Parametrizing Curves - Calculus 3

1. Consider the curve formed by the intersection of the cylinder $x^2 + y^2 = 4$ and the plane $z = x + y$ in the first octant. Which parametrization correctly represents this curve from the point $(2, 0, 2)$ to $(0, 2, 2)$?
   Answer: $\mathbf{r}(t) = (2\cos t, 2\sin t, 2\cos t + 2\sin t)$ for $t \in [0, \pi/2]$ The intersection curve lies on both surfaces, so we use $x = 2\cos t$ and $y = 2\sin t$ from the cylinder, and $z = x + y = 2\cos t + 2\sin t$ from the plane constraint. At $t = 0$: $(2, 0, 2)$ ✓. At $t = \pi/2$: $(0, 2, 2)$ ✓. Choice B has the wrong parameter interval direction. Choice C swaps sine and cosine, giving wrong start/end points. Choice D uses $z = 4\cos t\sin t$, which doesn't satisfy the plane equation $z = x + y$.
2. Consider the piecewise curve consisting of the line segment from $(0, 1)$ to $(2, 3)$ followed by the parabolic arc $y = x^2 - 1$ from $(2, 3)$ to $(0, -1)$. Which parametrization correctly represents the second piece of this curve?
   Answer: $\mathbf{r}(t) = (2 - 2t, (2 - 2t)^2 - 1)$ for $t \in [0, 1]$ When parametrizing a curve segment, you need to ensure your parameter values trace the path in the correct direction from start to finish. Here, you're looking for a parametrization of the parabola $y = x^2 - 1$ that goes from $(2, 3)$ to $(0, -1)$. The correct answer is C: $\mathbf{r}(t) = (2 - 2t, (2 - 2t)^2 - 1)$ for $t \in [0, 1]$. Let's verify: when $t = 0$, we get $(2 - 0, (2)^2 - 1) = (2, 3)$, which is our starting point. When $t = 1$, we get $(2 - 2, (0)^2 - 1) = (0, -1)$, which is our ending point. The $x$-coordinate is $2 - 2t$, and substituting this into $y = x^2 - 1$ gives $y = (2 - 2t)^2 - 1$, confirming we're on the parabola. Option A has the right curve equation but uses $t \in [2, 0]$, which is problematic since the interval goes backward. Option B uses $\mathbf{r}(t) = (2t, (2t)^2 - 1)$ for $t \in [1, 0]$, but this would start at $(2, 3)$ when $t = 1$ and end at $(0, -1)$ when $t = 0$—again, a backward parameter interval. Option D gives $\mathbf{r}(t) = (2 - t, (2 - t)^2 - 1)$ for $t \in [0, 2]$, which starts correctly at $(2, 3)$ but ends at $(0, -1)$ when $t = 2$, not matching the given interval structure. Always check both endpoints of your parametrization and ensure the parameter interval runs in the natural direction from smaller to larger values.
3. A curve $C$ is parametrized by $\mathbf{r}(t) = (2\cos t, 3\sin t, t)$ for $t \in [0, 2\pi]$. If we want to evaluate a line integral along $C$ but in the opposite direction, which parametrization should we use?
   Answer: $\mathbf{r}(t) = (2\cos(2\pi - t), 3\sin(2\pi - t), 2\pi - t)$ for $t \in [0, 2\pi]$ To reverse the orientation of a parametric curve, we need to traverse the same geometric path in the opposite direction. The substitution $u = 2\pi - t$ achieves this: when $t = 0$, $u = 2\pi$ (end point), and when $t = 2\pi$, $u = 0$ (start point). Choice A uses $t \mapsto -t$ which creates a different curve entirely. Choice C negates the position vector components, creating a different geometric path. Choice D is invalid notation since the interval $[2\pi, 0]$ is not properly ordered.
4. A helix is given by $\mathbf{r}(t) = (a\cos t, a\sin t, bt)$ where $a > 0$ and $b \neq 0$. If we want to parametrize the same geometric curve so that it makes exactly 3 complete revolutions as the parameter varies from 0 to 1, which parametrization is correct?
   Answer: $\mathbf{r}(s) = (a\cos(6\pi s), a\sin(6\pi s), 6\pi bs)$ for $s \in [0, 1]$ For 3 complete revolutions as $s$ goes from 0 to 1, the angular component must change by $3 \cdot 2\pi = 6\pi$. This requires the substitution $t = 6\pi s$. When $s = 0$, $t = 0$; when $s = 1$, $t = 6\pi$ (exactly 3 full revolutions). The $z$-component becomes $z = bt = b(6\pi s) = 6\pi bs$. Choice A only gives 1.5 revolutions. Choice C has the wrong $z$-component scaling. Choice D changes the radius and gives only 1 revolution.
