Calculus 2 Quiz: Arc Length And Distance Traveled
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Arc Length And Distance TraveledQuestion 1 of 20

The derivative of a function f(x)f(x) is given by f′(x)=x4−1f'(x) = \sqrt{x^4 - 1}. Which integral represents the arc length of y=f(x)y=f(x) from x=1x=1 to x=2x=2?

∫12(x2−1) dx\int_1^2 (x^2-1) \, dx
∫121+x4−1 dx\int_1^2 \sqrt{1 + \sqrt{x^4-1}} \, dx
∫12x4−2 dx\int_1^2 \sqrt{x^4-2} \, dx
∫12x2 dx\int_1^2 x^2 \, dx
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Calculus 2 Quiz

Calculus 2 Quiz: Arc Length And Distance Traveled

Practice Arc Length And Distance Traveled 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 And Distance Traveled, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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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.

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Question 1

The derivative of a function f(x)f(x) is given by f′(x)=x4−1f'(x) = \sqrt{x^4 - 1}. Which integral represents the arc length of y=f(x)y=f(x) from x=1x=1 to x=2x=2?

  1. ∫12(x2−1) dx\int_1^2 (x^2-1) \, dx
  2. ∫121+x4−1 dx\int_1^2 \sqrt{1 + \sqrt{x^4-1}} \, dx
  3. ∫12x4−2 dx\int_1^2 \sqrt{x^4-2} \, dx
  4. ∫12x2 dx\int_1^2 x^2 \, dx (correct answer)
Explanation: When you encounter an arc length problem, you need to recall the arc length formula for a function y=f(x)y = f(x): L=∫ab1+[f′(x)]2 dxL = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx. This formula measures the actual distance along the curve, accounting for both horizontal and vertical changes. Given that f′(x)=x4−1f'(x) = \sqrt{x^4 - 1}, you need to substitute this into the arc length formula. First, calculate [f′(x)]2=(x4−1)2=x4−1[f'(x)]^2 = (\sqrt{x^4 - 1})^2 = x^4 - 1. Then the arc length integral becomes: L=∫121+(x4−1) dx=∫12x4 dx=∫12x2 dxL = \int_1^2 \sqrt{1 + (x^4 - 1)} \, dx = \int_1^2 \sqrt{x^4} \, dx = \int_1^2 x^2 \, dx This matches answer choice D. Let's examine why the other answers are incorrect. Choice A gives ∫12(x2−1) dx\int_1^2 (x^2-1) \, dx, which completely omits the square root from the arc length formula—this would be incorrect even for a simple area calculation. Choice B shows ∫121+x4−1 dx\int_1^2 \sqrt{1 + \sqrt{x^4-1}} \, dx, which incorrectly treats f′(x)f'(x) as x4−1\sqrt{x^4-1} rather than squaring it first; this represents the common mistake of forgetting to square the derivative. Choice C gives ∫12x4−2 dx\int_1^2 \sqrt{x^4-2} \, dx, which appears to result from the error 1+(x4−1)=x4−21 + (x^4 - 1) = x^4 - 2 instead of the correct x4x^4. Remember: arc length problems always require the formula 1+[f′(x)]2\sqrt{1 + [f'(x)]^2}. Square the given derivative first, then add 1, then take the square root of the entire expression.

Question 2

A robot's path is defined by y=13(x2+2)3/2y = \frac{1}{3}(x^2+2)^{3/2}. What is the distance the robot travels along this path from x=0x=0 to x=3x=3?

  1. 9
  2. 12 (correct answer)
  3. 15
  4. 18
Explanation: The distance is the arc length. First, find the derivative: y′=13⋅32(x2+2)1/2⋅2x=xx2+2y' = \frac{1}{3} \cdot \frac{3}{2}(x^2+2)^{1/2} \cdot 2x = x\sqrt{x^2+2}. Next, compute 1+(y′)21+(y')^2: 1+(xx2+2)2=1+x2(x2+2)=1+x4+2x2=(x2+1)21 + (x\sqrt{x^2+2})^2 = 1 + x^2(x^2+2) = 1 + x^4 + 2x^2 = (x^2+1)^2. The arc length integral is L=∫03(x2+1)2 dx=∫03(x2+1) dxL = \int_0^3 \sqrt{(x^2+1)^2} \, dx = \int_0^3 (x^2+1) \, dx. Evaluating the integral gives [x33+x]03=(273+3)−0=9+3=12[\frac{x^3}{3} + x]_0^3 = (\frac{27}{3} + 3) - 0 = 9 + 3 = 12.

