Calculus 2 Quiz: Integrating Vector Valued Functions
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Integrating Vector Valued FunctionsQuestion 1 of 20

A particle's acceleration is a(t)=⟨1/t,0⟩\mathbf{a}(t) = \langle 1/t, 0 \rangle for t>0t > 0. Given v(1)=⟨1,2⟩\mathbf{v}(1) = \langle 1, 2 \rangle and r(1)=⟨1,1⟩\mathbf{r}(1) = \langle 1, 1 \rangle, find r(e)\mathbf{r}(e).

⟨e,2e⟩\langle e, 2e \rangle
⟨2,1⟩\langle 2, 1 \rangle
⟨1/e+e−1,2e−1⟩\langle 1/e+e-1, 2e-1 \rangle
⟨e+1,2e−1⟩\langle e+1, 2e-1 \rangle
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Calculus 2 Quiz

Calculus 2 Quiz: Integrating Vector Valued Functions

Practice Integrating Vector Valued Functions in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Integrating Vector Valued Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Question 1

A particle's acceleration is a(t)=⟨1/t,0⟩\mathbf{a}(t) = \langle 1/t, 0 \rangle for t>0t > 0. Given v(1)=⟨1,2⟩\mathbf{v}(1) = \langle 1, 2 \rangle and r(1)=⟨1,1⟩\mathbf{r}(1) = \langle 1, 1 \rangle, find r(e)\mathbf{r}(e).

  1. ⟨e,2e⟩\langle e, 2e \rangle
  2. ⟨2,1⟩\langle 2, 1 \rangle
  3. ⟨1/e+e−1,2e−1⟩\langle 1/e+e-1, 2e-1 \rangle
  4. ⟨e+1,2e−1⟩\langle e+1, 2e-1 \rangle (correct answer)
Explanation: First, integrate acceleration to find velocity:
v(t)=∫⟨1/t,0⟩ dt=⟨ln⁡(t)+C1,C2⟩\mathbf{v}(t) = \int \langle 1/t, 0 \rangle \,dt = \langle \ln(t) + C_1, C_2 \rangle for t>0t>0.
Use v(1)=⟨1,2⟩\mathbf{v}(1) = \langle 1, 2 \rangle to find the constants:
ln⁡(1)+C1=1⇒0+C1=1⇒C1=1\ln(1) + C_1 = 1 \Rightarrow 0 + C_1 = 1 \Rightarrow C_1 = 1.
C2=2C_2 = 2.
So, v(t)=⟨ln⁡(t)+1,2⟩\mathbf{v}(t) = \langle \ln(t) + 1, 2 \rangle.
Next, integrate velocity to find position:
r(t)=∫⟨ln⁡(t)+1,2⟩ dt\mathbf{r}(t) = \int \langle \ln(t) + 1, 2 \rangle \,dt. The integral of ln⁡(t)\ln(t) is tln⁡(t)−tt\ln(t) - t.
So, r(t)=⟨tln⁡(t)−t+t+D1,2t+D2⟩=⟨tln⁡(t)+D1,2t+D2⟩\mathbf{r}(t) = \langle t\ln(t) - t + t + D_1, 2t + D_2 \rangle = \langle t\ln(t) + D_1, 2t + D_2 \rangle.
Use r(1)=⟨1,1⟩\mathbf{r}(1) = \langle 1, 1 \rangle to find the constants:
1ln⁡(1)+D1=1⇒0+D1=1⇒D1=11\ln(1) + D_1 = 1 \Rightarrow 0 + D_1 = 1 \Rightarrow D_1 = 1.
2(1)+D2=1⇒D2=−12(1) + D_2 = 1 \Rightarrow D_2 = -1.
So, r(t)=⟨tln⁡(t)+1,2t−1⟩\mathbf{r}(t) = \langle t\ln(t) + 1, 2t - 1 \rangle.
Finally, evaluate at t=et=e:
r(e)=⟨eln⁡(e)+1,2e−1⟩=⟨e(1)+1,2e−1⟩=⟨e+1,2e−1⟩\mathbf{r}(e) = \langle e\ln(e) + 1, 2e - 1 \rangle = \langle e(1) + 1, 2e - 1 \rangle = \langle e+1, 2e-1 \rangle.

Question 2

Let R(t)\mathbf{R}(t) be a vector-valued function representing position. Its rate of change is given by R′(t)=⟨1t+1,2tt2+1⟩\mathbf{R}'(t) = \langle \frac{1}{t+1}, \frac{2t}{t^2+1} \rangle. If R(0)=⟨1,1⟩\mathbf{R}(0) = \langle 1, 1 \rangle, find the vector representing the net change in position from t=0t=0 to t=1t=1, which is R(1)−R(0)\mathbf{R}(1) - \mathbf{R}(0).

  1. ⟨ln⁡(2)+1,ln⁡(2)+1⟩\langle \ln(2)+1, \ln(2)+1 \rangle
  2. ⟨ln⁡(2),ln⁡(2)⟩\langle \ln(2), \ln(2) \rangle (correct answer)
  3. ⟨ln⁡(2),π/2⟩\langle \ln(2), \pi/2 \rangle
  4. ⟨1,1⟩\langle 1, 1 \rangle
Explanation: The question asks for the net change in position, R(1)−R(0)\mathbf{R}(1) - \mathbf{R}(0). By the Fundamental Theorem of Calculus for vector-valued functions, this is equal to the definite integral of the rate of change, ∫01R′(t) dt\int_0^1 \mathbf{R}'(t) \,dt. The initial condition R(0)=⟨1,1⟩\mathbf{R}(0) = \langle 1, 1 \rangle is extra information that is not needed to find the net change, but would be needed to find the final position R(1)\mathbf{R}(1).
We calculate ∫01⟨1t+1,2tt2+1⟩ dt\int_0^1 \langle \frac{1}{t+1}, \frac{2t}{t^2+1} \rangle \,dt.
For the x-component: ∫011t+1 dt=[ln⁡(t+1)]01=ln⁡(2)−ln⁡(1)=ln⁡(2)\int_0^1 \frac{1}{t+1} \,dt = [\ln(t+1)]_0^1 = \ln(2) - \ln(1) = \ln(2).
For the y-component, we use a u-substitution with u=t2+1u = t^2+1, so du=2t dtdu = 2t \,dt. The integral is ∫1udu=ln⁡(u)=ln⁡(t2+1)\int \frac{1}{u} du = \ln(u) = \ln(t^2+1).
∫012tt2+1 dt=[ln⁡(t2+1)]01=ln⁡(12+1)−ln⁡(02+1)=ln⁡(2)−ln⁡(1)=ln⁡(2)\int_0^1 \frac{2t}{t^2+1} \,dt = [\ln(t^2+1)]_0^1 = \ln(1^2+1) - \ln(0^2+1) = \ln(2) - \ln(1) = \ln(2).
The net change vector is ⟨ln⁡(2),ln⁡(2)⟩\langle \ln(2), \ln(2) \rangle.

