Arc Length of Parametric Curves - Calculus 2

1. A parametric curve is given by $x = e^t\cos(t)$ and $y = e^t\sin(t)$ for $0 \leq t \leq \ln(2)$. Which expression correctly represents the arc length?
   Answer: $\int_0^{\ln(2)} \sqrt{e^{2t}(\cos(t) - \sin(t))^2 + e^{2t}(\sin(t) + \cos(t))^2} \, dt$ We have $\frac{dx}{dt} = e^t(\cos(t) - \sin(t))$ and $\frac{dy}{dt} = e^t(\sin(t) + \cos(t))$. The arc length is $\int_0^{\ln(2)} \sqrt{[e^t(\cos(t) - \sin(t))]^2 + [e^t(\sin(t) + \cos(t))]^2} \, dt$, which simplifies to choice B. Choice A has incorrect signs in the derivative calculation. Choice C incorrectly factors the expressions. Choice D factors out $e^t$ incorrectly from the square root.
2. The length of the curve defined by $x(t) = \ln(\sec t + \tan t) - \sin t$ and $y(t) = \cos t$ from $t=0$ to $t=\pi/3$ is being calculated. What is the value of the integrand $\sqrt{(dx/dt)^2 + (dy/dt)^2}$ after simplification?
   Answer: $\tan t$ This question tests the setup and simplification, not the final integration. First, find the derivatives. $y'(t) = -\sin t$. For $x(t)$, recall $\frac{d}{dt}\ln(\sec t + \tan t) = \sec t$. So, $x'(t) = \sec t - \cos t$. Now, compute the sum of squares: $(x')^2 + (y')^2 = (\sec t - \cos t)^2 + (-\sin t)^2 = \sec^2 t - 2\sec t \cos t + \cos^2 t + \sin^2 t$. Since $\sec t \cos t = 1$ and $\cos^2 t + \sin^2 t = 1$, this simplifies to $\sec^2 t - 2(1) + 1 = \sec^2 t - 1$. Finally, using the identity $\tan^2 t = \sec^2 t - 1$, the expression becomes $\tan^2 t$. The integrand is $\sqrt{\tan^2 t} = |\tan t|$. Since the interval is $[0, \pi/3]$, $\tan t$ is non-negative, so the integrand is $\tan t$.
3. A line segment from (1, 2) to (5, 5) is parameterized by $x(t) = 1+4t$ and $y(t) = 2+3t$ for $0 \le t \le 1$. If the parameterization is changed to $x(u) = 1+4u^2$ and $y(u) = 2+3u^2$ for $0 \le u \le 1$, how does the arc length change?
   Answer: It stays the same. The arc length of a curve is a geometric property and does not depend on the specific parameterization, as long as the parameterization traces the curve exactly once without reversing direction. The first parameterization traces the line segment from (1,2) to (5,5). Its length can be found by the distance formula: $\sqrt{(5-1)^2 + (5-2)^2} = \sqrt{4^2+3^2} = \sqrt{16+9} = \sqrt{25} = 5$. The second parameterization has its starting point at $u=0$, where $x(0)=1, y(0)=2$, and its endpoint at $u=1$, where $x(1)=5, y(1)=5$. Since $u^2$ is monotonically increasing on $[0,1]$, this parameterization also traces the same line segment exactly once. Therefore, the arc length remains the same. This question tests the conceptual understanding that arc length is independent of parameterization.
4. The arc length of the parabola $y=x^2$ from $x=0$ to $x=1$ can be found by parameterizing the curve as $x(t)=t, y(t)=t^2$ for $0 \le t \le 1$. Which of the following integrals correctly represents this length?
   Answer: $\int_0^1 \sqrt{1+4t^2} \, dt$ This question connects the Cartesian form of arc length with the parametric form. Given the parameterization $x(t)=t$ and $y(t)=t^2$, we first find the derivatives: $x'(t) = 1$ and $y'(t) = 2t$. The formula for parametric arc length is $L = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2} dt$. Plugging in our derivatives, we get $L = \int_0^1 \sqrt{(1)^2 + (2t)^2} dt = \int_0^1 \sqrt{1+4t^2} dt$. This matches option A. Note that this is the same integral obtained from the Cartesian formula $L = \int_a^b \sqrt{1+(dy/dx)^2} dx$, since $dy/dx = 2x$, giving $\int_0^1 \sqrt{1+(2x)^2} dx$.
