AP Calculus BC Quiz: Arc Lengths Of Curves Parametric Equations
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Arc Lengths Of Curves Parametric EquationsQuestion 1 of 20

For x(t)=sec⁡tx(t)=\sec t, y(t)=tan⁡ty(t)=\tan t on 0≤t≤π40\le t\le\frac{\pi}{4}, which integral sets up the arc length?

∫0π/4(sec⁡ttan⁡t)2+(sec⁡2t)2 dt\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec t\tan t)^2+(\sec^2 t)^2}\,dt
∫0π/4(sec⁡t)2+(tan⁡t)2 dt\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec t)^2+(\tan t)^2}\,dt
∫0π/4(sec⁡2t)2+(sec⁡ttan⁡t)2 dt\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec^2 t)^2+(\sec t\tan t)^2}\,dt
∫0π/2(sec⁡ttan⁡t)2+(sec⁡2t)2 dt\displaystyle \int_{0}^{\pi/2}\sqrt{(\sec t\tan t)^2+(\sec^2 t)^2}\,dt
∫0π/4(sec⁡ttan⁡t)+(sec⁡2t) dt\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec t\tan t)+(\sec^2 t)}\,dt
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AP Calculus BC Quiz

AP Calculus BC Quiz: Arc Lengths Of Curves Parametric Equations

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

What this quiz covers

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

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

For x(t)=sec⁡tx(t)=\sec t, y(t)=tan⁡ty(t)=\tan t on 0≤t≤π40\le t\le\frac{\pi}{4}, which integral sets up the arc length?

  1. ∫0π/4(sec⁡ttan⁡t)2+(sec⁡2t)2 dt\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec t\tan t)^2+(\sec^2 t)^2}\,dt (correct answer)
  2. ∫0π/4(sec⁡t)2+(tan⁡t)2 dt\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec t)^2+(\tan t)^2}\,dt
  3. ∫0π/4(sec⁡2t)2+(sec⁡ttan⁡t)2 dt\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec^2 t)^2+(\sec t\tan t)^2}\,dt
  4. ∫0π/2(sec⁡ttan⁡t)2+(sec⁡2t)2 dt\displaystyle \int_{0}^{\pi/2}\sqrt{(\sec t\tan t)^2+(\sec^2 t)^2}\,dt
  5. ∫0π/4(sec⁡ttan⁡t)+(sec⁡2t) dt\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec t\tan t)+(\sec^2 t)}\,dt
Explanation: This question tests your understanding of finding the arc length of a curve defined by parametric equations. To find the arc length, compute the integral from t=0 to t=π/4 of the square root of (dx/dt)^2 plus (dy/dt)^2 dt. Here, differentiate x(t) = sec t to get dx/dt = sec t tan t, and y(t) = tan t to get dy/dt = sec² t. Plug these into the formula to obtain ∫ from 0 to π/4 of √[(sec t tan t)² + (sec² t)²] dt, which matches choice A. A tempting distractor is choice B, which uses the original functions sec t and tan t instead of their derivatives, failing to apply the arc length formula correctly. Always remember to differentiate the parametric functions to get the speed components before integrating for arc length.

Question 2

A drone follows x=t2+1x=t^2+1, y=3t−t3y=3t-t^3 for 0≤t≤20\le t\le 2; which integral gives the arc length?

  1. ∫02(2t)2+(3−3t2)2 dt\displaystyle \int_{0}^{2}\sqrt{(2t)^2+(3-3t^2)^2}\,dt (correct answer)
  2. ∫02(2t)+(3−3t2) dt\displaystyle \int_{0}^{2}\sqrt{(2t)+(3-3t^2)}\,dt
  3. ∫02((2t)2+(3−3t2)2) dt\displaystyle \int_{0}^{2}\left((2t)^2+(3-3t^2)^2\right)\,dt
  4. ∫151+(dydx)2 dx\displaystyle \int_{1}^{5}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx
  5. ∫02(t2+1)2+(3t−t3)2 dt\displaystyle \int_{0}^{2}\sqrt{(t^2+1)^2+(3t-t^3)^2}\,dt
Explanation: This problem asks for the arc length of a parametric curve, which requires the formula L=∫ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. For x=t2+1x = t^2 + 1 and y=3t−t3y = 3t - t^3, we compute derivatives: dxdt=2t\frac{dx}{dt} = 2t and dydt=3−3t2\frac{dy}{dt} = 3 - 3t^2. Substituting these into the arc length formula gives ∫02(2t)2+(3−3t2)2dt\int_0^2 \sqrt{(2t)^2 + (3-3t^2)^2} dt. Choice B incorrectly omits the squares inside the square root, which would give a nonsensical result since we need the magnitude of the velocity vector. The key strategy for parametric arc length is to always square the derivatives before adding them under the square root.

Question 3

A parametric curve is x=ln⁡tx=\ln t, y=ty=\sqrt{t} for 1≤t≤41\le t\le 4; which arc length setup is correct?