Question 3

Let LL be the arc length of the curve y=f(x)y=f(x) from x=ax=a to x=bx=b. What is the arc length of the curve y=cf(x/c)y=cf(x/c) from x=cax=ca to x=cbx=cb, where c>0c > 0 is a constant?

  1. LL
  2. cLcL (correct answer)
  3. c2Lc^2L
  4. L/cL/c
Explanation: Let the new curve be g(x)=cf(x/c)g(x) = cf(x/c). The arc length is given by the integral ∫cacb1+[g′(x)]2 dx\int_{ca}^{cb} \sqrt{1 + [g'(x)]^2} \, dx. First, we find the derivative of g(x)g(x) using the chain rule: g′(x)=c⋅f′(x/c)⋅1c=f′(x/c)g'(x) = c \cdot f'(x/c) \cdot \frac{1}{c} = f'(x/c). Substituting this into the arc length formula gives ∫cacb1+[f′(x/c)]2 dx\int_{ca}^{cb} \sqrt{1 + [f'(x/c)]^2} \, dx. We use a u-substitution with u=x/cu = x/c, which means x=cux = cu and dx=c dudx = c \, du. The limits of integration also change: when x=cax=ca, u=au=a; when x=cbx=cb, u=bu=b. The integral becomes ∫ab1+[f′(u)]2⋅c du=c∫ab1+[f′(u)]2 du\int_a^b \sqrt{1 + [f'(u)]^2} \cdot c \, du = c \int_a^b \sqrt{1 + [f'(u)]^2} \, du. This is exactly cc times the original arc length LL. Therefore, the new arc length is cLcL.

Question 4

Consider the curve given by y=x2y = x^2 for 0≤x≤10 \le x \le 1. The arc length is L1=∫011+4x2dxL_1 = \int_0^1 \sqrt{1+4x^2}dx. Now consider the curve x=y2x=y^2 for 0≤y≤10 \le y \le 1. Its arc length is L2=∫011+4y2dyL_2 = \int_0^1 \sqrt{1+4y^2}dy. Which of the following statements is true?

  1. L1>L2L_1 > L_2
  2. L1<L2L_1 < L_2
  3. L1=L2L_1 = L_2 (correct answer)
  4. The relationship cannot be determined without evaluating the integrals.
Explanation: The arc length for y=x2y=x^2 from x=0x=0 to x=1x=1 is given by L1=∫011+(2x)2dx=∫011+4x2dxL_1 = \int_0^1 \sqrt{1+(2x)^2}dx = \int_0^1 \sqrt{1+4x^2}dx. The arc length for x=y2x=y^2 from y=0y=0 to y=1y=1 is given by L2=∫011+(2y)2dy=∫011+4y2dyL_2 = \int_0^1 \sqrt{1+(2y)^2}dy = \int_0^1 \sqrt{1+4y^2}dy. The two integrals are definite integrals of the same function, f(z)=1+4z2f(z) = \sqrt{1+4z^2}, over the same interval [0,1][0,1]. The variable of integration (whether it's called xx or yy) is a dummy variable and does not affect the value of the definite integral. Therefore, the values of the two integrals are equal, and L1=L2L_1 = L_2. Geometrically, the curve x=y2x=y^2 for 0≤y≤10 \le y \le 1 is a reflection of the curve y=x2y=x^2 for 0≤x≤10 \le x \le 1 across the line y=xy=x, and reflection is a rigid transformation that preserves arc length.

Question 5

The astroid curve is given parametrically by x=acos⁡3(t)x = a\cos^3(t), y=asin⁡3(t)y = a\sin^3(t) where a>0a > 0 and t∈[0,2π]t \in [0, 2\pi]. If the total arc length of the astroid is 1212, what is the value of aa?