5. Consider the curve given by $x(t) = \int_0^t \cos(\frac{\pi}{2}u^2)du$ and $y(t) = \int_0^t \sin(\frac{\pi}{2}u^2)du$ (a Fresnel integral). What is the arc length of this curve from $t=0$ to $t=2$?
   Answer: 2 When you encounter parametric curves defined by integrals, the key insight is recognizing how the Fundamental Theorem of Calculus helps you find derivatives and compute arc length using the standard formula. To find the arc length from $t=0$ to $t=2$, you need $\int_0^2 \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$. Using the Fundamental Theorem of Calculus, the derivatives are: $\frac{dx}{dt} = \cos\left(\frac{\pi}{2}t^2\right)$ and $\frac{dy}{dt} = \sin\left(\frac{\pi}{2}t^2\right)$ The arc length becomes: $\int_0^2 \sqrt{\cos^2\left(\frac{\pi}{2}t^2\right) + \sin^2\left(\frac{\pi}{2}t^2\right)} dt$ Since $\cos^2(\theta) + \sin^2(\theta) = 1$ for any angle $\theta$, this simplifies to: $\int_0^2 1 \, dt = 2$ Choice A (1) might tempt you if you incorrectly integrated from 0 to 1 instead of 0 to 2. Choice B ($2\pi$) could result from mistakenly thinking the trigonometric functions contribute a factor of $2\pi$ without recognizing the Pythagorean identity. Choice C ($\pi$) might arise from confusing this with a circular arc or incorrectly handling the $\frac{\pi}{2}$ factor. Remember: For parametric curves, especially those involving trigonometric integrals, always check if $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2$ simplifies using trigonometric identities before attempting complex integration. The Fresnel spiral has the special property that its speed is always 1, making arc length calculations surprisingly simple.
6. A parametric curve is given by $x(t) = a(t-\sin t)$ and $y(t) = a(1-\cos t)$ for $a > 0$. The length of one arch of this cycloid (from $t=0$ to $t=2\pi$) is $8a$. What is the length of the curve from the point where it first touches the x-axis (after $t=0$) to the highest point of the first arch?
   Answer: $4a - 2a\sqrt{2}$ The problem gives the total length of one arch as $8a$. The first arch starts at $t=0$. The highest point of the arch occurs when $y(t)$ is maximum. $y(t) = a(1-\cos t)$ is max when $\cos t = -1$, which is at $t=\pi$. The curve touches the x-axis when $y(t)=0$, so $a(1-\cos t)=0 \implies \cos t = 1$. This occurs at $t=0, 2\pi, 4\pi, ...$. The first time it touches the x-axis after $t=0$ is at $t=2\pi$. The question asks for the length from where it first touches the x-axis (after t=0) to the highest point. This seems backward. Let's assume it means from the start $(t=0)$ to the highest point $(t=\pi)$, which is half the arch. By symmetry, this length is $8a/2=4a$. Let's assume it means from the highest point $(t=\pi)$ to the end of the arch $(t=2\pi)$. This is also $4a$. The question seems unusual. 'from the point where it first touches the x-axis (after t=0)'. This is at $(2\pi a, 0)$ corresponding to $t=2\pi$. 'to the highest point of the first arch'. This is at $(\pi a, 2a)$ corresponding to $t=\pi$. This asks for the length from $t=2\pi$ to $t=\pi$, which is the negative of the integral from $\pi$ to $2\pi$. The length is positive, so it's the integral from $\pi$ to $2\pi$. This is the second half of the arch, which has length $4a$. The wording is confusing. Let me re-read. Maybe it means a different point. Let's find the length from $t=\pi/2$ to $t=\pi$. The integrand is $2a|\sin(t/2)| = 2a\sin(t/2)$ on $[0, 2\pi]$. $\int_{\pi/2}^{\pi} 2a\sin(t/2) dt = [-4a\cos(t/2)]_{\pi/2}^{\pi} = -4a(\cos(\pi/2)-\cos(\pi/4)) = -4a(0-\frac{\sqrt{2}}{2}) = 2a\sqrt{2}$. The length from 0 to $\pi/2$ is $[-4a\cos(t/2)]_0^{\pi/2} = -4a(\cos(\pi/4)-\cos(0)) = -4a(\frac{\sqrt{2}}{2}-1) = 4a - 2a\sqrt{2}$. The length from 0 to $\pi$ is $4a$. The question is likely asking for something non-obvious. What if 'first touches the x-axis' means $t=0$, and 'highest point' is $t=\pi$? Length is $4a$. What if it means from $t=0$ to the point where the tangent is vertical? $dx/dt = a(1-\cos t)=0$ at $t=0, 2\pi$. Not helpful. Where tangent is horizontal? $dy/dt = a\sin t = 0$ at $t=0, \pi$. So from the start to the highest point. That's $4a$. Let's assume the question is asking for the arc length from $t=0$ to $t=\pi/2$. That length is $4a - 2a\sqrt{2}$. Let's re-word the stem slightly to make this clearer. 'What is the length of the curve from its starting point to the point where $y=a$ for the first time?'. $y(t)=a \implies a(1-\cos t)=a \implies 1-\cos t=1 \implies \cos t=0$. This is at $t=\pi/2$. The length from $t=0$ to $t=\pi/2$ is $4a - 2a\sqrt{2}$. This is a good multi-step question.