Question 3

A particle moves in such a way that its velocity v(t)\mathbf{v}(t) is always orthogonal to a constant vector k=⟨1,−2⟩\mathbf{k} = \langle 1, -2 \rangle. If the particle is at the origin at t=0t=0, which of the following must be true about its position r(t)\mathbf{r}(t)?

  1. r(t)\mathbf{r}(t) is always parallel to k\mathbf{k}.
  2. r(t)⋅k=0\mathbf{r}(t) \cdot \mathbf{k} = 0 for all tt. (correct answer)
  3. r(t)\mathbf{r}(t) is always a constant vector.
  4. The magnitude of r(t)\mathbf{r}(t) is constant.
Explanation: The condition that v(t)\mathbf{v}(t) is orthogonal to k\mathbf{k} means their dot product is zero for all tt: v(t)⋅k=0\mathbf{v}(t) \cdot \mathbf{k} = 0.
We know that v(t)=r′(t)\mathbf{v}(t) = \mathbf{r}'(t). So, r′(t)⋅k=0\mathbf{r}'(t) \cdot \mathbf{k} = 0.
Consider the derivative of the scalar function r(t)⋅k\mathbf{r}(t) \cdot \mathbf{k}:
ddt(r(t)⋅k)=r′(t)⋅k+r(t)⋅k′(t)\frac{d}{dt}(\mathbf{r}(t) \cdot \mathbf{k}) = \mathbf{r}'(t) \cdot \mathbf{k} + \mathbf{r}(t) \cdot \mathbf{k}'(t).
Since k\mathbf{k} is a constant vector, k′(t)=0\mathbf{k}'(t) = \mathbf{0}.
So, ddt(r(t)⋅k)=r′(t)⋅k=0\frac{d}{dt}(\mathbf{r}(t) \cdot \mathbf{k}) = \mathbf{r}'(t) \cdot \mathbf{k} = 0.
If the derivative of r(t)⋅k\mathbf{r}(t) \cdot \mathbf{k} is always zero, then r(t)⋅k\mathbf{r}(t) \cdot \mathbf{k} must be a constant. Let's call this constant CC.
We can find CC using the initial condition r(0)=⟨0,0⟩\mathbf{r}(0) = \langle 0, 0 \rangle.
r(0)⋅k=⟨0,0⟩⋅⟨1,−2⟩=0\mathbf{r}(0) \cdot \mathbf{k} = \langle 0, 0 \rangle \cdot \langle 1, -2 \rangle = 0.
So, the constant is 0. This means r(t)⋅k=0\mathbf{r}(t) \cdot \mathbf{k} = 0 for all tt, which implies the position vector is always orthogonal to k\mathbf{k}.

Question 4

The velocity of a particle is v(t)=⟨sec⁡2(t),2sin⁡(t)cos⁡(t)⟩\mathbf{v}(t) = \langle \sec^2(t), 2\sin(t)\cos(t) \rangle on the interval −π/2<t<π/2-\pi/2 < t < \pi/2. If r(0)=⟨1,0⟩\mathbf{r}(0) = \langle 1, 0 \rangle, find r(π/4)\mathbf{r}(\pi/4).

  1. ⟨2,1/2⟩\langle 2, 1/2 \rangle (correct answer)
  2. ⟨2,1⟩\langle 2, 1 \rangle
  3. ⟨1,1/2⟩\langle 1, 1/2 \rangle
  4. ⟨2,1⟩\langle \sqrt{2}, 1 \rangle
Explanation: To find the position, we integrate the velocity. We can use the identity 2sin⁡(t)cos⁡(t)=sin⁡(2t)2\sin(t)\cos(t) = \sin(2t) to simplify the y-component. So, v(t)=⟨sec⁡2(t),sin⁡(2t)⟩\mathbf{v}(t) = \langle \sec^2(t), \sin(2t) \rangle.
r(t)=∫⟨sec⁡2(t),sin⁡(2t)⟩ dt=⟨tan⁡(t)+C1,−12cos⁡(2t)+C2⟩\mathbf{r}(t) = \int \langle \sec^2(t), \sin(2t) \rangle \,dt = \langle \tan(t) + C_1, -\frac{1}{2}\cos(2t) + C_2 \rangle.
Use the initial condition r(0)=⟨1,0⟩\mathbf{r}(0) = \langle 1, 0 \rangle to find the constants.
x-comp: tan⁡(0)+C1=1⇒0+C1=1⇒C1=1\tan(0) + C_1 = 1 \Rightarrow 0 + C_1 = 1 \Rightarrow C_1 = 1.
y-comp: −12cos⁡(0)+C2=0⇒−12(1)+C2=0⇒C2=1/2-\frac{1}{2}\cos(0) + C_2 = 0 \Rightarrow -\frac{1}{2}(1) + C_2 = 0 \Rightarrow C_2 = 1/2.
So, r(t)=⟨tan⁡(t)+1,−12cos⁡(2t)+12⟩\mathbf{r}(t) = \langle \tan(t) + 1, -\frac{1}{2}\cos(2t) + \frac{1}{2} \rangle.
Now, evaluate at t=π/4t = \pi/4:
r(π/4)=⟨tan⁡(π/4)+1,−12cos⁡(2⋅π/4)+12⟩\mathbf{r}(\pi/4) = \langle \tan(\pi/4) + 1, -\frac{1}{2}\cos(2 \cdot \pi/4) + \frac{1}{2} \rangle.
=⟨1+1,−12cos⁡(π/2)+12⟩=⟨2,−12(0)+12⟩=⟨2,1/2⟩= \langle 1 + 1, -\frac{1}{2}\cos(\pi/2) + \frac{1}{2} \rangle = \langle 2, -\frac{1}{2}(0) + \frac{1}{2} \rangle = \langle 2, 1/2 \rangle.

