## 17Calculus - Calculus 1 - Practice Exam 1 (Midterm) (Semester A)

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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.

### Practice Exam Tips

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.
- 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).

Part A - Questions 1-11

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

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

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

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.

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

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

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$$\displaystyle{ \lim_{x\to2}{ \left[ \frac{x^2+x-6}{x-2} \right] } }$$

Problem Statement

$$\displaystyle{ \lim_{x\to2}{ \left[ \frac{x^2+x-6}{x-2} \right] } }$$

$$\displaystyle{ \lim_{x\to2}{ \left[ \frac{x^2+x-6}{x-2} \right] } = 5 }$$

Problem Statement

$$\displaystyle{ \lim_{x\to2}{ \left[ \frac{x^2+x-6}{x-2} \right] } }$$

Solution

Direct substitution yields $$0/0$$ which is indeterminate. Let's see if we can factor the numerator and, perhaps, cancel with the $$(x-2)$$ factor in the denominator.
$$\displaystyle{\lim_{x\to 2}{\left[ \frac{x^2+x-6}{x-2} \right]} = }$$ $$\displaystyle{\lim_{x \to 2}{\left[ \frac{(x+3)(x-2)}{(x-2)}\right]} = }$$ $$\displaystyle{\lim_{x \to 2}{(x+3)} = 2+3 = 5}$$

$$\displaystyle{ \lim_{x\to2}{ \left[ \frac{x^2+x-6}{x-2} \right] } = 5 }$$

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$$\displaystyle{ \lim_{x\to3}{ \left[ \frac{x^2-4x+6}{x^2-9} \right] } }$$

Problem Statement

$$\displaystyle{ \lim_{x\to3}{ \left[ \frac{x^2-4x+6}{x^2-9} \right] } }$$

$$\displaystyle{ \lim_{x\to 3}{ \left[ \frac{x^2-4x+6}{x^2-9} \right] } }$$ does not exist

Problem Statement

$$\displaystyle{ \lim_{x\to3}{ \left[ \frac{x^2-4x+6}{x^2-9} \right] } }$$

Solution

p>As usual, first try direct substitution.
$$\displaystyle{ \lim_{x\to3}{ \left[ \frac{x^2-4x+6}{x^2-9} \right] } = \frac{9-12+6}{9-9} = \frac{3}{0} }$$
So we have three possible situations here.
1. The limit goes to $$+\infty$$.
2. The limit goes to $$-\infty$$.
3. The limit does not exist.
We need to look at what happens on both sides of $$x=3$$ to determine which case we have.
Left: $$\displaystyle{ \lim_{x\to 3^-}{\left[\frac{x^2-4x+6}{x^2-9}\right]} }$$ $$\to$$ $$\displaystyle{ \frac{3}{negative} }$$ $$\to$$ $$-\infty$$
Right: $$\displaystyle{ \lim_{x\to 3^+}{\left[ \frac{x^2-4x+6}{x^2-9} \right]} }$$ $$\to$$ $$\displaystyle{ \frac{3}{positive} }$$ $$\to$$ $$+\infty$$
Since the limit approaches $$-\infty$$ on the left and $$+\infty$$ on the right, and the left and right limits are not equal, the limit does not exist.
Note - - Check the discussion here to understand the difference between when a limit goes to $$\pm \infty$$ and when a limit does not exist.

$$\displaystyle{ \lim_{x\to 3}{ \left[ \frac{x^2-4x+6}{x^2-9} \right] } }$$ does not exist

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$$\displaystyle{ \lim_{x\to1}{ \left[ \frac{x-1}{x-x^3} \right] } }$$

Problem Statement

$$\displaystyle{ \lim_{x\to1}{ \left[ \frac{x-1}{x-x^3} \right] } }$$

$$\displaystyle{ \lim_{x\to1}{ \left[ \frac{x-1}{x-x^3} \right] } = -1/2 }$$

Problem Statement

$$\displaystyle{ \lim_{x\to1}{ \left[ \frac{x-1}{x-x^3} \right] } }$$

Solution

Direct substitution yields $$0/0$$ which is indeterminate. So let's factor and see what we get.

 $$\displaystyle{\lim_{x\to 1}{\left[\frac{x-1}{x-x^3}\right]}}$$ $$\displaystyle{\lim_{x\to1}{\left[\frac{x-1}{x(1-x^2)}\right]}}$$ $$\displaystyle{\lim_{x\to1}{\left[\frac{x-1}{-x(x^2-1)}\right]}}$$ $$\displaystyle{\lim_{x\to1}{\left[\frac{x-1}{-x(x-1)(x+1)}\right]}}$$ $$\displaystyle{\lim_{x\to1}{\left[\frac{1}{-x(x+1)}\right]}}$$ $$\displaystyle{\frac{1}{-1(1+1)}=-1/2}$$

$$\displaystyle{ \lim_{x\to1}{ \left[ \frac{x-1}{x-x^3} \right] } = -1/2 }$$

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$$\displaystyle{ \lim_{x\to\infty}{ \left[ \frac{5x-1}{3x^2+1/4} \right] } }$$

Problem Statement

$$\displaystyle{ \lim_{x\to\infty}{ \left[ \frac{5x-1}{3x^2+1/4} \right] } }$$

$$\displaystyle{ \lim_{x\to\infty}{ \left[ \frac{5x-1}{3x^2+1/4} \right] } = 0 }$$

Problem Statement

$$\displaystyle{ \lim_{x\to\infty}{ \left[ \frac{5x-1}{3x^2+1/4} \right] } }$$

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.

