Calculus 2 Quiz: Arc Length Of Parametric Curves
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Arc Length Of Parametric CurvesQuestion 1 of 20

A path is defined by x(t)=et+e−tx(t) = e^t + e^{-t} and y(t)=5−2ty(t) = 5 - 2t. Find the arc length from t=0t=0 to t=1t=1.

e−e−1e - e^{-1}
e+e−1e + e^{-1}
e+e−1−2e + e^{-1} - 2
e−e−1+2e - e^{-1} + 2
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Calculus 2 Quiz

Calculus 2 Quiz: Arc Length Of Parametric Curves

Practice Arc Length Of Parametric Curves in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Arc Length Of Parametric Curves, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Question 1

A path is defined by x(t)=et+e−tx(t) = e^t + e^{-t} and y(t)=5−2ty(t) = 5 - 2t. Find the arc length from t=0t=0 to t=1t=1.

  1. e−e−1e - e^{-1} (correct answer)
  2. e+e−1e + e^{-1}
  3. e+e−1−2e + e^{-1} - 2
  4. e−e−1+2e - e^{-1} + 2
Explanation: First, find the derivatives: x′(t)=et−e−tx'(t) = e^t - e^{-t} and y′(t)=−2y'(t) = -2. Then, find the sum of their squares: (x′)2+(y′)2=(et−e−t)2+(−2)2=(e2t−2ete−t+e−2t)+4=e2t−2+e−2t+4=e2t+2+e−2t(x')^2 + (y')^2 = (e^t - e^{-t})^2 + (-2)^2 = (e^{2t} - 2e^t e^{-t} + e^{-2t}) + 4 = e^{2t} - 2 + e^{-2t} + 4 = e^{2t} + 2 + e^{-2t}. This expression is a perfect square: (et+e−t)2(e^t + e^{-t})^2. So, the integrand is (et+e−t)2=et+e−t\sqrt{(e^t + e^{-t})^2} = e^t + e^{-t} (since et+e−te^t+e^{-t} is always positive). The arc length is L=∫01(et+e−t)dt=[et−e−t]01=(e1−e−1)−(e0−e0)=e−e−1−0=e−e−1L = \int_0^1 (e^t + e^{-t}) dt = [e^t - e^{-t}]_0^1 = (e^1 - e^{-1}) - (e^0 - e^0) = e - e^{-1} - 0 = e - e^{-1}.

Question 2

Consider the parametric equations x=a(cos⁡(t)+tsin⁡(t))x = a(\cos(t) + t\sin(t)) and y=a(sin⁡(t)−tcos⁡(t))y = a(\sin(t) - t\cos(t)) where a>0a > 0 is a constant. This represents an involute of a circle. What is the arc length from t=0t = 0 to t=2πt = 2\pi?

  1. ∫02πat2+1 dt\int_0^{2\pi} a\sqrt{t^2 + 1} \, dt
  2. ∫02πat dt\int_0^{2\pi} at \, dt (correct answer)
  3. ∫02πa2+t2 dt\int_0^{2\pi} a\sqrt{2 + t^2} \, dt
  4. ∫02πa∣t∣ dt\int_0^{2\pi} a|t| \, dt
Explanation: We compute dxdt=a(−sin⁡(t)+sin⁡(t)+tcos⁡(t))=atcos⁡(t)\frac{dx}{dt} = a(-\sin(t) + \sin(t) + t\cos(t)) = at\cos(t) and dydt=a(cos⁡(t)−cos⁡(t)+tsin⁡(t))=atsin⁡(t)\frac{dy}{dt} = a(\cos(t) - \cos(t) + t\sin(t)) = at\sin(t). Therefore (dxdt)2+(dydt)2=a2t2cos⁡2(t)+a2t2sin⁡2(t)=atcos⁡2(t)+sin⁡2(t)=at\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≥0t \geq 0. Choice A incorrectly has t2+1\sqrt{t^2 + 1}. Choice C has 2+t2\sqrt{2 + t^2}. Choice D uses absolute value, but since t≥0t \geq 0 on [0,2π][0, 2\pi], this equals choice B anyway, but the reasoning would be different.

Question 3

A parametric curve is given by x=∫0tcos⁡(u2) dux = \int_0^t \cos(u^2) \, du and y=∫0tsin⁡(u2) duy = \int_0^t \sin(u^2) \, du for t∈[0,π]t \in [0, \sqrt{\pi}]. This is known as a Fresnel spiral. What integral represents its arc length?

