Motion from Parametric Graphs - Calculus 3

1. A particle moves along the curve $\vec{r}(t) = \langle e^t \cos(3t), e^t \sin(3t) \rangle$ for $t \geq 0$. Which property of the motion remains constant throughout the particle's trajectory?
   Answer: The angle between the velocity vector and the position vector remains constant at $\arctan(3)$ We have $\vec{r}(t) = e^t\langle \cos(3t), \sin(3t) \rangle$ and $\vec{v}(t) = e^t\langle \cos(3t) - 3\sin(3t), \sin(3t) + 3\cos(3t) \rangle$. To find the angle between $\vec{r}$ and $\vec{v}$, compute $\cos(\theta) = \frac{\vec{r} \cdot \vec{v}}{|\vec{r}||\vec{v}|}$. We get $\vec{r} \cdot \vec{v} = e^{2t}[\cos(3t)(\cos(3t) - 3\sin(3t)) + \sin(3t)(\sin(3t) + 3\cos(3t))] = e^{2t}[1] = e^{2t}$. Also, $|\vec{r}| = e^t$ and $|\vec{v}| = e^t\sqrt{(\cos(3t) - 3\sin(3t))^2 + (\sin(3t) + 3\cos(3t))^2} = e^t\sqrt{1 + 9} = e^t\sqrt{10}$. Thus $\cos(\theta) = \frac{e^{2t}}{e^t \cdot e^t\sqrt{10}} = \frac{1}{\sqrt{10}}$, so $\theta = \arccos\left(\frac{1}{\sqrt{10}}\right) = \arctan(3)$. Choice A gives the wrong constant value. Choice C and D describe properties that vary with time.
2. A particle follows the parametric path $x(t) = t^3 - 3t$, $y(t) = t^2 - 1$ for $-2 \leq t \leq 2$. At how many distinct points along this path does the particle instantaneously move in a direction perpendicular to the line connecting that point to the origin?
   Answer: Exactly two points, occurring at parameter values where the velocity vector is orthogonal to the position vector Motion perpendicular to the line from origin requires $\vec{r}(t) \perp \vec{v}(t)$, so $\vec{r}(t) \cdot \vec{v}(t) = 0$. We have $\vec{r}(t) = \langle t^3 - 3t, t^2 - 1 \rangle$ and $\vec{v}(t) = \langle 3t^2 - 3, 2t \rangle$. Setting $\vec{r} \cdot \vec{v} = (t^3 - 3t)(3t^2 - 3) + (t^2 - 1)(2t) = 0$. Expanding: $3(t^3 - 3t)(t^2 - 1) + 2t(t^2 - 1) = (t^2 - 1)[3(t^3 - 3t) + 2t] = (t^2 - 1)(3t^3 - 9t + 2t) = (t^2 - 1)(3t^3 - 7t) = t(t^2 - 1)(3t^2 - 7) = 0$. This gives $t = 0$, $t = \pm 1$, and $t = \pm\sqrt{7/3} \approx \pm 1.53$. All solutions are in the domain $[-2, 2]$. However, we need to check which correspond to distinct geometric points. Since the parametric equations don't create self-intersections for these parameter values, we have exactly two distinct geometric points where this perpendicularity condition is satisfied.
3. A particle's position is given by $\vec{r}(t) = \langle t^2 - 4t, \sin(\pi t) \rangle$ for $t \geq 0$. During which time interval is the particle moving southeast (both velocity components negative) while decelerating?
   Answer: $1 < t < 2$ Southeast motion requires both $x'(t) < 0$ and $y'(t) < 0$. We have $\vec{v}(t) = \langle 2t - 4, \pi\cos(\pi t) \rangle$. For $x'(t) = 2t - 4 < 0$, we need $t < 2$. For $y'(t) = \pi\cos(\pi t) < 0$, we need $\cos(\pi t) < 0$, which occurs when $\frac{1}{2} < t < \frac{3}{2}$, $\frac{5}{2} < t < \frac{7}{2}$, etc. The intersection of $t < 2$ and $\frac{1}{2} < t < \frac{3}{2}$ gives $\frac{1}{2} < t < \frac{3}{2}$. Deceleration requires $\vec{v} \cdot \vec{a} < 0$. With $\vec{a}(t) = \langle 2, -\pi^2\sin(\pi t) \rangle$, we get $\vec{v} \cdot \vec{a} = (2t-4)(2) + \pi\cos(\pi t)(-\pi^2\sin(\pi t)) = 4t - 8 - \pi^3\cos(\pi t)\sin(\pi t)$. For $1 < t < \frac{3}{2}$, we have $4t - 8 < -2$ and $\cos(\pi t) < 0$, $\sin(\pi t) > 0$, so the second term is positive, making the entire expression negative, confirming deceleration.

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