  1. 43\frac{4}{3}
  2. 22 (correct answer)
  3. 83\frac{8}{3}
  4. 33
Explanation: For the astroid, dxdt=−3acos⁡2(t)sin⁡(t)\frac{dx}{dt} = -3a\cos^2(t)\sin(t) and dydt=3asin⁡2(t)cos⁡(t)\frac{dy}{dt} = 3a\sin^2(t)\cos(t). So (dxdt)2+(dydt)2=9a2cos⁡4(t)sin⁡2(t)+9a2sin⁡4(t)cos⁡2(t)=9a2cos⁡2(t)sin⁡2(t)[cos⁡2(t)+sin⁡2(t)]=9a2cos⁡2(t)sin⁡2(t)(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 = 9a^2\cos^4(t)\sin^2(t) + 9a^2\sin^4(t)\cos^2(t) = 9a^2\cos^2(t)\sin^2(t)[\cos^2(t) + \sin^2(t)] = 9a^2\cos^2(t)\sin^2(t). The arc length for one complete cycle is L=∫02π3a∣cos⁡(t)sin⁡(t)∣dtL = \int_0^{2\pi} 3a|\cos(t)\sin(t)| dt. Due to symmetry, we can compute 4∫0π/23acos⁡(t)sin⁡(t)dt=4⋅3a∫0π/212sin⁡(2t)dt=6a[−12cos⁡(2t)]0π/2=6a[−12(−1−1)]=6a4\int_0^{\pi/2} 3a\cos(t)\sin(t) dt = 4 \cdot 3a \int_0^{\pi/2} \frac{1}{2}\sin(2t) dt = 6a[-\frac{1}{2}\cos(2t)]_0^{\pi/2} = 6a[-\frac{1}{2}(-1 - 1)] = 6a. Setting 6a=126a = 12 gives a=2a = 2. Choice A gives a=43a = \frac{4}{3}, which would yield arc length 88. Choice C gives a=83a = \frac{8}{3}, yielding arc length 1616. Choice D gives a=3a = 3, yielding arc length 1818.

Question 6

A particle moves along a path described by x(t)=etcos⁡(t)x(t) = e^t \cos(t), y(t)=etsin⁡(t)y(t) = e^t \sin(t) for t∈[0,π]t \in [0, \pi]. The particle's speed varies according to its position. What is the total distance traveled?

  1. 2(eπ−1)\sqrt{2}(e^{\pi} - 1) (correct answer)
  2. eπ−1e^{\pi} - 1
  3. 2eπ\sqrt{2}e^{\pi}
  4. 2(eπ−1)2(e^{\pi} - 1)
Explanation: The distance traveled is ∫0π(dxdt)2+(dydt)2dt\int_0^{\pi} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt. We have dxdt=etcos⁡(t)−etsin⁡(t)=et(cos⁡(t)−sin⁡(t))\frac{dx}{dt} = e^t \cos(t) - e^t \sin(t) = e^t(\cos(t) - \sin(t)) and dydt=etsin⁡(t)+etcos⁡(t)=et(sin⁡(t)+cos⁡(t))\frac{dy}{dt} = e^t \sin(t) + e^t \cos(t) = e^t(\sin(t) + \cos(t)). Therefore (dxdt)2+(dydt)2=e2t[(cos⁡(t)−sin⁡(t))2+(sin⁡(t)+cos⁡(t))2]=e2t[cos⁡2(t)−2cos⁡(t)sin⁡(t)+sin⁡2(t)+sin⁡2(t)+2sin⁡(t)cos⁡(t)+cos⁡2(t)]=e2t[2cos⁡2(t)+2sin⁡2(t)]=2e2t(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 = e^{2t}[(\cos(t) - \sin(t))^2 + (\sin(t) + \cos(t))^2] = e^{2t}[\cos^2(t) - 2\cos(t)\sin(t) + \sin^2(t) + \sin^2(t) + 2\sin(t)\cos(t) + \cos^2(t)] = e^{2t}[2\cos^2(t) + 2\sin^2(t)] = 2e^{2t}. So the distance is ∫0π2e2tdt=2∫0πetdt=2[et]0π=2(eπ−1)\int_0^{\pi} \sqrt{2e^{2t}} dt = \sqrt{2} \int_0^{\pi} e^t dt = \sqrt{2}[e^t]_0^{\pi} = \sqrt{2}(e^{\pi} - 1). Choice B omits the 2\sqrt{2} factor. Choice C forgets to subtract 1. Choice D incorrectly computes the coefficient.

Question 7

Which expression represents the arc length of the curve y=x∣x∣y = x|x| from x=−2x = -2 to x=2x = 2?