  1. ∫0πcos⁡2(t2)+sin⁡2(t2) dt\int_0^{\sqrt{\pi}} \sqrt{\cos^2(t^2) + \sin^2(t^2)} \, dt
  2. ∫0π∣cos⁡(t2)+sin⁡(t2)∣ dt\int_0^{\sqrt{\pi}} |\cos(t^2) + \sin(t^2)| \, dt
  3. ∫0πcos⁡4(t2)+sin⁡4(t2) dt\int_0^{\sqrt{\pi}} \sqrt{\cos^4(t^2) + \sin^4(t^2)} \, dt
  4. ∫0πdt\int_0^{\sqrt{\pi}} dt (correct answer)
Explanation: Using the Fundamental Theorem of Calculus, dxdt=cos⁡(t2)\frac{dx}{dt} = \cos(t^2) and dydt=sin⁡(t2)\frac{dy}{dt} = \sin(t^2). Therefore the arc length is ∫0πcos⁡2(t2)+sin⁡2(t2) dt=∫0π1 dt=∫0πdt\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.

Question 4

A point moves along the curve x(t)=2cos⁡(t)+cos⁡(2t)x(t) = 2\cos(t) + \cos(2t), y(t)=2sin⁡(t)−sin⁡(2t)y(t) = 2\sin(t) - \sin(2t). The speed of the point at time tt is given by (x′(t))2+(y′(t))2\sqrt{(x'(t))^2+(y'(t))^2}. Find the speed at t=π/2t = \pi/2.

  1. 2
  2. 4
  3. 222\sqrt{2} (correct answer)
  4. 424\sqrt{2}
Explanation: 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)=−2sin⁡(t)−2sin⁡(2t)x'(t) = -2\sin(t) - 2\sin(2t) and y′(t)=2cos⁡(t)−2cos⁡(2t)y'(t) = 2\cos(t) - 2\cos(2t). Now evaluate these derivatives at t=π/2t = \pi/2: x′(π/2)=−2sin⁡(π/2)−2sin⁡(π)=−2(1)−2(0)=−2x'(\pi/2) = -2\sin(\pi/2) - 2\sin(\pi) = -2(1) - 2(0) = -2. y′(π/2)=2cos⁡(π/2)−2cos⁡(π)=2(0)−2(−1)=2y'(\pi/2) = 2\cos(\pi/2) - 2\cos(\pi) = 2(0) - 2(-1) = 2. The speed at t=π/2t = \pi/2 is (x′(π/2))2+(y′(π/2))2=(−2)2+(2)2=4+4=8=22\sqrt{(x'(\pi/2))^2 + (y'(\pi/2))^2} = \sqrt{(-2)^2 + (2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}.

Question 5

The arc length of the curve x(t)=sin⁡(2t),y(t)=cos⁡(2t)x(t) = \sin(2t), y(t) = \cos(2t) from t=0t=0 to t=πt=\pi is found to be 2π2\pi. What is the arc length of the curve x(t)=sin⁡(kt),y(t)=cos⁡(kt)x(t) = \sin(kt), y(t) = \cos(kt) from t=0t=0 to t=πt=\pi for a constant k>0k>0?

  1. 2π2\pi
  2. 2kπ2k\pi
  3. kπk\pi (correct answer)
  4. 2π/k2\pi/k
Explanation: This question tests the understanding of how a parameter affects arc length. For the general curve, x′(t)=kcos⁡(kt)x'(t) = k\cos(kt) and y′(t)=−ksin⁡(kt)y'(t) = -k\sin(kt). The sum of squares of the derivatives is (kcos⁡(kt))2+(−ksin⁡(kt))2=k2cos⁡2(kt)+k2sin⁡2(kt)=k2(k\cos(kt))^2 + (-k\sin(kt))^2 = k^2\cos^2(kt) + k^2\sin^2(kt) = k^2. The integrand is k2=∣k∣\sqrt{k^2} = |k|. Since k>0k>0, this is simply kk. The arc length is L=∫0πk dt=[kt]0π=kπL = \int_0^{\pi} k \, dt = [kt]_0^{\pi} = k\pi. The initial information about the case k=2k=2 is consistent with this formula, as 2π2\pi is the result.

