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AP Calculus BC Quiz

AP Calculus BC Quiz: Arc Length Of Smooth Planar Curve

Practice Arc Length Of Smooth Planar Curve in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

A particle moves with x(t)=2sin⁡tx(t)=2\sin tx(t)=2sint and y(t)=2cos⁡ty(t)=2\cos ty(t)=2cost for 0≤t≤π20\le t\le\tfrac{\pi}{2}0≤t≤2π​. What distance is traveled?

Select an answer to continue

What this quiz covers

This quiz focuses on Arc Length Of Smooth Planar Curve, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

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

A particle moves with x(t)=2sin⁡tx(t)=2\sin tx(t)=2sint and y(t)=2cos⁡ty(t)=2\cos ty(t)=2cost for 0≤t≤π20\le t\le\tfrac{\pi}{2}0≤t≤2π​. What distance is traveled?

  1. π\piπ (correct answer)
  2. 2π2\pi2π
  3. π2\dfrac{\pi}{2}2π​
  4. π2\pi\sqrt{2}π2​
  5. 222\sqrt{2}22​

Explanation: This particle moves along a circle of radius 2, with velocity components dxdt=2cos⁡(t)\frac{dx}{dt} = 2\cos(t)dtdx​=2cos(t) and dydt=−2sin⁡(t)\frac{dy}{dt} = -2\sin(t)dtdy​=−2sin(t). The speed is (2cos⁡t)2+(−2sin⁡t)2=4cos⁡2t+4sin⁡2t=2cos⁡2t+sin⁡2t=2\sqrt{(2\cos t)^2 + (-2\sin t)^2} = \sqrt{4\cos^2 t + 4\sin^2 t} = 2\sqrt{\cos^2 t + \sin^2 t} = 2(2cost)2+(−2sint)2​=4cos2t+4sin2t​=2cos2t+sin2t​=2. With constant speed 2, the distance from t=0 to t=π/2 is ∫0π/22 dt=2(π/2)=π\int_0^{\pi/2} 2 \, dt = 2(\pi/2) = \pi∫0π/2​2dt=2(π/2)=π. Choice B (2π2\pi2π) would be a full circle, while we only travel a quarter circle. The key is recognizing that for circular motion with radius r, the constant speed equals r times the angular velocity (here r=2r=2r=2, ω=1\omega=1ω=1).

Question 2

A cyclist’s position is r(t)=⟨cos⁡t, sin⁡t⟩\mathbf r(t)=\langle \cos t,\,\sin t\rangler(t)=⟨cost,sint⟩ for 0≤t≤π0\le t\le\pi0≤t≤π. What is the total distance traveled?

  1. π\piπ (correct answer)
  2. 2π2\pi2π
  3. 2 π\sqrt{2}\,\pi2​π
  4. 222
  5. π2\dfrac{\pi}{2}2π​

Explanation: This problem involves finding the distance traveled along a circular path given by parametric equations. The velocity components are dx/dt = -sin(t) and dy/dt = cos(t), giving speed = √[(-sin t)² + (cos t)²] = √[sin²t + cos²t] = 1. Since the speed is constant at 1, the distance traveled from t=0 to t=π equals ∫₀^π 1 dt = π. Choice B (2π) would be the full circumference, but we only travel half the circle. The key insight is recognizing that for uniform circular motion, speed equals the radius (1 in this case), making distance calculations straightforward.

Question 3

For x(t)=ln⁡(t+1)x(t)=\ln(t+1)x(t)=ln(t+1) and y(t)=ty(t)=\sqrt ty(t)=t​ on 0≤t≤30\le t\le 30≤t≤3, which integral equals arc length?

  1. ∫03(1t+1)2+(12t)2 dt\displaystyle \int_{0}^{3}\sqrt{\left(\frac{1}{t+1}\right)^2+\left(\frac{1}{2\sqrt t}\right)^2}\,dt∫03​(t+11​)2+(2t​1​)2​dt (correct answer)
  2. ∫03(1t+1+12t)dt\displaystyle \int_{0}^{3}\left(\frac{1}{t+1}+\frac{1}{2\sqrt t}\right)dt∫03​(t+11​+2t​1​)dt
  3. ∫03(ln⁡(t+1))2+(t)2 dt\displaystyle \int_{0}^{3}\sqrt{(\ln(t+1))^2+(\sqrt t)^2}\,dt∫03​(ln(t+1))2+(t​)2​dt
  4. ∫30(1t+1)2+(12t)2 dt\displaystyle \int_{3}^{0}\sqrt{\left(\frac{1}{t+1}\right)^2+\left(\frac{1}{2\sqrt t}\right)^2}\,dt∫30​(t+11​)2+(2t​1​)2​dt
  5. ∫03((1t+1)2+(12t)2)dt\displaystyle \int_{0}^{3}\left(\left(\frac{1}{t+1}\right)^2+\left(\frac{1}{2\sqrt t}\right)^2\right)dt∫03​((t+11​)2+(2t​1​)2)dt

