Vector-Valued Functions & Derivatives - Calculus 2

1. Let $\mathbf{r}(t) = \langle t+1, t^2, 1 \rangle$ and $\mathbf{s}(t) = \langle t+1, \sin t, \cos t \rangle$. Find the derivative $\frac{d}{dt}[\mathbf{r}(t) \times \mathbf{s}(t)]$ at $t=0$.
   Answer: $\langle -1, 0, 1 \rangle$ Using the product rule for cross products, $(\mathbf{r} \times \mathbf{s})' = \mathbf{r}' \times \mathbf{s} + \mathbf{r} \times \mathbf{s}'$. First, find the derivatives: $\mathbf{r}'(t) = \langle 1, 2t, 0 \rangle$ and $\mathbf{s}'(t) = \langle 1, \cos t, -\sin t \rangle$. Evaluate all vectors at $t=0$: $\mathbf{r}(0)=\langle 1,0,1 \rangle$, $\mathbf{s}(0)=\langle 1,0,1 \rangle$, $\mathbf{r}'(0)=\langle 1,0,0 \rangle$, $\mathbf{s}'(0)=\langle 1,1,0 \rangle$. Now compute the cross products: $\mathbf{r}'(0) \times \mathbf{s}(0) = \langle 1,0,0 \rangle \times \langle 1,0,1 \rangle = \langle 0, -1, 0 \rangle$. And $\mathbf{r}(0) \times \mathbf{s}'(0) = \langle 1,0,1 \rangle \times \langle 1,1,0 \rangle = \langle -1, 1, 1 \rangle$. The sum is $\langle 0, -1, 0 \rangle + \langle -1, 1, 1 \rangle = \langle -1, 0, 1 \rangle$.
2. Find the unit tangent vector $\mathbf{T}(t)$ for the curve $\mathbf{r}(t) = \langle e^{2t}, e^{-2t}, 2\sqrt{2}t \rangle$ at $t=0$.
   Answer: $\langle 1/2, -1/2, \sqrt{2}/2 \rangle$ The unit tangent vector is $\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||}$. First, find the derivative: $\mathbf{r}'(t) = \langle 2e^{2t}, -2e^{-2t}, 2\sqrt{2} \rangle$. Evaluate at $t=0$: $\mathbf{r}'(0) = \langle 2e^0, -2e^0, 2\sqrt{2} \rangle = \langle 2, -2, 2\sqrt{2} \rangle$. Next, find the magnitude of this vector: $||\mathbf{r}'(0)|| = \sqrt{2^2 + (-2)^2 + (2\sqrt{2})^2} = \sqrt{4 + 4 + 8} = \sqrt{16} = 4$. Finally, divide the vector by its magnitude: $\mathbf{T}(0) = \frac{1}{4}\langle 2, -2, 2\sqrt{2} \rangle = \langle 1/2, -1/2, \sqrt{2}/2 \rangle$.
3. If $\mathbf{T}(t)$ is the unit tangent vector to a smooth curve, which of the following must be true for all $t$ where $\mathbf{T}'(t)$ exists?
   Answer: $\mathbf{T}(t) \cdot \mathbf{T}'(t) = 0$ The unit tangent vector $\mathbf{T}(t)$ has a constant magnitude of 1, i.e., $||\mathbf{T}(t)|| = 1$. As shown in a previous problem, if a vector function has a constant magnitude, it must be orthogonal to its derivative. We can prove this by differentiating $\mathbf{T}(t) \cdot \mathbf{T}(t) = 1$. Using the product rule, we get $\mathbf{T}'(t) \cdot \mathbf{T}(t) + \mathbf{T}(t) \cdot \mathbf{T}'(t) = 0$, which simplifies to $2\mathbf{T}(t) \cdot \mathbf{T}'(t) = 0$. Therefore, $\mathbf{T}(t) \cdot \mathbf{T}'(t) = 0$.
4. A particle moves along the path $\mathbf{r}(t) = \langle t, t^2 \rangle$. Find the cosine of the angle between the position vector and the velocity vector at $t = 2$.