Question 5

The acceleration of a particle is a(t)=⟨sin⁡(t),−cos⁡(t)⟩\mathbf{a}(t) = \langle \sin(t), -\cos(t) \rangle. The velocity of the particle at t=πt=\pi is v(π)=⟨1,1⟩\mathbf{v}(\pi) = \langle 1, 1 \rangle. What is the velocity vector v(t)\mathbf{v}(t)?

  1. ⟨−cos⁡(t),−sin⁡(t)⟩\langle -\cos(t), -\sin(t) \rangle
  2. ⟨−cos⁡(t)+2,−sin⁡(t)+1⟩\langle -\cos(t) + 2, -\sin(t) + 1 \rangle
  3. ⟨−cos⁡(t),1−sin⁡(t)⟩\langle -\cos(t), 1 - \sin(t) \rangle (correct answer)
  4. ⟨cos⁡(t)+2,sin⁡(t)+1⟩\langle \cos(t) + 2, \sin(t) + 1 \rangle
Explanation: To find the velocity vector v(t)\mathbf{v}(t), integrate the acceleration vector a(t)\mathbf{a}(t):
v(t)=∫⟨sin⁡(t),−cos⁡(t)⟩ dt=⟨−cos⁡(t)+C1,−sin⁡(t)+C2⟩\mathbf{v}(t) = \int \langle \sin(t), -\cos(t) \rangle \,dt = \langle -\cos(t) + C_1, -\sin(t) + C_2 \rangle.
Use the given condition v(π)=⟨1,1⟩\mathbf{v}(\pi) = \langle 1, 1 \rangle to solve for the constants C1C_1 and C2C_2.
For the x-component: −cos⁡(π)+C1=1⇒−(−1)+C1=1⇒1+C1=1⇒C1=0-\cos(\pi) + C_1 = 1 \Rightarrow -(-1) + C_1 = 1 \Rightarrow 1 + C_1 = 1 \Rightarrow C_1 = 0.
For the y-component: −sin⁡(π)+C2=1⇒0+C2=1⇒C2=1-\sin(\pi) + C_2 = 1 \Rightarrow 0 + C_2 = 1 \Rightarrow C_2 = 1.
Substituting the constants back into the expression for v(t)\mathbf{v}(t) gives v(t)=⟨−cos⁡(t),−sin⁡(t)+1⟩=⟨−cos⁡(t),1−sin⁡(t)⟩\mathbf{v}(t) = \langle -\cos(t), -\sin(t) + 1 \rangle = \langle -\cos(t), 1 - \sin(t) \rangle.

Question 6

A particle's acceleration is a(t)=⟨6t,−t−2⟩\mathbf{a}(t) = \langle 6t, -t^{-2} \rangle. At t=1t=1, its position is r(1)=⟨−2,0⟩\mathbf{r}(1) = \langle -2, 0 \rangle and at t=2t=2, its position is r(2)=⟨2,ln⁡(2)⟩\mathbf{r}(2) = \langle 2, \ln(2) \rangle. Find the velocity v(1)\mathbf{v}(1).

  1. ⟨4,ln⁡(2)⟩\langle 4, \ln(2) \rangle
  2. ⟨0,1⟩\langle 0, 1 \rangle (correct answer)
  3. ⟨−3,1⟩\langle -3, 1 \rangle
  4. ⟨3,0⟩\langle 3, 0 \rangle
Explanation: This problem requires two integrations and solving a system of equations for the constants.
Integrate a(t)\mathbf{a}(t) to get v(t)=⟨3t2+C1,t−1+C2⟩=⟨3t2+C1,1/t+C2⟩\mathbf{v}(t) = \langle 3t^2 + C_1, t^{-1} + C_2 \rangle = \langle 3t^2 + C_1, 1/t + C_2 \rangle.
Integrate v(t)\mathbf{v}(t) to get r(t)=⟨t3+C1t+D1,ln⁡(t)+C2t+D2⟩\mathbf{r}(t) = \langle t^3 + C_1 t + D_1, \ln(t) + C_2 t + D_2 \rangle.
Use the two position data points:
r(1)=⟨1+C1+D1,ln⁡(1)+C2+D2⟩=⟨1+C1+D1,C2+D2⟩=⟨−2,0⟩\mathbf{r}(1) = \langle 1+C_1+D_1, \ln(1)+C_2+D_2 \rangle = \langle 1+C_1+D_1, C_2+D_2 \rangle = \langle -2, 0 \rangle.
r(2)=⟨8+2C1+D1,ln⁡(2)+2C2+D2⟩=⟨2,ln⁡(2)⟩\mathbf{r}(2) = \langle 8+2C_1+D_1, \ln(2)+2C_2+D_2 \rangle = \langle 2, \ln(2) \rangle.
From the y-components: C2+D2=0C_2+D_2=0 and ln⁡(2)+2C2+D2=ln⁡(2)⇒2C2+D2=0\ln(2)+2C_2+D_2=\ln(2) \Rightarrow 2C_2+D_2=0. Subtracting these gives C2=0C_2=0, and so D2=0D_2=0.
From the x-components: 1+C1+D1=−2⇒C1+D1=−31+C_1+D_1=-2 \Rightarrow C_1+D_1=-3 and 8+2C1+D1=2⇒2C1+D1=−68+2C_1+D_1=2 \Rightarrow 2C_1+D_1=-6. Subtracting the first from the second gives C1=−3C_1=-3, and so D1=0D_1=0.
We need the constants for velocity, which are C1=−3C_1=-3 and C2=0C_2=0.
So, v(t)=⟨3t2−3,1/t⟩\mathbf{v}(t) = \langle 3t^2 - 3, 1/t \rangle.
Finally, v(1)=⟨3(1)2−3,1/1⟩=⟨0,1⟩\mathbf{v}(1) = \langle 3(1)^2 - 3, 1/1 \rangle = \langle 0, 1 \rangle.