Explanation: The skill tested is formulating the arc length integral for distance traveled with non-polynomial parametrics. Distance traveled integrates the magnitude of velocity, distinguishing it from signed displacement which could net zero for closed paths. For x(t) = ln(t+1) and y(t) = √t, dx/dt = 1/(t+1) and dy/dt = 1/(2√t), yielding ∫ √[(1/(t+1))² + (1/(2√t))²] dt from 0 to 3. Choice C is tempting, using positions ln(t+1) and √t inside the square root instead of their derivatives. Adopt the magnitude-first approach: calculate derivatives for velocity, find the magnitude via square root of summed squares, and integrate to get arc length.

Question 4

A particle moves with x(t)=t2−1x(t)=t^2-1x(t)=t2−1 and y(t)=3ty(t)=3ty(t)=3t for 0≤t≤20\le t\le20≤t≤2. Which integral gives total distance traveled?

  1. ∫02(2t+3) dt\displaystyle \int_{0}^{2}\left(2t+3\right)\,dt∫02​(2t+3)dt
  2. ∫02(2t)2+32 dt\displaystyle \int_{0}^{2}\sqrt{(2t)^2+3^2}\,dt∫02​(2t)2+32​dt (correct answer)
  3. ∫02((2t)2+32) dt\displaystyle \int_{0}^{2}\left((2t)^2+3^2\right)\,dt∫02​((2t)2+32)dt
  4. ∫−13(2t)2+32 dt\displaystyle \int_{-1}^{3}\sqrt{(2t)^2+3^2}\,dt∫−13​(2t)2+32​dt
  5. ∫022t+3 dt\displaystyle \int_{0}^{2}\sqrt{2t+3}\,dt∫02​2t+3​dt

Explanation: This problem asks for the total distance traveled, which requires finding the arc length of the parametric curve. For parametric equations x(t) and y(t), distance is calculated using the speed formula: ∫√[(dx/dt)² + (dy/dt)²]dt. Here, dx/dt = 2t and dy/dt = 3, so we need ∫₀² √[(2t)² + 3²]dt. Choice A incorrectly adds the derivatives instead of using the Pythagorean formula for speed. Remember: distance traveled always uses the magnitude of velocity, found by taking the square root of the sum of squared component derivatives.

Question 5

Two particles move for 0≤t≤10\le t\le10≤t≤1: r1=⟨t,t⟩\mathbf r_1=\langle t,t\rangler1​=⟨t,t⟩, r2=⟨t,2t⟩\mathbf r_2=\langle t,2t\rangler2​=⟨t,2t⟩. Which travels farther?

  1. Particle 1, because its velocity is larger for all ttt.
  2. Particle 2, because its speed is larger for all ttt. (correct answer)
  3. They travel the same distance, because both have constant speed.
  4. Particle 1, because its xxx-component of velocity is larger.
  5. Cannot be determined without finding acceleration.

Explanation: Particle 1 has velocity v₁ = ⟨1,1⟩ with speed |v₁| = √[1² + 1²] = √2. Particle 2 has velocity v₂ = ⟨1,2⟩ with speed |v₂| = √[1² + 2²] = √5. Since √5 > √2 and both speeds are constant, particle 2 travels farther: distance₂ = √5 × 1 > √2 × 1 = distance₁. Choice A confuses velocity magnitude with individual components. Choice C incorrectly assumes equal speeds. For distance comparisons, always calculate speed as the magnitude of the velocity vector, not just looking at components.

Question 6

For x(t)=t3x(t)=t^3x(t)=t3 and y(t)=t2y(t)=t^2y(t)=t2 on −1≤t≤1-1\le t\le1−1≤t≤1, which expression equals the distance traveled?