   Answer: $\frac{9}{\sqrt{85}}$ The cosine of the angle $\theta$ between two vectors $\mathbf{u}$ and $\mathbf{v}$ is given by $\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| ||\mathbf{v}||}$. Here, the vectors are the position vector $\mathbf{r}(t)$ and the velocity vector $\mathbf{r}'(t)$. First, $\mathbf{r}'(t) = \langle 1, 2t \rangle$. At $t=2$, we have $\mathbf{r}(2) = \langle 2, 4 \rangle$ and $\mathbf{r}'(2) = \langle 1, 4 \rangle$. The dot product is $\mathbf{r}(2) \cdot \mathbf{r}'(2) = (2)(1) + (4)(4) = 18$. The magnitudes are $||\mathbf{r}(2)|| = \sqrt{2^2+4^2} = \sqrt{20}$ and $||\mathbf{r}'(2)|| = \sqrt{1^2+4^2} = \sqrt{17}$. Therefore, $\cos\theta = \frac{18}{\sqrt{20}\sqrt{17}} = \frac{18}{\sqrt{340}} = \frac{18}{2\sqrt{85}} = \frac{9}{\sqrt{85}}$.
5. Let $\mathbf{r}(t) = \langle \cos(2t), \sin(2t), 3t \rangle$. What is the dot product of the velocity and acceleration vectors, $\mathbf{r}'(\pi/4) \cdot \mathbf{r}''(\pi/4)$?
   Answer: 0 First, find the velocity vector: $\mathbf{r}'(t) = \langle -2\sin(2t), 2\cos(2t), 3 \rangle$. Then, find the acceleration vector: $\mathbf{r}''(t) = \langle -4\cos(2t), -4\sin(2t), 0 \rangle$. Now evaluate both at $t = \pi/4$. $\mathbf{r}'(\pi/4) = \langle -2\sin(\pi/2), 2\cos(\pi/2), 3 \rangle = \langle -2, 0, 3 \rangle$. And $\mathbf{r}''(\pi/4) = \langle -4\cos(\pi/2), -4\sin(\pi/2), 0 \rangle = \langle 0, -4, 0 \rangle$. The dot product is $\langle -2, 0, 3 \rangle \cdot \langle 0, -4, 0 \rangle = (-2)(0) + (0)(-4) + (3)(0) = 0$. Alternatively, one can notice the speed, $||\mathbf{r}'(t)|| = \sqrt{4\sin^2(2t)+4\cos^2(2t)+9} = \sqrt{13}$, 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.
6. A particle moves on the surface of a sphere centered at the origin given by $x^2+y^2+z^2 = 14$. At a moment when the particle is at the point $P(1, 2, 3)$, its velocity components are $dx/dt = 2$ and $dy/dt = 1$. What is $dz/dt$ at this moment?
   Answer: $-4/3$ 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 $x^2 + y^2 + z^2 = 14$ at all times, you can differentiate both sides with respect to time: $\frac{d}{dt}(x^2 + y^2 + z^2) = \frac{d}{dt}(14)$ This gives you: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} + 2z\frac{dz}{dt} = 0$ Dividing by 2: $x\frac{dx}{dt} + y\frac{dy}{dt} + z\frac{dz}{dt} = 0$ Now substitute the given values. At point $P(1, 2, 3)$ with $\frac{dx}{dt} = 2$ and $\frac{dy}{dt} = 1$: $(1)(2) + (2)(1) + (3)\frac{dz}{dt} = 0$ $2 + 2 + 3\frac{dz}{dt} = 0$ $4 + 3\frac{dz}{dt} = 0$ $\frac{dz}{dt} = -\frac{4}{3}$ Answer D is correct: $-\frac{4}{3}$. Answer A ($\frac{4}{3}$) 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.
7. Let $\mathbf{F}(t) = \langle e^t \cos(t), e^t \sin(t), t^2 \rangle$. If $\mathbf{G}(t) = \mathbf{F}(t^2)$, what is $\mathbf{G}'(1)$?