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

$$\displaystyle{ \lim_{x\to\infty}{ \left[ \frac{5x-1}{3x^2+1/4} \right] } = 0 }$$

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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

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

$$2010 (6x+1) (3x^2+x+1)^{2009}$$

Problem Statement

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

Solution

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

$$2010 (6x+1) (3x^2+x+1)^{2009}$$

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

Problem Statement

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

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

Problem Statement

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

Solution

Use the quotient rule.

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

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

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

Problem Statement

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

$$8 \sec^2(4x) \tan(4x)$$

Problem Statement

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

Solution

Use the chain rule.

 $$\displaystyle{ D_x[\sec^2 (4x)] }$$ $$\displaystyle{ 2\sec(4x)\frac{d}{dx}[\sec(4x)] }$$ $$2\sec(4x)[ \sec(4x)\tan(4x) ](4)$$ $$8\sec^2(4x)\tan(4x)$$

$$8 \sec^2(4x) \tan(4x)$$

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

Problem Statement

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

$$f'(3) = 8(11)^{1/3}$$

Problem Statement

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

Solution

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

$$f'(3) = 8(11)^{1/3}$$

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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 $$2x - y^2 = \cos(xy) + 5$$

$$\displaystyle{ \frac{dy}{dx} = \frac{2+y\sin(xy)}{2y-x\sin(xy)} }$$

Problem Statement

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

Solution

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

$$\displaystyle{ \frac{dy}{dx} = \frac{2+y\sin(xy)}{2y-x\sin(xy)} }$$

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

Problem Statement

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

$$D_x^{(101)}[\cos(x)]=-\sin(x)$$

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.
$$D_x^{(101)}[\cos(x)]=D_x^1[\cos(x)]=-\sin(x)$$

$$D_x^{(101)}[\cos(x)]=-\sin(x)$$

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Part B - Questions 12-17

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 slope-intercept 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 slope-intercept form.

$$\displaystyle{ y = \frac{-x}{16} + \frac{1}{4} }$$

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 slope-intercept 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 y-value 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.
$$\begin{array}{rcl} \displaystyle{y-\frac{1}{8}} & = & \displaystyle{\frac{-1}{16}(x-2)} \\ y & = & \displaystyle{\frac{-x}{16}+\frac{1}{8}+\frac{1}{8}} \\ y & = & \displaystyle{\frac{-x}{16}+\frac{1}{4}} \end{array}$$
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.

$$\displaystyle{ y = \frac{-x}{16} + \frac{1}{4} }$$

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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 right-handed 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 $$S = [-1,4]$$.

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 $$f'(x)=0$$.
$$f'(x) = 3x^2 - 6x = 0$$
$$3x(x-2)=0$$
$$3x = 0 ~~ \to ~~ x=0$$
$$x-2 = 0 ~~ \to ~~ x=2$$
So, $$x=0$$ and $$x=2$$ are the critical numbers. Since the question used the term critical 'points', we need to find each y-value for each corresponding critical number.
$$f(0) = 0^3-3(0)^2+1 = 1$$
$$f(2) = 2^3 - 3(2)^2+1 = 8 -3(4)+1 = 8-12+1 = -3$$
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
$$f(0) = 1$$
$$f(2) = -3$$
$$f(-1) = (-1)^3-3(-1)^2+1 = -1-3+1 = -3$$
$$f(4) = 4^3 -3(4)^2_1 = 64-48+1 = 17$$
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.

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 $$\pi/6$$?

$$\displaystyle{ \frac{dA}{dt} = \frac{45}{8} }$$ m2/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 $$\theta=\pi/6 ~~ d\theta/dt=2.5 rad/min$$

 $$\displaystyle{\frac{dA}{dt}}$$ $$\displaystyle{\frac{9}{2}\left[\cos^2(\pi/6)-\sin^2(\pi/6)\right] (m^2)(2.5 rad/min)}$$ $$\displaystyle{\frac{9}{2}\left[\frac{3}{4}-\frac{1}{4}\right] 2.5 m^2/min}$$ $$\displaystyle{\left[\frac{9}{4}\cdot\frac{5}{2}\right] m^2/min}$$ $$\displaystyle{\frac{45}{8} m^2/min}$$

$$\displaystyle{ \frac{dA}{dt} = \frac{45}{8} }$$ m2/min

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Section 4

For the questions in this section, consider the function $$f(x) = x^4-4x^3-10$$.

[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.

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.
$$f'(x) = 4x^3 - 12x^2$$
$$4x^3 - 12x^2 = 0$$
$$4x^2(x-3) = 0$$
$$4x^2 = 0 ~~ \to ~~ x = 0$$
$$x-3=0 ~~ \to ~~ x=3$$
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 x-value

$$x=-1$$$$x=1$$$$x=4$$

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.

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.

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 $$f'(x) = 4x^3 - 12x^2$$.
$$f''(x) = 12x^2-24x$$
$$12x^2-24x = 0$$
$$12x(x-2) = 0$$
$$12x=0 ~~ \to ~~ x = 0$$
$$x-2 = 0 ~~ \to ~~ x=2$$
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 x-value

$$x=-1$$$$x=1$$$$x=3$$

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.

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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 limits Most limit topics are required for this exam except for L'Hôpital's Rule which is usually covered in calculus 2. all derivative topics

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other exams from calculus 1

### Trig Formulas

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 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle 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

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

### Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

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