  1. ∫−11∣3t2+2t∣ dt\displaystyle \int_{-1}^{1}\left|3t^2+2t\right|\,dt∫−11​​3t2+2t​dt
  2. ∫−11(3t2)2+(2t)2 dt\displaystyle \int_{-1}^{1}\sqrt{(3t^2)^2+(2t)^2}\,dt∫−11​(3t2)2+(2t)2​dt (correct answer)
  3. ∫01(3t2)2+(2t)2 dt\displaystyle \int_{0}^{1}\sqrt{(3t^2)^2+(2t)^2}\,dt∫01​(3t2)2+(2t)2​dt
  4. ∫−11((3t2)2+(2t)2) dt\displaystyle \int_{-1}^{1}\left((3t^2)^2+(2t)^2\right)\,dt∫−11​((3t2)2+(2t)2)dt
  5. ∫−113t2+2t dt\displaystyle \int_{-1}^{1}\sqrt{3t^2+2t}\,dt∫−11​3t2+2t​dt

Explanation: To find distance traveled for the parametric curve x(t)=t³, y(t)=t², we need the arc length formula with derivatives dx/dt = 3t² and dy/dt = 2t. The speed is √[(3t²)² + (2t)²] = √[9t⁴ + 4t²] = |t|√[9t² + 4]. The integral becomes ∫₋₁¹ |t|√[9t² + 4] dt, which equals ∫₋₁¹ √[(3t²)² + (2t)²] dt since the square root handles the absolute value. Choice A incorrectly tries to use absolute value of a sum rather than the magnitude formula. Always use the Pythagorean theorem for speed: √[(dx/dt)² + (dy/dt)²], not |dx/dt + dy/dt|.

Question 7

For x(t)=etx(t)=e^tx(t)=et, y(t)=e−ty(t)=e^{-t}y(t)=e−t on 0≤t≤10\le t\le10≤t≤1, which integral represents the arc length?

  1. ∫01e2t+e−2t dt\displaystyle \int_{0}^{1}\sqrt{e^{2t}+e^{-2t}}\,dt∫01​e2t+e−2t​dt (correct answer)
  2. ∫01(et−e−t) dt\displaystyle \int_{0}^{1}\left(e^t-e^{-t}\right)\,dt∫01​(et−e−t)dt
  3. ∫01e2t−e−2t dt\displaystyle \int_{0}^{1}\sqrt{e^{2t}-e^{-2t}}\,dt∫01​e2t−e−2t​dt
  4. ∫10e2t+e−2t dt\displaystyle \int_{1}^{0}\sqrt{e^{2t}+e^{-2t}}\,dt∫10​e2t+e−2t​dt
  5. ∫01(e2t+e−2t) dt\displaystyle \int_{0}^{1}\left(e^{2t}+e^{-2t}\right)\,dt∫01​(e2t+e−2t)dt

Explanation: For x(t) = e^t and y(t) = e^(-t), we calculate x'(t) = e^t and y'(t) = -e^(-t). The arc length integrand is √[(e^t)² + (-e^(-t))²] = √[e^(2t) + e^(-2t)]. Therefore, the correct integral is ∫₀¹ √[e^(2t) + e^(-2t)]dt. Choice B shows e^t - e^(-t), which incorrectly subtracts the derivatives without squaring and taking the square root—this would give displacement, not distance. Remember that arc length always uses the speed formula √[(x'(t))² + (y'(t))²], ensuring all contributions are positive.

Question 8

A particle has r(t)=⟨t2, t2⟩\mathbf r(t)=\langle t^2,\,t^2\rangler(t)=⟨t2,t2⟩ for −1≤t≤2-1 \leq t \leq 2−1≤t≤2; which setup gives distance traveled?

  1. ∫−12(2t)2+(2t)2 dt\displaystyle \int_{-1}^{2}\sqrt{(2t)^2+(2t)^2}\,dt∫−12​(2t)2+(2t)2​dt (correct answer)
  2. ∫−12(2t+2t) dt\displaystyle \int_{-1}^{2}(2t+2t)\,dt∫−12​(2t+2t)dt
  3. ∫−12((2t)2+(2t)2) dt\displaystyle \int_{-1}^{2}\big((2t)^2+(2t)^2\big)\,dt∫−12​((2t)2+(2t)2)dt
  4. ∫2−1(2t)2+(2t)2 dt\displaystyle \int_{2}^{-1}\sqrt{(2t)^2+(2t)^2}\,dt∫2−1​(2t)2+(2t)2​dt
  5. ∫−12(t2)2+(t2)2 dt\displaystyle \int_{-1}^{2}\sqrt{(t^2)^2+(t^2)^2}\,dt∫−12​(t2)2+(t2)2​dt

Explanation: This question evaluates computing the distance traveled via arc length for a parametric curve with negative t values. Distance is the integral of speed, the magnitude of velocity, not signed displacement, to account for the entire path regardless of direction. For r(t)=⟨t2,t2⟩r(t) = \langle t^2, t^2 \rangler(t)=⟨t2,t2⟩, dxdt=2t\frac{dx}{dt} = 2tdtdx​=2t and dydt=2t\frac{dy}{dt} = 2tdtdy​=2t, so ∫−12(2t)2+(2t)2 dt\int_{-1}^{2} \sqrt{(2t)^2 + (2t)^2} \, dt∫−12​(2t)2+(2t)2​dt gives the distance. A tempting distractor is choice E, which uses positions t2t^2t2 instead of derivatives 2t2t2t in the integrand. The magnitude-first strategy is key: always derive to get velocity, compute its magnitude with the square root, then integrate over the interval for total distance.

