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This is the first exam for third semester, multi-variable calculus. The exam covers the basics of vectors and vector functions and applications including curvature, arc length and tangent and normal vectors.

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

2 hours

Questions

5

Total Points

75

Tools

Calculator

no

Formula Sheet(s)

1 page, 8.5x11 or A4

Other Tools

none

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.

(20 points) For \( \vec{r}(t) = 2\cos(t^2) \hat{i} + 2\sin(t^2)\hat{j} \), \( t \ge 0 \), compute
a. \(\vec{T}(t)\); b. \(\vec{N}(t)\); c. curvature; d. \(a_T\) and \(a_N\); e. \(\vec{r}(s)\)

Problem Statement

(20 points) For \( \vec{r}(t) = 2\cos(t^2) \hat{i} + 2\sin(t^2)\hat{j} \), \( t \ge 0 \), compute
a. \(\vec{T}(t)\); b. \(\vec{N}(t)\); c. curvature; d. \(a_T\) and \(a_N\); e. \(\vec{r}(s)\)

Solution

a. \(\displaystyle{\vec{T} = \frac{\vec{v}(t)}{\|\vec{v}(t)\|}}\)
\(\vec{r}'(t)=\vec{v}(t)=-2\sin(t^2)(2t)\hat{i}+2\cos(t^2)(2t)\hat{j}\)
\(\|\vec{v}(t)\| = \sqrt{ [4t\sin(t^2)]^2+[4t\cos(t^2)]^2} = \) \(\sqrt{ 16t^2\sin^2(t^2) + 16t^2\cos^2(t^2) } = \) \(\sqrt{ 16t^2[\sin^2(t^2) + \cos^2(t^2)]} = \sqrt{16t^2} = 4t\)
\(\displaystyle{ \vec{T} = \frac{\vec{v}(t)}{\|\vec{v}(t)\|} = \frac{-4t\sin(t^2)\hat{i}+4t\cos(t^2)\hat{j}}{4t}}\)
\(\boxed{ \vec{T} = -\sin(t^2)\hat{i}+\cos(t^2)\hat{j}}\)

b. \(\displaystyle{\vec{N}(t) = \frac{ d\vec{T}/dt }{ \|d\vec{T}/dt\|}}\)
\(\displaystyle{ \frac{d\vec{T}}{dt} = -\cos(t^2)(2t)\hat{i} + (-\sin(t^2))(2t)\hat{j}}\)
\(\| d\vec{T}/dt \| = \sqrt{ 4t^2\cos^2(t^2) + 4t^2\sin^2(t^2)} = \) \(\sqrt{4t^2} = 2t\)
\(\displaystyle{\vec{N}(t) = \frac{-2t\cos(t^2)\hat{i}-2t\sin(t^2)\hat{j}}{2t}}\)
\(\boxed{ \vec{N}(t) = -\cos(t^2)\hat{i} -\sin(t^2)\hat{j}}\)

c. \(\displaystyle{ K=\frac{1}{\|\vec{v}\|} \left\|\frac{d\vec{T}}{dt}\right\| = }\) \(\displaystyle{\frac{1}{4t}(2t) = \frac{1}{2}}\)     \(\boxed{K=1/2}\)

d. \(a_N = k \|\vec{v}\|^2 = (1/2)(4t)^2 = 8t^2\)     \(\boxed{a_N=8t^2}\)
You may be tempted to use the equation \(\displaystyle{a_T = \frac{\vec{v} \cdot \vec{a}}{\|\vec{v}\|}}\), which will work but it is the hard way. However, if you look ahead at the next part of this problem, you are asked to find \(\vec{r}(s)\), the arc length parameterization of \(\vec{r}(t)\). During the course of solving that, you will find that \(s=t^2\). So, to find \(a_T\), we can use \(\displaystyle{a_T=\frac{d^2s}{dt^2}}\).
\(\displaystyle{a_T=\frac{d^2s}{dt^2} = \frac{d}{dt}[4t] = 4}\)    \(\boxed{a_T=4}\)

e. \(\displaystyle{s(t) = \int_{a}^{t}{\|\vec{v}(u)\|~du}}\)
Since \(\|\vec{v}(u)\| = 4u\), we have \(\displaystyle{ s(t)=\int_{0}^{t}{4u~du} = \left. 2u^2\right|_0^t = 2t^2}\)
So \(s=2t^2\). Solving for t, we have \(s=2t^2 \to s/2=t^2 \to t=\pm\sqrt{s/2}\). Since \(t\ge0\), we choose the positive square root, giving us \(t=\sqrt{s/2}\). Substituting for t in the original position function, we get \(\vec{r}\) in terms of s.
\(\boxed{\vec{r}(s) = 2\cos(s/2)\hat{i} + 2\sin(s/2)\hat{j} ~~~ s \ge 0}\)

close solution

(5 points) Compute the indefinite integral of \(\vec{r}(t)= (5t^{-4}-t^2)\hat{i} + (t^6-4t^3)\hat{j} + (2/t)\hat{k}\)

