Calculus 2 Quiz: Vector Valued Functions And Derivatives
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Vector Valued Functions And DerivativesQuestion 1 of 20

Let f(t)=e2tf(t) = e^{2t} and r(t)=⟨cos⁡(t),sin⁡(t),t⟩\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle. Find the derivative of the scalar multiple g(t)=f(t)r(t)g(t) = f(t)\mathbf{r}(t) at t=0t=0.

⟨2,1,0⟩\langle 2, 1, 0 \rangle
⟨2,0,0⟩\langle 2, 0, 0 \rangle
⟨1,0,1⟩\langle 1, 0, 1 \rangle
⟨2,1,1⟩\langle 2, 1, 1 \rangle
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Calculus 2 Quiz

Calculus 2 Quiz: Vector Valued Functions And Derivatives

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

What this quiz covers

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

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let f(t)=e2tf(t) = e^{2t} and r(t)=⟨cos⁡(t),sin⁡(t),t⟩\mathbf{r}(t) = \langle \cos(t), \sin(t), t \rangle. Find the derivative of the scalar multiple g(t)=f(t)r(t)g(t) = f(t)\mathbf{r}(t) at t=0t=0.

  1. ⟨2,1,0⟩\langle 2, 1, 0 \rangle
  2. ⟨2,0,0⟩\langle 2, 0, 0 \rangle
  3. ⟨1,0,1⟩\langle 1, 0, 1 \rangle
  4. ⟨2,1,1⟩\langle 2, 1, 1 \rangle (correct answer)
Explanation: When you encounter the derivative of a scalar function multiplied by a vector function, you need to apply the product rule for vector calculus. Just like with regular functions, the product rule states that (f(t)r(t))′=f′(t)r(t)+f(t)r′(t)(f(t)\mathbf{r}(t))' = f'(t)\mathbf{r}(t) + f(t)\mathbf{r}'(t) Let's find each component. First, f′(t)=ddt[e2t]=2e2tf'(t) = \frac{d}{dt}[e^{2t}] = 2e^{2t}, so f′(0)=2e0=2f'(0) = 2e^0 = 2. Next, r′(t)=⟨−sin⁡(t),cos⁡(t),1⟩\mathbf{r}'(t) = \langle -\sin(t), \cos(t), 1 \rangle, so r′(0)=⟨0,1,1⟩\mathbf{r}'(0) = \langle 0, 1, 1 \rangle. At t=0t = 0: f(0)=e0=1f(0) = e^0 = 1 and r(0)=⟨1,0,0⟩\mathbf{r}(0) = \langle 1, 0, 0 \rangle. Applying the product rule: g′(0)=f′(0)r(0)+f(0)r′(0)=2⟨1,0,0⟩+1⟨0,1,1⟩=⟨2,0,0⟩+⟨0,1,1⟩=⟨2,1,1⟩g'(0) = f'(0)\mathbf{r}(0) + f(0)\mathbf{r}'(0) = 2\langle 1, 0, 0 \rangle + 1\langle 0, 1, 1 \rangle = \langle 2, 0, 0 \rangle + \langle 0, 1, 1 \rangle = \langle 2, 1, 1 \rangle This confirms answer D is correct. A ⟨2,1,0⟩\langle 2, 1, 0 \rangle incorrectly omits the contribution from the zz-component of r′(0)\mathbf{r}'(0). B ⟨2,0,0⟩\langle 2, 0, 0 \rangle only includes the first term of the product rule, forgetting f(0)r′(0)f(0)\mathbf{r}'(0) entirely. C ⟨1,0,1⟩\langle 1, 0, 1 \rangle appears to confuse the scalar multiples, possibly using f(0)f(0) instead of f′(0)f'(0) in the first term. Remember: vector product rule problems require computing both terms completely. Don't skip the second term f(t)r′(t)f(t)\mathbf{r}'(t)—it's often where students make mistakes by forgetting to include all vector components.

Question 2

The position of a particle is given by r(t)=⟨t2,at,a−10t⟩\mathbf{r}(t) = \langle t^2, at, a-10t \rangle, where aa is a constant. The velocity vector of the particle at t=2t=2 is orthogonal to the vector v=⟨1,1,1⟩\mathbf{v} = \langle 1, 1, 1 \rangle. Find the value of aa.

  1. 6 (correct answer)
  2. 10
  3. -4
  4. 14
Explanation: First, find the velocity vector by differentiating the position vector: r′(t)=⟨2t,a,−10⟩\mathbf{r}'(t) = \langle 2t, a, -10 \rangle. Next, evaluate the velocity vector at t=2t=2: r′(2)=⟨2(2),a,−10⟩=⟨4,a,−10⟩\mathbf{r}'(2) = \langle 2(2), a, -10 \rangle = \langle 4, a, -10 \rangle. Two vectors are orthogonal if their dot product is zero. So, we set r′(2)⋅v=0\mathbf{r}'(2) \cdot \mathbf{v} = 0. ⟨4,a,−10⟩⋅⟨1,1,1⟩=4(1)+a(1)+(−10)(1)=4+a−10=a−6\langle 4, a, -10 \rangle \cdot \langle 1, 1, 1 \rangle = 4(1) + a(1) + (-10)(1) = 4 + a - 10 = a - 6. Setting the dot product to zero gives a−6=0a - 6 = 0, which implies a=6a = 6.