Question 9

A particle moves on r(t)=⟨et, e−t⟩\mathbf{r}(t)=\langle e^t,\,e^{-t}\rangler(t)=⟨et,e−t⟩ for −1≤t≤1-1\le t\le1−1≤t≤1; which integral equals its distance traveled?

  1. ∫−11(et−e−t)dt\displaystyle \int_{-1}^{1}\left(e^t-e^{-t}\right)dt∫−11​(et−e−t)dt
  2. ∫−11e2t+e−2t dt\displaystyle \int_{-1}^{1}\sqrt{e^{2t}+e^{-2t}}\,dt∫−11​e2t+e−2t​dt
  3. ∫−11(et)2+(−e−t)2 dt\displaystyle \int_{-1}^{1}\sqrt{\left(e^t\right)^2+\left(-e^{-t}\right)^2}\,dt∫−11​(et)2+(−e−t)2​dt (correct answer)
  4. ∫1−1(et)2+(−e−t)2 dt\displaystyle \int_{1}^{-1}\sqrt{\left(e^t\right)^2+\left(-e^{-t}\right)^2}\,dt∫1−1​(et)2+(−e−t)2​dt
  5. ∫−11((et)2+(−e−t)2)dt\displaystyle \int_{-1}^{1}\left(\left(e^t\right)^2+\left(-e^{-t}\right)^2\right)dt∫−11​((et)2+(−e−t)2)dt

Explanation: The skill being tested is calculating the arc length or distance traveled by a particle along a parametric curve. The distance traveled is found by integrating the speed, which is the magnitude of the velocity vector, over the time interval. This differs from displacement, which is the straight-line distance from start to end and can be signed. For r(t) = <e^t, e^{-t}>, the velocity components are e^t and -e^{-t}, so the integral is ∫ from -1 to 1 of sqrt((e^t)² + (-e^{-t})²) dt, matching choice C. A tempting distractor is choice D, which reverses the limits, yielding a negative value unsuitable for distance. Always compute the magnitude inside the integral first to capture the instantaneous speed. This magnitude-first strategy ensures accurate distance for any parametric path.

Question 10

Two particles move for 0≤t≤10\le t\le10≤t≤1: r1(t)=⟨t,0⟩\mathbf{r}_1(t)=\langle t,0\rangler1​(t)=⟨t,0⟩ and r2(t)=⟨t,t2⟩\mathbf{r}_2(t)=\langle t,t^2\rangler2​(t)=⟨t,t2⟩; which travels farther?

  1. Particle 1, because ∫011 dt>∫011+4t2 dt\int_0^1 1\,dt>\int_0^1 \sqrt{1+4t^2}\,dt∫01​1dt>∫01​1+4t2​dt.
  2. Particle 2, because ∫011+4t2 dt>∫011 dt\int_0^1 \sqrt{1+4t^2}\,dt>\int_0^1 1\,dt∫01​1+4t2​dt>∫01​1dt. (correct answer)
  3. They travel the same distance, because both have x(1)−x(0)=1x(1)-x(0)=1x(1)−x(0)=1.
  4. Particle 1, because ∫01(1+2t) dt>∫011 dt\int_0^1 (1+2t)\,dt>\int_0^1 1\,dt∫01​(1+2t)dt>∫01​1dt.
  5. They travel the same distance, because ∫01(1+4t2) dt=∫011 dt\int_0^1 (1+4t^2)\,dt=\int_0^1 1\,dt∫01​(1+4t2)dt=∫01​1dt.

Explanation: The skill being tested is calculating the arc length or distance traveled by a particle along a parametric curve. The distance traveled is found by integrating the speed, which is the magnitude of the velocity vector, over the time interval. This differs from displacement, which is the straight-line distance from start to end and can be signed. For the two particles, particle 2's path along the parabola has greater length since ∫ sqrt(1 + 4t²) dt > ∫ 1 dt, matching choice B. A tempting distractor is choice C, which equates distances based on x-displacement alone, ignoring the y-component. Always compute the magnitude inside the integral first to capture the instantaneous speed. This magnitude-first strategy ensures accurate distance for any parametric path.