Question 6

A line segment from (1, 2) to (5, 5) is parameterized by x(t)=1+4tx(t) = 1+4t and y(t)=2+3ty(t) = 2+3t for 0≤t≤10 \le t \le 1. If the parameterization is changed to x(u)=1+4u2x(u) = 1+4u^2 and y(u)=2+3u2y(u) = 2+3u^2 for 0≤u≤10 \le u \le 1, how does the arc length change?

  1. It stays the same. (correct answer)
  2. It increases.
  3. It decreases.
  4. The change cannot be determined without more information.
Explanation: 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: (5−1)2+(5−2)2=42+32=16+9=25=5\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=0u=0, where x(0)=1,y(0)=2x(0)=1, y(0)=2, and its endpoint at u=1u=1, where x(1)=5,y(1)=5x(1)=5, y(1)=5. Since u2u^2 is monotonically increasing on [0,1][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.

Question 7

The arc length of the parabola y=x2y=x^2 from x=0x=0 to x=1x=1 can be found by parameterizing the curve as x(t)=t,y(t)=t2x(t)=t, y(t)=t^2 for 0≤t≤10 \le t \le 1. Which of the following integrals correctly represents this length?

  1. ∫011+4t2 dt\int_0^1 \sqrt{1+4t^2} \, dt (correct answer)
  2. ∫011+t2 dt\int_0^1 \sqrt{1+t^2} \, dt
  3. ∫01(1+2t) dt\int_0^1 (1+2t) \, dt
  4. ∫011+4t dt\int_0^1 \sqrt{1+4t} \, dt
Explanation: This question connects the Cartesian form of arc length with the parametric form. Given the parameterization x(t)=tx(t)=t and y(t)=t2y(t)=t^2, we first find the derivatives: x′(t)=1x'(t) = 1 and y′(t)=2ty'(t) = 2t. The formula for parametric arc length is L=∫ab(x′(t))2+(y′(t))2dtL = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2} dt. Plugging in our derivatives, we get L=∫01(1)2+(2t)2dt=∫011+4t2dtL = \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=∫ab1+(dy/dx)2dxL = \int_a^b \sqrt{1+(dy/dx)^2} dx, since dy/dx=2xdy/dx = 2x, giving ∫011+(2x)2dx\int_0^1 \sqrt{1+(2x)^2} dx.

Question 8

The length of the curve defined by x(t)=ln⁡(sec⁡t+tan⁡t)−sin⁡tx(t) = \ln(\sec t + \tan t) - \sin t and y(t)=cos⁡ty(t) = \cos t from t=0t=0 to t=π/3t=\pi/3 is being calculated. What is the value of the integrand (dx/dt)2+(dy/dt)2\sqrt{(dx/dt)^2 + (dy/dt)^2} after simplification?

  1. sec⁡t\sec t
  2. tan⁡t\tan t (correct answer)
  3. sin⁡t\sin t
  4. cos⁡t\cos t
Explanation: This question tests the setup and simplification, not the final integration. First, find the derivatives. y′(t)=−sin⁡ty'(t) = -\sin t. For x(t)x(t), recall ddtln⁡(sec⁡t+tan⁡t)=sec⁡t\frac{d}{dt}\ln(\sec t + \tan t) = \sec t. So, x′(t)=sec⁡t−cos⁡tx'(t) = \sec t - \cos t. Now, compute the sum of squares: (x′)2+(y′)2=(sec⁡t−cos⁡t)2+(−sin⁡t)2=sec⁡2t−2sec⁡tcos⁡t+cos⁡2t+sin⁡2t(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⁡tcos⁡t=1\sec t \cos t = 1 and cos⁡2t+sin⁡2t=1\cos^2 t + \sin^2 t = 1, this simplifies to sec⁡2t−2(1)+1=sec⁡2t−1\sec^2 t - 2(1) + 1 = \sec^2 t - 1. Finally, using the identity tan⁡2t=sec⁡2t−1\tan^2 t = \sec^2 t - 1, the expression becomes tan⁡2t\tan^2 t. The integrand is tan⁡2t=∣tan⁡t∣\sqrt{\tan^2 t} = |\tan t|. Since the interval is [0,π/3][0, \pi/3], tan⁡t\tan t is non-negative, so the integrand is tan⁡t\tan t.