  1. ∫−221+4x2 dx\int_{-2}^2 \sqrt{1 + 4x^2} \, dx (correct answer)
  2. ∫−221+x4 dx\int_{-2}^2 \sqrt{1 + x^4} \, dx
  3. 0
  4. ∫−201−4x2 dx+∫021+4x2 dx\int_{-2}^0 \sqrt{1 - 4x^2} \, dx + \int_0^2 \sqrt{1 + 4x^2} \, dx
Explanation: The function can be written piecewise: y=x2y = x^2 for x≥0x \ge 0 and y=−x2y = -x^2 for x<0x < 0. The derivatives are y′=2xy' = 2x for x>0x > 0 and y′=−2xy' = -2x for x<0x < 0. The arc length integrand is 1+(y′)2\sqrt{1+(y')^2}. For x>0x > 0, this is 1+(2x)2=1+4x2\sqrt{1+(2x)^2} = \sqrt{1+4x^2}. For x<0x < 0, this is 1+(−2x)2=1+4x2\sqrt{1+(-2x)^2} = \sqrt{1+4x^2}. Since the integrand is the same for both pieces, the total arc length is the sum of the integrals over the two parts: ∫−201+4x2 dx+∫021+4x2 dx\int_{-2}^0 \sqrt{1+4x^2} \, dx + \int_0^2 \sqrt{1+4x^2} \, dx. This can be combined into a single integral: ∫−221+4x2 dx\int_{-2}^2 \sqrt{1+4x^2} \, dx.

Question 8

Find the arc length of the polar curve r=e2θr = e^{2\theta} from θ=0\theta=0 to θ=ln⁡(2)\theta=\ln(2).

  1. 352\frac{3\sqrt{5}}{2} (correct answer)
  2. 453\frac{4\sqrt{5}}{3}
  3. 232\sqrt{3}
  4. 323\sqrt{2}
Explanation: The arc length formula for a polar curve is L=∫abr2+(dr/dθ)2 dθL = \int_a^b \sqrt{r^2 + (dr/d\theta)^2} \, d\theta. Here, r=e2θr = e^{2\theta}, so dr/dθ=2e2θdr/d\theta = 2e^{2\theta}. The term under the square root is r2+(dr/dθ)2=(e2θ)2+(2e2θ)2=e4θ+4e4θ=5e4θr^2 + (dr/d\theta)^2 = (e^{2\theta})^2 + (2e^{2\theta})^2 = e^{4\theta} + 4e^{4\theta} = 5e^{4\theta}. The integrand is 5e4θ=5e2θ\sqrt{5e^{4\theta}} = \sqrt{5}e^{2\theta}. The integral for the arc length is L=∫0ln⁡(2)5e2θ dθ=5[12e2θ]0ln⁡(2)L = \int_0^{\ln(2)} \sqrt{5}e^{2\theta} \, d\theta = \sqrt{5}[\frac{1}{2}e^{2\theta}]_0^{\ln(2)}. Evaluating this gives 52(e2ln⁡(2)−e0)=52(eln⁡(4)−1)=52(4−1)=352\frac{\sqrt{5}}{2}(e^{2\ln(2)} - e^0) = \frac{\sqrt{5}}{2}(e^{\ln(4)} - 1) = \frac{\sqrt{5}}{2}(4-1) = \frac{3\sqrt{5}}{2}.

Question 9

A particle's position is described by the parametric equations x(t)=etcos⁡(t)x(t) = e^t \cos(t) and y(t)=etsin⁡(t)y(t) = e^t \sin(t). What is the total distance the particle travels from t=0t=0 to t=ln⁡(3)t=\ln(3)?