Question 11

A particle follows r(t)=⟨3cos⁡t, 4sin⁡t⟩\mathbf{r}(t)=\langle 3\cos t,\,4\sin t\rangler(t)=⟨3cost,4sint⟩ for 0≤t≤π20\le t\le\frac{\pi}{2}0≤t≤2π​; which integral gives distance?

  1. ∫0π/29cos⁡2t+16sin⁡2t dt\displaystyle \int_0^{\pi/2}\sqrt{9\cos^2 t+16\sin^2 t}\,dt∫0π/2​9cos2t+16sin2t​dt
  2. ∫0π/29sin⁡2t+16cos⁡2t dt\displaystyle \int_0^{\pi/2}\sqrt{9\sin^2 t+16\cos^2 t}\,dt∫0π/2​9sin2t+16cos2t​dt (correct answer)
  3. ∫0π/2(3sin⁡t+4cos⁡t)dt\displaystyle \int_0^{\pi/2}\left(3\sin t+4\cos t\right)dt∫0π/2​(3sint+4cost)dt
  4. ∫π/209sin⁡2t+16cos⁡2t dt\displaystyle \int_{\pi/2}^{0}\sqrt{9\sin^2 t+16\cos^2 t}\,dt∫π/20​9sin2t+16cos2t​dt
  5. ∫0π/2(9sin⁡2t+16cos⁡2t)dt\displaystyle \int_0^{\pi/2}\left(9\sin^2 t+16\cos^2 t\right)dt∫0π/2​(9sin2t+16cos2t)dt

Explanation: The skill being tested is calculating the arc length or distance traveled by a particle along a parametric curve. The distance traveled is found by integrating the speed, which is the magnitude of the velocity vector, over the time interval. This differs from displacement, which is the straight-line distance from start to end and can be signed. For r(t) = <3 cos t, 4 sin t>, the velocity components are -3 sin t and 4 cos t, so the integral is ∫ from 0 to π/2 of sqrt(9 sin² t + 16 cos² t) dt, matching choice B. A tempting distractor is choice A, which swaps the coefficients, altering the ellipse axes. Always compute the magnitude inside the integral first to capture the instantaneous speed. This magnitude-first strategy ensures accurate distance for any parametric path.

Question 12

A particle has x(t)=t3−3t, y(t)=4tx(t)=t^3-3t,\ y(t)=4tx(t)=t3−3t, y(t)=4t on −1≤t≤2-1\le t\le2−1≤t≤2; which is the correct distance integral?

  1. ∫−12(3t2−3)2+42 dt\displaystyle \int_{-1}^{2}\sqrt{(3t^2-3)^2+4^2}\,dt∫−12​(3t2−3)2+42​dt (correct answer)
  2. ∫−12((3t2−3)2+42) dt\displaystyle \int_{-1}^{2}\big((3t^2-3)^2+4^2\big)\,dt∫−12​((3t2−3)2+42)dt
  3. ∫2−1(3t2−3)2+42 dt\displaystyle \int_{2}^{-1}\sqrt{(3t^2-3)^2+4^2}\,dt∫2−1​(3t2−3)2+42​dt
  4. ∫−12((3t2−3)+4) dt\displaystyle \int_{-1}^{2}\big((3t^2-3)+4\big)\,dt∫−12​((3t2−3)+4)dt
  5. (∫−12(3t2−3) dt)2+(∫−124 dt)2\displaystyle \sqrt{\left(\int_{-1}^{2}(3t^2-3)\,dt\right)^2+\left(\int_{-1}^{2}4\,dt\right)^2}(∫−12​(3t2−3)dt)2+(∫−12​4dt)2​

Explanation: The skill being tested is calculating the arc length or distance traveled by a particle along a parametric curve. The distance traveled is found by integrating the speed, which is the magnitude of the velocity vector, over the time interval. This differs from displacement, which is the straight-line distance from start to end and can be signed. For x=t³-3t, y=4t, the velocity components are 3t²-3 and 4, so the integral is ∫ from -1 to 2 of sqrt((3t²-3)² + 4²) dt, matching choice A. A tempting distractor is choice E, which computes the magnitude of the integrals separately, giving displacement instead. Always compute the magnitude inside the integral first to capture the instantaneous speed. This magnitude-first strategy ensures accurate distance for any parametric path.

Question 13

A drone follows r(t)=⟨et, e−t⟩\mathbf r(t)=\langle e^t,\,e^{-t}\rangler(t)=⟨et,e−t⟩ for 0≤t≤10\le t\le10≤t≤1. Which integral gives its distance traveled?