Question 9

Consider the curve given by x(t)=∫0tcos⁡(π2u2)dux(t) = \int_0^t \cos(\frac{\pi}{2}u^2)du and y(t)=∫0tsin⁡(π2u2)duy(t) = \int_0^t \sin(\frac{\pi}{2}u^2)du (a Fresnel integral). What is the arc length of this curve from t=0t=0 to t=2t=2?

  1. 1
  2. 2π2\pi
  3. π\pi
  4. 2 (correct answer)
Explanation: 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=0t=0 to t=2t=2, you need ∫02(dxdt)2+(dydt)2dt\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: dxdt=cos⁡(π2t2)\frac{dx}{dt} = \cos\left(\frac{\pi}{2}t^2\right) and dydt=sin⁡(π2t2)\frac{dy}{dt} = \sin\left(\frac{\pi}{2}t^2\right) The arc length becomes: ∫02cos⁡2(π2t2)+sin⁡2(π2t2)dt\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(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1 for any angle θ\theta, this simplifies to: ∫021 dt=2\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 (2π2\pi) could result from mistakenly thinking the trigonometric functions contribute a factor of 2π2\pi without recognizing the Pythagorean identity. Choice C (π\pi) might arise from confusing this with a circular arc or incorrectly handling the π2\frac{\pi}{2} factor. Remember: For parametric curves, especially those involving trigonometric integrals, always check if (dxdt)2+(dydt)2\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.

Question 10

A particle moves on a circle of radius 5 centered at the origin, with position given by x(t)=5cos⁡(ωt)x(t) = 5\cos(\omega t) and y(t)=5sin⁡(ωt)y(t) = 5\sin(\omega t). The particle travels a distance of 10π10\pi in 2 seconds. What is the value of the angular frequency ω>0\omega > 0?

  1. π/2\pi/2
  2. 2π2\pi
  3. π\pi (correct answer)
  4. 1
Explanation: 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)=5cos⁡(ωt)x(t) = 5\cos(\omega t) and y(t)=5sin⁡(ωt)y(t) = 5\sin(\omega t), you need to find the arc length traveled. The arc length formula for parametric curves is s=∫ab(dxdt)2+(dydt)2dts = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt. Taking derivatives: dxdt=−5ωsin⁡(ωt)\frac{dx}{dt} = -5\omega\sin(\omega t) and dydt=5ωcos⁡(ωt)\frac{dy}{dt} = 5\omega\cos(\omega t). Therefore: (dxdt)2+(dydt)2=25ω2sin⁡2(ωt)+25ω2cos⁡2(ωt)=5ω\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=∫025ω dt=10ωs = \int_0^2 5\omega \, dt = 10\omega. Since this equals 10π10\pi, we have 10ω=10π10\omega = 10\pi, so ω=π\omega = \pi. Answer A (π/2\pi/2) gives a distance of only 5π5\pi, half the required distance. Answer B (2π2\pi) would result in 20π20\pi distance traveled, which is twice too much. Answer D (1) yields a distance of 10, missing the factor of π\pi entirely. The correct answer is C: ω=π\omega = \pi. Study tip: For circular motion problems, remember that the speed is constant and equals the radius times the angular frequency (v=rωv = r\omega). Distance equals speed times time, so d=rωtd = r\omega t provides a quick check for your answer.

Question 11

A parametric curve is given by x(t)=a(t−sin⁡t)x(t) = a(t-\sin t) and y(t)=a(1−cos⁡t)y(t) = a(1-\cos t) for a>0a > 0. The length of one arch of this cycloid (from t=0t=0 to t=2πt=2\pi) is 8a8a. What is the length of the curve from the point where it first touches the x-axis (after t=0t=0) to the highest point of the first arch?

