Parametrizing Curves - Multivariable Calculus

The curve $C$ is the intersection of the sphere $x^2 + y^2 + z^2 = 9$ and the cylinder $x^2 + y^2 = 4$ . To parametrize this curve, which approach correctly accounts for the geometric constraints?
Answer: $\mathbf{r}(t) = (2\cos t, 2\sin t, \pm\sqrt{5})$ for $t \in [0, 2\pi]$ , giving two separate circles When finding the intersection of two surfaces, you need to satisfy both constraints simultaneously to understand the resulting curve's geometry. The intersection occurs where both $x^2 + y^2 + z^2 = 9$ (sphere of radius 3) and $x^2 + y^2 = 4$ (cylinder of radius 2) are satisfied. Since the cylinder constrains $x^2 + y^2 = 4$ , substituting into the sphere equation gives: $4 + z^2 = 9$ , so $z^2 = 5$ and $z = \pm\sqrt{5}$ . This means the intersection consists of two horizontal circles at heights $z = \sqrt{5}$ and $z = -\sqrt{5}$ , both with radius 2 in the $xy$ -plane. Option C correctly captures this geometry: $\mathbf{r}(t) = (2\cos t, 2\sin t, \pm\sqrt{5})$ parametrizes circles of radius 2 at the two fixed $z$ -values, requiring separate parametrizations for each circle. Option A uses radius 3 from the sphere, ignoring the cylinder's radius-2 constraint. The $z$ -component simplifies to 0, missing the actual intersection geometry entirely. Option B uses the correct radius 2 but has the wrong interval $[0, 4\pi]$ and incorrectly varies the $z$ -coordinate, which should be constant for each circle. Option D attempts to connect both circles with $\sqrt{5}\cos(2t)$ , but this creates a single continuous curve oscillating between the circles—geometrically impossible since the actual intersection is two separate, disconnected circles. Key takeaway: When parametrizing surface intersections, first solve the constraint equations algebraically to understand the geometric shape, then choose parametrization variables that respect all constraints simultaneously.
A curve $C$ is given by $\mathbf{r}(t) = (t^2, t^3, t)$ for $t \in [-1, 2]$ . To evaluate a line integral over this curve using a different parameter $u$ where $u = t + 1$ , what is the correct reparametrization?
Answer: $\mathbf{s}(u) = ((u-1)^2, (u-1)^3, u-1)$ for $u \in [0, 3]$ When reparametrizing a curve for line integrals, you need to express the old parameter in terms of the new one, then substitute systematically. This preserves the geometric path while changing how you traverse it. Given $u = t + 1$ , you can solve for $t$ in terms of $u$ : $t = u - 1$ . Now substitute this relationship into every component of the original parametrization $\mathbf{r}(t) = (t^2, t^3, t)$ . This gives you $\mathbf{s}(u) = ((u-1)^2, (u-1)^3, u-1)$ . For the parameter bounds, when $t = -1$ , we have $u = -1 + 1 = 0$ , and when $t = 2$ , we have $u = 2 + 1 = 3$ . So the new interval is $u \in [0, 3]$ . Choice A incorrectly uses $u$ directly instead of substituting $t = u - 1$ , creating a completely different curve. The mention of "Jacobian adjustment" is also misleading—while the derivative $\frac{dt}{du}$ matters for line integrals, it doesn't change the reparametrization itself. Choice B attempts some adjustment with $u^2 - 1$ and $u^3 - 1$ for the first two components but leaves the third as $u$ , creating an inconsistent and incorrect parametrization. Choice C correctly substitutes $(u+1)$ into each component but uses the wrong parameter bounds $[-2, 1]$ , which would correspond to a different relationship between $u$ and $t$ . Study tip: Always work backwards from the parameter relationship: solve for the old parameter, substitute completely, then transform the bounds using the same relationship. Check your work by verifying that endpoint values match.
Let $C$ be the line segment from the point of intersection of the lines $y=x+1$ and $y=7-2x$ to the point $(5, 8)$ . Which of the following is a parametrization of the curve $C$ ?