Question 3

If r(t)\mathbf{r}(t) is a differentiable vector function such that r′(t)=cr(t)\mathbf{r}'(t) = c \mathbf{r}(t) for some scalar function c(t)c(t), which of the following best describes the path of the particle?

  1. A circle centered at the origin.
  2. A line passing through the origin. (correct answer)
  3. A parabola with its vertex at the origin.
  4. The particle is stationary at the origin.
Explanation: The condition r′(t)=c(t)r(t)\mathbf{r}'(t) = c(t) \mathbf{r}(t) means that the velocity vector r′(t)\mathbf{r}'(t) is always parallel to the position vector r(t)\mathbf{r}(t). This implies that the direction of motion is always directly away from or towards the origin. Such motion is constrained to a straight line that passes through the origin. For example, if r(t)=ektv\mathbf{r}(t) = e^{kt}\mathbf{v} for a constant vector v\mathbf{v}, then r′(t)=kektv=kr(t)\mathbf{r}'(t) = ke^{kt}\mathbf{v} = k\mathbf{r}(t), and the path is a ray from the origin.

Question 4

For a differentiable vector function r(t)\mathbf{r}(t), the derivative of its unit vector, ddt(r(t)∣∣r(t)∣∣)\frac{d}{dt} \left( \frac{\mathbf{r}(t)}{||\mathbf{r}(t)||} \right), represents the rate of change of the particle's:

  1. Speed
  2. Position
  3. Direction (correct answer)
  4. Acceleration
Explanation: The vector r(t)\mathbf{r}(t) represents the position of a particle. The vector r(t)∣∣r(t)∣∣\frac{\mathbf{r}(t)}{||\mathbf{r}(t)||} is a unit vector, meaning it has a magnitude of 1. Its only property that can change is its direction. Therefore, this unit vector points in the direction of the particle from the origin. The derivative of this unit vector with respect to time represents the rate at which this direction is changing.

Question 5

The position of a particle is described by the piecewise function r(t)={⟨t2,t3⟩t≤1⟨at−1,bt−2⟩t>1\mathbf{r}(t) = \begin{cases} \langle t^2, t^3 \rangle & t \le 1 \\ \langle at-1, bt-2 \rangle & t > 1 \end{cases}. For the function to be differentiable at t=1t=1, what must be the value of a+ba+b?

  1. 5 (correct answer)
  2. 3
  3. 1
  4. -1
Explanation: For r(t)\mathbf{r}(t) to be differentiable at t=1t=1, it must first be continuous at t=1t=1. This means lim⁡t→1−r(t)=lim⁡t→1+r(t)\lim_{t \to 1^-} \mathbf{r}(t) = \lim_{t \to 1^+} \mathbf{r}(t). So, ⟨12,13⟩=⟨a(1)−1,b(1)−2⟩\langle 1^2, 1^3 \rangle = \langle a(1)-1, b(1)-2 \rangle, which gives ⟨1,1⟩=⟨a−1,b−2⟩\langle 1, 1 \rangle = \langle a-1, b-2 \rangle. This yields a=2a=2 and b=3b=3. Next, the derivative must exist and be equal from both sides. For t<1t < 1, r′(t)=⟨2t,3t2⟩\mathbf{r}'(t) = \langle 2t, 3t^2 \rangle, so lim⁡t→1−r′(t)=⟨2,3⟩\lim_{t \to 1^-} \mathbf{r}'(t) = \langle 2, 3 \rangle. For t>1t > 1, r′(t)=⟨a,b⟩\mathbf{r}'(t) = \langle a, b \rangle, so lim⁡t→1+r′(t)=⟨a,b⟩\lim_{t \to 1^+} \mathbf{r}'(t) = \langle a, b \rangle. For differentiability, we must have ⟨2,3⟩=⟨a,b⟩\langle 2, 3 \rangle = \langle a, b \rangle, so a=2a=2 and b=3b=3. Both conditions give the same values for a and b. The question asks for a+ba+b, which is 2+3=52+3=5.

Question 6

A particle moves on the path r(t)=⟨t2,sin⁡(πt),cos⁡(πt)⟩\mathbf{r}(t) = \langle t^2, \sin(\pi t), \cos(\pi t) \rangle. At what positive time tt is the particle's velocity vector parallel to the vector v=⟨4,π,0⟩\mathbf{v} = \langle 4, \pi, 0 \rangle?