  1. ∫14(1t)2+(12t)2 dt\displaystyle \int_{1}^{4}\sqrt{\left(\frac{1}{t}\right)^2+\left(\frac{1}{2\sqrt{t}}\right)^2}\,dt (correct answer)
  2. ∫141t+12t dt\displaystyle \int_{1}^{4}\sqrt{\frac{1}{t}+\frac{1}{2\sqrt{t}}}\,dt
  3. ∫14(ln⁡t)2+(t)2 dt\displaystyle \int_{1}^{4}\sqrt{(\ln t)^2+(\sqrt{t})^2}\,dt
  4. ∫04(1t)2+(12t)2 dt\displaystyle \int_{0}^{4}\sqrt{\left(\frac{1}{t}\right)^2+\left(\frac{1}{2\sqrt{t}}\right)^2}\,dt
  5. ∫14((1t)2+(12t)2) dt\displaystyle \int_{1}^{4}\left(\left(\frac{1}{t}\right)^2+\left(\frac{1}{2\sqrt{t}}\right)^2\right)\,dt
Explanation: For parametric arc length, we apply L=∫ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. With x=ln⁡tx = \ln t and y=ty = \sqrt{t}, we find derivatives: dxdt=1t\frac{dx}{dt} = \frac{1}{t} and dydt=12t\frac{dy}{dt} = \frac{1}{2\sqrt{t}}. The arc length integral is ∫14(1t)2+(12t)2dt\int_1^4 \sqrt{(\frac{1}{t})^2 + (\frac{1}{2\sqrt{t}})^2} dt. Choice C mistakenly uses the original functions ln⁡t\ln t and t\sqrt{t} instead of their derivatives, which is a common error. Remember that arc length measures how fast the curve is traced out, requiring derivatives not the position functions.

Question 4

A parametric curve is x=tx=\sqrt{t}, y=13t3/2y=\frac{1}{3}t^{3/2} for 0≤t≤90\le t\le 9; which arc length setup is correct?

  1. ∫09(12t)2+(12t)2 dt\displaystyle \int_{0}^{9}\sqrt{\left(\frac{1}{2\sqrt{t}}\right)^2+\left(\frac{1}{2}\sqrt{t}\right)^2}\,dt (correct answer)
  2. ∫0912t+12t dt\displaystyle \int_{0}^{9}\sqrt{\frac{1}{2\sqrt{t}}+\frac{1}{2}\sqrt{t}}\,dt
  3. ∫09(t)2+(13t3/2)2 dt\displaystyle \int_{0}^{9}\sqrt{(\sqrt{t})^2+\left(\frac{1}{3}t^{3/2}\right)^2}\,dt
  4. ∫09((12t)2+(12t)2) dt\displaystyle \int_{0}^{9}\left(\left(\frac{1}{2\sqrt{t}}\right)^2+\left(\frac{1}{2}\sqrt{t}\right)^2\right)\,dt
  5. ∫19(12t)2+(12t)2 dt\displaystyle \int_{1}^{9}\sqrt{\left(\frac{1}{2\sqrt{t}}\right)^2+\left(\frac{1}{2}\sqrt{t}\right)^2}\,dt
Explanation: For parametric arc length, we apply L=∫ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. With x=tx = \sqrt{t} and y=13t3/2y = \frac{1}{3}t^{3/2}, the derivatives are dxdt=12t and dydt=12t1/2=12t\frac{dx}{dt} = \frac{1}{2\sqrt{t}} \text{ and } \frac{dy}{dt} = \frac{1}{2}t^{1/2} = \frac{1}{2}\sqrt{t}. This yields ∫09(12t)2+(12t)2dt\int_0^9 \sqrt{(\frac{1}{2\sqrt{t}})^2 + (\frac{1}{2}\sqrt{t})^2} dt. Choice E starts at t=1t = 1 instead of t=0t = 0, which would miss part of the curve and give an incorrect arc length. Always use the exact parameter bounds given in the problem.

Question 5

A curve is given by x=etx=e^t, y=t2y=t^2 for 1≤t≤31\le t\le 3; which integral sets up its arc length?

  1. ∫13(et)2+(2t)2 dt\displaystyle \int_{1}^{3}\sqrt{(e^t)^2+(2t)^2}\,dt (correct answer)
  2. ∫13et+2t dt\displaystyle \int_{1}^{3}\sqrt{e^t+2t}\,dt
  3. ∫13(et)+(2t)2 dt\displaystyle \int_{1}^{3}\sqrt{(e^t)+(2t)^2}\,dt
  4. ∫ee31+(dydx)2 dx\displaystyle \int_{e}^{e^3}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx
  5. ∫02(et)2+(2t)2 dt\displaystyle \int_{0}^{2}\sqrt{(e^t)^2+(2t)^2}\,dt
Explanation: To find arc length of a parametric curve, we use L=∫ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. For x=etx = e^t and y=t2y = t^2, the derivatives are dxdt=et\frac{dx}{dt} = e^t and dydt=2t\frac{dy}{dt} = 2t. Substituting into the formula gives ∫13(et)2+(2t)2dt\int_1^3 \sqrt{(e^t)^2 + (2t)^2} dt, using the given bounds 1≤t≤31 \leq t \leq 3. Choice E uses incorrect bounds 00 to 22 instead of the given 11 to 33, which would calculate arc length for a different portion of the curve. Always verify that the integration bounds match the given parameter interval.