Answer: $\vec{r}(t) = \langle 2+3t, 3+5t \rangle$ , for $0 \le t \le 1$ First, find the starting point by setting the two line equations equal to each other: $x+1 = 7-2x$ , which gives $3x=6$ , so $x=2$ . The corresponding $y$ -value is $y=2+1=3$ . The starting point is $P_0 = (2, 3)$ . The ending point is $P_1 = (5, 8)$ . A standard parametrization for a line segment from $P_0$ to $P_1$ is $\vec{r}(t) = (1-t)P_0 + tP_1$ for $0 \le t \le 1$ . This becomes $\vec{r}(t) = (1-t)\langle 2, 3 \rangle + t\langle 5, 8 \rangle = \langle 2(1-t)+5t, 3(1-t)+8t \rangle = \langle 2-2t+5t, 3-3t+8t \rangle = \langle 2+3t, 3+5t \rangle$ .
Let $C$ be the curve parametrized by $\vec{r}(t) = \langle e^t, t^3, \cos(\pi t) \rangle$ for $0 \le t \le 1$ . Let $-C$ be the same curve with the opposite orientation. Which of the following parametrizes $-C$ ?
Answer: $\vec{s}(t) = \langle e^{1-t}, (1-t)^3, \cos(\pi(1-t)) \rangle$ , for $0 \le t \le 1$ When you encounter a question about curve orientation, you're dealing with the direction of traversal along a parametric curve. The key insight is that reversing orientation means starting at the endpoint and moving backward to the starting point. The original curve $\vec{r}(t) = \langle e^t, t^3, \cos(\pi t) \rangle$ for $0 \le t \le 1$ starts at $\vec{r}(0) = \langle 1, 0, 1 \rangle$ and ends at $\vec{r}(1) = \langle e, 1, -1 \rangle$ . To reverse the orientation, you need a new parametrization that starts at $\langle e, 1, -1 \rangle$ when $t = 0$ and ends at $\langle 1, 0, 1 \rangle$ when $t = 1$ . Option D accomplishes this perfectly using the substitution $u = 1-t$ . When $t = 0$ , we get $\vec{s}(0) = \langle e^1, 1^3, \cos(\pi) \rangle = \langle e, 1, -1 \rangle$ . When $t = 1$ , we get $\vec{s}(1) = \langle e^0, 0^3, \cos(0) \rangle = \langle 1, 0, 1 \rangle$ . This traces the same geometric path but in reverse. Option A creates an entirely different curve by negating components rather than reversing direction. Option B shifts the parameter but doesn't reverse orientation—it still goes from the original starting point to endpoint. Option C multiplies the entire vector by $-1$ , which reflects the curve through the origin rather than reversing its direction of traversal. Remember: to reverse curve orientation, substitute $t$ with $(1-t)$ or similar transformations that flip the parameter direction. Don't confuse this with negating vector components, which creates a different curve entirely.
A particle moves on the cylinder $x^2+y^2=1$ . Its vertical velocity is given by $dz/dt = x(t)$ . If the particle's horizontal motion is counter-clockwise when viewed from above, which of the following could be a parametrization of its path, $\vec{r}(t)$ ?
Answer: $\vec{r}(t) = \langle \cos(t), \sin(t), \sin(t) \rangle$ The particle's motion on the cylinder $x^2+y^2=1$ in a counter-clockwise direction can be parametrized by $x(t) = \cos(t)$ and $y(t) = \sin(t)$ . The vertical velocity is given as $dz/dt = x(t) = \cos(t)$ . To find the vertical position $z(t)$ , we must integrate the vertical velocity with respect to time: $z(t) = \int \frac{dz}{dt} dt = \int \cos(t) dt = \sin(t) + C$ , where $C$ is a constant of integration. A possible parametrization is therefore $\vec{r}(t) = \langle \cos(t), \sin(t), \sin(t) + C \rangle$ . Option C matches this form with $C=0$ .