  1. t=1/2t = 1/2
  2. t=1t = 1
  3. t=2t = 2 (correct answer)
  4. t=3/2t = 3/2
Explanation: The velocity vector is found by differentiating the position vector: r′(t)=⟨2t,πcos⁡(πt),−πsin⁡(πt)⟩\mathbf{r}'(t) = \langle 2t, \pi\cos(\pi t), -\pi\sin(\pi t) \rangle. We want to find a positive time tt such that r′(t)\mathbf{r}'(t) is parallel to v=⟨4,π,0⟩\mathbf{v} = \langle 4, \pi, 0 \rangle. This means there exists a scalar cc such that r′(t)=cv\mathbf{r}'(t) = c\mathbf{v}. Let's test the given options. At t=2t=2, the velocity vector is r′(2)=⟨2(2),πcos⁡(2π),−πsin⁡(2π)⟩=⟨4,π(1),−π(0)⟩=⟨4,π,0⟩\mathbf{r}'(2) = \langle 2(2), \pi\cos(2\pi), -\pi\sin(2\pi) \rangle = \langle 4, \pi(1), -\pi(0) \rangle = \langle 4, \pi, 0 \rangle. This is exactly the vector v\mathbf{v}, so they are parallel (with c=1c=1). Thus, t=2t=2 is the correct answer.

Question 7

Let r(t)=⟨2cos⁡(t),2sin⁡(t)⟩\mathbf{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle. The vector r′(t)\mathbf{r}'(t) is the velocity. The vector r′′(t)\mathbf{r}''(t) is the acceleration. Which statement correctly describes the relationship between r(t)\mathbf{r}(t) and r′′(t)\mathbf{r}''(t)?

  1. r′′(t)\mathbf{r}''(t) is orthogonal to r(t)\mathbf{r}(t).
  2. r′′(t)\mathbf{r}''(t) is in the same direction as r(t)\mathbf{r}(t).
  3. r′′(t)\mathbf{r}''(t) is in the opposite direction of r(t)\mathbf{r}(t). (correct answer)
  4. r′′(t)\mathbf{r}''(t) is a zero vector.
Explanation: First, we find the first and second derivatives of r(t)\mathbf{r}(t). The velocity is r′(t)=⟨−2sin⁡(t),2cos⁡(t)⟩\mathbf{r}'(t) = \langle -2\sin(t), 2\cos(t) \rangle. The acceleration is r′′(t)=⟨−2cos⁡(t),−2sin⁡(t)⟩\mathbf{r}''(t) = \langle -2\cos(t), -2\sin(t) \rangle. We can factor out -1 from the acceleration vector: r′′(t)=−1⟨2cos⁡(t),2sin⁡(t)⟩\mathbf{r}''(t) = -1 \langle 2\cos(t), 2\sin(t) \rangle. Notice that the vector ⟨2cos⁡(t),2sin⁡(t)⟩\langle 2\cos(t), 2\sin(t) \rangle is the original position vector r(t)\mathbf{r}(t). Therefore, r′′(t)=−r(t)\mathbf{r}''(t) = -\mathbf{r}(t). This means the acceleration vector is always pointing in the direction opposite to the position vector. This is characteristic of uniform circular motion where the acceleration (centripetal) points towards the center of the circle.

Question 8

Let r(t)=⟨t+1,t2,1⟩\mathbf{r}(t) = \langle t+1, t^2, 1 \rangle and s(t)=⟨t+1,sin⁡t,cos⁡t⟩\mathbf{s}(t) = \langle t+1, \sin t, \cos t \rangle. Find the derivative ddt[r(t)×s(t)]\frac{d}{dt}[\mathbf{r}(t) \times \mathbf{s}(t)] at t=0t=0.

  1. ⟨−1,0,1⟩\langle -1, 0, 1 \rangle (correct answer)
  2. ⟨0,−1,0⟩\langle 0, -1, 0 \rangle
  3. ⟨−1,1,1⟩\langle -1, 1, 1 \rangle
  4. ⟨1,0,−1⟩\langle 1, 0, -1 \rangle
Explanation: Using the product rule for cross products, (r×s)′=r′×s+r×s′(\mathbf{r} \times \mathbf{s})' = \mathbf{r}' \times \mathbf{s} + \mathbf{r} \times \mathbf{s}'. First, find the derivatives: r′(t)=⟨1,2t,0⟩\mathbf{r}'(t) = \langle 1, 2t, 0 \rangle and s′(t)=⟨1,cos⁡t,−sin⁡t⟩\mathbf{s}'(t) = \langle 1, \cos t, -\sin t \rangle. Evaluate all vectors at t=0t=0: r(0)=⟨1,0,1⟩\mathbf{r}(0)=\langle 1,0,1 \rangle, s(0)=⟨1,0,1⟩\mathbf{s}(0)=\langle 1,0,1 \rangle, r′(0)=⟨1,0,0⟩\mathbf{r}'(0)=\langle 1,0,0 \rangle, s′(0)=⟨1,1,0⟩\mathbf{s}'(0)=\langle 1,1,0 \rangle. Now compute the cross products: r′(0)×s(0)=⟨1,0,0⟩×⟨1,0,1⟩=⟨0,−1,0⟩\mathbf{r}'(0) \times \mathbf{s}(0) = \langle 1,0,0 \rangle \times \langle 1,0,1 \rangle = \langle 0, -1, 0 \rangle. And r(0)×s′(0)=⟨1,0,1⟩×⟨1,1,0⟩=⟨−1,1,1⟩\mathbf{r}(0) \times \mathbf{s}'(0) = \langle 1,0,1 \rangle \times \langle 1,1,0 \rangle = \langle -1, 1, 1 \rangle. The sum is ⟨0,−1,0⟩+⟨−1,1,1⟩=⟨−1,0,1⟩\langle 0, -1, 0 \rangle + \langle -1, 1, 1 \rangle = \langle -1, 0, 1 \rangle.