Question 6

A particle travels along x=t3−2tx=t^3-2t, y=4ty=4t for −1≤t≤2-1\le t\le 2; which arc length integral is correct?

  1. ∫−12(3t2−2)2+42 dt\displaystyle \int_{-1}^{2}\sqrt{(3t^2-2)^2+4^2}\,dt (correct answer)
  2. ∫−12(3t2−2)+4 dt\displaystyle \int_{-1}^{2}\sqrt{(3t^2-2)+4}\,dt
  3. ∫−12((3t2−2)2+42) dt\displaystyle \int_{-1}^{2}\left((3t^2-2)^2+4^2\right)\,dt
  4. ∫−12(t3−2t)2+(4t)2 dt\displaystyle \int_{-1}^{2}\sqrt{(t^3-2t)^2+(4t)^2}\,dt
  5. ∫02(3t2−2)2+42 dt\displaystyle \int_{0}^{2}\sqrt{(3t^2-2)^2+4^2}\,dt
Explanation: Arc length for parametric curves requires the formula L=∫ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. Given x=t3−2tx = t^3 - 2t and y=4ty = 4t, we compute derivatives: dxdt=3t2−2\frac{dx}{dt} = 3t^2 - 2 and dydt=4\frac{dy}{dt} = 4. The arc length integral becomes ∫−12(3t2−2)2+42dt\int_{-1}^2 \sqrt{(3t^2-2)^2 + 4^2} dt. Choice D incorrectly uses the original functions instead of their derivatives, showing (t3−2t)2+(4t)2(t^3-2t)^2 + (4t)^2 under the square root. The fundamental principle is that arc length depends on the rate of change (derivatives), not the position functions themselves.

Question 7

A curve is x=t−sin⁡tx=t-\sin t, y=1−cos⁡ty=1-\cos t for 0≤t≤2π0\le t\le 2\pi; which integral gives its arc length?

  1. ∫02π(1−cos⁡t)2+(sin⁡t)2 dt\displaystyle \int_{0}^{2\pi}\sqrt{(1-\cos t)^2+(\sin t)^2}\,dt (correct answer)
  2. ∫02π(1+cos⁡t)2+(sin⁡t)2 dt\displaystyle \int_{0}^{2\pi}\sqrt{(1+\cos t)^2+(\sin t)^2}\,dt
  3. ∫02π(t−sin⁡t)2+(1−cos⁡t)2 dt\displaystyle \int_{0}^{2\pi}\sqrt{(t-\sin t)^2+(1-\cos t)^2}\,dt
  4. ∫02π((1−cos⁡t)2+(sin⁡t)2) dt\displaystyle \int_{0}^{2\pi}\left((1-\cos t)^2+(\sin t)^2\right)\,dt
  5. ∫0π(1−cos⁡t)2+(sin⁡t)2 dt\displaystyle \int_{0}^{\pi}\sqrt{(1-\cos t)^2+(\sin t)^2}\,dt
Explanation: Arc length for parametric curves uses L=∫ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. Given x=t−sin⁡tx = t - \sin t and y=1−cos⁡ty = 1 - \cos t, we compute derivatives: dxdt=1−cos⁡t\frac{dx}{dt} = 1 - \cos t and dydt=sin⁡t\frac{dy}{dt} = \sin t. The arc length integral becomes ∫02π(1−cos⁡t)2+(sin⁡t)2dt\int_0^{2\pi} \sqrt{(1-\cos t)^2 + (\sin t)^2} dt. Choice C incorrectly uses the original parametric functions (t−sin⁡t)2+(1−cos⁡t)2(t-\sin t)^2 + (1-\cos t)^2 instead of their derivatives. The key insight is that arc length depends on instantaneous velocity (derivatives), not position.

Question 8

For x=1+t2x=1+t^2 and y=t3−1y=t^3-1 on 1≤t≤31\le t\le3, which integral represents the arc length?

