Vector-Valued Functions - Calculus 3

1. Consider the vector-valued function $\vec{F}(t) = \langle t\cos(t), t\sin(t), t^2 \rangle$. The curve traced by this function from $t = 0$ to $t = 2\pi$ most closely resembles which geometric shape?
   Answer: A spiral cone with increasing radius and increasing pitch The parametric equations show that in the xy-plane, we have $x = t\cos(t)$ and $y = t\sin(t)$, which gives $\sqrt{x^2 + y^2} = t$. This means the distance from the z-axis increases linearly with the parameter $t$, creating an expanding spiral. Meanwhile, $z = t^2$ increases quadratically, so the vertical spacing between successive turns increases as we move up. This describes a spiral cone with increasing radius and increasing pitch. Choice A is wrong because the radius increases (not constant). Choice B is wrong because the pitch increases ($z = t^2$, not linear). Choice D is wrong because the radius increases (not decreases) and the height increment increases (not constant).
2. A parametric curve is defined by $\vec{r}(t) = \langle t^3 - 6t, t^2 \rangle$ for $t \in \mathbb{R}$. At how many distinct points does this curve intersect itself?
   Answer: 1 Self-intersections occur when $\vec{r}(t_1) = \vec{r}(t_2)$ for $t_1 \neq t_2$. This means $t_1^3 - 6t_1 = t_2^3 - 6t_2$ and $t_1^2 = t_2^2$. From the second equation, either $t_1 = t_2$ (which we exclude) or $t_1 = -t_2$. Substituting $t_2 = -t_1$ into the first equation: $t_1^3 - 6t_1 = (-t_1)^3 - 6(-t_1) = -t_1^3 + 6t_1$. This gives $2t_1^3 - 12t_1 = 0$, so $2t_1(t_1^2 - 6) = 0$. Thus $t_1 = 0, \pm\sqrt{6}$. For $t_1 = 0$, we get $t_2 = 0$ (same point). For $t_1 = \sqrt{6}$, we get $t_2 = -\sqrt{6}$, giving the intersection point $(\sqrt{6}^3 - 6\sqrt{6}, 6) = (-5\sqrt{6}, 6)$. Similarly for $t_1 = -\sqrt{6}$, we get the same point. Therefore, there is exactly 1 self-intersection.
3. The vector-valued function $\vec{r}(t) = \langle \ln(t^2 + 1), \arctan(t), \frac{t}{\sqrt{t^2 + 1}} \rangle$ is defined for all real $t$. Which of the following statements about this curve is true?
   Answer: The curve is unbounded in the $x$-direction but bounded in both $y$ and $z$ directions Let's analyze each component: For the $x$-component: $\ln(t^2 + 1) \to \infty$ as $t \to \pm\infty$, so it's unbounded above and bounded below by $\ln(1) = 0$. For the $y$-component: $\arctan(t)$ has range $(-\frac{\pi}{2}, \frac{\pi}{2})$, so it's bounded. For the $z$-component: $\frac{t}{\sqrt{t^2 + 1}}$ approaches $\pm 1$ as $t \to \pm\infty$ (since $\frac{t}{\sqrt{t^2 + 1}} = \frac{t/|t|}{\sqrt{1 + 1/t^2}} \to \pm 1$), so it has range $(-1, 1)$ and is bounded. Therefore, the curve is unbounded only in the $x$-direction. Choice A is wrong because it's not bounded in all directions. Choice C is wrong because the $z$-component is bounded. Choice D is wrong because only the $x$-component is unbounded.
4. Consider the parametric curve $\vec{r}(t) = \langle a\cos^3(t), a\sin^3(t) \rangle$ where $a > 0$. This curve is known as an astroid. What is the total arc length of one complete astroid ($0 \leq t \leq 2\pi$)?
   Answer: $6a$ When you encounter arc length problems for parametric curves, you need to use the formula $L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$. For this astroid, let's find the derivatives first. Given $\vec{r}(t) = \langle a\cos^3(t), a\sin^3(t) \rangle$, we have: - $\frac{dx}{dt} = -3a\cos^2(t)\sin(t)$ - $\frac{dy}{dt} = 3a\sin^2(t)\cos(t)$ The integrand becomes: $\sqrt{9a^2\cos^4(t)\sin^2(t) + 9a^2\sin^4(t)\cos^2(t)} = 3a|\cos(t)\sin(t)|\sqrt{\cos^2(t) + \sin^2(t)} = 3a|\cos(t)\sin(t)|$ Since the astroid has four-fold symmetry, you can calculate the arc length for one quadrant ($0 \leq t \leq \frac{\pi}{2}$) and multiply by 4. In this interval, both $\cos(t)$ and $\sin(t)$ are non-negative, so: $L = 4 \int_0^{\pi/2} 3a\cos(t)\sin(t) \, dt = 12a \int_0^{\pi/2} \cos(t)\sin(t) \, dt$ Using substitution $u = \sin(t)$, $du = \cos(t)dt$: $L = 12a \int_0^1 u \, du = 12a \cdot \frac{1}{2} = 6a$ This confirms answer A is correct. Answer B ($4\pi a$) and C ($3\pi a$) incorrectly assume the arc length involves $\pi$, likely from confusing this with circular arc formulas. Answer D ($8a$) might result from computational errors in the integration or symmetry analysis. Study tip: For parametric curves with clear symmetry like astroids, always exploit that symmetry to simplify your integration limits and avoid sign complications with absolute values.
5. Consider the space curve $\vec{r}(t) = \langle 2\cos(3t), 2\sin(3t), 4\sin(t) \rangle$. What is the period of this parametric curve (i.e., the smallest positive value $T$ such that $\vec{r}(t+T) = \vec{r}(t)$ for all $t$)?
   Answer: $2\pi$ When finding the period of a parametric curve, you need to determine when all components return to their starting values simultaneously. This requires finding the least common multiple of the individual periods of each trigonometric function. Let's analyze each component of $\vec{r}(t) = \langle 2\cos(3t), 2\sin(3t), 4\sin(t) \rangle$: - The $x$-component $2\cos(3t)$ has period $\frac{2\pi}{3}$ (since $\cos(3t)$ completes one cycle when $3t = 2\pi$) - The $y$-component $2\sin(3t)$ also has period $\frac{2\pi}{3}$ - The $z$-component $4\sin(t)$ has period $2\pi$ For the entire curve to repeat, all components must return to their starting values. This happens at the least common multiple of $\frac{2\pi}{3}$ and $2\pi$. Since $2\pi = 3 \cdot \frac{2\pi}{3}$, the LCM is $2\pi$. We can verify: at $t + 2\pi$, we have $\cos(3(t + 2\pi)) = \cos(3t + 6\pi) = \cos(3t)$ and $\sin(t + 2\pi) = \sin(t)$. Answer choice A ($\frac{2\pi}{3}$) is just the period of the $x$ and $y$ components, but ignores the $z$-component. Choice B ($\pi$) doesn't work since $\cos(3t + 3\pi) = -\cos(3t)$. Choice C ($6\pi$) is a multiple of the true period but not the smallest one. Study tip: For parametric curves with multiple trigonometric functions, always find the LCM of all individual periods—don't just focus on one component.

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