5. A curve $C$ is parametrized as $\mathbf{r}(t) = (e^t \cos t, e^t \sin t, e^t)$ for $t \in [0, \ln 2]$. If we want to reparametrize this curve using arc length $s$ as the parameter, which relationship between $s$ and $t$ is correct?
   Answer: $s = \sqrt{3} \int_0^t e^u \, du = \sqrt{3}(e^t - 1)$ Arc length parametrization requires $s = \int_0^t |\mathbf{r}'(u)| \, du$. We have $\mathbf{r}'(t) = (e^t(\cos t - \sin t), e^t(\sin t + \cos t), e^t)$. Computing the magnitude: $|\mathbf{r}'(t)|^2 = e^{2t}[(\cos t - \sin t)^2 + (\sin t + \cos t)^2 + 1] = e^{2t}[2 + 1] = 3e^{2t}$. Therefore $|\mathbf{r}'(t)| = \sqrt{3}e^t$ and $s = \sqrt{3} \int_0^t e^u \, du = \sqrt{3}(e^t - 1)$. Choice B has the wrong coefficient. Choice C treats $\sqrt{3}$ as constant outside the integral incorrectly. Choice D makes an error in expanding the derivative magnitude.
6. A curve is defined implicitly by the equation $x^2 + y^2 + z^2 = 25$ and $z = 3$. To parametrize this curve for a line integral, we want the parameter to represent the angle measured from the positive $x$-axis in the $xy$-plane. Which parametrization achieves this with the standard domain $[0, 2\pi]$?
   Answer: $\mathbf{r}(\theta) = (4\cos\theta, 4\sin\theta, 3)$ for $\theta \in [0, 2\pi]$ When you encounter a curve defined by the intersection of two surfaces, you need to find which points satisfy both equations simultaneously. Here, you have a sphere $x^2 + y^2 + z^2 = 25$ (radius 5, centered at origin) intersected with the plane $z = 3$. This intersection creates a circle in the plane $z = 3$. To find the radius of this circle, substitute $z = 3$ into the sphere equation: $x^2 + y^2 + 3^2 = 25$, which gives $x^2 + y^2 = 16$. This is a circle with radius 4 in the plane $z = 3$. Since you want the parameter $\theta$ to represent the angle from the positive $x$-axis, you use the standard parametric form for a circle: $x = r\cos\theta$, $y = r\sin\theta$, where $r = 4$. Therefore, the correct parametrization is $\mathbf{r}(\theta) = (4\cos\theta, 4\sin\theta, 3)$, which is answer C. Answer A uses radius 5, which would be correct if the intersection were at $z = 0$, but ignores the constraint $z = 3$. Answer B incorrectly uses radius 3 (confusing the $z$-coordinate with the radius) and places the curve at $z = 4$. Answer D uses $2\sqrt{2} \approx 2.83$ as the radius, which doesn't correspond to any geometric relationship in this problem. Study tip: For intersection problems, always substitute one constraint into the other to find the actual dimensions of the resulting curve. Don't confuse given coordinates with radii.
7. Consider the curve formed by the intersection of the elliptic paraboloid $z = x^2 + 4y^2$ and the plane $z = 4$. If we parametrize this curve and want the $x$-coordinate to achieve its maximum value when the parameter $t = 0$, which parametrization is appropriate?
   Answer: $\mathbf{r}(t) = (2\cos t, \sin t, 4)$ for $t \in [0, 2\pi]$ When you encounter problems involving the intersection of surfaces, start by finding the curve equation, then analyze how different parametrizations behave at specific parameter values. To find the intersection curve, substitute $z = 4$ into the paraboloid equation: $4 = x^2 + 4y^2$, which simplifies to $\frac{x^2}{4} + y^2 = 1$. This is an ellipse with semi-major axis 2 along the x-direction and semi-minor axis 1 along the y-direction. The standard parametrization for this ellipse is $x = 2\cos t$ and $y = \sin t$. Since we want the x-coordinate to be maximum when $t = 0$, we need $x(0) = 2$ (the maximum possible x-value). Using $x = 2\cos t$, we get $x(0) = 2\cos(0) = 2$, which is indeed the maximum. Choice D gives $\mathbf{r}(t) = (2\cos t, \sin t, 4)$, which satisfies our requirement perfectly. Choice A has $x = \cos t$, so $x(0) = 1$, but the maximum x-value should be 2, not 1. Choice B has $x = 2\sin t$, so $x(0) = 0$, which is far from the maximum. Choice C has $x = 2\cos(t + \pi) = -2\cos t$, so $x(0) = -2$, which is actually the minimum x-value. Study tip: When parametrizing ellipses, remember that $\cos t$ and $\sin t$ achieve their maximum values of 1 when $t = 0$ and $t = \pi/2$ respectively. Match the trigonometric function to the desired behavior at your chosen parameter value.

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