   Answer: $\langle 2e(\cos(1) - \sin(1)), 2e(\sin(1) + \cos(1)), 4 \rangle$ Using the chain rule, $\mathbf{G}'(t) = \mathbf{F}'(t^2) \cdot \frac{d}{dt}[t^2] = \mathbf{F}'(t^2) \cdot 2t$. First find $\mathbf{F}'(t) = \langle e^t(\cos(t) - \sin(t)), e^t(\sin(t) + \cos(t)), 2t \rangle$. At $t = 1$, we need $\mathbf{F}'(1^2) \cdot 2(1) = 2\mathbf{F}'(1) = 2\langle e(\cos(1) - \sin(1)), e(\sin(1) + \cos(1)), 2 \rangle = \langle 2e(\cos(1) - \sin(1)), 2e(\sin(1) + \cos(1)), 4 \rangle$. 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.
8. For a vector function $\mathbf{r}(t)$ representing the position of a particle, which expression represents the rate of change of the particle's speed?
   Answer: The tangential component of acceleration, $\frac{\mathbf{r}'(t) \cdot \mathbf{r}''(t)}{||\mathbf{r}'(t)||}$ The speed of the particle is given by $s(t) = ||\mathbf{r}'(t)||$. To find the rate of change of the speed, we must differentiate $s(t)$ with respect to $t$. We can write $s(t) = (\mathbf{r}'(t) \cdot \mathbf{r}'(t))^{1/2}$. Differentiating using the chain rule gives $s'(t) = \frac{1}{2}(\mathbf{r}'(t) \cdot \mathbf{r}'(t))^{-1/2} \cdot (2 \mathbf{r}'(t) \cdot \mathbf{r}''(t)) = \frac{\mathbf{r}'(t) \cdot \mathbf{r}''(t)}{(\mathbf{r}'(t) \cdot \mathbf{r}'(t))^{1/2}} = \frac{\mathbf{r}'(t) \cdot \mathbf{r}''(t)}{||\mathbf{r}'(t)||}$. This quantity is also known as the tangential component of acceleration.
9. If $\mathbf{r}(t)$ is a differentiable vector function such that $\mathbf{r}'(t) = c \mathbf{r}(t)$ for some scalar function $c(t)$, which of the following best describes the path of the particle?
   Answer: A line passing through the origin. The condition $\mathbf{r}'(t) = c(t) \mathbf{r}(t)$ means that the velocity vector $\mathbf{r}'(t)$ is always parallel to the position vector $\mathbf{r}(t)$. 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 $\mathbf{r}(t) = e^{kt}\mathbf{v}$ for a constant vector $\mathbf{v}$, then $\mathbf{r}'(t) = ke^{kt}\mathbf{v} = k\mathbf{r}(t)$, and the path is a ray from the origin.
10. The position of a particle is given by $\mathbf{r}(t) = \langle t^2, at, a-10t \rangle$, where $a$ is a constant. The velocity vector of the particle at $t=2$ is orthogonal to the vector $\mathbf{v} = \langle 1, 1, 1 \rangle$. Find the value of $a$.
    Answer: 6 First, find the velocity vector by differentiating the position vector: $\mathbf{r}'(t) = \langle 2t, a, -10 \rangle$. Next, evaluate the velocity vector at $t=2$: $\mathbf{r}'(2) = \langle 2(2), a, -10 \rangle = \langle 4, a, -10 \rangle$. Two vectors are orthogonal if their dot product is zero. So, we set $\mathbf{r}'(2) \cdot \mathbf{v} = 0$. $\langle 4, a, -10 \rangle \cdot \langle 1, 1, 1 \rangle = 4(1) + a(1) + (-10)(1) = 4 + a - 10 = a - 6$. Setting the dot product to zero gives $a - 6 = 0$, which implies $a = 6$.