A line integral is to be evaluated over the line segment from $P=(0, 1, 5)$ to $Q=(3, 5, 2)$ . Which parametrization describes this path as being traced at a constant speed of 1 (i.e., is an arc-length parametrization)?
Answer: $\vec{r}(t) = \langle \frac{3t}{\sqrt{34}}, 1+\frac{4t}{\sqrt{34}}, 5-\frac{3t}{\sqrt{34}} \rangle$ , for $0 \le t \le \sqrt{34}$ When evaluating line integrals, understanding arc-length parametrization is crucial. An arc-length parametrization traces a curve at constant speed 1, meaning the magnitude of the velocity vector $|\vec{r}'(t)| = 1$ for all values of $t$ . To find the correct parametrization, you need to first determine the direction vector from $P$ to $Q$ : $\vec{PQ} = (3,5,2) - (0,1,5) = (3,4,-3)$ . The length of this segment is $|\vec{PQ}| = \sqrt{3^2 + 4^2 + (-3)^2} = \sqrt{34}$ . For arc-length parametrization, you need the unit direction vector: $\frac{\vec{PQ}}{|\vec{PQ}|} = \frac{(3,4,-3)}{\sqrt{34}}$ . The parametrization becomes $\vec{r}(t) = P + t \cdot \text{(unit vector)} = (0,1,5) + t \cdot \frac{(3,4,-3)}{\sqrt{34}}$ , which gives option D. Since the total length is $\sqrt{34}$ , the parameter runs from $0$ to $\sqrt{34}$ . Option A parametrizes the correct path but at speed $|\vec{r}'(t)| = \sqrt{34}$ , not 1. Option B attempts to normalize but uses the wrong denominator (34 instead of $\sqrt{34}$ ), creating speed $\frac{\sqrt{34}}{34}$ . Option C doesn't even pass through the starting point $P$ since $\vec{r}(0) = (0,0,0) \neq (0,1,5)$ . Study tip: For arc-length parametrization, always normalize the direction vector by dividing by its magnitude, and let the parameter run from 0 to the total arc length.
A curve $C$ is the portion of the ellipse $9x^2 + y^2 = 9$ in the first quadrant, traversed from the point $(0, 3)$ to $(1, 0)$ . Which of the following parametrizes $C$ ?
Answer: $\vec{r}(t) = \langle \sin(t), 3\cos(t) \rangle$ , for $0 \le t \le \pi/2$ The equation of the ellipse can be written in standard form as $x^2/1^2 + y^2/3^2 = 1$ . The semi-axes are $a=1$ (in the $x$ -direction) and $b=3$ (in the $y$ -direction). A standard counter-clockwise parametrization is $\vec{r}(t) = \langle a\cos(t), b\sin(t) \rangle = \langle \cos(t), 3\sin(t) \rangle$ . For this parametrization, $t=0$ gives $(1, 0)$ and $t=\pi/2$ gives $(0, 3)$ . This traces the first quadrant arc from the $x$ -axis to the $y$ -axis. The question asks for the reverse direction, from $(0, 3)$ to $(1, 0)$ . To reverse the direction, we can swap the roles of sine and cosine with appropriate adjustments, or change the parameter. Let's test option B: $\vec{r}(t) = \langle \sin(t), 3\cos(t) \rangle$ . At $t=0$ , $\vec{r}(0) = \langle \sin(0), 3\cos(0) \rangle = \langle 0, 3 \rangle$ . This is the correct starting point. At $t=\pi/2$ , $\vec{r}(\pi/2) = \langle \sin(\pi/2), 3\cos(\pi/2) \rangle = \langle 1, 0 \rangle$ . This is the correct ending point. This parametrization correctly describes the path.
Consider the curve of intersection of the two cylinders $x^2+z^2=4$ and $y^2+z^2=4$ . Which of the following parametrizes the portion of this curve that lies in the first octant, starting from the $xy$ -plane and ending at the point of maximum height?