7. Consider the parametric curve $x = t^3 - 3t$ and $y = t^2$ for $t \in [-2, 2]$. The curve has a self-intersection at the origin. What is the total arc length of both branches of the curve?
   Answer: $\int_{-2}^2 \sqrt{(3t^2 - 3)^2 + 4t^2} \, dt$ The arc length formula for parametric curves is $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt$. Here $\frac{dx}{dt} = 3t^2 - 3$ and $\frac{dy}{dt} = 2t$. The curve self-intersects but we still integrate over the full parameter interval to get the total arc length of both branches. Choice A incorrectly assumes symmetry and doubles half the integral. Choice C uses incorrect bounds. Choice D incorrectly splits the integral, missing the portions from $t \in [-1,1]$.
8. A particle moves on a circle of radius 5 centered at the origin, with position given by $x(t) = 5\cos(\omega t)$ and $y(t) = 5\sin(\omega t)$. The particle travels a distance of $10\pi$ in 2 seconds. What is the value of the angular frequency $\omega > 0$?
   Answer: $\pi$ This problem tests your understanding of parametric motion and the relationship between arc length, angular velocity, and time for circular motion. When a particle moves in a circle with parametric equations $x(t) = 5\cos(\omega t)$ and $y(t) = 5\sin(\omega t)$, you need to find the arc length traveled. The arc length formula for parametric curves is $s = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$. Taking derivatives: $\frac{dx}{dt} = -5\omega\sin(\omega t)$ and $\frac{dy}{dt} = 5\omega\cos(\omega t)$. Therefore: $\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{25\omega^2\sin^2(\omega t) + 25\omega^2\cos^2(\omega t)} = 5\omega$ The distance traveled in 2 seconds is $s = \int_0^2 5\omega \, dt = 10\omega$. Since this equals $10\pi$, we have $10\omega = 10\pi$, so $\omega = \pi$. Answer A ($\pi/2$) gives a distance of only $5\pi$, half the required distance. Answer B ($2\pi$) would result in $20\pi$ distance traveled, which is twice too much. Answer D (1) yields a distance of 10, missing the factor of $\pi$ entirely. The correct answer is C: $\omega = \pi$. Study tip: For circular motion problems, remember that the speed is constant and equals the radius times the angular frequency ($v = r\omega$). Distance equals speed times time, so $d = r\omega t$ provides a quick check for your answer.
9. The position of a particle is given by $x(t) = t^2$ and $y(t) = \frac{1}{3}t^3 - t$. Find the total distance traveled by the particle during the time interval $-1 \le t \le 1$.