Question 7

A projectile is launched from the origin with an initial velocity vector v(0)=⟨10,50⟩\mathbf{v}(0) = \langle 10, 50 \rangle. The acceleration due to gravity is a(t)=⟨0,−10⟩\mathbf{a}(t) = \langle 0, -10 \rangle. What is the position vector r(t)\mathbf{r}(t) of the projectile?

  1. ⟨10t,−5t2+50t⟩\langle 10t, -5t^2 + 50t \rangle (correct answer)
  2. ⟨10,−10t+50⟩\langle 10, -10t + 50 \rangle
  3. ⟨0,−5t2⟩\langle 0, -5t^2 \rangle
  4. ⟨10t+10,−5t2+50t+50⟩\langle 10t + 10, -5t^2 + 50t + 50 \rangle
Explanation: We are given acceleration and need to find position, which requires two integrations.
First, integrate a(t)\mathbf{a}(t) to find v(t)\mathbf{v}(t):
v(t)=∫⟨0,−10⟩ dt=⟨C1,−10t+C2⟩\mathbf{v}(t) = \int \langle 0, -10 \rangle \,dt = \langle C_1, -10t + C_2 \rangle.
Use the initial velocity v(0)=⟨10,50⟩\mathbf{v}(0) = \langle 10, 50 \rangle to find the constants: C1=10C_1 = 10 and C2=50C_2 = 50.
So, v(t)=⟨10,−10t+50⟩\mathbf{v}(t) = \langle 10, -10t + 50 \rangle.
Next, integrate v(t)\mathbf{v}(t) to find r(t)\mathbf{r}(t):
r(t)=∫⟨10,−10t+50⟩ dt=⟨10t+D1,−5t2+50t+D2⟩\mathbf{r}(t) = \int \langle 10, -10t + 50 \rangle \,dt = \langle 10t + D_1, -5t^2 + 50t + D_2 \rangle.
Since the projectile is launched from the origin, r(0)=⟨0,0⟩\mathbf{r}(0) = \langle 0, 0 \rangle. Use this to find the constants: D1=0D_1 = 0 and D2=0D_2 = 0.
Therefore, the position vector is r(t)=⟨10t,−5t2+50t⟩\mathbf{r}(t) = \langle 10t, -5t^2 + 50t \rangle.

Question 8

The velocity of a particle is v(t)=⟨2t,3t2⟩\mathbf{v}(t) = \langle 2t, 3t^2 \rangle. Its position vector r(t)\mathbf{r}(t) is such that r(1)\mathbf{r}(1) is orthogonal to v(1)\mathbf{v}(1). If the x-component of r(1)\mathbf{r}(1) is 3, find the position vector r(t)\mathbf{r}(t).

  1. ⟨t2+2,t3−3⟩\langle t^2 + 2, t^3 - 3 \rangle (correct answer)
  2. ⟨t2,t3⟩\langle t^2, t^3 \rangle
  3. ⟨t2+3,t3−2⟩\langle t^2 + 3, t^3 - 2 \rangle
  4. ⟨t2+2,t3+2⟩\langle t^2 + 2, t^3 + 2 \rangle
Explanation: First, integrate v(t)\mathbf{v}(t) to find the general form of r(t)\mathbf{r}(t):
r(t)=∫⟨2t,3t2⟩ dt=⟨t2+C1,t3+C2⟩\mathbf{r}(t) = \int \langle 2t, 3t^2 \rangle \,dt = \langle t^2 + C_1, t^3 + C_2 \rangle.
Next, use the given information to find the specific vector r(1)\mathbf{r}(1). We have v(1)=⟨2(1),3(1)2⟩=⟨2,3⟩\mathbf{v}(1) = \langle 2(1), 3(1)^2 \rangle = \langle 2, 3 \rangle.
Let r(1)=⟨3,y1⟩\mathbf{r}(1) = \langle 3, y_1 \rangle. The condition that r(1)\mathbf{r}(1) is orthogonal to v(1)\mathbf{v}(1) means their dot product is zero:
r(1)⋅v(1)=⟨3,y1⟩⋅⟨2,3⟩=0\mathbf{r}(1) \cdot \mathbf{v}(1) = \langle 3, y_1 \rangle \cdot \langle 2, 3 \rangle = 0.
3(2)+y1(3)=0⇒6+3y1=0⇒y1=−23(2) + y_1(3) = 0 \Rightarrow 6 + 3y_1 = 0 \Rightarrow y_1 = -2.
So, we have the specific position r(1)=⟨3,−2⟩\mathbf{r}(1) = \langle 3, -2 \rangle.
Now use this to find the constants of integration:
From r(t)=⟨t2+C1,t3+C2⟩\mathbf{r}(t) = \langle t^2 + C_1, t^3 + C_2 \rangle, we have r(1)=⟨12+C1,13+C2⟩=⟨1+C1,1+C2⟩\mathbf{r}(1) = \langle 1^2 + C_1, 1^3 + C_2 \rangle = \langle 1+C_1, 1+C_2 \rangle.
Set this equal to ⟨3,−2⟩\langle 3, -2 \rangle:
1+C1=3⇒C1=21 + C_1 = 3 \Rightarrow C_1 = 2.
1+C2=−2⇒C2=−31 + C_2 = -2 \Rightarrow C_2 = -3.
Therefore, the position vector is r(t)=⟨t2+2,t3−3⟩\mathbf{r}(t) = \langle t^2 + 2, t^3 - 3 \rangle.

Question 9

Let w(t)=⟨cos⁡(t2),sin⁡(t2),t⟩\mathbf{w}(t) = \langle \cos(t^2), \sin(t^2), t \rangle. If the arc length of w(t)\mathbf{w}(t) from t=0t = 0 to t=at = a is given by L=∫0a4t2sin⁡2(t2)+4t2cos⁡2(t2)+1 dtL = \int_0^a \sqrt{4t^2 \sin^2(t^2) + 4t^2 \cos^2(t^2) + 1} \, dt, what can you conclude about the relationship between this curve and the standard helix?