Answer: $\vec{r}(t) = \langle 2\cos(t), 2\cos(t), 2\sin(t) \rangle$ , for $0 \le t \le \pi/2$ From the two cylinder equations, we have $x^2+z^2 = y^2+z^2$ , which implies $x^2 = y^2$ . In the first octant ( $x, y, z \ge 0$ ), this simplifies to $x=y$ . Substituting $x=y$ into the first cylinder equation gives $x^2+z^2=4$ . This is the equation of a circle of radius 2 in the plane $x=y$ (projected onto the $xz$ -plane). We can parametrize this circular arc in the first octant using $x(t)=2\cos(t)$ and $z(t)=2\sin(t)$ for $0 \le t \le \pi/2$ . Since $y=x$ , we have $y(t)=2\cos(t)$ . This gives the parametrization $\vec{r}(t) = \langle 2\cos(t), 2\cos(t), 2\sin(t) \rangle$ . Let's check the start and end points. The path starts on the $xy$ -plane, where $z=0$ . This corresponds to $2\sin(t)=0$ , so $t=0$ . At $t=0$ , $\vec{r}(0) = \langle 2, 2, 0 \rangle$ . The maximum height occurs when $z$ is maximum. $z=2\sin(t)$ is maximum when $t=\pi/2$ , giving $z=2$ . At $t=\pi/2$ , $\vec{r}(\pi/2) = \langle 0, 0, 2 \rangle$ . The range $0 \le t \le \pi/2$ correctly traces the curve from the $xy$ -plane to its highest point in the first octant.
A particle follows a path along the arc of the circle $x^2 + y^2 = 25$ from the point $(0, -5)$ to the point $(5, 0)$ in a clockwise direction. Which of the following parametrizations describes this path?
Answer: $\vec{r}(t) = \langle 5\sin(t), -5\cos(t) \rangle$ , for $0 \le t \le \pi/2$ The path is clockwise. A standard counter-clockwise (CCW) parametrization is $\langle 5\cos(t), 5\sin(t) \rangle$ . To achieve clockwise motion, we can, for example, replace $t$ with $-t$ to get $\langle 5\cos(-t), 5\sin(-t) \rangle = \langle 5\cos(t), -5\sin(t) \rangle$ , or swap sine and cosine to get $\langle 5\sin(t), 5\cos(t) \rangle$ or $\langle 5\sin(t), -5\cos(t) \rangle$ . Let's test choice B: $\vec{r}(t) = \langle 5\sin(t), -5\cos(t) \rangle$ . At $t=0$ , $\vec{r}(0) = \langle 0, -5 \rangle$ , which is the correct starting point. At $t=\pi/2$ , $\vec{r}(\pi/2) = \langle 5\sin(\pi/2), -5\cos(\pi/2) \rangle = \langle 5, 0 \rangle$ , which is the correct ending point. The interval $0 \le t \le \pi/2$ correctly traces the desired arc.
Which of the following parametrizes the curve of intersection of the parabolic cylinder $z=x^2$ and the plane $y=2x$ , from the point $(0,0,0)$ to $(1,2,1)$ ?
Answer: $\vec{r}(t) = \langle t, 2t, t^2 \rangle$ , for $0 \le t \le 1$ To find a parametrization for the curve of intersection, we can use one variable as a parameter and express the other variables in terms of it. Let's use $x$ as the parameter, so let $x=t$ . From the plane equation, we have $y = 2x = 2t$ . From the parabolic cylinder equation, we have $z = x^2 = t^2$ . Combining these gives the vector function $\vec{r}(t) = \langle t, 2t, t^2 \rangle$ . Now we determine the range for the parameter $t$ . The curve starts at $(0,0,0)$ , which corresponds to $x=0$ , so $t=0$ . It ends at $(1,2,1)$ , which corresponds to $x=1$ , so $t=1$ . Thus, the range is $0 \le t \le 1$ .
The curve $C$ is defined by the intersection of the cylinder $x^2 + y^2 = 1$ and the plane $z = x + y$ . Which parametrization correctly represents this curve?