   Answer: $8/3$ When you encounter parametric equations describing a particle's motion, finding total distance requires calculating arc length, not just displacement. This is a crucial distinction that separates many students. To find the total distance traveled, you need the arc length formula: $s = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$ First, find the derivatives: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = t^2 - 1$ Next, compute the integrand: $\sqrt{(2t)^2 + (t^2-1)^2} = \sqrt{4t^2 + t^4 - 2t^2 + 1} = \sqrt{t^4 + 2t^2 + 1} = \sqrt{(t^2+1)^2} = |t^2+1|$ Since $t^2 + 1 > 0$ for all real $t$, we have $|t^2+1| = t^2+1$. The total distance is: $\int_{-1}^1 (t^2 + 1) \, dt = \left[\frac{t^3}{3} + t\right]_{-1}^1 = \left(\frac{1}{3} + 1\right) - \left(\frac{-1}{3} - 1\right) = \frac{4}{3} + 2 = \frac{8}{3}$ Answer choice A ($\frac{4}{3}$) likely comes from forgetting the "+1" term in the integrand. Choice B (4) might result from computational errors in the integration bounds. Choice C (2) could arise from calculating displacement magnitude instead of total distance traveled. Remember: for parametric motion problems, always use the arc length formula for distance. Don't confuse distance traveled with displacement—distance is always non-negative and accounts for the actual path taken.
10. The arc length of the curve $x(t) = \sin(2t), y(t) = \cos(2t)$ from $t=0$ to $t=\pi$ is found to be $2\pi$. What is the arc length of the curve $x(t) = \sin(kt), y(t) = \cos(kt)$ from $t=0$ to $t=\pi$ for a constant $k>0$?
    Answer: $k\pi$ This question tests the understanding of how a parameter affects arc length. For the general curve, $x'(t) = k\cos(kt)$ and $y'(t) = -k\sin(kt)$. The sum of squares of the derivatives is $(k\cos(kt))^2 + (-k\sin(kt))^2 = k^2\cos^2(kt) + k^2\sin^2(kt) = k^2$. The integrand is $\sqrt{k^2} = |k|$. Since $k>0$, this is simply $k$. The arc length is $L = \int_0^{\pi} k \, dt = [kt]_0^{\pi} = k\pi$. The initial information about the case $k=2$ is consistent with this formula, as $2\pi$ is the result.
11. A parametric curve is given by $x = \int_0^t \cos(u^2) \, du$ and $y = \int_0^t \sin(u^2) \, du$ for $t \in [0, \sqrt{\pi}]$. This is known as a Fresnel spiral. What integral represents its arc length?
    Answer: $\int_0^{\sqrt{\pi}} dt$ Using the Fundamental Theorem of Calculus, $\frac{dx}{dt} = \cos(t^2)$ and $\frac{dy}{dt} = \sin(t^2)$. Therefore the arc length is $\int_0^{\sqrt{\pi}} \sqrt{\cos^2(t^2) + \sin^2(t^2)} \, dt = \int_0^{\sqrt{\pi}} \sqrt{1} \, dt = \int_0^{\sqrt{\pi}} dt$. Choice A doesn't simplify the trigonometric identity. Choice B incorrectly adds the derivatives before taking absolute value. Choice C squares the derivatives incorrectly.
12. A point moves along the curve $x(t) = 2\cos(t) + \cos(2t)$, $y(t) = 2\sin(t) - \sin(2t)$. The speed of the point at time $t$ is given by $\sqrt{(x'(t))^2+(y'(t))^2}$. Find the speed at $t = \pi/2$.
    Answer: $2\sqrt{2}$ This question asks for the speed (the integrand of the arc length formula) at a specific point, not the full arc length. First, find the derivatives: $x'(t) = -2\sin(t) - 2\sin(2t)$ and $y'(t) = 2\cos(t) - 2\cos(2t)$. Now evaluate these derivatives at $t = \pi/2$: $x'(\pi/2) = -2\sin(\pi/2) - 2\sin(\pi) = -2(1) - 2(0) = -2$. $y'(\pi/2) = 2\cos(\pi/2) - 2\cos(\pi) = 2(0) - 2(-1) = 2$. The speed at $t = \pi/2$ is $\sqrt{(x'(\pi/2))^2 + (y'(\pi/2))^2} = \sqrt{(-2)^2 + (2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$.
13. A particle moves along a path such that its position is given by $x(t) = 6t$ and $y(t) = 4t^{3/2}$. How long does it take for the particle to travel a distance of 28 units, starting from $t=0$?
