is zero or the derivative doesn't exist. So our first step is to find the critical points.
\(f'(x) = 4x^3  12x^2\)
\(4x^3  12x^2 = 0\)
\(4x^2(x3) = 0\)
\(4x^2 = 0 ~~ \to ~~ x = 0\)
\(x3=0 ~~ \to ~~ x=3\)
So our critical values are \(x=0\) and \(x=3\). Now let's build a table to determine increasing and decreasing intervals.
Interval  \(\infty < x < 0\)  \(0 < x < 3\)  \(3 < x < \infty\) 

Test xvalue  \(x=1\)  \(x=1\)  \(x=4\) 
Sign of \(f'(x)\)  \(f'(1)=16<0\)  \(f'(1)=8<0\)  \(f'(4)=64>0\) 
Conclusion  decreasing  decreasing  increasing 
Although we were not asked to graph the function, it is always good to quickly plot it on your calculator to check your answers if there is adequate time.
Final Answer 

The function is decreasing on the intervals \((\infty,0)\) and \((0,3)\) and increasing on the interval \((3,\infty)\). 
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[7 points] Find the intervals on which \(f(x)\) is concave up and the intervals where \(f(x)\) is concave down.
Problem Statement 

[7 points] Find the intervals on which \(f(x)\) is concave up and the intervals where \(f(x)\) is concave down.
Final Answer 

The function is concave up on the intervals \((\infty,0)\) and \((2,\infty)\) and concave down on the interval \((0,2)\).
Problem Statement 

[7 points] Find the intervals on which \(f(x)\) is concave up and the intervals where \(f(x)\) is concave down.
Solution 

Change in concavity is determined by inflection points. Inflection points occur where the second derivative is zero or the second derivative does not exist. From the previous question, we know that \(f'(x) = 4x^3  12x^2\).
\(f''(x) = 12x^224x\)
\(12x^224x = 0\)
\(12x(x2) = 0\)
\(12x=0 ~~ \to ~~ x = 0\)
\(x2 = 0 ~~ \to ~~ x=2\)
Since the second derivative is continuous for all real numbers, the only inflection points we have are \( x=0\) and \(x=2\).
So, now we will build a table to determine concavity.
Interval  \(\infty < x < 0\)  \(0 < x < 2\)  \(2 < x < \infty\) 

Test xvalue  \(x=1\)  \(x=1\)  \(x=3\) 
Sign of \(f''(x)\)  \(f''(1)=36>0\)  \(f''(1)=12<0\)  \(f''(3)=36>0\) 
Conclusion  concave up  concave down  concave up 
Although we were not asked to graph the function, it is always good to quickly plot it on your calculator to check your answers if there is adequate time.
Final Answer 

The function is concave up on the intervals \((\infty,0)\) and \((2,\infty)\) and concave down on the interval \((0,2)\). 
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You CAN Ace Calculus
Most limit topics are required for this exam except for L'Hôpital's Rule which is usually covered in calculus 2. 
all derivative topics 
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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