  1. ∫13(2t)2+(3t2)2 dt\displaystyle \int_{1}^{3}\sqrt{(2t)^2+(3t^2)^2}\,dt (correct answer)
  2. ∫13(2t)+(3t2) dt\displaystyle \int_{1}^{3}\sqrt{(2t)+(3t^2)}\,dt
  3. ∫131+(3t22t)2 dt\displaystyle \int_{1}^{3}\sqrt{1+\left(\frac{3t^2}{2t}\right)^2}\,dt
  4. ∫13((2t)2+(3t2)2) dt\displaystyle \int_{1}^{3}\left((2t)^2+(3t^2)^2\right)\,dt
  5. ∫13(1+t2)2+(t3−1)2 dt\displaystyle \int_{1}^{3}\sqrt{(1+t^2)^2+(t^3-1)^2}\,dt
Explanation: To find the arc length of this parametric curve, we use the formula L=∫ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. For x=1+t2x = 1 + t^2, differentiating gives dxdt=2t\frac{dx}{dt} = 2t. For y=t3−1y = t^3 - 1, differentiating gives dydt=3t2\frac{dy}{dt} = 3t^2. Substituting these derivatives into the arc length formula yields ∫13(2t)2+(3t2)2dt\int_1^3 \sqrt{(2t)^2 + (3t^2)^2} dt. Choice E incorrectly uses the original parametric functions (1+t2)(1+t^2) and (t3−1)(t^3-1) instead of their derivatives, which is a fundamental misunderstanding of how arc length measures distance along a curve. The essential principle is that arc length integrals always contain squared derivatives, not the original functions.

Question 9

For x=ln⁡(1+t)x=\ln(1+t) and y=t2y=t^2 on 0≤t≤30\le t\le3, which integral sets up the arc length?

  1. ∫03(11+t)2+(2t)2 dt\displaystyle \int_{0}^{3}\sqrt{\left(\frac{1}{1+t}\right)^2+(2t)^2}\,dt (correct answer)
  2. ∫03(11+t)+(2t) dt\displaystyle \int_{0}^{3}\sqrt{\left(\frac{1}{1+t}\right)+(2t)}\,dt
  3. ∫031+(2t1/(1+t))2 dt\displaystyle \int_{0}^{3}\sqrt{1+\left(\frac{2t}{1/(1+t)}\right)^2}\,dt
  4. ∫03((11+t)2+(2t)2) dt\displaystyle \int_{0}^{3}\left(\left(\frac{1}{1+t}\right)^2+(2t)^2\right)\,dt
  5. ∫03(ln⁡(1+t))2+(t2)2 dt\displaystyle \int_{0}^{3}\sqrt{\left(\ln(1+t)\right)^2+(t^2)^2}\,dt
Explanation: To find the arc length of this parametric curve, we apply the formula L=∫ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. For x=ln⁡(1+t)x = \ln(1+t), we use the chain rule to get dxdt=11+t\frac{dx}{dt} = \frac{1}{1+t}. For y=t2y = t^2, we have dydt=2t\frac{dy}{dt} = 2t. Substituting these derivatives gives ∫03(11+t)2+(2t)2dt\int_0^3 \sqrt{\left(\frac{1}{1+t}\right)^2 + (2t)^2} dt. Choice E incorrectly uses the original functions ln⁡(1+t)\ln(1+t) and t2t^2 instead of their derivatives, which is a common error when students forget that arc length measures the rate of change along the curve. The essential principle is that arc length integrals always contain derivatives, not the original parametric functions.

Question 10

A parametric curve is x=t−1tx=t-\frac{1}{t} and y=t+1ty=t+\frac{1}{t} for 1≤t≤21\le t\le2. Which integral gives arc length?

  1. ∫12(1+1t2)2+(1−1t2)2 dt\displaystyle \int_{1}^{2}\sqrt{\left(1+\frac{1}{t^2}\right)^2+\left(1-\frac{1}{t^2}\right)^2}\,dt (correct answer)
  2. ∫12(1+1t2)+(1−1t2) dt\displaystyle \int_{1}^{2}\sqrt{\left(1+\frac{1}{t^2}\right)+\left(1-\frac{1}{t^2}\right)}\,dt
  3. ∫121+(1−1/t21+1/t2)2 dt\displaystyle \int_{1}^{2}\sqrt{1+\left(\frac{1-1/t^2}{1+1/t^2}\right)^2}\,dt
  4. ∫12((1+1t2)2+(1−1t2)2) dt\displaystyle \int_{1}^{2}\left(\left(1+\frac{1}{t^2}\right)^2+\left(1-\frac{1}{t^2}\right)^2\right)\,dt
  5. ∫12(t−1t)2+(t+1t)2 dt\displaystyle \int_{1}^{2}\sqrt{\left(t-\frac{1}{t}\right)^2+\left(t+\frac{1}{t}\right)^2}\,dt
Explanation: This problem requires finding the arc length using the parametric formula L=∫ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. For x=t−1tx = t - \frac{1}{t}, we differentiate to get dxdt=1+1t2\frac{dx}{dt} = 1 + \frac{1}{t^2}. For y=t+1ty = t + \frac{1}{t}, we differentiate to get dydt=1−1t2\frac{dy}{dt} = 1 - \frac{1}{t^2}. Substituting these derivatives gives ∫12(1+1t2)2+(1−1t2)2dt\int_1^2 \sqrt{\left(1+\frac{1}{t^2}\right)^2 + \left(1-\frac{1}{t^2}\right)^2} dt. Choice E incorrectly shows the original parametric expressions rather than their derivatives, which would not measure the instantaneous velocity needed for arc length. The key strategy is to always differentiate parametric equations before constructing the arc length integral.