    Answer: 3 units of time First, find the derivatives: $x'(t) = 6$ and $y'(t) = 4 \cdot \frac{3}{2} t^{1/2} = 6t^{1/2}$. The sum of their squares is $(x')^2 + (y')^2 = 6^2 + (6t^{1/2})^2 = 36 + 36t = 36(1+t)$. The integrand is $\sqrt{36(1+t)} = 6\sqrt{1+t}$. The arc length as a function of time $T$ is $L(T) = \int_0^T 6\sqrt{1+t} \, dt$. Using substitution $u=1+t, du=dt$: $\int_1^{1+T} 6u^{1/2} du = 6[\frac{2}{3}u^{3/2}]_1^{1+T} = 4[(1+T)^{3/2} - 1]$. Setting this equal to 28: $4[(1+T)^{3/2} - 1] = 28 \implies (1+T)^{3/2} - 1 = 7 \implies (1+T)^{3/2} = 8 \implies 1+T = 8^{2/3} = 4 \implies T = 3$.
14. Consider the parametric equations $x = a(\cos(t) + t\sin(t))$ and $y = a(\sin(t) - t\cos(t))$ where $a > 0$ is a constant. This represents an involute of a circle. What is the arc length from $t = 0$ to $t = 2\pi$?
    Answer: $\int_0^{2\pi} at \, dt$ We compute $\frac{dx}{dt} = a(-\sin(t) + \sin(t) + t\cos(t)) = at\cos(t)$ and $\frac{dy}{dt} = a(\cos(t) - \cos(t) + t\sin(t)) = at\sin(t)$. Therefore $\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{a^2t^2\cos^2(t) + a^2t^2\sin^2(t)} = at\sqrt{\cos^2(t) + \sin^2(t)} = at$ for $t \geq 0$. Choice A incorrectly has $\sqrt{t^2 + 1}$. Choice C has $\sqrt{2 + t^2}$. Choice D uses absolute value, but since $t \geq 0$ on $[0, 2\pi]$, this equals choice B anyway, but the reasoning would be different.
15. The curve $x(t) = 2\cos t, y(t) = \sin(2t)$ creates a Lissajous figure. Find the length of the path from $t=0$ to $t=\pi/2$.
    Answer: $\int_0^{\pi/2} \sqrt{4\sin^2 t + 4\cos^2(2t)} \, dt$ When you encounter a parametric curve and need to find arc length, you're applying the fundamental arc length formula for parametric equations. The key insight is that you need to account for how both x and y change with respect to the parameter t. For a parametric curve with $x(t)$ and $y(t)$, the arc length from $t = a$ to $t = b$ is given by $\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$. Let's find the derivatives: $\frac{dx}{dt} = \frac{d}{dt}(2\cos t) = -2\sin t$ and $\frac{dy}{dt} = \frac{d}{dt}(\sin(2t)) = 2\cos(2t)$. Substituting into the arc length formula: $\sqrt{(-2\sin t)^2 + (2\cos(2t))^2} = \sqrt{4\sin^2 t + 4\cos^2(2t)}$. This matches answer choice C exactly. Looking at the wrong answers: A gives you $2\sin t + 2\cos(2t)$, which incorrectly adds the derivatives instead of using the Pythagorean relationship. B uses $4\sin^2(2t)$ instead of $4\cos^2(2t)$, incorrectly applying the derivative of $\sin(2t)$. D uses $4\cos^2 t$ instead of $4\sin^2 t$, which would come from forgetting the negative sign when squaring $\frac{dx}{dt}$. Remember: Arc length always involves the square root of the sum of squares of the derivatives. Don't let the trigonometric complexity distract you from applying this fundamental pattern correctly.
16. Find the arc length of the curve given by $x(t) = t - \sin t$ and $y(t) = 1 - \cos t$ for $0 \le t \le \pi$.
    Answer: 4 First, find the derivatives: $dx/dt = 1 - \cos t$ and $dy/dt = \sin t$. Then, compute the sum of their squares: $(dx/dt)^2 + (dy/dt)^2 = (1 - \cos t)^2 + (\sin t)^2 = 1 - 2\cos t + \cos^2 t + \sin^2 t = 2 - 2\cos t$. The arc length integral is $L = \int_0^{\pi} \sqrt{2 - 2\cos t} \, dt$. Using the half-angle identity $1 - \cos t = 2\sin^2(t/2)$, the integrand becomes $\sqrt{2(2\sin^2(t/2))} = \sqrt{4\sin^2(t/2)} = 2|\sin(t/2)|$. For $t$ in $[0, \pi]$, $t/2$ is in $[0, \pi/2]$, so $\sin(t/2) \ge 0$. The absolute value can be dropped. The integral is $L = \int_0^{\pi} 2\sin(t/2) dt = [-4\cos(t/2)]_0^{\pi} = -4(\cos(\pi/2) - \cos(0)) = -4(0 - 1) = 4$.