Question 9

A particle moves on the surface of a sphere centered at the origin given by x2+y2+z2=14x^2+y^2+z^2 = 14. At a moment when the particle is at the point P(1,2,3)P(1, 2, 3), its velocity components are dx/dt=2dx/dt = 2 and dy/dt=1dy/dt = 1. What is dz/dtdz/dt at this moment?

  1. 4/34/3
  2. 1
  3. −2-2
  4. −4/3-4/3 (correct answer)
Explanation: When you see a particle moving on a constrained surface like a sphere, you're dealing with implicit differentiation and constraint relationships. The key insight is that the particle must always satisfy the sphere equation, so the rates of change of its coordinates are linked. Since the particle stays on the sphere x2+y2+z2=14x^2 + y^2 + z^2 = 14 at all times, you can differentiate both sides with respect to time: ddt(x2+y2+z2)=ddt(14)\frac{d}{dt}(x^2 + y^2 + z^2) = \frac{d}{dt}(14) This gives you: 2xdxdt+2ydydt+2zdzdt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} + 2z\frac{dz}{dt} = 0 Dividing by 2: xdxdt+ydydt+zdzdt=0x\frac{dx}{dt} + y\frac{dy}{dt} + z\frac{dz}{dt} = 0 Now substitute the given values. At point P(1,2,3)P(1, 2, 3) with dxdt=2\frac{dx}{dt} = 2 and dydt=1\frac{dy}{dt} = 1: (1)(2)+(2)(1)+(3)dzdt=0(1)(2) + (2)(1) + (3)\frac{dz}{dt} = 0 2+2+3dzdt=02 + 2 + 3\frac{dz}{dt} = 0 4+3dzdt=04 + 3\frac{dz}{dt} = 0 dzdt=−43\frac{dz}{dt} = -\frac{4}{3} Answer D is correct: −43-\frac{4}{3}. Answer A (43\frac{4}{3}) has the right magnitude but wrong sign—this happens if you forget that the sum must equal zero. Answer B (1) ignores the constraint relationship entirely. Answer C (-2) likely comes from incorrectly setting up the constraint equation or making arithmetic errors. Study tip: For constrained motion problems, always differentiate the constraint equation implicitly with respect to time. The constraint forces the velocity components to be related—they can't all be independent.

Question 10

A particle moves along a path described by the vector function r(t)\mathbf{r}(t) such that its distance from the origin is always 5. Which of the following statements must be true for all tt for which r(t)\mathbf{r}(t) is differentiable?

  1. The velocity vector r′(t)\mathbf{r}'(t) is the zero vector.
  2. The speed ∣∣r′(t)∣∣||\mathbf{r}'(t)|| is constant.
  3. The position vector r(t)\mathbf{r}(t) is orthogonal to the velocity vector r′(t)\mathbf{r}'(t). (correct answer)
  4. The position vector r(t)\mathbf{r}(t) is parallel to the velocity vector r′(t)\mathbf{r}'(t).
Explanation: The condition that the distance from the origin is always 5 means ∣∣r(t)∣∣=5||\mathbf{r}(t)|| = 5. Squaring both sides gives ∣∣r(t)∣∣2=25||\mathbf{r}(t)||^2 = 25, which can be written as r(t)⋅r(t)=25\mathbf{r}(t) \cdot \mathbf{r}(t) = 25. Differentiating both sides with respect to tt yields ddt[r(t)⋅r(t)]=0\frac{d}{dt}[\mathbf{r}(t) \cdot \mathbf{r}(t)] = 0. Applying the product rule for dot products, we get r′(t)⋅r(t)+r(t)⋅r′(t)=0\mathbf{r}'(t) \cdot \mathbf{r}(t) + \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0, which simplifies to 2r(t)⋅r′(t)=02\mathbf{r}(t) \cdot \mathbf{r}'(t) = 0. Therefore, r(t)⋅r′(t)=0\mathbf{r}(t) \cdot \mathbf{r}'(t) = 0, which means the position vector is orthogonal to the velocity vector.