  1. ∫01(et−e−t) dt\displaystyle \int_{0}^{1}\left(e^t-e^{-t}\right)\,dt∫01​(et−e−t)dt
  2. ∫01et+e−t dt\displaystyle \int_{0}^{1}\sqrt{e^t+e^{-t}}\,dt∫01​et+e−t​dt
  3. ∫01(et)2+(e−t)2 dt\displaystyle \int_{0}^{1}\sqrt{(e^t)^2+(e^{-t})^2}\,dt∫01​(et)2+(e−t)2​dt
  4. ∫10(et)2+(e−t)2 dt\displaystyle \int_{1}^{0}\sqrt{(e^t)^2+(e^{-t})^2}\,dt∫10​(et)2+(e−t)2​dt
  5. ∫01(et)2+(−e−t)2 dt\displaystyle \int_{0}^{1}\sqrt{(e^t)^2+(-e^{-t})^2}\,dt∫01​(et)2+(−e−t)2​dt (correct answer)

Explanation: For the parametric curve r(t)=⟨et,e−t⟩r(t) = \langle e^t, e^{-t} \rangler(t)=⟨et,e−t⟩, we calculate distance using derivatives dxdt=et\frac{dx}{dt} = e^tdtdx​=et and dydt=−e−t\frac{dy}{dt} = -e^{-t}dtdy​=−e−t. The speed formula gives [(et)2+(−e−t)2]=[e2t+e−2t]\sqrt{[(e^t)^2 + (-e^{-t})^2]} = \sqrt{[e^{2t} + e^{-2t}]}[(et)2+(−e−t)2]​=[e2t+e−2t]​. Choice C omits the negative sign on the y-derivative, but since we square it, (−e−t)2=e−2t(-e^{-t})^2 = e^{-2t}(−e−t)2=e−2t, making choice E correct. Choice B incorrectly adds inside the square root instead of squaring first. Remember: when finding speed, always square the individual derivatives before adding—the negative sign on dydt\frac{dy}{dt}dtdy​ matters for direction but not for speed magnitude.

Question 14

Two particles: r1(t)=⟨t,2t⟩\mathbf r_1(t)=\langle t,2t\rangler1​(t)=⟨t,2t⟩, r2(t)=⟨2t,t⟩\mathbf r_2(t)=\langle 2t,t\rangler2​(t)=⟨2t,t⟩ for 0≤t≤10\le t\le10≤t≤1; compare distances traveled.​

  1. Particle 1 travels farther.
  2. Particle 2 travels farther.
  3. They travel the same distance. (correct answer)
  4. Cannot be determined without positions at t=0t=0t=0.
  5. Cannot be determined without computing displacement vectors only.

Explanation: To compare distances traveled, we need to compute arc length for each particle. For r₁(t) = ⟨t, 2t⟩: velocity is ⟨1, 2⟩, speed is √[1² + 2²] = √5. Distance traveled: ∫₀¹ √5 dt = √5. For r₂(t) = ⟨2t, t⟩: velocity is ⟨2, 1⟩, speed is √[2² + 1²] = √5. Distance traveled: ∫₀¹ √5 dt = √5. Both particles travel the same distance √5, despite following different paths. This illustrates that arc length depends on speed magnitude, not on the specific path direction. When comparing distances, always compute the integral of speed for each particle.

Question 15

A particle has velocity v(t)=⟨6t, −8⟩\mathbf v(t)=\langle 6t,\,-8\ranglev(t)=⟨6t,−8⟩ for 0≤t≤10\le t\le10≤t≤1; which integral gives distance traveled?​

  1. ∫01(6t)2+(−8)2 dt\displaystyle \int_{0}^{1}\sqrt{(6t)^2+(-8)^2}\,dt∫01​(6t)2+(−8)2​dt (correct answer)
  2. ∫01(6t−8) dt\displaystyle \int_{0}^{1}(6t-8)\,dt∫01​(6t−8)dt
  3. ∫01((6t)2+(−8)2) dt\displaystyle \int_{0}^{1}\left((6t)^2+(-8)^2\right)\,dt∫01​((6t)2+(−8)2)dt
  4. ∫10(6t)2+(−8)2 dt\displaystyle \int_{1}^{0}\sqrt{(6t)^2+(-8)^2}\,dt∫10​(6t)2+(−8)2​dt
  5. ∫01⟨6t, −8⟩ dt\displaystyle \int_{0}^{1}\left\langle 6t,\,-8\right\rangle\,dt∫01​⟨6t,−8⟩dt

Explanation: Given velocity v(t) = ⟨6t, -8⟩, distance traveled is found by integrating speed (magnitude of velocity). The speed is |v(t)| = √[(6t)² + (-8)²] = √[36t² + 64]. The correct integral is ∫₀¹ √[(6t)² + (-8)²] dt. Choice B shows 6t - 8, which incorrectly subtracts components—this would give a signed quantity that could be negative, but distance is always non-negative. When given velocity directly, distance traveled equals ∫|v(t)| dt, where |v(t)| is the magnitude of the velocity vector.