Question 11

A bead moves on x=sin⁡tx=\sin t, y=cos⁡(2t)y=\cos(2t) for 0≤t≤π0\le t\le \pi; which arc length integral is correct?

  1. ∫0π(cos⁡t)2+(−2sin⁡(2t))2 dt\displaystyle \int_{0}^{\pi}\sqrt{(\cos t)^2+(-2\sin(2t))^2}\,dt (correct answer)
  2. ∫0π(sin⁡t)2+(cos⁡(2t))2 dt\displaystyle \int_{0}^{\pi}\sqrt{(\sin t)^2+(\cos(2t))^2}\,dt
  3. ∫0π(cos⁡t)+(−2sin⁡(2t)) dt\displaystyle \int_{0}^{\pi}\sqrt{(\cos t)+(-2\sin(2t))}\,dt
  4. ∫0π((cos⁡t)2+(−2sin⁡(2t))2) dt\displaystyle \int_{0}^{\pi}\left((\cos t)^2+(-2\sin(2t))^2\right)\,dt
  5. ∫02π(cos⁡t)2+(−2sin⁡(2t))2 dt\displaystyle \int_{0}^{2\pi}\sqrt{(\cos t)^2+(-2\sin(2t))^2}\,dt
Explanation: This problem requires finding the arc length of a parametric curve using L=∫ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. Given x=sin⁡tx = \sin t and y=cos⁡(2t)y = \cos(2t), we find derivatives: dxdt=cos⁡t\frac{dx}{dt} = \cos t and dydt=−2sin⁡(2t)\frac{dy}{dt} = -2\sin(2t). Plugging into the arc length formula yields ∫0π(cos⁡t)2+(−2sin⁡(2t))2dt\int_0^\pi \sqrt{(\cos t)^2 + (-2\sin(2t))^2} dt. Choice C incorrectly adds the derivatives without squaring them, which violates the Pythagorean nature of arc length. Remember that arc length measures the magnitude of velocity, requiring squared derivatives under the square root.

Question 12

Given x=2t−sin⁡tx=2t-\sin t and y=1−cos⁡ty=1-\cos t, 0≤t≤2π0\le t\le 2\pi, which integral sets up the arc length?

  1. ∫02π(2−cos⁡t)2+(sin⁡t)2 dt\displaystyle \int_{0}^{2\pi}\sqrt{(2-\cos t)^2+(\sin t)^2}\,dt (correct answer)
  2. ∫02π(2t−sin⁡t)2+(1−cos⁡t)2 dt\displaystyle \int_{0}^{2\pi}\sqrt{(2t-\sin t)^2+(1-\cos t)^2}\,dt
  3. ∫02π(2−cos⁡t)+(sin⁡t) dt\displaystyle \int_{0}^{2\pi}\sqrt{(2-\cos t)+(\sin t)}\,dt
  4. ∫02π((2−cos⁡t)2+(sin⁡t)2) dt\displaystyle \int_{0}^{2\pi}\big((2-\cos t)^2+(\sin t)^2\big)\,dt
  5. ∫−2π2π(2−cos⁡t)2+(sin⁡t)2 dt\displaystyle \int_{-2\pi}^{2\pi}\sqrt{(2-\cos t)^2+(\sin t)^2}\,dt
Explanation: Finding the arc length of this parametric curve requires the formula ∫ab(dxdt)2+(dydt)2dt\int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. From x=2t−sin⁡tx = 2t - \sin t and y=1−cos⁡ty = 1 - \cos t, we compute dxdt=2−cos⁡t\frac{dx}{dt} = 2 - \cos t and dydt=sin⁡t\frac{dy}{dt} = \sin t. The arc length integral becomes ∫02π(2−cos⁡t)2+(sin⁡t)2dt\int_0^{2\pi} \sqrt{(2-\cos t)^2 + (\sin t)^2} dt. Choice C incorrectly omits the squares, which would give a meaningless expression since we need the magnitude of the velocity vector, not a signed quantity. The fundamental strategy for parametric arc length is: differentiate, square both derivatives, add, then take the square root.

Question 13

A drone traces x(t)=t2+1x(t)=t^2+1, y(t)=3t−t3y(t)=3t-t^3 for 0≤t≤10\le t\le1; which integral gives the arc length?

  1. ∫01(2t)2+(3−3t2)2 dt\displaystyle \int_{0}^{1}\sqrt{(2t)^2+(3-3t^2)^2}\,dt (correct answer)
  2. ∫01(t2+1)2+(3t−t3)2 dt\displaystyle \int_{0}^{1}\sqrt{(t^2+1)^2+(3t-t^3)^2}\,dt
  3. ∫01(2t)+(3−3t2) dt\displaystyle \int_{0}^{1}\sqrt{(2t)+(3-3t^2)}\,dt
  4. ∫01(2t)2+(3t−t3)2 dt\displaystyle \int_{0}^{1}\sqrt{(2t)^2+(3t-t^3)^2}\,dt
  5. ∫10(2t)2+(3−3t2)2 dt\displaystyle \int_{1}^{0}\sqrt{(2t)^2+(3-3t^2)^2}\,dt
Explanation: This question tests your understanding of finding the arc length of a curve defined by parametric equations. To find the arc length, compute the integral from t=0 to t=1 of the square root of (dx/dt)^2 plus (dy/dt)^2 dt. Here, differentiate x(t) = t² + 1 to get dx/dt = 2t, and y(t) = 3t - t³ to get dy/dt = 3 - 3t². Plug these into the formula to obtain ∫ from 0 to 1 of √[(2t)² + (3 - 3t²)²] dt, which matches choice A. A tempting distractor is choice D, which incorrectly uses the original y(t) = 3t - t³ inside the square root instead of its derivative, leading to an invalid setup. Always remember to differentiate the parametric functions to get the speed components before integrating for arc length.