Answer: $\mathbf{r}(t) = (\cos t, \sin t, \cos t + \sin t)$ for $t \in [0, 2\pi]$ The intersection curve lies on both surfaces simultaneously. Points on the cylinder satisfy x² + y² = 1, which can be parametrized as x = cos t, y = sin t. Points on the plane satisfy z = x + y. Combining these constraints gives z = cos t + sin t. We can verify: x² + y² = cos²t + sin²t = 1 ✓, and z = cos t + sin t satisfies the plane equation. Choice B uses z = cos t sin t, which doesn't satisfy z = x + y. Choice C parametrizes the cylinder differently but gives different points on the curve. Choice D only covers half the circle (t ∈ [-1,1] doesn't give the full parameter range for the unit circle).
Find a parametrization for the curve defined by the parabola $x = 1 - 2y^2$ that starts at the point $(-1, 1)$ and ends at the point $(-7, 2)$ .
Answer: $\vec{r}(t) = \langle 1 - 2t^2, t \rangle$ , for $1 \le t \le 2$ Since $x$ is given as a function of $y$ , it is natural to choose $y$ as the parameter. Let $y=t$ . Then $x = 1 - 2t^2$ . The parametrization is $\vec{r}(t) = \langle 1 - 2t^2, t \rangle$ . The path starts at the point where $y=1$ and ends at the point where $y=2$ . Therefore, the parameter $t$ must range from 1 to 2. This gives the interval $1 \le t \le 2$ .
A curve $C$ consists of the line segment from $(0,1)$ to $(2,3)$ followed by the parabolic arc $y = x^2 - 1$ from $(2,3)$ to $(3,8)$ . Which parametrization correctly represents the entire curve $C$ for $t \in [0,2]$ ?
Answer: $\mathbf{r}(t) = \begin{cases} (2t, 1+2t) & \text{if } 0 \leq t \leq 1 \\ (1+t, (1+t)^2-1) & \text{if } 1 < t \leq 2 \end{cases}$ For the line segment from (0,1) to (2,3), we need x to go from 0 to 2 and y to go from 1 to 3 as t goes from 0 to 1. This gives x = 2t and y = 1+2t. At t=1, this gives (2,3) correctly. For the parabolic arc from (2,3) to (3,8), we need x to go from 2 to 3 as t goes from 1 to 2, so x = 1+t. Since the arc follows y = x² - 1, we have y = (1+t)² - 1. At t=1, this gives (2,3), and at t=2, this gives (3,8). Choice A has incorrect parametrization for the second piece. Choice C doesn't maintain continuity at the junction point. Choice D uses the wrong parameter intervals.
To evaluate a line integral over the cardioid given by the polar equation $r = 1 + \cos(\theta)$ , for $0 \le \theta \le \pi$ , a Cartesian parametrization is required. Which of the following is a correct parametrization?
Answer: $\vec{r}(t) = \langle (1+\cos(t))\cos(t), (1+\cos(t))\sin(t) \rangle$ , for $0 \le t \le \pi$ The standard conversion from polar coordinates $(r, \theta)$ to Cartesian coordinates $(x, y)$ is given by the equations $x = r\cos(\theta)$ and $y = r\sin(\theta)$ . In this problem, the radius $r$ is a function of the angle $\theta$ : $r = 1 + \cos(\theta)$ . To parametrize the curve, we can use the angle $\theta$ as the parameter, which we can call $t$ . So, let $\theta=t$ . Substituting the expression for $r$ into the conversion formulas, we get: $x(t) = (1 + \cos(t))\cos(t)$ and $y(t) = (1 + \cos(t))\sin(t)$ . The given range for the angle is $0 \le \theta \le \pi$ , so the parameter range is $0 \le t \le \pi$ . This gives the vector parametrization $\vec{r}(t) = \langle (1+\cos(t))\cos(t), (1+\cos(t))\sin(t) \rangle$ .
To compute a line integral over the curve $C$ , which is the intersection of the cylinder $x^2 + y^2 = 9$ and the plane $z = 1 - x/3$ , a parametrization is needed. The curve is oriented counter-clockwise when viewed from above. Which of the following is a suitable parametrization?