  1. The curve has constant curvature equal to 14t2+1\frac{1}{\sqrt{4t^2 + 1}}
  2. The arc length simplifies to L=∫0a4t2+1 dtL = \int_0^a \sqrt{4t^2 + 1} \, dt (correct answer)
  3. The curve lies on a circular cylinder of radius 1
  4. The torsion of the curve is constant and equal to 14t2+1\frac{1}{\sqrt{4t^2 + 1}}
Explanation: To find the arc length, we need ∣∣w′(t)∣∣||\mathbf{w}'(t)||. Computing: w′(t)=⟨−2tsin⁡(t2),2tcos⁡(t2),1⟩\mathbf{w}'(t) = \langle -2t\sin(t^2), 2t\cos(t^2), 1 \rangle. Then ∣∣w′(t)∣∣2=(−2tsin⁡(t2))2+(2tcos⁡(t2))2+12=4t2sin⁡2(t2)+4t2cos⁡2(t2)+1=4t2(sin⁡2(t2)+cos⁡2(t2))+1=4t2+1||\mathbf{w}'(t)||^2 = (-2t\sin(t^2))^2 + (2t\cos(t^2))^2 + 1^2 = 4t^2\sin^2(t^2) + 4t^2\cos^2(t^2) + 1 = 4t^2(\sin^2(t^2) + \cos^2(t^2)) + 1 = 4t^2 + 1. Therefore ∣∣w′(t)∣∣=4t2+1||\mathbf{w}'(t)|| = \sqrt{4t^2 + 1}, and the arc length is L=∫0a4t2+1 dtL = \int_0^a \sqrt{4t^2 + 1} \, dt. The given expression in the problem statement correctly simplifies to this form using the trigonometric identity sin⁡2(t2)+cos⁡2(t2)=1\sin^2(t^2) + \cos^2(t^2) = 1. Options A, C, and D make claims about curvature, geometry, and torsion that are not directly supported by the arc length calculation.

Question 10

Consider p(t)=⟨t2sin⁡(t3),et2cos⁡(t),t1−t2⟩\mathbf{p}(t) = \langle t^2 \sin(t^3), e^{t^2} \cos(t), \frac{t}{\sqrt{1-t^2}} \rangle. If Q(x)=∫0xp(t) dt\mathbf{Q}(x) = \int_0^x \mathbf{p}(t) \, dt, what is d2dx2[Q(x)⋅Q(x)]\frac{d^2}{dx^2}[\mathbf{Q}(x) \cdot \mathbf{Q}(x)] at x=0x = 0?

  1. 22 (correct answer)
  2. 00
  3. 44
  4. 2∣∣p(0)∣∣22||\mathbf{p}(0)||^2
Explanation: By the Fundamental Theorem of Calculus, Q′(x)=p(x)\mathbf{Q}'(x) = \mathbf{p}(x) and Q(0)=0\mathbf{Q}(0) = \mathbf{0}. Let f(x)=Q(x)⋅Q(x)=∣∣Q(x)∣∣2f(x) = \mathbf{Q}(x) \cdot \mathbf{Q}(x) = ||\mathbf{Q}(x)||^2. Then f′(x)=2Q(x)⋅Q′(x)=2Q(x)⋅p(x)f'(x) = 2\mathbf{Q}(x) \cdot \mathbf{Q}'(x) = 2\mathbf{Q}(x) \cdot \mathbf{p}(x). At x=0x = 0: f′(0)=2Q(0)⋅p(0)=20⋅p(0)=0f'(0) = 2\mathbf{Q}(0) \cdot \mathbf{p}(0) = 2\mathbf{0} \cdot \mathbf{p}(0) = 0. Taking the second derivative: f′′(x)=2Q′(x)⋅p(x)+2Q(x)⋅p′(x)=2p(x)⋅p(x)+2Q(x)⋅p′(x)=2∣∣p(x)∣∣2+2Q(x)⋅p′(x)f''(x) = 2\mathbf{Q}'(x) \cdot \mathbf{p}(x) + 2\mathbf{Q}(x) \cdot \mathbf{p}'(x) = 2\mathbf{p}(x) \cdot \mathbf{p}(x) + 2\mathbf{Q}(x) \cdot \mathbf{p}'(x) = 2||\mathbf{p}(x)||^2 + 2\mathbf{Q}(x) \cdot \mathbf{p}'(x). At x=0x = 0: f′′(0)=2∣∣p(0)∣∣2+20⋅p′(0)=2∣∣p(0)∣∣2f''(0) = 2||\mathbf{p}(0)||^2 + 2\mathbf{0} \cdot \mathbf{p}'(0) = 2||\mathbf{p}(0)||^2. Since p(0)=⟨0,1,0⟩\mathbf{p}(0) = \langle 0, 1, 0 \rangle, we have ∣∣p(0)∣∣2=1||\mathbf{p}(0)||^2 = 1, so f′′(0)=2f''(0) = 2.

Question 11

A particle's acceleration is a(t)=⟨6t,2⟩\mathbf{a}(t) = \langle 6t, 2 \rangle. Given the initial conditions v(1)=⟨3,0⟩\mathbf{v}(1) = \langle 3, 0 \rangle and r(1)=⟨0,1⟩\mathbf{r}(1) = \langle 0, 1 \rangle, find the particle's position r(2)\mathbf{r}(2).