11. A particle moves on the path $\mathbf{r}(t) = \langle t^2, \sin(\pi t), \cos(\pi t) \rangle$. At what positive time $t$ is the particle's velocity vector parallel to the vector $\mathbf{v} = \langle 4, \pi, 0 \rangle$?
    Answer: $t = 2$ The velocity vector is found by differentiating the position vector: $\mathbf{r}'(t) = \langle 2t, \pi\cos(\pi t), -\pi\sin(\pi t) \rangle$. We want to find a positive time $t$ such that $\mathbf{r}'(t)$ is parallel to $\mathbf{v} = \langle 4, \pi, 0 \rangle$. This means there exists a scalar $c$ such that $\mathbf{r}'(t) = c\mathbf{v}$. Let's test the given options. At $t=2$, the velocity vector is $\mathbf{r}'(2) = \langle 2(2), \pi\cos(2\pi), -\pi\sin(2\pi) \rangle = \langle 4, \pi(1), -\pi(0) \rangle = \langle 4, \pi, 0 \rangle$. This is exactly the vector $\mathbf{v}$, so they are parallel (with $c=1$). Thus, $t=2$ is the correct answer.
12. The position of a particle is described by the piecewise function $\mathbf{r}(t) = \begin{cases} \langle t^2, t^3 \rangle & t \le 1 \\ \langle at-1, bt-2 \rangle & t > 1 \end{cases}$. For the function to be differentiable at $t=1$, what must be the value of $a+b$?
    Answer: 5 For $\mathbf{r}(t)$ to be differentiable at $t=1$, it must first be continuous at $t=1$. This means $\lim_{t \to 1^-} \mathbf{r}(t) = \lim_{t \to 1^+} \mathbf{r}(t)$. So, $\langle 1^2, 1^3 \rangle = \langle a(1)-1, b(1)-2 \rangle$, which gives $\langle 1, 1 \rangle = \langle a-1, b-2 \rangle$. This yields $a=2$ and $b=3$. Next, the derivative must exist and be equal from both sides. For $t < 1$, $\mathbf{r}'(t) = \langle 2t, 3t^2 \rangle$, so $\lim_{t \to 1^-} \mathbf{r}'(t) = \langle 2, 3 \rangle$. For $t > 1$, $\mathbf{r}'(t) = \langle a, b \rangle$, so $\lim_{t \to 1^+} \mathbf{r}'(t) = \langle a, b \rangle$. For differentiability, we must have $\langle 2, 3 \rangle = \langle a, b \rangle$, so $a=2$ and $b=3$. Both conditions give the same values for a and b. The question asks for $a+b$, which is $2+3=5$.
13. Let $\mathbf{r}(u) = \langle u^3, 1/u, \ln u \rangle$ and let $u(t) = e^{2t}$. Find the derivative of the composite function $\mathbf{p}(t) = \mathbf{r}(u(t))$ at $t=0$.
    Answer: $\langle 6, -2, 2 \rangle$ By the chain rule for vector-valued functions, $\mathbf{p}'(t) = \mathbf{r}'(u(t)) \cdot u'(t)$. First, we compute the derivatives: $\mathbf{r}'(u) = \langle 3u^2, -1/u^2, 1/u \rangle$ and $u'(t) = 2e^{2t}$. Next, we evaluate $u(t)$ and $u'(t)$ at $t=0$: $u(0) = e^0 = 1$ and $u'(0) = 2e^0 = 2$. Now we evaluate $\mathbf{r}'(u)$ at $u=u(0)=1$: $\mathbf{r}'(1) = \langle 3(1)^2, -1/(1)^2, 1/1 \rangle = \langle 3, -1, 1 \rangle$. Finally, we apply the chain rule: $\mathbf{p}'(0) = \mathbf{r}'(1) \cdot u'(0) = \langle 3, -1, 1 \rangle \cdot 2 = \langle 6, -2, 2 \rangle$.