Question 14

A particle moves on x(t)=t3−1x(t) = t^3 - 1, y(t)=4ty(t) = 4\sqrt{t} for 1≤t≤91 \le t \le 9; which integral gives arc length?

  1. ∫19(3t2)2+(2t)2 dt\displaystyle \int_{1}^{9}\sqrt{(3t^2)^2+\left(\frac{2}{\sqrt{t}}\right)^2}\,dt (correct answer)
  2. ∫19(t3−1)2+(4t)2 dt\displaystyle \int_{1}^{9}\sqrt{(t^3-1)^2+(4\sqrt{t})^2}\,dt
  3. ∫19(3t2)2+(4t)2 dt\displaystyle \int_{1}^{9}\sqrt{(3t^2)^2+\left(4\sqrt{t}\right)^2}\,dt
  4. ∫19[(3t2)2+(2t)2]dt\displaystyle \int_{1}^{9}\left[(3t^2)^2+\left(\frac{2}{\sqrt{t}}\right)^2\right]dt
  5. ∫08(3t2)2+(2t)2 dt\displaystyle \int_{0}^{8}\sqrt{(3t^2)^2+\left(\frac{2}{\sqrt{t}}\right)^2}\,dt
Explanation: This question tests your understanding of finding the arc length of a curve defined by parametric equations. To find the arc length, compute the integral from t=1 to t=9 of the square root of (dx/dt)2(dx/dt)^2 plus (dy/dt)2(dy/dt)^2 dt. Here, differentiate x(t)=t3−1x(t) = t^3 - 1 to get dx/dt=3t2dx/dt = 3t^2, and y(t)=4ty(t) = 4\sqrt{t} to get dy/dt=2/tdy/dt = 2 / \sqrt{t}. Plug these into the formula to obtain ∫19[(3t2)2+(2/t)2 dt\int_{1}^{9} \sqrt{[(3t^2)^2 + (2/\sqrt{t})^2} \, dt, which matches choice A. A tempting distractor is choice C, which incorrectly uses 4t4\sqrt{t} instead of 2/t2 / \sqrt{t} for dy/dtdy/dt, leading to an erroneous integrand. Always remember to differentiate the parametric functions to get the speed components before integrating for arc length.

Question 15

A curve is x(t)=ln⁡(t+1)x(t)=\ln(t+1), y(t)=ty(t)=\sqrt{t} for 0≤t≤40\le t\le4; select the correct arc length integral.

  1. ∫04(1t+1)2+(12t)2 dt\displaystyle \int_{0}^{4}\sqrt{\left(\frac{1}{t+1}\right)^2+\left(\frac{1}{2\sqrt{t}}\right)^2}\,dt (correct answer)
  2. ∫04(ln⁡(t+1))2+(t)2 dt\displaystyle \int_{0}^{4}\sqrt{\left(\ln(t+1)\right)^2+\left(\sqrt{t}\right)^2}\,dt
  3. ∫04(1t+1)+(12t) dt\displaystyle \int_{0}^{4}\sqrt{\left(\frac{1}{t+1}\right)+\left(\frac{1}{2\sqrt{t}}\right)}\,dt
  4. ∫04(1t+1)2+(t)2 dt\displaystyle \int_{0}^{4}\sqrt{\left(\frac{1}{t+1}\right)^2+\left(\sqrt{t}\right)^2}\,dt
  5. ∫15(1t+1)2+(12t)2 dt\displaystyle \int_{1}^{5}\sqrt{\left(\frac{1}{t+1}\right)^2+\left(\frac{1}{2\sqrt{t}}\right)^2}\,dt
Explanation: This question tests your understanding of finding the arc length of a curve defined by parametric equations. To find the arc length, compute the integral from t=0 to t=4 of the square root of (dx/dt)^2 plus (dy/dt)^2 dt. Here, differentiate x(t) = ln(t+1) to get dx/dt = 1/(t+1), and y(t) = √t to get dy/dt = 1/(2√t). Plug these into the formula to obtain ∫ from 0 to 4 of √[(1/(t+1))² + (1/(2√t))²] dt, which matches choice A. A tempting distractor is choice D, which incorrectly uses √t instead of its derivative 1/(2√t) for y, resulting in a wrong setup. Always remember to differentiate the parametric functions to get the speed components before integrating for arc length.