  1. ⟨8,4⟩\langle 8, 4 \rangle
  2. ⟨7,2⟩\langle 7, 2 \rangle (correct answer)
  3. ⟨7,4⟩\langle 7, 4 \rangle
  4. ⟨8,0⟩\langle 8, 0 \rangle
Explanation: First, integrate the acceleration vector a(t)\mathbf{a}(t) to find the velocity vector v(t)\mathbf{v}(t).
v(t)=∫⟨6t,2⟩ dt=⟨3t2+C1,2t+C2⟩\mathbf{v}(t) = \int \langle 6t, 2 \rangle \,dt = \langle 3t^2 + C_1, 2t + C_2 \rangle.
Use the condition v(1)=⟨3,0⟩\mathbf{v}(1) = \langle 3, 0 \rangle to find the constants.
For the x-component: 3(1)2+C1=3⇒C1=03(1)^2 + C_1 = 3 \Rightarrow C_1 = 0.
For the y-component: 2(1)+C2=0⇒C2=−22(1) + C_2 = 0 \Rightarrow C_2 = -2.
So, v(t)=⟨3t2,2t−2⟩\mathbf{v}(t) = \langle 3t^2, 2t - 2 \rangle.
Next, integrate the velocity vector v(t)\mathbf{v}(t) to find the position vector r(t)\mathbf{r}(t).
r(t)=∫⟨3t2,2t−2⟩ dt=⟨t3+D1,t2−2t+D2⟩\mathbf{r}(t) = \int \langle 3t^2, 2t - 2 \rangle \,dt = \langle t^3 + D_1, t^2 - 2t + D_2 \rangle.
Use the condition r(1)=⟨0,1⟩\mathbf{r}(1) = \langle 0, 1 \rangle to find the constants.
For the x-component: (1)3+D1=0⇒D1=−1(1)^3 + D_1 = 0 \Rightarrow D_1 = -1.
For the y-component: (1)2−2(1)+D2=1⇒−1+D2=1⇒D2=2(1)^2 - 2(1) + D_2 = 1 \Rightarrow -1 + D_2 = 1 \Rightarrow D_2 = 2.
So, r(t)=⟨t3−1,t2−2t+2⟩\mathbf{r}(t) = \langle t^3 - 1, t^2 - 2t + 2 \rangle.
Finally, evaluate r(2)\mathbf{r}(2):
r(2)=⟨(2)3−1,(2)2−2(2)+2⟩=⟨8−1,4−4+2⟩=⟨7,2⟩\mathbf{r}(2) = \langle (2)^3 - 1, (2)^2 - 2(2) + 2 \rangle = \langle 8 - 1, 4 - 4 + 2 \rangle = \langle 7, 2 \rangle.

Question 12

The velocity vector of a particle moving in a plane is v(t)\mathbf{v}(t). Which of the following expressions represents the total distance traveled by the particle from t=at=a to t=bt=b?

  1. ∫ab∥v(t)∥ dt\int_a^b \|\mathbf{v}(t)\| \,dt (correct answer)
  2. ∥∫abv(t) dt∥\| \int_a^b \mathbf{v}(t) \,dt \|
  3. ∫abv(t) dt\int_a^b \mathbf{v}(t) \,dt
  4. r(b)−r(a)b−a\frac{\mathbf{r}(b) - \mathbf{r}(a)}{b-a}
Explanation: The total distance traveled by a particle is the integral of its speed over the time interval. The speed is the magnitude of the velocity vector, ∥v(t)∥\|\mathbf{v}(t)\|. Therefore, the total distance is given by the arc length formula, ∫ab∥v(t)∥ dt\int_a^b \|\mathbf{v}(t)\| \,dt.
Choice B, ∥∫abv(t) dt∥\| \int_a^b \mathbf{v}(t) \,dt \|, represents the magnitude of the displacement vector, which is the straight-line distance between the start and end points, not the total distance traveled along the path.
Choice C is the displacement vector itself.
Choice D is the average velocity vector over the interval.

Question 13

A particle's velocity is v(t)=⟨3t2−1,2e2t⟩\mathbf{v}(t) = \langle 3t^2 - 1, 2e^{2t} \rangle. If the particle is at r(0)=⟨2,0⟩\mathbf{r}(0) = \langle 2, 0 \rangle, what is its position at t=1t=1?

  1. ⟨0,e2−1⟩\langle 0, e^2-1 \rangle
  2. ⟨2,e2⟩\langle 2, e^2 \rangle
  3. ⟨2,e2−1⟩\langle 2, e^2 - 1 \rangle (correct answer)
  4. ⟨0,e2+1⟩\langle 0, e^2+1 \rangle
Explanation: The final position r(1)\mathbf{r}(1) can be found using the relationship r(b)=r(a)+∫abv(t) dt\mathbf{r}(b) = \mathbf{r}(a) + \int_a^b \mathbf{v}(t) \,dt.
Here, a=0,b=1a=0, b=1. First, calculate the displacement by integrating the velocity from 0 to 1:
∫01⟨3t2−1,2e2t⟩ dt=⟨[t3−t]01,[e2t]01⟩\int_0^1 \langle 3t^2 - 1, 2e^{2t} \rangle \,dt = \left\langle [t^3 - t]_0^1, [e^{2t}]_0^1 \right\rangle.
Evaluating the definite integrals:
For the x-component: (13−1)−(03−0)=0(1^3 - 1) - (0^3 - 0) = 0.
For the y-component: e2(1)−e2(0)=e2−1e^{2(1)} - e^{2(0)} = e^2 - 1.
The displacement is ⟨0,e2−1⟩\langle 0, e^2 - 1 \rangle.
Now, add the initial position to the displacement:
r(1)=r(0)+displacement=⟨2,0⟩+⟨0,e2−1⟩=⟨2,e2−1⟩\mathbf{r}(1) = \mathbf{r}(0) + \text{displacement} = \langle 2, 0 \rangle + \langle 0, e^2 - 1 \rangle = \langle 2, e^2 - 1 \rangle.

Question 14

The velocity of a particle is given by v(t)=⟨1t+1,e−t⟩\mathbf{v}(t) = \langle \frac{1}{t+1}, e^{-t} \rangle. What is the displacement of the particle from t=0t=0 to t=1t=1?

  1. ⟨ln⁡(2),−e−1⟩\langle \ln(2), -e^{-1} \rangle
  2. ⟨ln⁡(2),1−e−1⟩\langle \ln(2), 1 - e^{-1} \rangle (correct answer)
  3. ⟨−ln⁡(2),e−1−1⟩\langle -\ln(2), e^{-1} - 1 \rangle
  4. ⟨ln⁡(2),1+e−1⟩\langle \ln(2), 1 + e^{-1} \rangle
Explanation: Displacement is the definite integral of the velocity vector over the given interval. We need to calculate ∫01v(t) dt\int_0^1 \mathbf{v}(t) \,dt.
∫01⟨1t+1,e−t⟩ dt=⟨∫011t+1 dt,∫01e−t dt⟩\int_0^1 \langle \frac{1}{t+1}, e^{-t} \rangle \,dt = \left\langle \int_0^1 \frac{1}{t+1} \,dt, \int_0^1 e^{-t} \,dt \right\rangle For the x-component: ∫011t+1 dt=[ln⁡∣t+1∣]01=ln⁡(2)−ln⁡(1)=ln⁡(2)\int_0^1 \frac{1}{t+1} \,dt = [\ln|t+1|]_0^1 = \ln(2) - \ln(1) = \ln(2).
For the y-component: ∫01e−t dt=[−e−t]01=(−e−1)−(−e0)=−e−1+1=1−e−1\int_0^1 e^{-t} \,dt = [-e^{-t}]_0^1 = (-e^{-1}) - (-e^0) = -e^{-1} + 1 = 1 - e^{-1}.
Therefore, the displacement vector is ⟨ln⁡(2),1−e−1⟩\langle \ln(2), 1 - e^{-1} \rangle.

