Vector-Valued Function Derivatives - Calculus 3

1. Consider the vector function $\vec{r}(t) = \langle t\cos(t), t\sin(t), \sqrt{t} \rangle$ for $t > 0$. At $t = \pi$, which statement about the relationship between velocity and acceleration vectors is correct?
   Answer: The acceleration vector has both tangential and normal components relative to the velocity vector direction We need to analyze $\vec{v}(\pi)$ and $\vec{a}(\pi)$. First: $\vec{v}(t) = \langle \cos(t) - t\sin(t), \sin(t) + t\cos(t), \frac{1}{2\sqrt{t}} \rangle$. At $t = \pi$: $\vec{v}(\pi) = \langle -1 - \pi(0), 0 + \pi(-1), \frac{1}{2\sqrt{\pi}} \rangle = \langle -1, -\pi, \frac{1}{2\sqrt{\pi}} \rangle$. Next: $\vec{a}(t) = \langle -\sin(t) - \sin(t) - t\cos(t), \cos(t) + \cos(t) - t\sin(t), \frac{-1}{4t^{3/2}} \rangle = \langle -2\sin(t) - t\cos(t), 2\cos(t) - t\sin(t), \frac{-1}{4t^{3/2}} \rangle$. At $t = \pi$: $\vec{a}(\pi) = \langle 0 - \pi(-1), 2(-1) - \pi(0), \frac{-1}{4\pi^{3/2}} \rangle = \langle \pi, -2, \frac{-1}{4\pi^{3/2}} \rangle$. Computing $\vec{v}(\pi) \cdot \vec{a}(\pi) = (-1)(\pi) + (-\pi)(-2) + (\frac{1}{2\sqrt{\pi}})(\frac{-1}{4\pi^{3/2}}) = -\pi + 2\pi - \frac{1}{8\pi^2} = \pi - \frac{1}{8\pi^2} \neq 0$. Since the dot product is neither zero nor makes the vectors parallel/antiparallel, the acceleration has both tangential and normal components.
2. For the parametric curve $\vec{r}(t) = \langle t^2, t^3, t^4 \rangle$, at what point does the velocity vector first become parallel to the vector $\langle 2, 6, 12 \rangle$?
   Answer: $(4, 8, 16)$, corresponding to the parameter value that creates vector parallelism conditions Two vectors are parallel when one is a scalar multiple of the other. We have $\vec{v}(t) = \langle 2t, 3t^2, 4t^3 \rangle$. For parallelism with $\langle 2, 6, 12 \rangle$, we need $\langle 2t, 3t^2, 4t^3 \rangle = k\langle 2, 6, 12 \rangle$ for some scalar $k$. This gives us: $2t = 2k$, $3t^2 = 6k$, $4t^3 = 12k$. From the first equation: $k = t$. Substituting into the second: $3t^2 = 6t$, so $t = 2$ (excluding $t = 0$). Check with the third: $4t^3 = 12t$ gives $4(8) = 24$ and $12(2) = 24$ ✓. At $t = 2$: $\vec{r}(2) = \langle 4, 8, 16 \rangle$. The other choices correspond to different parameter values that don't satisfy the parallelism condition.
3. A particle's position is given by $\vec{r}(t) = \langle 3\cos(t), 3\sin(t), 4t \rangle$. The particle experiences a force $\vec{F}(t) = m\vec{a}(t)$ where $m = 2$ kg. At $t = \frac{\pi}{2}$, what is the component of the force in the direction of motion?
   Answer: $0$ N, since acceleration is always perpendicular to velocity for circular motion projected components When analyzing force components along a particle's path, you need to understand the relationship between velocity, acceleration, and the resulting force directions. First, find the velocity and acceleration vectors by differentiating the position function: $\vec{v}(t) = \langle -3\sin(t), 3\cos(t), 4 \rangle$ and $\vec{a}(t) = \langle -3\cos(t), -3\sin(t), 0 \rangle$. At $t = \frac{\pi}{2}$: $\vec{v}(\frac{\pi}{2}) = \langle -3, 0, 4 \rangle$ and $\vec{a}(\frac{\pi}{2}) = \langle 0, -3, 0 \rangle$. The component of force in the direction of motion is found using the dot product: $\vec{F} \cdot \hat{v} = m\vec{a} \cdot \hat{v}$. However, notice that $\vec{v} \cdot \vec{a} = (-3)(0) + (0)(-3) + (4)(0) = 0$. Since the dot product is zero, the vectors are perpendicular, meaning there's no force component in the direction of motion. Choice A incorrectly assumes you multiply the total force magnitude by direction components. Choice B misinterprets this as having a tangential component along the helical path—while the path is helical, the acceleration has no component parallel to velocity. Choice C incorrectly projects radial acceleration onto velocity, but this projection yields zero since they're perpendicular. Choice D is correct: the acceleration (and thus force) is entirely perpendicular to velocity, resulting in zero component along the direction of motion. Study tip: For any motion problem, always check if $\vec{v} \cdot \vec{a} = 0$. When this occurs, all acceleration is centripetal (changing direction, not speed), so no force acts along the motion direction.