  1. 22
  2. 222\sqrt{2} (correct answer)
  3. 33
  4. 323\sqrt{2}
Explanation: The distance traveled is the integral of the speed from t=0t=0 to t=ln⁡(3)t=\ln(3). The speed is s(t)=(x′(t))2+(y′(t))2s(t) = \sqrt{(x'(t))^2 + (y'(t))^2}. First, we find the derivatives: x′(t)=etcos⁡(t)−etsin⁡(t)x'(t) = e^t \cos(t) - e^t \sin(t) and y′(t)=etsin⁡(t)+etcos⁡(t)y'(t) = e^t \sin(t) + e^t \cos(t). Next, we square and add them: (x′(t))2=e2t(cos⁡2(t)−2sin⁡(t)cos⁡(t)+sin⁡2(t))=e2t(1−2sin⁡(t)cos⁡(t))(x'(t))^2 = e^{2t}(\cos^2(t) - 2\sin(t)\cos(t) + \sin^2(t)) = e^{2t}(1 - 2\sin(t)\cos(t)). (y′(t))2=e2t(sin⁡2(t)+2sin⁡(t)cos⁡(t)+cos⁡2(t))=e2t(1+2sin⁡(t)cos⁡(t))(y'(t))^2 = e^{2t}(\sin^2(t) + 2\sin(t)\cos(t) + \cos^2(t)) = e^{2t}(1 + 2\sin(t)\cos(t)). The sum is (x′)2+(y′)2=e2t(1−2sin⁡(t)cos⁡(t)+1+2sin⁡(t)cos⁡(t))=2e2t(x')^2 + (y')^2 = e^{2t}(1 - 2\sin(t)\cos(t) + 1 + 2\sin(t)\cos(t)) = 2e^{2t}. The speed is s(t)=2e2t=2ets(t) = \sqrt{2e^{2t}} = \sqrt{2}e^t. The distance is ∫0ln⁡(3)2et dt=[2et]0ln⁡(3)=2(eln⁡(3)−e0)=2(3−1)=22\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 10

The arc length of the curve y=ln⁡(cos⁡x)y = \ln(\cos x) from x=0x=0 to x=kx=k, where 0<k<π/20 < k < \pi/2, is found to be ln⁡(2+3)\ln(2+\sqrt{3}). What is the value of kk?

  1. π/6\pi/6
  2. π/4\pi/4
  3. π/3\pi/3 (correct answer)
  4. π/2\pi/2
Explanation: First, we set up the arc length integral. The derivative is y′=1cos⁡x(−sin⁡x)=−tan⁡xy' = \frac{1}{\cos x}(-\sin x) = -\tan x. The arc length integrand is 1+(y′)2=1+tan⁡2x=sec⁡2x=∣sec⁡x∣\sqrt{1+(y')^2} = \sqrt{1+\tan^2 x} = \sqrt{\sec^2 x} = |\sec x|. Since 0<k<π/20 < k < \pi/2, sec⁡x\sec x is positive, so the integrand is sec⁡x\sec x. The arc length is L=∫0ksec⁡x dx=[ln⁡∣sec⁡x+tan⁡x∣]0kL = \int_0^k \sec x \, dx = [\ln|\sec x + \tan x|]_0^k. Evaluating the integral gives ln⁡∣sec⁡k+tan⁡k∣−ln⁡∣sec⁡0+tan⁡0∣=ln⁡(sec⁡k+tan⁡k)−ln⁡(1+0)=ln⁡(sec⁡k+tan⁡k)\ln|\sec k + \tan k| - \ln|\sec 0 + \tan 0| = \ln(\sec k + \tan k) - \ln(1+0) = \ln(\sec k + \tan k). We are given that this length is ln⁡(2+3)\ln(2+\sqrt{3}). Therefore, sec⁡k+tan⁡k=2+3\sec k + \tan k = 2+\sqrt{3}. We can test the given angle values. For k=π/3k=\pi/3, sec⁡(π/3)=2\sec(\pi/3) = 2 and tan⁡(π/3)=3\tan(\pi/3) = \sqrt{3}. Their sum is 2+32+\sqrt{3}, so k=π/3k=\pi/3.

Question 11

Which of the following integrals correctly represents the arc length of the astroid x2/3+y2/3=a2/3x^{2/3} + y^{2/3} = a^{2/3} in the first quadrant?