Question 15

The velocity of a particle is given by v(t)=⟨4t,3t2⟩\mathbf{v}(t) = \langle 4t, 3t^2 \rangle. What is the average velocity of the particle from t=0t=0 to t=2t=2?

  1. ⟨4,6⟩\langle 4, 6 \rangle
  2. ⟨8,8⟩\langle 8, 8 \rangle
  3. ⟨4,3⟩\langle 4, 3 \rangle
  4. ⟨4,4⟩\langle 4, 4 \rangle (correct answer)
Explanation: The average value of a vector function f(t)\mathbf{f}(t) over an interval [a,b][a, b] is given by 1b−a∫abf(t) dt\frac{1}{b-a} \int_a^b \mathbf{f}(t) \,dt.
For this problem, the average velocity is 12−0∫02⟨4t,3t2⟩ dt\frac{1}{2-0} \int_0^2 \langle 4t, 3t^2 \rangle \,dt.
First, compute the definite integral:
∫02⟨4t,3t2⟩ dt=⟨[2t2]02,[t3]02⟩=⟨2(2)2−0,23−0⟩=⟨8,8⟩\int_0^2 \langle 4t, 3t^2 \rangle \,dt = \left\langle [2t^2]_0^2, [t^3]_0^2 \right\rangle = \langle 2(2)^2 - 0, 2^3 - 0 \rangle = \langle 8, 8 \rangle.
Then, multiply by 1b−a=12\frac{1}{b-a} = \frac{1}{2}:
Average velocity = 12⟨8,8⟩=⟨4,4⟩\frac{1}{2} \langle 8, 8 \rangle = \langle 4, 4 \rangle.
Choice A is the average of the initial and final velocities. Choice B is the total displacement, not the average velocity. Choice C is the velocity at the midpoint of the interval.

Question 16

The rate of change of the velocity of a particle is given by a(t)=⟨6,6t⟩\mathbf{a}(t) = \langle 6, 6t \rangle. What is the net change in velocity from t=1t=1 to t=3t=3?

  1. ⟨18,27⟩\langle 18, 27 \rangle
  2. ⟨12,24⟩\langle 12, 24 \rangle (correct answer)
  3. ⟨12,26⟩\langle 12, 26 \rangle
  4. Cannot be determined without an initial velocity.
Explanation: The net change in a quantity is the definite integral of its rate of change. The net change in velocity from t=1t=1 to t=3t=3 is given by Δv=v(3)−v(1)=∫13a(t) dt\Delta \mathbf{v} = \mathbf{v}(3) - \mathbf{v}(1) = \int_1^3 \mathbf{a}(t) \,dt.
∫13⟨6,6t⟩ dt=⟨[6t]13,[3t2]13⟩\int_1^3 \langle 6, 6t \rangle \,dt = \left\langle [6t]_1^3, [3t^2]_1^3 \right\rangle.
Evaluating the x-component: 6(3)−6(1)=18−6=126(3) - 6(1) = 18 - 6 = 12.
Evaluating the y-component: 3(3)2−3(1)2=3(9)−3(1)=27−3=243(3)^2 - 3(1)^2 = 3(9) - 3(1) = 27 - 3 = 24.
Therefore, the net change in velocity is ⟨12,24⟩\langle 12, 24 \rangle. The constants of integration for velocity cancel out when computing a definite integral, so the initial velocity is not needed.

Question 17

A particle's velocity is v(t)=⟨11+t2,11−t2⟩\mathbf{v}(t) = \langle \frac{1}{1+t^2}, \frac{1}{\sqrt{1-t^2}} \rangle for −1<t<1-1 < t < 1. If r(0)=⟨1,−1⟩\mathbf{r}(0) = \langle 1, -1 \rangle, find r(t)\mathbf{r}(t).

  1. ⟨arctan⁡(t),arcsin⁡(t)⟩\langle \arctan(t), \arcsin(t) \rangle
  2. ⟨arctan⁡(t)+1,arcsin⁡(t)−1⟩\langle \arctan(t) + 1, \arcsin(t) - 1 \rangle (correct answer)
  3. ⟨ln⁡(1+t2)+1,21−t2−3⟩\langle \ln(1+t^2) + 1, 2\sqrt{1-t^2} - 3 \rangle
  4. ⟨arctan⁡(t)−1,arcsin⁡(t)+1⟩\langle \arctan(t) - 1, \arcsin(t) + 1 \rangle
Explanation: To find the position vector r(t)\mathbf{r}(t), we must integrate the velocity vector v(t)\mathbf{v}(t).
r(t)=∫⟨11+t2,11−t2⟩ dt\mathbf{r}(t) = \int \langle \frac{1}{1+t^2}, \frac{1}{\sqrt{1-t^2}} \rangle \,dt.
The antiderivatives are standard inverse trigonometric functions:
r(t)=⟨arctan⁡(t)+C1,arcsin⁡(t)+C2⟩\mathbf{r}(t) = \langle \arctan(t) + C_1, \arcsin(t) + C_2 \rangle.
Now, use the initial condition r(0)=⟨1,−1⟩\mathbf{r}(0) = \langle 1, -1 \rangle to find the constants.
For the x-component: arctan⁡(0)+C1=1⇒0+C1=1⇒C1=1\arctan(0) + C_1 = 1 \Rightarrow 0 + C_1 = 1 \Rightarrow C_1 = 1.
For the y-component: arcsin⁡(0)+C2=−1⇒0+C2=−1⇒C2=−1\arcsin(0) + C_2 = -1 \Rightarrow 0 + C_2 = -1 \Rightarrow C_2 = -1.
Therefore, the position vector is r(t)=⟨arctan⁡(t)+1,arcsin⁡(t)−1⟩\mathbf{r}(t) = \langle \arctan(t) + 1, \arcsin(t) - 1 \rangle.