  1. 2a2a
  2. 4a4a
  3. 2a22a\sqrt{2}
  4. 4a−2a24a - 2a\sqrt{2} (correct answer)
Explanation: The problem gives the total length of one arch as 8a8a. The first arch starts at t=0t=0. The highest point of the arch occurs when y(t)y(t) is maximum. y(t)=a(1−cos⁡t)y(t) = a(1-\cos t) is max when cos⁡t=−1\cos t = -1, which is at t=πt=\pi. The curve touches the x-axis when y(t)=0y(t)=0, so a(1−cos⁡t)=0  ⟹  cos⁡t=1a(1-\cos t)=0 \implies \cos t = 1. This occurs at t=0,2π,4π,...t=0, 2\pi, 4\pi, .... The first time it touches the x-axis after t=0t=0 is at t=2πt=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)(t=0) to the highest point (t=π)(t=\pi), which is half the arch. By symmetry, this length is 8a/2=4a8a/2=4a. Let's assume it means from the highest point (t=π)(t=\pi) to the end of the arch (t=2π)(t=2\pi). This is also 4a4a. The question seems unusual. 'from the point where it first touches the x-axis (after t=0)'. This is at (2πa,0)(2\pi a, 0) corresponding to t=2πt=2\pi. 'to the highest point of the first arch'. This is at (πa,2a)(\pi a, 2a) corresponding to t=πt=\pi. This asks for the length from t=2πt=2\pi to t=πt=\pi, which is the negative of the integral from π\pi to 2π2\pi. The length is positive, so it's the integral from π\pi to 2π2\pi. This is the second half of the arch, which has length 4a4a. The wording is confusing. Let me re-read. Maybe it means a different point. Let's find the length from t=π/2t=\pi/2 to t=πt=\pi. The integrand is 2a∣sin⁡(t/2)∣=2asin⁡(t/2)2a|\sin(t/2)| = 2a\sin(t/2) on [0,2π][0, 2\pi]. ∫π/2π2asin⁡(t/2)dt=[−4acos⁡(t/2)]π/2π=−4a(cos⁡(π/2)−cos⁡(π/4))=−4a(0−22)=2a2\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 π/2\pi/2 is [−4acos⁡(t/2)]0π/2=−4a(cos⁡(π/4)−cos⁡(0))=−4a(22−1)=4a−2a2[-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 π\pi is 4a4a. The question is likely asking for something non-obvious. What if 'first touches the x-axis' means t=0t=0, and 'highest point' is t=πt=\pi? Length is 4a4a. What if it means from t=0t=0 to the point where the tangent is vertical? dx/dt=a(1−cos⁡t)=0dx/dt = a(1-\cos t)=0 at t=0,2πt=0, 2\pi. Not helpful. Where tangent is horizontal? dy/dt=asin⁡t=0dy/dt = a\sin t = 0 at t=0,πt=0, \pi. So from the start to the highest point. That's 4a4a. Let's assume the question is asking for the arc length from t=0t=0 to t=π/2t=\pi/2. That length is 4a−2a24a - 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=ay=a for the first time?'. y(t)=a  ⟹  a(1−cos⁡t)=a  ⟹  1−cos⁡t=1  ⟹  cos⁡t=0y(t)=a \implies a(1-\cos t)=a \implies 1-\cos t=1 \implies \cos t=0. This is at t=π/2t=\pi/2. The length from t=0t=0 to t=π/2t=\pi/2 is 4a−2a24a - 2a\sqrt{2}. This is a good multi-step question.

Question 12

Find the arc length of the curve given by x(t)=t−sin⁡tx(t) = t - \sin t and y(t)=1−cos⁡ty(t) = 1 - \cos t for 0≤t≤π0 \le t \le \pi.

  1. 2
  2. 4 (correct answer)
  3. 8
  4. 222\sqrt{2}
Explanation: First, find the derivatives: dx/dt=1−cos⁡tdx/dt = 1 - \cos t and dy/dt=sin⁡tdy/dt = \sin t. Then, compute the sum of their squares: (dx/dt)2+(dy/dt)2=(1−cos⁡t)2+(sin⁡t)2=1−2cos⁡t+cos⁡2t+sin⁡2t=2−2cos⁡t(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=∫0π2−2cos⁡t dtL = \int_0^{\pi} \sqrt{2 - 2\cos t} \, dt. Using the half-angle identity 1−cos⁡t=2sin⁡2(t/2)1 - \cos t = 2\sin^2(t/2), the integrand becomes 2(2sin⁡2(t/2))=4sin⁡2(t/2)=2∣sin⁡(t/2)∣\sqrt{2(2\sin^2(t/2))} = \sqrt{4\sin^2(t/2)} = 2|\sin(t/2)|. For tt in [0,π][0, \pi], t/2t/2 is in [0,π/2][0, \pi/2], so sin⁡(t/2)≥0\sin(t/2) \ge 0. The absolute value can be dropped. The integral is L=∫0π2sin⁡(t/2)dt=[−4cos⁡(t/2)]0π=−4(cos⁡(π/2)−cos⁡(0))=−4(0−1)=4L = \int_0^{\pi} 2\sin(t/2) dt = [-4\cos(t/2)]_0^{\pi} = -4(\cos(\pi/2) - \cos(0)) = -4(0 - 1) = 4.