Question 16

For x(t)=2tx(t)=2t, y(t)=t2y(t)=t^2 from t=0t=0 to t=3t=3, which integral computes the arc length?

  1. ∫03(2)2+(2t)2 dt\displaystyle \int_{0}^{3}\sqrt{(2)^2+(2t)^2}\,dt (correct answer)
  2. ∫03(2t)2+(t2)2 dt\displaystyle \int_{0}^{3}\sqrt{(2t)^2+(t^2)^2}\,dt
  3. ∫03(2)2+(t2)2 dt\displaystyle \int_{0}^{3}\sqrt{(2)^2+(t^2)^2}\,dt
  4. ∫03[(2)2+(2t)2]dt\displaystyle \int_{0}^{3}\left[(2)^2+(2t)^2\right]dt
  5. ∫06(2)2+(2t)2 dt\displaystyle \int_{0}^{6}\sqrt{(2)^2+(2t)^2}\,dt
Explanation: This question tests your understanding of finding the arc length of a curve defined by parametric equations. To find the arc length, compute the integral from t=0 to t=3 of the square root of (dx/dt)^2 plus (dy/dt)^2 dt. Here, differentiate x(t) = 2t to get dx/dt = 2, and y(t) = t² to get dy/dt = 2t. Plug these into the formula to obtain ∫ from 0 to 3 of √[(2)² + (2t)²] dt, which matches choice A. A tempting distractor is choice C, which incorrectly uses t² instead of 2t for dy/dt, leading to an invalid integrand. Always remember to differentiate the parametric functions to get the speed components before integrating for arc length.

Question 17

A curve is defined by x(t)=1+t4x(t)=1+t^4, y(t)=2t2y(t)=2t^2 for −1≤t≤1-1\le t\le1; which integral gives arc length?

  1. ∫−11(4t3)2+(4t)2 dt\displaystyle \int_{-1}^{1}\sqrt{(4t^3)^2+(4t)^2}\,dt (correct answer)
  2. ∫−11(1+t4)2+(2t2)2 dt\displaystyle \int_{-1}^{1}\sqrt{(1+t^4)^2+(2t^2)^2}\,dt
  3. ∫−11(4t3)2+(2t2)2 dt\displaystyle \int_{-1}^{1}\sqrt{(4t^3)^2+(2t^2)^2}\,dt
  4. ∫01(4t3)2+(4t)2 dt\displaystyle \int_{0}^{1}\sqrt{(4t^3)^2+(4t)^2}\,dt
  5. ∫−11(4t3)+(4t) dt\displaystyle \int_{-1}^{1}\sqrt{(4t^3)+(4t)}\,dt
Explanation: This question tests your understanding of finding the arc length of a curve defined by parametric equations. To find the arc length, compute the integral from t=-1 to t=1 of the square root of (dx/dt)^2 plus (dy/dt)^2 dt. Here, differentiate x(t) = 1 + t⁴ to get dx/dt = 4t³, and y(t) = 2t² to get dy/dt = 4t. Plug these into the formula to obtain ∫ from -1 to 1 of √[(4t³)² + (4t)²] dt, which matches choice A. A tempting distractor is choice C, which incorrectly uses 2t² instead of 4t for dy/dt, resulting in a wrong integrand. Always remember to differentiate the parametric functions to get the speed components before integrating for arc length.

Question 18

A curve is x(t)=tcos⁡tx(t)=t\cos t, y(t)=tsin⁡ty(t)=t\sin t for 0≤t≤20\le t\le2; choose the correct arc length integral.

  1. ∫02(cos⁡t−tsin⁡t)2+(sin⁡t+tcos⁡t)2 dt\displaystyle \int_{0}^{2}\sqrt{(\cos t-t\sin t)^2+(\sin t+t\cos t)^2}\,dt (correct answer)
  2. ∫02(tcos⁡t)2+(tsin⁡t)2 dt\displaystyle \int_{0}^{2}\sqrt{(t\cos t)^2+(t\sin t)^2}\,dt
  3. ∫02(cos⁡t−tsin⁡t)+(sin⁡t+tcos⁡t) dt\displaystyle \int_{0}^{2}\sqrt{(\cos t-t\sin t)+(\sin t+t\cos t)}\,dt
  4. ∫02[(cos⁡t−tsin⁡t)2+(sin⁡t+tcos⁡t)2]dt\displaystyle \int_{0}^{2}\left[(\cos t-t\sin t)^2+(\sin t+t\cos t)^2\right]dt
  5. ∫02π(cos⁡t−tsin⁡t)2+(sin⁡t+tcos⁡t)2 dt\displaystyle \int_{0}^{2\pi}\sqrt{(\cos t-t\sin t)^2+(\sin t+t\cos t)^2}\,dt
Explanation: This question tests your understanding of finding the arc length of a curve defined by parametric equations. To find the arc length, compute ∫02(dxdt)2+(dydt)2 dt\int_{0}^{2} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt. Here, differentiate x(t) = t cos t using product rule to get dx/dt=cos⁡t−tsin⁡tdx/dt = \cos t - t \sin t, and y(t) = t sin t to get dy/dt=sin⁡t+tcos⁡tdy/dt = \sin t + t \cos t. Plug these into the formula to obtain ∫02(cos⁡t−tsin⁡t)2+(sin⁡t+tcos⁡t)2 dt\int_{0}^{2} \sqrt{(\cos t - t \sin t)^2 + (\sin t + t \cos t)^2} \, dt, which matches choice A. A tempting distractor is choice B, which incorrectly uses t cos t and t sin t without differentiating properly, leading to an invalid setup. Always remember to differentiate the parametric functions to get the speed components before integrating for arc length.