Problem Statement

(5 points) Compute the indefinite integral of \(\vec{r}(t)= (5t^{-4}-t^2)\hat{i} + (t^6-4t^3)\hat{j} + (2/t)\hat{k}\)

Final Answer

\(\displaystyle{ \int{ \vec{r}(t)~dt } = \left( \frac{-5}{3}t^{-3}-\frac{t^3}{3} \right)\hat{i} + \left( \frac{t^7}{7}-t^4 \right) \hat{j} + \left( 2\ln \abs{t} \right)\hat{k} + \vec{C} }\)

Problem Statement

(5 points) Compute the indefinite integral of \(\vec{r}(t)= (5t^{-4}-t^2)\hat{i} + (t^6-4t^3)\hat{j} + (2/t)\hat{k}\)

Solution

Evaluating each integral separately, gives us
x-component: \(\displaystyle{\int{ 5t^{-4}-t^2~dt } = \frac{5t^{-3}}{-3} - \frac{t^3}{3} + c_1}\)
y-component: \(\displaystyle{\int{ t^6-4t^3 ~dt} = \frac{t^7}{7}-\frac{4t^4}{4} + c_2}\)
z-component: \(\displaystyle{\int{ \frac{2}{t}~dt} = 2\ln \abs{t} + c_3}\)
Let \(\vec{C}= c_1\hat{i} + c_2 \hat{j} + c_3 \hat{k} \).

Final Answer

\(\displaystyle{ \int{ \vec{r}(t)~dt } = \left( \frac{-5}{3}t^{-3}-\frac{t^3}{3} \right)\hat{i} + \left( \frac{t^7}{7}-t^4 \right) \hat{j} + \left( 2\ln \abs{t} \right)\hat{k} + \vec{C} }\)

close solution

(10 points) For the trajectory \(\vec{r}(t) = \langle e^t\sin(t), e^t\cos(t), e^t\rangle\), find
a. the speed associated with the trajectory;
b. the length of the trajectory on the interval \(0\leq t \leq \ln(2)\)

Problem Statement

(10 points) For the trajectory \(\vec{r}(t) = \langle e^t\sin(t), e^t\cos(t), e^t\rangle\), find
a. the speed associated with the trajectory;
b. the length of the trajectory on the interval \(0\leq t \leq \ln(2)\)

Solution

a. speed \(\|\vec{v}(t)\|\) \(\vec{v}(t) = \vec{r}'(t) = \langle e^t\cos(t)+e^t\sin(t), -e^t\sin(t)+e^t\cos(t), e^t \rangle\)
\(\|\vec{v}(t)\| = \sqrt{ (e^t\cos t + e^t \sin t)^2 + (-e^t\sin t+e^t\cos t)^2 + e^{2t} } = \)
\(e^t\sqrt{ \cos^2 t + 2\sin t \cos t + \sin^2 t + \sin^2 t - 2\sin t \cos t + \cos^2 t + 1 } = \)
\(e^t \sqrt{1+1+1} \)
\(\boxed{ \|\vec{v}(t) \| = e^t\sqrt{3}}\)

b. \(\displaystyle{L = \int_{a}^{b}{ \|\vec{v}(t)\| } = \int_{0}^{\ln 2}{ e^t\sqrt{3}~dt} = }\)
\(\displaystyle{\left. e^t\sqrt{3}\right|_{0}^{\ln 2} = }\)
\(\sqrt{3}e^{\ln 2} - \sqrt{3}e^0 = 2\sqrt{3}-\sqrt{3} = \sqrt{3}\)
\(\boxed{L=\sqrt{3}}\)

close solution

(15 points) For the vectors \(\vec{u}=\langle 1,3,-2 \rangle\) and \(\vec{v}=\langle 4,2,1 \rangle\), find
a. the cosine of the angle \(\theta\) between the vectors
b. the projection of \(\vec{v}\) onto \(\vec{u}\); c. \(\vec{u} \times \vec{v}\)