Question 11

A particle has velocity v(t)=⟨2,3t,−t2⟩\mathbf{v}(t) = \langle 2, 3t, -t^2 \rangle. The particle's speed is stationary (at a local extremum) when its velocity and acceleration vectors are orthogonal. Find the acceleration vector at this time.

  1. ⟨0,3,−2⟩\langle 0, 3, -2 \rangle
  2. ⟨2,0,0⟩\langle 2, 0, 0 \rangle
  3. ⟨0,0,0⟩\langle 0, 0, 0 \rangle
  4. ⟨0,3,0⟩\langle 0, 3, 0 \rangle (correct answer)
Explanation: When you see a problem about particle motion asking for stationary speed, you're dealing with the relationship between velocity, acceleration, and the rate of change of speed. The key insight is that speed is stationary (has zero derivative) precisely when velocity and acceleration are orthogonal. Start by finding the acceleration vector: a(t)=v′(t)=⟨0,3,−2t⟩\mathbf{a}(t) = \mathbf{v}'(t) = \langle 0, 3, -2t \rangle For orthogonal vectors, their dot product equals zero. Set up the orthogonality condition: v(t)⋅a(t)=0\mathbf{v}(t) \cdot \mathbf{a}(t) = 0 ⟨2,3t,−t2⟩⋅⟨0,3,−2t⟩=0\langle 2, 3t, -t^2 \rangle \cdot \langle 0, 3, -2t \rangle = 0 2(0)+3t(3)+(−t2)(−2t)=02(0) + 3t(3) + (-t^2)(-2t) = 0 9t+2t3=09t + 2t^3 = 0 t(9+2t2)=0t(9 + 2t^2) = 0 This gives us t=0t = 0 (since 9+2t2>09 + 2t^2 > 0 for all real tt). At t=0t = 0, the acceleration is a(0)=⟨0,3,0⟩\mathbf{a}(0) = \langle 0, 3, 0 \rangle. Choice A gives ⟨0,3,−2⟩\langle 0, 3, -2 \rangle, which would correspond to t=1t = 1, but at t=1t = 1 the vectors aren't orthogonal. Choice B gives ⟨2,0,0⟩\langle 2, 0, 0 \rangle, which isn't even a possible acceleration value since the first component is always zero. Choice C gives the zero vector, which would require all acceleration components to be zero simultaneously—impossible here. Remember: when speed is stationary, velocity and acceleration are perpendicular. Always find when their dot product equals zero, then evaluate the acceleration at that time.

Question 12

Let r(t)=⟨2t,t2,sin⁡(t)⟩\mathbf{r}(t) = \langle 2t, t^2, \sin(t) \rangle and s(t)=⟨et,cos⁡(t),3⟩\mathbf{s}(t) = \langle e^t, \cos(t), 3 \rangle. Find the value of ddt[r(t)⋅s(t)]\frac{d}{dt}[\mathbf{r}(t) \cdot \mathbf{s}(t)] at t=0t=0.

  1. 5 (correct answer)
  2. 0
  3. 2
  4. -1
Explanation: To find the derivative of the dot product, we use the product rule: ddt[r(t)⋅s(t)]=r′(t)⋅s(t)+r(t)⋅s′(t)\frac{d}{dt}[\mathbf{r}(t) \cdot \mathbf{s}(t)] = \mathbf{r}'(t) \cdot \mathbf{s}(t) + \mathbf{r}(t) \cdot \mathbf{s}'(t). First, we find the derivatives: r′(t)=⟨2,2t,cos⁡(t)⟩\mathbf{r}'(t) = \langle 2, 2t, \cos(t) \rangle and s′(t)=⟨et,−sin⁡(t),0⟩\mathbf{s}'(t) = \langle e^t, -\sin(t), 0 \rangle. Next, we evaluate the functions and their derivatives at t=0t=0: r(0)=⟨0,0,0⟩\mathbf{r}(0) = \langle 0, 0, 0 \rangle, s(0)=⟨1,1,3⟩\mathbf{s}(0) = \langle 1, 1, 3 \rangle, r′(0)=⟨2,0,1⟩\mathbf{r}'(0) = \langle 2, 0, 1 \rangle, and s′(0)=⟨1,0,0⟩\mathbf{s}'(0) = \langle 1, 0, 0 \rangle. Finally, we substitute these into the product rule formula: r′(0)⋅s(0)+r(0)⋅s′(0)=(⟨2,0,1⟩⋅⟨1,1,3⟩)+(⟨0,0,0⟩⋅⟨1,0,0⟩)=(2+0+3)+0=5\mathbf{r}'(0) \cdot \mathbf{s}(0) + \mathbf{r}(0) \cdot \mathbf{s}'(0) = (\langle 2, 0, 1 \rangle \cdot \langle 1, 1, 3 \rangle) + (\langle 0, 0, 0 \rangle \cdot \langle 1, 0, 0 \rangle) = (2+0+3) + 0 = 5.