Answer: $\vec{r}(t) = \langle 3\cos(t), 3\sin(t), 1 - \cos(t) \rangle$ , for $0 \le t \le 2\pi$ The projection of the curve onto the $xy$ -plane is the circle $x^2 + y^2 = 9$ . A standard counter-clockwise parametrization for a circle of radius 3 is $x(t) = 3\cos(t)$ and $y(t) = 3\sin(t)$ , for $0 \le t \le 2\pi$ . To find the $z$ -component, substitute the expression for $x$ into the equation of the plane: $z = 1 - x/3 = 1 - (3\cos(t))/3 = 1 - \cos(t)$ . Combining these components gives the parametrization $\vec{r}(t) = \langle 3\cos(t), 3\sin(t), 1 - \cos(t) \rangle$ .
Consider parametrizing the boundary of the region $R = \{(x,y) : x^2 + y^2 \leq 4, x \geq 0, y \geq 0\}$ for a line integral. The boundary consists of three pieces: two line segments and one circular arc. If we traverse the boundary counterclockwise starting from $(2,0)$ , which represents the correct parametrization of the circular arc portion?
Answer: $\mathbf{r}_1(t) = (2\cos t, 2\sin t)$ for $t \in [0, \pi/2]$ When parametrizing boundaries for line integrals, you need to carefully match the direction of traversal with your parameter values. The region R is a quarter-circle of radius 2 in the first quadrant, and its boundary includes the arc from (2,0) to (0,2). For counterclockwise traversal starting at (2,0), you want the standard parametrization $\mathbf{r}(t) = (2\cos t, 2\sin t)$ with appropriate parameter bounds. At $t = 0$ : $\mathbf{r}(0) = (2\cos 0, 2\sin 0) = (2,0)$ ✓. At $t = \pi/2$ : $\mathbf{r}(\pi/2) = (2\cos(\pi/2), 2\sin(\pi/2)) = (0,2)$ ✓. This matches the desired counterclockwise path, making D correct. Choice A uses $\cos(\pi/2 - t) = \sin t$ and $\sin(\pi/2 - t) = \cos t$ , giving $\mathbf{r}_1(t) = (2\sin t, 2\cos t)$ . At $t = 0$ , this yields (0,2), not (2,0), so it starts at the wrong point. Choice B has the correct parametric equations but uses interval $[\pi/2, 0]$ . Since $\pi/2 > 0$ , this means $t$ decreases from $\pi/2$ to 0, creating clockwise motion instead of counterclockwise. Choice C gives $\mathbf{r}_1(t) = (2\sin t, 2\cos t)$ , which also starts at (0,2) when $t = 0$ , not the required starting point (2,0). Strategy tip: Always check your parametrization at the endpoints of your parameter interval to verify you start and end at the correct points, and ensure the direction matches the problem's requirements.
A particle moves along the curve $\mathbf{r}(t) = (e^t \cos t, e^t \sin t, e^t)$ for $t \in [0, \ln 2]$ . To parametrize the same geometric curve with constant speed, which of the following represents the correct approach?
Answer: Use arc length $s = \sqrt{3}(e^t - 1)$ and reparametrize with $t = \ln\left(\frac{s}{\sqrt{3}} + 1\right)$ First, find $\mathbf{r}'(t) = (e^t(\cos t - \sin t), e^t(\sin t + \cos t), e^t)$ . Then $\|\mathbf{r}'(t)\|^2 = e^{2t}[(\cos t - \sin t)^2 + (\sin t + \cos t)^2 + 1] = e^{2t}[2 + 1] = 3e^{2t}$ . So $\|\mathbf{r}'(t)\| = \sqrt{3}e^t$ . The arc length from 0 to t is $s = \int_0^t \sqrt{3}e^u du = \sqrt{3}(e^t - 1)$ . Solving for t: $e^t = \frac{s}{\sqrt{3}} + 1$ , so $t = \ln\left(\frac{s}{\sqrt{3}} + 1\right)$ . Choice A has the wrong arc length formula. Choice C doesn't give constant speed parametrization. Choice D has incorrect speed calculation.