Question 19

For x=1+sin⁡tx=1+\sin t and y=1−cos⁡ty=1-\cos t on 0≤t≤π20\le t\le \frac{\pi}{2}, which integral gives arc length?

  1. ∫0π/2(cos⁡t)2+(sin⁡t)2 dt\displaystyle \int_{0}^{\pi/2}\sqrt{\left(\cos t\right)^2+\left(\sin t\right)^2}\,dt (correct answer)
  2. ∫0π/2(sin⁡t)2+(−cos⁡t)2 dt\displaystyle \int_{0}^{\pi/2}\sqrt{\left(\sin t\right)^2+\left(-\cos t\right)^2}\,dt
  3. ∫0π/2(1+sin⁡t)2+(1−cos⁡t)2 dt\displaystyle \int_{0}^{\pi/2}\sqrt{\left(1+\sin t\right)^2+\left(1-\cos t\right)^2}\,dt
  4. ∫0π/21+(sin⁡tcos⁡t)2 dt\displaystyle \int_{0}^{\pi/2}\sqrt{1+\left(\frac{\sin t}{\cos t}\right)^2}\,dt
  5. ∫0π(cos⁡t)2+(sin⁡t)2 dt\displaystyle \int_{0}^{\pi}\sqrt{\left(\cos t\right)^2+\left(\sin t\right)^2}\,dt
Explanation: To find the arc length of this parametric curve, we use L=∫ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. Given x=1+sin⁡tx = 1 + \sin t and y=1−cos⁡ty = 1 - \cos t, we compute dxdt=cos⁡t\frac{dx}{dt} = \cos t and dydt=sin⁡t\frac{dy}{dt} = \sin t. Substituting these derivatives gives ∫0π/2(cos⁡t)2+(sin⁡t)2dt=∫0π/21dt=∫0π/21dt\int_0^{\pi/2} \sqrt{(\cos t)^2 + (\sin t)^2} dt = \int_0^{\pi/2} \sqrt{1} dt = \int_0^{\pi/2} 1 dt. Choice C incorrectly uses the original parametric functions instead of their derivatives, missing the fundamental requirement of measuring instantaneous rate of change. Notice that this particular curve has constant speed 1, making the arc length simply equal to the parameter interval length.

Question 20

For x=etx=e^{t} and y=e−ty=e^{-t} on 0≤t≤ln⁡20\le t\le \ln 2, which integral represents the arc length?

  1. ∫0ln⁡2(et)2+(−e−t)2 dt\displaystyle \int_{0}^{\ln 2}\sqrt{\left(e^{t}\right)^2+\left(-e^{-t}\right)^2}\,dt (correct answer)
  2. ∫0ln⁡2(et)2+(e−t)2 dt\displaystyle \int_{0}^{\ln 2}\sqrt{\left(e^{t}\right)^2+\left(e^{-t}\right)^2}\,dt
  3. ∫0ln⁡2(et)+(−e−t) dt\displaystyle \int_{0}^{\ln 2}\sqrt{\left(e^{t}\right)+\left(-e^{-t}\right)}\,dt
  4. ∫02(et)2+(−e−t)2 dt\displaystyle \int_{0}^{2}\sqrt{\left(e^{t}\right)^2+\left(-e^{-t}\right)^2}\,dt
  5. ∫0ln⁡21+(−e−tet)2 dt\displaystyle \int_{0}^{\ln 2}\sqrt{1+\left(\frac{-e^{-t}}{e^{t}}\right)^2}\,dt
Explanation: To find the arc length, we apply L=∫ab(dxdt)2+(dydt)2dtL = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. Given x=etx = e^t and y=e−ty = e^{-t}, we differentiate to get dxdt=et\frac{dx}{dt} = e^t and dydt=−e−t\frac{dy}{dt} = -e^{-t}. Substituting these derivatives gives ∫0ln⁡2(et)2+(−e−t)2dt\int_0^{\ln 2} \sqrt{(e^t)^2 + (-e^{-t})^2} dt. Choice B incorrectly uses e−te^{-t} instead of −e−t-e^{-t} for the derivative of yy, forgetting the chain rule's negative sign when differentiating e−te^{-t}. Remember that the derivative of e−te^{-t} is −e−t-e^{-t}, and this sign matters when squaring in the arc length formula.