14. For a differentiable vector function $\mathbf{r}(t)$, the derivative of its unit vector, $\frac{d}{dt} \left( \frac{\mathbf{r}(t)}{||\mathbf{r}(t)||} \right)$, represents the rate of change of the particle's:
    Answer: Direction The vector $\mathbf{r}(t)$ represents the position of a particle. The vector $\frac{\mathbf{r}(t)}{||\mathbf{r}(t)||}$ 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.
15. Let $\mathbf{r}(t) = \langle 2t, t^2, \sin(t) \rangle$ and $\mathbf{s}(t) = \langle e^t, \cos(t), 3 \rangle$. Find the value of $\frac{d}{dt}[\mathbf{r}(t) \cdot \mathbf{s}(t)]$ at $t=0$.
    Answer: 5 To find the derivative of the dot product, we use the product rule: $\frac{d}{dt}[\mathbf{r}(t) \cdot \mathbf{s}(t)] = \mathbf{r}'(t) \cdot \mathbf{s}(t) + \mathbf{r}(t) \cdot \mathbf{s}'(t)$. First, we find the derivatives: $\mathbf{r}'(t) = \langle 2, 2t, \cos(t) \rangle$ and $\mathbf{s}'(t) = \langle e^t, -\sin(t), 0 \rangle$. Next, we evaluate the functions and their derivatives at $t=0$: $\mathbf{r}(0) = \langle 0, 0, 0 \rangle$, $\mathbf{s}(0) = \langle 1, 1, 3 \rangle$, $\mathbf{r}'(0) = \langle 2, 0, 1 \rangle$, and $\mathbf{s}'(0) = \langle 1, 0, 0 \rangle$. Finally, we substitute these into the product rule formula: $\mathbf{r}'(0) \cdot \mathbf{s}(0) + \mathbf{r}(0) \cdot \mathbf{s}'(0) = (\langle 2, 0, 1 \rangle \cdot \langle 1, 1, 3 \rangle) + (\langle 0, 0, 0 \rangle \cdot \langle 1, 0, 0 \rangle) = (2+0+3) + 0 = 5$.
16. Let $\mathbf{r}(t) = \langle t e^{-t}, t^2, \ln(1+t) \rangle$. Find the velocity vector $\mathbf{r}'(0)$.
    Answer: $\langle 1, 0, 1 \rangle$ When you encounter a vector-valued function like $\mathbf{r}(t) = \langle t e^{-t}, t^2, \ln(1+t) \rangle$, finding the velocity vector means taking the derivative of each component separately, then evaluating at the given point. To find $\mathbf{r}'(t)$, you'll differentiate each component: For the first component $t e^{-t}$, use the product rule: $\frac{d}{dt}(t e^{-t}) = 1 \cdot e^{-t} + t \cdot (-e^{-t}) = e^{-t} - t e^{-t} = e^{-t}(1-t)$ For the second component $t^2$: $\frac{d}{dt}(t^2) = 2t$ For the third component $\ln(1+t)$: $\frac{d}{dt}(\ln(1+t)) = \frac{1}{1+t}$ So $\mathbf{r}'(t) = \langle e^{-t}(1-t), 2t, \frac{1}{1+t} \rangle$ Now evaluate at $t = 0$: - First component: $e^{0}(1-0) = 1 \cdot 1 = 1$ - Second component: $2(0) = 0$ - Third component: $\frac{1}{1+0} = 1$ Therefore $\mathbf{r}'(0) = \langle 1, 0, 1 \rangle$, which is answer D. Answer A gives $\langle 1, 0, 0 \rangle$, incorrectly making the third component 0 instead of 1. Answer B gives $\langle 0, 0, 1 \rangle$, incorrectly making the first component 0. Answer C gives $\langle -1, 0, 1 \rangle$, correctly getting the second and third components but missing the sign on the first component. Remember: when differentiating vector functions, work component by component and always double-check your derivative rules, especially the product rule for expressions like $t e^{-t}$.
17. A particle moves along a path described by the vector function $\mathbf{r}(t)$ such that its distance from the origin is always 5. Which of the following statements must be true for all $t$ for which $\mathbf{r}(t)$ is differentiable?