Question 13

The position of a particle is given by x(t)=t2x(t) = t^2 and y(t)=13t3−ty(t) = \frac{1}{3}t^3 - t. Find the total distance traveled by the particle during the time interval −1≤t≤1-1 \le t \le 1.

  1. 4/34/3
  2. 4
  3. 2
  4. 8/38/3 (correct answer)
Explanation: 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=∫ab(dxdt)2+(dydt)2 dts = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt First, find the derivatives: dxdt=2t\frac{dx}{dt} = 2t and dydt=t2−1\frac{dy}{dt} = t^2 - 1 Next, compute the integrand: (2t)2+(t2−1)2=4t2+t4−2t2+1=t4+2t2+1=(t2+1)2=∣t2+1∣\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 t2+1>0t^2 + 1 > 0 for all real tt, we have ∣t2+1∣=t2+1|t^2+1| = t^2+1. The total distance is: ∫−11(t2+1) dt=[t33+t]−11=(13+1)−(−13−1)=43+2=83\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 (43\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.

Question 14

A parametric curve is given by x=etcos⁡(t)x = e^t\cos(t) and y=etsin⁡(t)y = e^t\sin(t) for 0≤t≤ln⁡(2)0 \leq t \leq \ln(2). Which expression correctly represents the arc length?

  1. ∫0ln⁡(2)e2t(cos⁡2(t)−sin⁡2(t))+e2t(sin⁡2(t)+cos⁡2(t)) dt\int_0^{\ln(2)} \sqrt{e^{2t}(\cos^2(t) - \sin^2(t)) + e^{2t}(\sin^2(t) + \cos^2(t))} \, dt
  2. ∫0ln⁡(2)e2t(cos⁡(t)−sin⁡(t))2+e2t(sin⁡(t)+cos⁡(t))2 dt\int_0^{\ln(2)} \sqrt{e^{2t}(\cos(t) - \sin(t))^2 + e^{2t}(\sin(t) + \cos(t))^2} \, dt (correct answer)
  3. ∫0ln⁡(2)2e2tcos⁡2(t)+2e2tsin⁡2(t) dt\int_0^{\ln(2)} \sqrt{2e^{2t}\cos^2(t) + 2e^{2t}\sin^2(t)} \, dt
  4. ∫0ln⁡(2)et(cos⁡(t)−sin⁡(t))2+(sin⁡(t)+cos⁡(t))2 dt\int_0^{\ln(2)} e^t\sqrt{(\cos(t) - \sin(t))^2 + (\sin(t) + \cos(t))^2} \, dt
Explanation: We have dxdt=et(cos⁡(t)−sin⁡(t))\frac{dx}{dt} = e^t(\cos(t) - \sin(t)) and dydt=et(sin⁡(t)+cos⁡(t))\frac{dy}{dt} = e^t(\sin(t) + \cos(t)). The arc length is ∫0ln⁡(2)[et(cos⁡(t)−sin⁡(t))]2+[et(sin⁡(t)+cos⁡(t))]2 dt\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 ete^t incorrectly from the square root.

Question 15

Consider the parametric curve x=t3−3tx = t^3 - 3t and y=t2y = t^2 for t∈[−2,2]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?

  1. 2∫02(3t2−3)2+4t2 dt2\int_0^2 \sqrt{(3t^2 - 3)^2 + 4t^2} \, dt
  2. ∫−22(3t2−3)2+4t2 dt\int_{-2}^2 \sqrt{(3t^2 - 3)^2 + 4t^2} \, dt (correct answer)
  3. ∫−33(3t2−3)2+4t2 dt\int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{(3t^2 - 3)^2 + 4t^2} \, dt
  4. ∫−2−1(3t2−3)2+4t2 dt+∫12(3t2−3)2+4t2 dt\int_{-2}^{-1} \sqrt{(3t^2 - 3)^2 + 4t^2} \, dt + \int_1^2 \sqrt{(3t^2 - 3)^2 + 4t^2} \, dt
Explanation: The arc length formula for parametric curves is L=∫ab(dxdt)2+(dydt)2 dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt. Here dxdt=3t2−3\frac{dx}{dt} = 3t^2 - 3 and dydt=2t\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∈[−1,1]t \in [-1,1].