Problem Statement

(15 points) For the vectors \(\vec{u}=\langle 1,3,-2 \rangle\) and \(\vec{v}=\langle 4,2,1 \rangle\), find
a. the cosine of the angle \(\theta\) between the vectors
b. the projection of \(\vec{v}\) onto \(\vec{u}\); c. \(\vec{u} \times \vec{v}\)

Solution

a. To find the cosine of the angle, use \(\vec{u}\cdot\vec{v}= \|\vec{u}\|~\|\vec{v}\| \cos \theta \)
\(\vec{u}\cdot\vec{v}=1(4)+3(2)+(-2)(1)=4+6-2 = 8\)
\(\|\vec{u}\| = \sqrt{1^2+3^2+(-2)^2} = \sqrt{14}\)
\(\|\vec{v}\| = \sqrt{4^2+2^2+1^2} = \sqrt{21}\)
\(\displaystyle{ \cos \theta = \frac{8}{\sqrt{14}\sqrt{21}}}\)     \(\boxed{ \displaystyle{ \cos \theta = \frac{8}{7\sqrt{6}}}}\)

b. \(\displaystyle{proj_{\vec{u}}\vec{v} = \|\vec{v}\| \cos\theta\left( \frac{\vec{u}}{\|\vec{u}\|} \right) = }\) \(\displaystyle{\sqrt{21}\frac{8}{7\sqrt{6}} \frac{\langle1,3,-2\rangle}{\sqrt{14}} = }\) \(\displaystyle{\frac{4}{7}\langle 1,3,-2 \rangle}\)

c. \(\vec{u}\times\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ 4 & 2 & 1 \end{vmatrix} = \) \(\displaystyle{\hat{i}(3+4) - \hat{j}(1+8) + \hat{k}(2-12)}\)
\(\boxed{\vec{u}\times\vec{v}=\langle 7,-9,-10\rangle}\)

close solution

(25 points) A baseball is hit 3 ft above home plate with an initial velocity of \(\langle 65,85,85\rangle\) ft/sec. The spin on the baseball produces a horizontal acceleration of the ball of 12 ft/sec2 in the eastward direction. Assume the x-axis points east, the y-axis points north and the z-axis is vertical with g=32 ft/sec2.
a. Find the velocity vector for \(t\ge0\).     b. Find the position vector for \(t\ge0\).
c. Clearly explain how you would determine the time of flight and range of the ball (do not calculate).
d. Clearly explain how you would determine the maximum height of the ball (do not calculate).

Problem Statement

(25 points) A baseball is hit 3 ft above home plate with an initial velocity of \(\langle 65,85,85\rangle\) ft/sec. The spin on the baseball produces a horizontal acceleration of the ball of 12 ft/sec2 in the eastward direction. Assume the x-axis points east, the y-axis points north and the z-axis is vertical with g=32 ft/sec2.
a. Find the velocity vector for \(t\ge0\).     b. Find the position vector for \(t\ge0\).
c. Clearly explain how you would determine the time of flight and range of the ball (do not calculate).
d. Clearly explain how you would determine the maximum height of the ball (do not calculate).

Solution

a. \(\vec{a}=\langle 12,0,-32\rangle\) ft/sec2
\(\vec{v}=\int{\vec{a}~dt}=\langle 12t+c_1, c_2, -32t+c_3 \rangle\)
\(\vec{v}(0)=\langle c_1,c_2,c_3 \rangle = \langle 65,85,85 \rangle\)
\(\boxed{ \vec{v}(t)=\langle 12t+65, 85, -32t+85 \rangle ~ ft/sec }\)

b. \( \vec{r}(t) = \int{\vec{v}~dt} = \) \( \langle 6t^2+65t+k_1, 85t+k_2, -16t^2+85t+k_3 \rangle \)
\(\vec{r}(0) = \langle k_1, k_2, k_3 \rangle = \langle 0,0,3 \rangle\)
\(\boxed{\vec{r}(t)=\langle 6t^2+65t, 85t, -16t^2+85t+3 \rangle ~ ft }\)

c. To calculate the time of flight, set the z-component of the position vector to zero an solve for t. You will get two values since the equation is a quadratic. The small value is negative, so it has no significance. The large value will be the time when the ball hits the ground, which is the time of flight.
To find the range, plug the value of t found for the time of flight into the position and find the magnitude of that vector to get the range.

d. To calculate the maximum height, we take the z-component of the velocity vector and set it equal to zero. This is equivalent to taking the derivative of the position of vector and set it equal to zero that you are familiar with from calculus 1 to find the critical value. Solving for t gives us the time when the ball is at it's maximum value. We take that value of t and plug it into the z-component of the position vector to get the maximum height.

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