A helical curve in 3D space makes exactly 2 complete revolutions around the z-axis while rising from $z = 0$ to $z = 4\pi$ , with radius 3. If the curve is parametrized for $t \in [0, 1]$ , which parametrization is correct?
Answer: $\mathbf{r}(t) = (3\cos(4\pi t), 3\sin(4\pi t), 4\pi t)$ For 2 complete revolutions as t goes from 0 to 1, the angular parameter must go from 0 to 4π, so we need 4πt in the trigonometric functions. For z to go from 0 to 4π as t goes from 0 to 1, we need z = 4πt. The radius is constant at 3. Choice A only makes 1 revolution (2πt gives one full cycle). Choice C has the wrong z-component (only rises to 2π instead of 4π). Choice D only makes 0.5 revolutions (πt gives half a cycle).
Consider the piecewise smooth curve $C$ that follows the parabola $y = x^2$ from $(-1,1)$ to $(1,1)$ , then follows the line segment from $(1,1)$ to $(1,3)$ . If this curve is parametrized continuously with parameter $t \in [0,3]$ such that the parabolic portion corresponds to $t \in [0,2]$ , which parametrization ensures continuity at the junction point?
Answer: $\mathbf{r}(t) = \begin{cases} (t-1, (t-1)^2) & \text{if } 0 \leq t \leq 2 \\ (1, 1+2(t-2)) & \text{if } 2 < t \leq 3 \end{cases}$ When dealing with piecewise parametric curves, continuity at junction points is crucial. A parametric curve $\mathbf{r}(t)$ is continuous at $t = a$ if $\lim_{t \to a^-} \mathbf{r}(t) = \lim_{t \to a^+} \mathbf{r}(t) = \mathbf{r}(a)$ . Since the parabolic portion corresponds to $t \in [0,2]$ and the line segment starts at $t = 2$ , we need continuity at $t = 2$ . The junction point is $(1,1)$ , so both pieces of the parametrization must give this coordinate when $t = 2$ . Let's check each option: For choice C: At $t = 2$ , the parabolic piece gives $(2-1, (2-1)^2) = (1,1)$ . The linear piece at $t = 2$ gives $(1, 1+2(2-2)) = (1,1)$ . Perfect continuity! Choice A fails because at $t = 2$ , the parabolic piece gives $(-1+2, (-1+2)^2) = (1,1)$ , but the linear piece gives $(1, 2(2)-3) = (1,1)$ . Wait, this actually works too, but let's verify the parameter ranges match the problem constraints. Choice B has the parabolic portion only going to $t = 1$ , violating the given constraint that the parabola should correspond to $t \in [0,2]$ . Choice D has the linear piece giving $(1, 1+2(2)) = (1,5)$ at $t = 2$ , creating a discontinuous jump from $(1,1)$ to $(1,5)$ . Study tip: Always check continuity by evaluating both pieces of a piecewise function at the transition point. The coordinates must match exactly, and verify that the parameter intervals align with the problem's specifications.
Consider the closed curve formed by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ traversed counterclockwise. If this curve is parametrized as $\mathbf{r}(t) = (2\cos(t), 3\sin(t))$ for $t \in [0, 2\pi]$ , what is the orientation of the parametrization?
Answer: Counterclockwise, and the parametrization matches the desired orientation exactly To determine orientation, we check the sign of the cross product $\mathbf{r}'(t) \times \mathbf{k}$ where $\mathbf{k}$ is the unit vector in the z-direction. Here $\mathbf{r}'(t) = (-2\sin(t), 3\cos(t))$ . The 'signed area' element is $x\frac{dy}{dt} - y\frac{dx}{dt} = (2\cos(t))(3\cos(t)) - (3\sin(t))(-2\sin(t)) = 6\cos^2(t) + 6\sin^2(t) = 6 > 0$ . Since this is positive, the orientation is counterclockwise, matching the desired direction. Choice B is incorrect about the orientation. Choice C is wrong because starting point doesn't affect orientation. Choice D is incorrect because orientation is determined by the parametric form, not the axis lengths.

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