Question 16

For the parametric curve x=ln⁡(sec⁡(t)+tan⁡(t))x = \ln(\sec(t) + \tan(t)) and y=sec⁡(t)y = \sec(t) where 0≤t≤π40 \leq t \leq \frac{\pi}{4}, which integral gives the arc length?

  1. ∫0π/4sec⁡2(t)+sec⁡2(t)tan⁡2(t) dt\int_0^{\pi/4} \sqrt{\sec^2(t) + \sec^2(t)\tan^2(t)} \, dt (correct answer)
  2. ∫0π/41+sec⁡2(t)tan⁡2(t) dt\int_0^{\pi/4} \sqrt{1 + \sec^2(t)\tan^2(t)} \, dt
  3. ∫0π/4sec⁡2(t)+sec⁡4(t) dt\int_0^{\pi/4} \sqrt{\sec^2(t) + \sec^4(t)} \, dt
  4. ∫0π/4sec⁡(t)1+tan⁡2(t) dt\int_0^{\pi/4} \sec(t)\sqrt{1 + \tan^2(t)} \, dt
Explanation: We need dxdt=ddt[ln⁡(sec⁡(t)+tan⁡(t))]=sec⁡(t)tan⁡(t)+sec⁡2(t)sec⁡(t)+tan⁡(t)=sec⁡(t)\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 dydt=sec⁡(t)tan⁡(t)\frac{dy}{dt} = \sec(t)\tan(t). So the arc length is ∫0π/4sec⁡2(t)+sec⁡2(t)tan⁡2(t) dt\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)\sec^2(t). Choice C has sec⁡4(t)\sec^4(t) instead of sec⁡2(t)tan⁡2(t)\sec^2(t)\tan^2(t). Choice D factors incorrectly.

Question 17

Find the arc length of the curve defined by x(t)=etcos⁡tx(t) = e^t \cos t and y(t)=etsin⁡ty(t) = e^t \sin t for 0≤t≤ln⁡30 \le t \le \ln 3.

  1. 22
  2. 222\sqrt{2} (correct answer)
  3. 22(ln⁡3−1)2\sqrt{2}(\ln 3 - 1)
  4. 32−23\sqrt{2} - \sqrt{2}
Explanation: First, find the derivatives using the product rule: x′(t)=etcos⁡t−etsin⁡t=et(cos⁡t−sin⁡t)x'(t) = e^t \cos t - e^t \sin t = e^t(\cos t - \sin t) and y′(t)=etsin⁡t+etcos⁡t=et(sin⁡t+cos⁡t)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=e2t(cos⁡t−sin⁡t)2+e2t(sin⁡t+cos⁡t)2=e2t[(cos⁡2t−2sin⁡tcos⁡t+sin⁡2t)+(sin⁡2t+2sin⁡tcos⁡t+cos⁡2t)]=e2t[2(cos⁡2t+sin⁡2t)]=2e2t(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 2e2t=2et\sqrt{2e^{2t}} = \sqrt{2}e^t. Finally, integrate: L=∫0ln⁡32et dt=2[et]0ln⁡3=2(eln⁡3−e0)=2(3−1)=22L = \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}.

Question 18

Let s(t)s(t) be the arc length of the curve x(u)=u3,y(u)=u2x(u) = u^3, y(u) = u^2 from u=0u=0 to u=tu=t. Find the rate of change of the arc length at t=1t=1, i.e., s′(1)s'(1).

  1. 13(1313−8)\frac{1}{3}(13\sqrt{13}-8)
  2. 11
  3. 55
  4. 13\sqrt{13} (correct answer)
Explanation: The arc length function is defined as s(t)=∫0t(x′(u))2+(y′(u))2 dus(t) = \int_0^t \sqrt{(x'(u))^2 + (y'(u))^2} \, du. By the Fundamental Theorem of Calculus, Part 1, the derivative of this integral with respect to tt is simply the integrand evaluated at tt. So, s′(t)=(x′(t))2+(y′(t))2s'(t) = \sqrt{(x'(t))^2 + (y'(t))^2}, which is the speed. First, we find the derivatives of the given parametric equations: x′(t)=3t2x'(t) = 3t^2 and y′(t)=2ty'(t) = 2t. Then, s′(t)=(3t2)2+(2t)2=9t4+4t2s'(t) = \sqrt{(3t^2)^2 + (2t)^2} = \sqrt{9t^4 + 4t^2}. We need to evaluate this at t=1t=1: s′(1)=9(1)4+4(1)2=9+4=13s'(1) = \sqrt{9(1)^4 + 4(1)^2} = \sqrt{9+4} = \sqrt{13}. Distractor A represents the actual arc length from 0 to 2, not its derivative.