Question 13

The position vector of a particle is given by r(t)=⟨3cos⁡(2t),3sin⁡(2t),4t⟩\mathbf{r}(t) = \langle 3\cos(2t), 3\sin(2t), 4t \rangle. If the particle's speed at time t=π4t = \frac{\pi}{4} is kk, what is the particle's speed at time t=π2t = \frac{\pi}{2}?

  1. kk (correct answer)
  2. 2k2k
  3. k2\frac{k}{2}
  4. 4k4k
Explanation: The velocity vector is r′(t)=⟨−6sin⁡(2t),6cos⁡(2t),4⟩\mathbf{r}'(t) = \langle -6\sin(2t), 6\cos(2t), 4 \rangle. The speed is ∣∣r′(t)∣∣=36sin⁡2(2t)+36cos⁡2(2t)+16=36+16=52=213||\mathbf{r}'(t)|| = \sqrt{36\sin^2(2t) + 36\cos^2(2t) + 16} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}. This speed is constant for all values of tt because the trigonometric terms combine to give a constant. Therefore, the speed at t=π2t = \frac{\pi}{2} equals the speed at t=π4t = \frac{\pi}{4}, which is kk. Choice B assumes the speed doubles, C assumes it halves, and D assumes it quadruples, all reflecting misunderstanding of the constant nature of speed for this helical motion.

Question 14

The vector-valued function R(s)=⟨sln⁡(s),s2ln⁡(s),s3⟩\mathbf{R}(s) = \langle s\ln(s), s^2\ln(s), s^3 \rangle is defined for s>0s > 0. If S(t)=R(et)\mathbf{S}(t) = \mathbf{R}(e^t), what is the coefficient of e3te^{3t} in the third component of S′(t)\mathbf{S}'(t)?

  1. 11
  2. 33 (correct answer)
  3. ee
  4. 3e3e
Explanation: First, S(t)=R(et)=⟨etln⁡(et),e2tln⁡(et),e3t⟩=⟨tet,te2t,e3t⟩\mathbf{S}(t) = \mathbf{R}(e^t) = \langle e^t \ln(e^t), e^{2t}\ln(e^t), e^{3t} \rangle = \langle te^t, te^{2t}, e^{3t} \rangle. Taking the derivative: S′(t)=⟨et+tet,e2t+2te2t,3e3t⟩\mathbf{S}'(t) = \langle e^t + te^t, e^{2t} + 2te^{2t}, 3e^{3t} \rangle. The third component is 3e3t3e^{3t}, so the coefficient of e3te^{3t} is 3. Choice A would result from forgetting the chain rule, choice C might come from confusing the base of the exponential, and choice D might result from incorrectly including an extra factor of ee in the computation.

Question 15

Let F(t)=⟨etcos⁡(t),etsin⁡(t),t2⟩\mathbf{F}(t) = \langle e^t \cos(t), e^t \sin(t), t^2 \rangle. If G(t)=F(t2)\mathbf{G}(t) = \mathbf{F}(t^2), what is G′(1)\mathbf{G}'(1)?

  1. ⟨2e(cos⁡(1)−sin⁡(1)),2e(sin⁡(1)+cos⁡(1)),4⟩\langle 2e(\cos(1) - \sin(1)), 2e(\sin(1) + \cos(1)), 4 \rangle (correct answer)
  2. ⟨e(cos⁡(1)−sin⁡(1)),e(sin⁡(1)+cos⁡(1)),2⟩\langle e(\cos(1) - \sin(1)), e(\sin(1) + \cos(1)), 2 \rangle
  3. ⟨2ecos⁡(1),2esin⁡(1),4⟩\langle 2e\cos(1), 2e\sin(1), 4 \rangle
  4. ⟨4e(cos⁡(1)−sin⁡(1)),4e(sin⁡(1)+cos⁡(1)),8⟩\langle 4e(\cos(1) - \sin(1)), 4e(\sin(1) + \cos(1)), 8 \rangle
Explanation: Using the chain rule, G′(t)=F′(t2)⋅ddt[t2]=F′(t2)⋅2t\mathbf{G}'(t) = \mathbf{F}'(t^2) \cdot \frac{d}{dt}[t^2] = \mathbf{F}'(t^2) \cdot 2t. First find F′(t)=⟨et(cos⁡(t)−sin⁡(t)),et(sin⁡(t)+cos⁡(t)),2t⟩\mathbf{F}'(t) = \langle e^t(\cos(t) - \sin(t)), e^t(\sin(t) + \cos(t)), 2t \rangle. At t=1t = 1, we need F′(12)⋅2(1)=2F′(1)=2⟨e(cos⁡(1)−sin⁡(1)),e(sin⁡(1)+cos⁡(1)),2⟩=⟨2e(cos⁡(1)−sin⁡(1)),2e(sin⁡(1)+cos⁡(1)),4⟩\mathbf{F}'(1^2) \cdot 2(1) = 2\mathbf{F}'(1) = 2\langle e(\cos(1) - \sin(1)), e(\sin(1) + \cos(1)), 2 \rangle = \langle 2e(\cos(1) - \sin(1)), 2e(\sin(1) + \cos(1)), 4 \rangle. Choice B omits the factor of 2 from the chain rule, choice C incorrectly differentiates the trigonometric parts, and choice D incorrectly applies an extra factor of 2.