Question 16

A particle moves on r(t)=⟨t2, t3⟩\mathbf r(t)=\langle t^2,\,t^3\rangler(t)=⟨t2,t3⟩ for 0≤t≤10\le t\le10≤t≤1. Which integral gives the distance traveled?

  1. ∫01(2t+3t2) dt\displaystyle \int_{0}^{1}\left(2t+3t^2\right)\,dt∫01​(2t+3t2)dt
  2. ∫01(2t)2+(3t2)2 dt\displaystyle \int_{0}^{1}\sqrt{(2t)^2+(3t^2)^2}\,dt∫01​(2t)2+(3t2)2​dt (correct answer)
  3. (∫012t dt)2+(∫013t2 dt)2\displaystyle \sqrt{\left(\int_{0}^{1}2t\,dt\right)^2+\left(\int_{0}^{1}3t^2\,dt\right)^2}(∫01​2tdt)2+(∫01​3t2dt)2​
  4. ∫01((2t)2+(3t2)2) dt\displaystyle \int_{0}^{1}\left((2t)^2+(3t^2)^2\right)\,dt∫01​((2t)2+(3t2)2)dt
  5. ∫10(2t)2+(3t2)2 dt\displaystyle \int_{1}^{0}\sqrt{(2t)^2+(3t^2)^2}\,dt∫10​(2t)2+(3t2)2​dt

Explanation: This problem asks for the distance traveled by a particle, which requires finding the arc length of the parametric curve. For parametric equations x(t) = t² and y(t) = t³, we find the derivatives: x'(t) = 2t and y'(t) = 3t². The arc length formula for parametric curves is ∫√[(x'(t))² + (y'(t))²]dt, which gives us ∫₀¹√[(2t)² + (3t²)²]dt. Choice A incorrectly adds the derivatives directly without taking the magnitude, which would give displacement rather than distance. The key strategy is to always use the speed formula √[(dx/dt)² + (dy/dt)²] inside the integral when finding distance traveled.

Question 17

For x(t)=cos⁡tx(t)=\cos tx(t)=cost, y(t)=sin⁡ty(t)=\sin ty(t)=sint on 0≤t≤π0\le t\le\pi0≤t≤π, what is the distance traveled?

  1. π\displaystyle \piπ (correct answer)
  2. 2π\displaystyle 2\pi2π
  3. 2\displaystyle 22
  4. ∫0π( ⁣−sin⁡t+cos⁡t) dt\displaystyle \int_{0}^{\pi}(\!-\sin t+\cos t)\,dt∫0π​(−sint+cost)dt
  5. ∫0π(( ⁣−sin⁡t)2+(cos⁡t)2) dt\displaystyle \int_{0}^{\pi}\big((\!-\sin t)^2+(\cos t)^2\big)\,dt∫0π​((−sint)2+(cost)2)dt

Explanation: This problem involves finding the distance traveled along a parametric curve representing a semicircle. Given x(t) = cos t and y(t) = sin t, we compute x'(t) = -sin t and y'(t) = cos t. The speed is √[(-sin t)² + (cos t)²] = √[sin²t + cos²t] = √1 = 1. Therefore, the distance traveled is ∫₀^π 1 dt = π. Choice D temptingly shows the derivatives but incorrectly adds them without squaring and taking the square root first. Remember that distance traveled always uses the magnitude of velocity, found by √[(x'(t))² + (y'(t))²], integrated over the time interval.

Question 18

For r(t)=⟨ln⁡t, t2⟩\mathbf{r}(t)=\langle \ln t,\,t^2\rangler(t)=⟨lnt,t2⟩ on 1≤t≤e1\le t\le e1≤t≤e, which integral equals the arc length?

  1. ∫1e(1t)2+(2t)2 dt\displaystyle \int_1^{e}\sqrt{\left(\frac{1}{t}\right)^2+(2t)^2}\,dt∫1e​(t1​)2+(2t)2​dt (correct answer)
  2. ∫1e(1t+2t)dt\displaystyle \int_1^{e}\left(\frac{1}{t}+2t\right)dt∫1e​(t1​+2t)dt
  3. ∫e1(1t)2+(2t)2 dt\displaystyle \int_e^{1}\sqrt{\left(\frac{1}{t}\right)^2+(2t)^2}\,dt∫e1​(t1​)2+(2t)2​dt
  4. ∫1e1t+4t2 dt\displaystyle \int_1^{e}\sqrt{\frac{1}{t}+4t^2}\,dt∫1e​t1​+4t2​dt
  5. ∫1e((1t)2+(2t)2)dt\displaystyle \int_1^{e}\left(\left(\frac{1}{t}\right)^2+(2t)^2\right)dt∫1e​((t1​)2+(2t)2)dt