Question 19

The curve x(t)=2cos⁡t,y(t)=sin⁡(2t)x(t) = 2\cos t, y(t) = \sin(2t) creates a Lissajous figure. Find the length of the path from t=0t=0 to t=π/2t=\pi/2.

  1. ∫0π/2(2sin⁡t+2cos⁡(2t)) dt\int_0^{\pi/2} (2\sin t + 2\cos(2t)) \, dt
  2. ∫0π/24sin⁡2t+4sin⁡2(2t) dt\int_0^{\pi/2} \sqrt{4\sin^2 t + 4\sin^2(2t)} \, dt
  3. ∫0π/24sin⁡2t+4cos⁡2(2t) dt\int_0^{\pi/2} \sqrt{4\sin^2 t + 4\cos^2(2t)} \, dt (correct answer)
  4. ∫0π/24cos⁡2t+4sin⁡2(2t) dt\int_0^{\pi/2} \sqrt{4\cos^2 t + 4\sin^2(2t)} \, dt
Explanation: 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)x(t) and y(t)y(t), the arc length from t=at = a to t=bt = b is given by ∫ab(dxdt)2+(dydt)2 dt\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt. Let's find the derivatives: dxdt=ddt(2cos⁡t)=−2sin⁡t\frac{dx}{dt} = \frac{d}{dt}(2\cos t) = -2\sin t and dydt=ddt(sin⁡(2t))=2cos⁡(2t)\frac{dy}{dt} = \frac{d}{dt}(\sin(2t)) = 2\cos(2t). Substituting into the arc length formula: (−2sin⁡t)2+(2cos⁡(2t))2=4sin⁡2t+4cos⁡2(2t)\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 2sin⁡t+2cos⁡(2t)2\sin t + 2\cos(2t), which incorrectly adds the derivatives instead of using the Pythagorean relationship. B uses 4sin⁡2(2t)4\sin^2(2t) instead of 4cos⁡2(2t)4\cos^2(2t), incorrectly applying the derivative of sin⁡(2t)\sin(2t). D uses 4cos⁡2t4\cos^2 t instead of 4sin⁡2t4\sin^2 t, which would come from forgetting the negative sign when squaring dxdt\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.

Question 20

A particle follows a path described by x(t)=13(t2+2)3/2x(t) = \frac{1}{3}(t^2+2)^{3/2} and y(t)=ty(t) = t. What is the distance traveled by the particle from t=0t=0 to t=3t=3?

  1. 9
  2. 12 (correct answer)
  3. 13(1111−22)\frac{1}{3}(11\sqrt{11}-2\sqrt{2})
  4. 15
Explanation: First, compute the derivatives with respect to tt. x′(t)=13⋅32(t2+2)1/2⋅2t=tt2+2x'(t) = \frac{1}{3} \cdot \frac{3}{2}(t^2+2)^{1/2} \cdot 2t = t\sqrt{t^2+2}. And y′(t)=1y'(t) = 1. Next, find the sum of their squares: (x′(t))2+(y′(t))2=(tt2+2)2+12=t2(t2+2)+1=t4+2t2+1=(t2+1)2(x'(t))^2 + (y'(t))^2 = (t\sqrt{t^2+2})^2 + 1^2 = t^2(t^2+2) + 1 = t^4 + 2t^2 + 1 = (t^2+1)^2. The integrand for the arc length is (t2+1)2=t2+1\sqrt{(t^2+1)^2} = t^2+1. The arc length is the integral of this expression from t=0t=0 to t=3t=3: L=∫03(t2+1)dt=[t33+t]03=(333+3)−(0)=(9+3)=12L = \int_0^3 (t^2+1) dt = [\frac{t^3}{3} + t]_0^3 = (\frac{3^3}{3} + 3) - (0) = (9+3) = 12.