Question 16

A particle's position is given by r(t)=⟨t2−8,t3−12t⟩\mathbf{r}(t) = \langle t^2 - 8, t^3 - 12t \rangle. For which positive value of tt is the tangent line to the particle's path horizontal?

  1. t=2t = 2 (correct answer)
  2. t=0t = 0
  3. t=23t = 2\sqrt{3}
  4. t=22t = 2\sqrt{2}
Explanation: The tangent line is horizontal when the vertical component of the velocity vector is zero and the horizontal component is non-zero. The velocity vector is r′(t)=⟨2t,3t2−12⟩\mathbf{r}'(t) = \langle 2t, 3t^2 - 12 \rangle. The vertical component is zero when 3t2−12=03t^2 - 12 = 0, which gives t2=4t^2 = 4, so t=±2t = \pm 2. The question asks for the positive value, which is t=2t=2. At t=2t=2, the horizontal component of the velocity is 2(2)=42(2)=4, which is non-zero. Therefore, the tangent line is horizontal at t=2t=2.

Question 17

Let r(t)=⟨cos⁡(2t),sin⁡(2t),3t⟩\mathbf{r}(t) = \langle \cos(2t), \sin(2t), 3t \rangle. What is the dot product of the velocity and acceleration vectors, r′(π/4)⋅r′′(π/4)\mathbf{r}'(\pi/4) \cdot \mathbf{r}''(\pi/4)?

  1. 0 (correct answer)
  2. 9π/49\pi/4
  3. 13
  4. 16
Explanation: First, find the velocity vector: r′(t)=⟨−2sin⁡(2t),2cos⁡(2t),3⟩\mathbf{r}'(t) = \langle -2\sin(2t), 2\cos(2t), 3 \rangle. Then, find the acceleration vector: r′′(t)=⟨−4cos⁡(2t),−4sin⁡(2t),0⟩\mathbf{r}''(t) = \langle -4\cos(2t), -4\sin(2t), 0 \rangle. Now evaluate both at t=π/4t = \pi/4. r′(π/4)=⟨−2sin⁡(π/2),2cos⁡(π/2),3⟩=⟨−2,0,3⟩\mathbf{r}'(\pi/4) = \langle -2\sin(\pi/2), 2\cos(\pi/2), 3 \rangle = \langle -2, 0, 3 \rangle. And r′′(π/4)=⟨−4cos⁡(π/2),−4sin⁡(π/2),0⟩=⟨0,−4,0⟩\mathbf{r}''(\pi/4) = \langle -4\cos(\pi/2), -4\sin(\pi/2), 0 \rangle = \langle 0, -4, 0 \rangle. The dot product is ⟨−2,0,3⟩⋅⟨0,−4,0⟩=(−2)(0)+(0)(−4)+(3)(0)=0\langle -2, 0, 3 \rangle \cdot \langle 0, -4, 0 \rangle = (-2)(0) + (0)(-4) + (3)(0) = 0. Alternatively, one can notice the speed, ∣∣r′(t)∣∣=4sin⁡2(2t)+4cos⁡2(2t)+9=13||\mathbf{r}'(t)|| = \sqrt{4\sin^2(2t)+4\cos^2(2t)+9} = \sqrt{13}, is constant. For any object moving at a constant speed, its velocity and acceleration vectors are always orthogonal, so their dot product is always zero.

Question 18

Let r(u)=⟨u3,1/u,ln⁡u⟩\mathbf{r}(u) = \langle u^3, 1/u, \ln u \rangle and let u(t)=e2tu(t) = e^{2t}. Find the derivative of the composite function p(t)=r(u(t))\mathbf{p}(t) = \mathbf{r}(u(t)) at t=0t=0.

  1. ⟨3,−1,1⟩\langle 3, -1, 1 \rangle
  2. ⟨6,−2,2⟩\langle 6, -2, 2 \rangle (correct answer)
  3. ⟨6,2,2⟩\langle 6, 2, 2 \rangle
  4. ⟨0,0,0⟩\langle 0, 0, 0 \rangle
Explanation: By the chain rule for vector-valued functions, p′(t)=r′(u(t))⋅u′(t)\mathbf{p}'(t) = \mathbf{r}'(u(t)) \cdot u'(t). First, we compute the derivatives: r′(u)=⟨3u2,−1/u2,1/u⟩\mathbf{r}'(u) = \langle 3u^2, -1/u^2, 1/u \rangle and u′(t)=2e2tu'(t) = 2e^{2t}. Next, we evaluate u(t)u(t) and u′(t)u'(t) at t=0t=0: u(0)=e0=1u(0) = e^0 = 1 and u′(0)=2e0=2u'(0) = 2e^0 = 2. Now we evaluate r′(u)\mathbf{r}'(u) at u=u(0)=1u=u(0)=1: r′(1)=⟨3(1)2,−1/(1)2,1/1⟩=⟨3,−1,1⟩\mathbf{r}'(1) = \langle 3(1)^2, -1/(1)^2, 1/1 \rangle = \langle 3, -1, 1 \rangle. Finally, we apply the chain rule: p′(0)=r′(1)⋅u′(0)=⟨3,−1,1⟩⋅2=⟨6,−2,2⟩\mathbf{p}'(0) = \mathbf{r}'(1) \cdot u'(0) = \langle 3, -1, 1 \rangle \cdot 2 = \langle 6, -2, 2 \rangle.