17. A curve is given parametrically by $x = 3\cos^2(t)$ and $y = 2\sin^3(t)$ for $0 \leq t \leq \pi$. What is the arc length of this curve?
    Answer: $\int_0^\pi \sqrt{36\sin^2(t)\cos^2(t) + 36\sin^4(t)\cos^2(t)} \, dt$ For parametric curves, arc length is $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt$. Here, $\frac{dx}{dt} = -6\cos(t)\sin(t)$ and $\frac{dy}{dt} = 6\sin^2(t)\cos(t)$. So $(\frac{dx}{dt})^2 = 36\cos^2(t)\sin^2(t)$ and $(\frac{dy}{dt})^2 = 36\sin^4(t)\cos^2(t)$. Choice A has the wrong coefficient for the second term. Choice B uses incorrect derivatives. Choice D has an error in the first derivative calculation.
18. The curve $x = 2t - \sin(2t)$ and $y = 2 - 2\cos(2t)$ represents one arch of a cycloid for $0 \leq t \leq \pi$. What is the length of this arch?
    Answer: $\int_0^\pi \sqrt{8 - 8\cos(2t)} \, dt$ We have $\frac{dx}{dt} = 2 - 2\cos(2t)$ and $\frac{dy}{dt} = 4\sin(2t)$. So $(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 = (2 - 2\cos(2t))^2 + 16\sin^2(2t) = 4 - 8\cos(2t) + 4\cos^2(2t) + 16\sin^2(2t) = 4 + 12\sin^2(2t) - 8\cos(2t) + 4\cos^2(2t) = 8 - 8\cos(2t)$. Choice A omits terms in the expansion. Choice B has the wrong coefficient. Choice C is missing a factor of 2 under the square root.
19. For the parametric curve $x = \ln(\sec(t) + \tan(t))$ and $y = \sec(t)$ where $0 \leq t \leq \frac{\pi}{4}$, which integral gives the arc length?
    Answer: $\int_0^{\pi/4} \sqrt{\sec^2(t) + \sec^2(t)\tan^2(t)} \, dt$ We need $\frac{dx}{dt} = \frac{d}{dt}[\ln(\sec(t) + \tan(t))] = \frac{\sec(t)\tan(t) + \sec^2(t)}{\sec(t) + \tan(t)} = \sec(t)$ and $\frac{dy}{dt} = \sec(t)\tan(t)$. So the arc length is $\int_0^{\pi/4} \sqrt{\sec^2(t) + \sec^2(t)\tan^2(t)} \, dt$. Choice B incorrectly has 1 instead of $\sec^2(t)$. Choice C has $\sec^4(t)$ instead of $\sec^2(t)\tan^2(t)$. Choice D factors incorrectly.
20. Find the arc length of the curve defined by $x(t) = e^t \cos t$ and $y(t) = e^t \sin t$ for $0 \le t \le \ln 3$. Give your answer in simplest form.
    Answer: $2\sqrt{2}$ First, find the derivatives using the product rule: $x'(t) = e^t \cos t - e^t \sin t = e^t(\cos t - \sin t)$ and $y'(t) = e^t \sin t + e^t \cos t = e^t(\sin t + \cos t)$. Next, compute the sum of squares: $(x'(t))^2 + (y'(t))^2 = e^{2t}(\cos t - \sin t)^2 + e^{2t}(\sin t + \cos t)^2 = e^{2t}[(\cos^2 t - 2\sin t \cos t + \sin^2 t) + (\sin^2 t + 2\sin t \cos t + \cos^2 t)] = e^{2t}[2(\cos^2 t + \sin^2 t)] = 2e^{2t}$. The integrand is $\sqrt{2e^{2t}} = \sqrt{2}e^t$. Finally, integrate: $L = \int_0^{\ln 3} \sqrt{2}e^t \, dt = \sqrt{2}[e^t]_0^{\ln 3} = \sqrt{2}(e^{\ln 3} - e^0) = \sqrt{2}(3 - 1) = 2\sqrt{2}$.

Start this quiz