Speed & Arc Length - Calculus 3

1. A particle moves along the helix $\mathbf{r}(t) = \langle 3\cos t, 3\sin t, 4t \rangle$. If the particle travels a distance of $25\pi$ units, through what angle (in radians) has it rotated around the z-axis?
   Answer: $5\pi$ First find the speed: $\mathbf{r}'(t) = \langle -3\sin t, 3\cos t, 4 \rangle$, so $\|\mathbf{r}'(t)\| = \sqrt{9\sin^2 t + 9\cos^2 t + 16} = \sqrt{9 + 16} = 5$. Since speed is constant at 5 units per unit time, traveling distance $25\pi$ takes time $\Delta t = \frac{25\pi}{5} = 5\pi$. The angle rotated around the z-axis equals the change in the parameter $t$, which is $5\pi$ radians. Choice A uses radius 3, choice B uses the z-component coefficient 4, and choice D incorrectly uses the z-component in the calculation.
2. A particle moves along the curve $\mathbf{r}(t) = \langle t^2, 2t, t^3 \rangle$ for $t \in [0, 2]$. At what time $t$ is the particle moving fastest?
   Answer: $t = 2$ Speed is $\|\mathbf{r}'(t)\| = \|\langle 2t, 2, 3t^2 \rangle\| = \sqrt{4t^2 + 4 + 9t^4} = \sqrt{9t^4 + 4t^2 + 4}$. To find when speed is maximum, we take the derivative and set equal to zero: $\frac{d}{dt}\sqrt{9t^4 + 4t^2 + 4} = \frac{36t^3 + 8t}{2\sqrt{9t^4 + 4t^2 + 4}} = \frac{4t(9t^2 + 2)}{2\sqrt{9t^4 + 4t^2 + 4}}$. This equals zero only when $t = 0$. Since the numerator $4t(9t^2 + 2)$ is positive for all $t > 0$, the speed function is strictly increasing on $(0, 2]$. Therefore, maximum speed occurs at the right endpoint $t = 2$.
3. The position of a particle is given by $\mathbf{r}(t) = \langle t^3 - 3t, t^2, 2t \rangle$. At which value(s) of $t$ is the particle momentarily at rest?
   Answer: No values of $t$ A particle is at rest when its velocity vector $\mathbf{r}'(t) = \mathbf{0}$. We have $\mathbf{r}'(t) = \langle 3t^2 - 3, 2t, 2 \rangle$. For this to equal the zero vector, we need $3t^2 - 3 = 0$, $2t = 0$, and $2 = 0$ simultaneously. The third equation $2 = 0$ is never true, so there are no values of $t$ where the particle is at rest. Students might incorrectly solve only $3t^2 - 3 = 0$ (giving $t = \pm 1$) or only $2t = 0$ (giving $t = 0$), leading to choices A, B, or C.
4. The arc length of the curve $\mathbf{r}(t) = \langle \cos(t^2), \sin(t^2), 2t \rangle$ from $t = 0$ to $t = a$ is given by the integral $\int_0^a f(t) \, dt$. What is $f(t)$? Give your answer in simplest form.
   Answer: $2\sqrt{t^2 + 1}$ Arc length is given by $\int_0^a \|\mathbf{r}'(t)\| \, dt$. First, find $\mathbf{r}'(t) = \langle -2t\sin(t^2), 2t\cos(t^2), 2 \rangle$. Then $\|\mathbf{r}'(t)\|^2 = 4t^2\sin^2(t^2) + 4t^2\cos^2(t^2) + 4 = 4t^2(\sin^2(t^2) + \cos^2(t^2)) + 4 = 4t^2 + 4$. Therefore $f(t) = \sqrt{4t^2 + 4} = 2\sqrt{t^2 + 1}$. Choice A incorrectly expands the trigonometric terms, choice B forgets the factor of 2, and choice D incorrectly places the coefficient.

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