Explanation: The skill being tested is calculating the arc length or distance traveled by a particle along a parametric curve. The distance traveled is found by integrating the speed, which is the magnitude of the velocity vector, over the time interval. This differs from displacement, which is the straight-line distance from start to end and can be signed. For r(t) = <ln t, t²>, the velocity components are 1/t and 2t, so the integral is ∫ from 1 to e of sqrt((1/t)² + (2t)²) dt, matching choice A. A tempting distractor is choice B, which adds the components without the square root, giving a different quantity. Always compute the magnitude inside the integral first to capture the instantaneous speed. This magnitude-first strategy ensures accurate distance for any parametric path.

Question 19

A particle follows x(t)=tx(t)=tx(t)=t, y(t)=ln⁡ty(t)=\ln ty(t)=lnt for 1≤t≤e1\le t\le e1≤t≤e. Which is the correct distance traveled setup?

  1. ∫1e1+(1t)2 dt\displaystyle \int_{1}^{e}\sqrt{1+\left(\frac1t\right)^2}\,dt∫1e​1+(t1​)2​dt (correct answer)
  2. ∫1e(1+1t) dt\displaystyle \int_{1}^{e}\left(1+\frac1t\right)\,dt∫1e​(1+t1​)dt
  3. ∫e11+(1t)2 dt\displaystyle \int_{e}^{1}\sqrt{1+\left(\frac1t\right)^2}\,dt∫e1​1+(t1​)2​dt
  4. ∫1e(1+(1t)2) dt\displaystyle \int_{1}^{e}\left(1+\left(\frac1t\right)^2\right)\,dt∫1e​(1+(t1​)2)dt
  5. (∫1e1 dt)2+(∫1e1t dt)2\displaystyle \sqrt{\left(\int_{1}^{e}1\,dt\right)^2+\left(\int_{1}^{e}\frac1t\,dt\right)^2}(∫1e​1dt)2+(∫1e​t1​dt)2​

Explanation: To find the distance traveled by a particle with position x(t) = t and y(t) = ln t, we need the arc length formula. Taking derivatives: x'(t) = 1 and y'(t) = 1/t. The arc length integral becomes ∫₁^e √[1² + (1/t)²]dt = ∫₁^e √[1 + 1/t²]dt. Choice B incorrectly adds the derivatives directly (1 + 1/t) without squaring and taking the square root, which would not give the correct speed. The fundamental principle is that arc length requires integrating the speed √[(dx/dt)² + (dy/dt)²], not the sum of derivatives.

Question 20

For x(t)=tx(t)=tx(t)=t, y(t)=ln⁡ty(t)=\ln ty(t)=lnt on 1≤t≤e1\le t\le e1≤t≤e, which is the correct arc length integral?​

  1. ∫1e1+(1t)2 dt\displaystyle \int_{1}^{e}\sqrt{1+\left(\frac{1}{t}\right)^2}\,dt∫1e​1+(t1​)2​dt (correct answer)
  2. ∫1e(1+1t) dt\displaystyle \int_{1}^{e}\left(1+\frac{1}{t}\right)\,dt∫1e​(1+t1​)dt
  3. ∫1e(1+(1t)2) dt\displaystyle \int_{1}^{e}\left(1+\left(\frac{1}{t}\right)^2\right)\,dt∫1e​(1+(t1​)2)dt
  4. ∫e11+(1t)2 dt\displaystyle \int_{e}^{1}\sqrt{1+\left(\frac{1}{t}\right)^2}\,dt∫e1​1+(t1​)2​dt
  5. ∫0e1+(1t)2 dt\displaystyle \int_{0}^{e}\sqrt{1+\left(\frac{1}{t}\right)^2}\,dt∫0e​1+(t1​)2​dt

Explanation: For x(t) = t and y(t) = ln t on [1, e], we need the arc length integral to find distance traveled. Computing derivatives: dx/dt = 1 and dy/dt = 1/t. The speed is √[1² + (1/t)²] = √[1 + 1/t²]. The correct arc length integral is ∫₁ᵉ √[1 + (1/t)²] dt. Choice B shows 1 + 1/t without the square root, which would incorrectly add the derivatives algebraically rather than finding their magnitude. Remember that arc length always uses the formula ∫√[(dx/dt)² + (dy/dt)²] dt, requiring the square root of the sum of squares.