Question 19

Find the unit tangent vector T(t)\mathbf{T}(t) for the curve r(t)=⟨e2t,e−2t,22t⟩\mathbf{r}(t) = \langle e^{2t}, e^{-2t}, 2\sqrt{2}t \rangle at t=0t=0.

  1. ⟨2,−2,22⟩\langle 2, -2, 2\sqrt{2} \rangle
  2. ⟨1/2,−1/2,2/2⟩\langle 1/2, -1/2, \sqrt{2}/2 \rangle (correct answer)
  3. ⟨1/2,1/2,0⟩\langle 1/\sqrt{2}, 1/\sqrt{2}, 0 \rangle
  4. ⟨1/3,−1/3,2/3⟩\langle 1/\sqrt{3}, -1/\sqrt{3}, \sqrt{2}/\sqrt{3} \rangle
Explanation: The unit tangent vector is T(t)=r′(t)∣∣r′(t)∣∣\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||}. First, find the derivative: r′(t)=⟨2e2t,−2e−2t,22⟩\mathbf{r}'(t) = \langle 2e^{2t}, -2e^{-2t}, 2\sqrt{2} \rangle. Evaluate at t=0t=0: r′(0)=⟨2e0,−2e0,22⟩=⟨2,−2,22⟩\mathbf{r}'(0) = \langle 2e^0, -2e^0, 2\sqrt{2} \rangle = \langle 2, -2, 2\sqrt{2} \rangle. Next, find the magnitude of this vector: ∣∣r′(0)∣∣=22+(−2)2+(22)2=4+4+8=16=4||\mathbf{r}'(0)|| = \sqrt{2^2 + (-2)^2 + (2\sqrt{2})^2} = \sqrt{4 + 4 + 8} = \sqrt{16} = 4. Finally, divide the vector by its magnitude: T(0)=14⟨2,−2,22⟩=⟨1/2,−1/2,2/2⟩\mathbf{T}(0) = \frac{1}{4}\langle 2, -2, 2\sqrt{2} \rangle = \langle 1/2, -1/2, \sqrt{2}/2 \rangle.

Question 20

For a vector function r(t)\mathbf{r}(t) representing the position of a particle, which expression represents the rate of change of the particle's speed?

  1. The magnitude of the acceleration, ∣∣r′′(t)∣∣||\mathbf{r}''(t)||
  2. The tangential component of acceleration, r′(t)⋅r′′(t)∣∣r′(t)∣∣\frac{\mathbf{r}'(t) \cdot \mathbf{r}''(t)}{||\mathbf{r}'(t)||} (correct answer)
  3. The speed of the particle, ∣∣r′(t)∣∣||\mathbf{r}'(t)||
  4. The derivative of the direction vector, ddt(r(t)∣∣r(t)∣∣)\frac{d}{dt} \left( \frac{\mathbf{r}(t)}{||\mathbf{r}(t)||} \right)
Explanation: The speed of the particle is given by s(t)=∣∣r′(t)∣∣s(t) = ||\mathbf{r}'(t)||. To find the rate of change of the speed, we must differentiate s(t)s(t) with respect to tt. We can write s(t)=(r′(t)⋅r′(t))1/2s(t) = (\mathbf{r}'(t) \cdot \mathbf{r}'(t))^{1/2}. Differentiating using the chain rule gives s′(t)=12(r′(t)⋅r′(t))−1/2⋅(2r′(t)⋅r′′(t))=r′(t)⋅r′′(t)(r′(t)⋅r′(t))1/2=r′(t)⋅r′′(t)∣∣r′(t)∣∣s'(t) = \frac{1}{2}(\mathbf{r}'(t) \cdot \mathbf{r}'(t))^{-1/2} \cdot (2 \mathbf{r}'(t) \cdot \mathbf{r}''(t)) = \frac{\mathbf{r}'(t) \cdot \mathbf{r}''(t)}{(\mathbf{r}'(t) \cdot \mathbf{r}'(t))^{1/2}} = \frac{\mathbf{r}'(t) \cdot \mathbf{r}''(t)}{||\mathbf{r}'(t)||}. This quantity is also known as the tangential component of acceleration.