Question 18

A particle's velocity is v(t)=⟨e−t2,2t+1⟩\mathbf{v}(t) = \langle e^{-t^2}, 2t+1 \rangle. If r(0)=⟨3,4⟩\mathbf{r}(0) = \langle 3, 4 \rangle, which expression represents r(2)\mathbf{r}(2)?

  1. ⟨∫02e−t2dt,6⟩\langle \int_0^2 e^{-t^2} dt, 6 \rangle
  2. ⟨3+∫02e−t2dt,10⟩\langle 3 + \int_0^2 e^{-t^2} dt, 10 \rangle (correct answer)
  3. ⟨3+∫02e−t2dt,8⟩\langle 3 + \int_0^2 e^{-t^2} dt, 8 \rangle
  4. ⟨3−14e−4,10⟩\langle 3 - \frac{1}{4}e^{-4}, 10 \rangle
Explanation: The position r(2)\mathbf{r}(2) is given by the initial position r(0)\mathbf{r}(0) plus the displacement from t=0t=0 to t=2t=2. The displacement is ∫02v(t) dt\int_0^2 \mathbf{v}(t) \,dt.
r(2)=r(0)+∫02⟨e−t2,2t+1⟩ dt\mathbf{r}(2) = \mathbf{r}(0) + \int_0^2 \langle e^{-t^2}, 2t+1 \rangle \,dt.
The integral of the x-component, ∫02e−t2dt\int_0^2 e^{-t^2} dt, cannot be expressed in terms of elementary functions, so it must be left in its integral form.
The integral of the y-component can be calculated: ∫02(2t+1) dt=[t2+t]02=(22+2)−(02+0)=6\int_0^2 (2t+1) \,dt = [t^2+t]_0^2 = (2^2+2) - (0^2+0) = 6.
So the displacement vector is ⟨∫02e−t2dt,6⟩\langle \int_0^2 e^{-t^2} dt, 6 \rangle.
Now, we add the initial position:
r(2)=⟨3,4⟩+⟨∫02e−t2dt,6⟩=⟨3+∫02e−t2dt,4+6⟩=⟨3+∫02e−t2dt,10⟩\mathbf{r}(2) = \langle 3, 4 \rangle + \langle \int_0^2 e^{-t^2} dt, 6 \rangle = \langle 3 + \int_0^2 e^{-t^2} dt, 4+6 \rangle = \langle 3 + \int_0^2 e^{-t^2} dt, 10 \rangle.

Question 19

The velocity of a particle is v(t)\mathbf{v}(t). It is known that ∫13v(t) dt=⟨4,−2⟩\int_1^3 \mathbf{v}(t) \,dt = \langle 4, -2 \rangle. If r(1)=⟨1,5⟩\mathbf{r}(1) = \langle 1, 5 \rangle, what is r(3)\mathbf{r}(3)?

  1. ⟨4,−2⟩\langle 4, -2 \rangle
  2. ⟨−3,7⟩\langle -3, 7 \rangle
  3. ⟨5,3⟩\langle 5, 3 \rangle (correct answer)
  4. ⟨5,7⟩\langle 5, 7 \rangle
Explanation: The Fundamental Theorem of Calculus for vector-valued functions states that ∫abv(t) dt=r(b)−r(a)\int_a^b \mathbf{v}(t) \,dt = \mathbf{r}(b) - \mathbf{r}(a).
In this case, a=1a=1 and b=3b=3. So, ∫13v(t) dt=r(3)−r(1)\int_1^3 \mathbf{v}(t) \,dt = \mathbf{r}(3) - \mathbf{r}(1).
We are given ∫13v(t) dt=⟨4,−2⟩\int_1^3 \mathbf{v}(t) \,dt = \langle 4, -2 \rangle and r(1)=⟨1,5⟩\mathbf{r}(1) = \langle 1, 5 \rangle.
We can rearrange the formula to solve for r(3)\mathbf{r}(3): r(3)=r(1)+∫13v(t) dt\mathbf{r}(3) = \mathbf{r}(1) + \int_1^3 \mathbf{v}(t) \,dt.
r(3)=⟨1,5⟩+⟨4,−2⟩=⟨1+4,5−2⟩=⟨5,3⟩\mathbf{r}(3) = \langle 1, 5 \rangle + \langle 4, -2 \rangle = \langle 1+4, 5-2 \rangle = \langle 5, 3 \rangle.

Question 20

A particle has velocity v(t)=⟨2t,1⟩\mathbf{v}(t) = \langle 2t, 1 \rangle. If its initial position is r(0)=⟨0,0⟩\mathbf{r}(0) = \langle 0, 0 \rangle, what is the magnitude of its position vector at t=2t=2?

  1. 252\sqrt{5} (correct answer)
  2. 17\sqrt{17}
  3. 66
  4. 44
Explanation: First, find the position vector r(t)\mathbf{r}(t) by integrating the velocity vector v(t)\mathbf{v}(t).
r(t)=∫⟨2t,1⟩ dt=⟨t2+C1,t+C2⟩\mathbf{r}(t) = \int \langle 2t, 1 \rangle \,dt = \langle t^2 + C_1, t + C_2 \rangle.
Given the initial position r(0)=⟨0,0⟩\mathbf{r}(0) = \langle 0, 0 \rangle, we find the constants:
02+C1=0⇒C1=00^2 + C_1 = 0 \Rightarrow C_1 = 0.
0+C2=0⇒C2=00 + C_2 = 0 \Rightarrow C_2 = 0.
So, the position vector is r(t)=⟨t2,t⟩\mathbf{r}(t) = \langle t^2, t \rangle.
Next, find the position vector at t=2t=2:
r(2)=⟨22,2⟩=⟨4,2⟩\mathbf{r}(2) = \langle 2^2, 2 \rangle = \langle 4, 2 \rangle.
Finally, calculate the magnitude of this vector:
∥r(2)∥=∥⟨4,2⟩∥=42+22=16+4=20=4⋅5=25\|\mathbf{r}(2)\| = \|\langle 4, 2 \rangle\| = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}.