4. A projectile follows the path $\vec{r}(t) = \langle 20t, 30t - 5t^2, 0 \rangle$ where distance is in meters and time in seconds. At what time does the projectile reach its maximum height, and what is the magnitude of its acceleration at that time?
   Answer: Maximum height at $t = 3$ s with acceleration magnitude of $10$ m/s², considering gravitational effects only Maximum height occurs when the vertical velocity component equals zero. We have $\vec{v}(t) = \langle 20, 30 - 10t, 0 \rangle$. Setting the y-component to zero: $30 - 10t = 0$, so $t = 3$ s. The acceleration is $\vec{a}(t) = \vec{r}''(t) = \langle 0, -10, 0 \rangle$, which is constant. Therefore $|\vec{a}(t)| = \sqrt{0^2 + (-10)^2 + 0^2} = 10$ m/s² at all times. Choice B has incorrect acceleration magnitude (5 instead of 10). Choices C and D incorrectly identify the time as 6 s instead of 3 s (perhaps confusing when the projectile returns to ground level).
5. Consider $\vec{r}(t) = \langle f(t), g(t), h(t) \rangle$ where $f(t) = t^2\cos(t)$, $g(t) = e^t\sin(t)$, and $h(t) = \frac{t}{t^2+1}$. What is the magnitude of the velocity vector at $t = 0$?
   Answer: $\sqrt{2}$, calculated from the sum of squares of all three velocity component derivatives When you encounter a vector function and need to find the magnitude of its velocity vector, you're looking for $|\vec{r}'(t)|$, which requires finding the derivative of each component function and then computing the magnitude. First, let's find the derivatives of each component at $t = 0$: For $f(t) = t^2\cos(t)$: Using the product rule, $f'(t) = 2t\cos(t) - t^2\sin(t)$, so $f'(0) = 0$. For $g(t) = e^t\sin(t)$: Using the product rule, $g'(t) = e^t\sin(t) + e^t\cos(t)$, so $g'(0) = 0 + 1 = 1$. For $h(t) = \frac{t}{t^2+1}$: Using the quotient rule, $h'(t) = \frac{(t^2+1) \cdot 1 - t \cdot 2t}{(t^2+1)^2} = \frac{1-t^2}{(t^2+1)^2}$, so $h'(0) = 1$. Therefore, $\vec{r}'(0) = \langle 0, 1, 1 \rangle$, and $|\vec{r}'(0)| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$. Answer A suggests the result is 2, but this would come from incorrectly computing the magnitude. Answer B gives 1, which might result from forgetting one component or assuming unit speed without calculation. Answer C suggests $\sqrt{5}$, which would arise from computational errors in the derivatives. The correct answer is D: $\sqrt{2}$. Study tip: Always compute derivatives carefully using appropriate rules (product, quotient, chain), evaluate at the given point, then use the 3D distance formula for magnitude: $\sqrt{x^2 + y^2 + z^2}$.
6. For the vector function $\vec{r}(t) = \langle te^t, t^2\ln(t), \arctan(t) \rangle$ where $t > 0$, which expression correctly represents the second derivative $\vec{r}''(t)$?
   Answer: $\langle (2+t)e^t, 2\ln(t) + 3, \frac{-2t}{(1+t^2)^2} \rangle$ obtained through careful product and chain rule application We need to find $\vec{r}''(t)$ by differentiating twice. First: $\vec{r}'(t) = \langle e^t + te^t, 2t\ln(t) + t, \frac{1}{1+t^2} \rangle = \langle (1+t)e^t, 2t\ln(t) + t, \frac{1}{1+t^2} \rangle$. Then: $\vec{r}''(t) = \langle \frac{d}{dt}[(1+t)e^t], \frac{d}{dt}[2t\ln(t) + t], \frac{d}{dt}[\frac{1}{1+t^2}] \rangle$. For the first component: $\frac{d}{dt}[(1+t)e^t] = e^t + (1+t)e^t = (2+t)e^t$. For the second: $\frac{d}{dt}[2t\ln(t) + t] = 2\ln(t) + 2 + 1 = 2\ln(t) + 3$. For the third: $\frac{d}{dt}[\frac{1}{1+t^2}] = \frac{-2t}{(1+t^2)^2}$. Choice A is $\vec{r}'(t)$, not $\vec{r}''(t)$. Choices B and D have errors in the second component.

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