    Answer: The position vector $\mathbf{r}(t)$ is orthogonal to the velocity vector $\mathbf{r}'(t)$. The condition that the distance from the origin is always 5 means $||\mathbf{r}(t)|| = 5$. Squaring both sides gives $||\mathbf{r}(t)||^2 = 25$, which can be written as $\mathbf{r}(t) \cdot \mathbf{r}(t) = 25$. Differentiating both sides with respect to $t$ yields $\frac{d}{dt}[\mathbf{r}(t) \cdot \mathbf{r}(t)] = 0$. Applying the product rule for dot products, we get $\mathbf{r}'(t) \cdot \mathbf{r}(t) + \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0$, which simplifies to $2\mathbf{r}(t) \cdot \mathbf{r}'(t) = 0$. Therefore, $\mathbf{r}(t) \cdot \mathbf{r}'(t) = 0$, which means the position vector is orthogonal to the velocity vector.
18. The vector-valued function $\mathbf{R}(s) = \langle s\ln(s), s^2\ln(s), s^3 \rangle$ is defined for $s > 0$. If $\mathbf{S}(t) = \mathbf{R}(e^t)$, what is the coefficient of $e^{3t}$ in the third component of $\mathbf{S}'(t)$?
    Answer: $3$ First, $\mathbf{S}(t) = \mathbf{R}(e^t) = \langle e^t \ln(e^t), e^{2t}\ln(e^t), e^{3t} \rangle = \langle te^t, te^{2t}, e^{3t} \rangle$. Taking the derivative: $\mathbf{S}'(t) = \langle e^t + te^t, e^{2t} + 2te^{2t}, 3e^{3t} \rangle$. The third component is $3e^{3t}$, so the coefficient of $e^{3t}$ 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 $e$ in the computation.
19. Let $\mathbf{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle$. The vector $\mathbf{r}'(t)$ is the velocity. The vector $\mathbf{r}''(t)$ is the acceleration. Which statement correctly describes the relationship between $\mathbf{r}(t)$ and $\mathbf{r}''(t)$?
    Answer: $\mathbf{r}''(t)$ is in the opposite direction of $\mathbf{r}(t)$. First, we find the first and second derivatives of $\mathbf{r}(t)$. The velocity is $\mathbf{r}'(t) = \langle -2\sin(t), 2\cos(t) \rangle$. The acceleration is $\mathbf{r}''(t) = \langle -2\cos(t), -2\sin(t) \rangle$. We can factor out -1 from the acceleration vector: $\mathbf{r}''(t) = -1 \langle 2\cos(t), 2\sin(t) \rangle$. Notice that the vector $\langle 2\cos(t), 2\sin(t) \rangle$ is the original position vector $\mathbf{r}(t)$. Therefore, $\mathbf{r}''(t) = -\mathbf{r}(t)$. 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.
20. A particle has velocity $\mathbf{v}(t) = \langle 2, 3t, -t^2 \rangle$. 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.
    Answer: $\langle 0, 3, 0 \rangle$ 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: $\mathbf{a}(t) = \mathbf{v}'(t) = \langle 0, 3, -2t \rangle$ For orthogonal vectors, their dot product equals zero. Set up the orthogonality condition: $\mathbf{v}(t) \cdot \mathbf{a}(t) = 0$ $\langle 2, 3t, -t^2 \rangle \cdot \langle 0, 3, -2t \rangle = 0$ $2(0) + 3t(3) + (-t^2)(-2t) = 0$ $9t + 2t^3 = 0$ $t(9 + 2t^2) = 0$ This gives us $t = 0$ (since $9 + 2t^2 > 0$ for all real $t$). At $t = 0$, the acceleration is $\mathbf{a}(0) = \langle 0, 3, 0 \rangle$. Choice A gives $\langle 0, 3, -2 \rangle$, which would correspond to $t = 1$, but at $t = 1$ the vectors aren't orthogonal. Choice B gives $\langle 2, 0, 0 \rangle$, 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.

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