  1. ∫0a1+y2/3x2/3 dx\int_0^a \sqrt{1 + \frac{y^{2/3}}{x^{2/3}}} \, dx
  2. ∫0aa1/3x−1/3 dx\int_0^a a^{1/3}x^{-1/3} \, dx (correct answer)
  3. ∫0a1+x−2/3y2/3 dx\int_0^a \sqrt{1 + x^{-2/3}y^{2/3}} \, dx
  4. ∫0aa2/3x−1/3 dx\int_0^a a^{2/3}x^{-1/3} \, dx
Explanation: To find the arc length, we first need dy/dxdy/dx. Using implicit differentiation on x2/3+y2/3=a2/3x^{2/3} + y^{2/3} = a^{2/3}: 23x−1/3+23y−1/3dydx=0\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0, which gives dydx=−y1/3x1/3\frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}}. The arc length integrand requires 1+(dy/dx)21+(dy/dx)^2. So, 1+(−y1/3x1/3)2=1+y2/3x2/3=x2/3+y2/3x2/31 + (-\frac{y^{1/3}}{x^{1/3}})^2 = 1 + \frac{y^{2/3}}{x^{2/3}} = \frac{x^{2/3} + y^{2/3}}{x^{2/3}}. From the original equation, x2/3+y2/3=a2/3x^{2/3} + y^{2/3} = a^{2/3}, so this simplifies to a2/3x2/3\frac{a^{2/3}}{x^{2/3}}. The arc length element is ds=a2/3x2/3 dx=a1/3x1/3 dx=a1/3x−1/3 dxds = \sqrt{\frac{a^{2/3}}{x^{2/3}}} \, dx = \frac{a^{1/3}}{x^{1/3}} \, dx = a^{1/3}x^{-1/3} \, dx. In the first quadrant, xx ranges from 0 to aa. So the integral is ∫0aa1/3x−1/3 dx\int_0^a a^{1/3}x^{-1/3} \, dx.

Question 12

Which integral represents the distance traveled by a particle from t=0t=0 to t=2t=2 whose position is given by x(t)=t2x(t) = t^2 and y(t)=t3−3ty(t) = t^3 - 3t?

  1. ∫024t2+(3t2−3)2 dt\int_0^2 \sqrt{4t^2 + (3t^2-3)^2} \, dt (correct answer)
  2. ∫02t4+(t3−3t)2 dt\int_0^2 \sqrt{t^4 + (t^3-3t)^2} \, dt
  3. ∫02(2t+3t2−3) dt\int_0^2 (2t + 3t^2 - 3) \, dt
  4. ∫022t+3t2−3 dt\int_0^2 \sqrt{2t + 3t^2 - 3} \, dt
Explanation: The distance traveled is the integral of the speed, which is s(t)=(x′(t))2+(y′(t))2s(t) = \sqrt{(x'(t))^2 + (y'(t))^2}. First, we find the derivatives of the position functions: x′(t)=2tx'(t) = 2t and y′(t)=3t2−3y'(t) = 3t^2 - 3. Next, we square these derivatives: (x′(t))2=(2t)2=4t2(x'(t))^2 = (2t)^2 = 4t^2 and (y′(t))2=(3t2−3)2(y'(t))^2 = (3t^2 - 3)^2. The speed is s(t)=4t2+(3t2−3)2s(t) = \sqrt{4t^2 + (3t^2-3)^2}. The total distance traveled is the integral of the speed over the given interval, which is ∫024t2+(3t2−3)2 dt\int_0^2 \sqrt{4t^2 + (3t^2-3)^2} \, dt. Note that the expression under the radical simplifies to 9t4−14t2+9\sqrt{9t^4 - 14t^2 + 9}, which is not a perfect square, so the question only asks for the setup.

Question 13

A particle's position is given by x(t)=∫0tcos⁡(u2) dux(t) = \int_0^t \cos(u^2) \, du and y(t)=∫0tsin⁡(u2) duy(t) = \int_0^t \sin(u^2) \, du. What is the total distance traveled by the particle from t=0t=0 to t=5t=5?

  1. 0
  2. 1
  3. 5 (correct answer)
  4. 525\sqrt{2}
Explanation: To find the distance traveled, we need to integrate the particle's speed. The speed is s(t)=(x′(t))2+(y′(t))2s(t) = \sqrt{(x'(t))^2 + (y'(t))^2}. Using the Fundamental Theorem of Calculus Part 1, we can find the derivatives of the position functions: x′(t)=cos⁡(t2)x'(t) = \cos(t^2) and y′(t)=sin⁡(t2)y'(t) = \sin(t^2). Now, we calculate the speed: s(t)=(cos⁡(t2))2+(sin⁡(t2))2=cos⁡2(t2)+sin⁡2(t2)s(t) = \sqrt{(\cos(t^2))^2 + (\sin(t^2))^2} = \sqrt{\cos^2(t^2) + \sin^2(t^2)}. Using the trigonometric identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1, the speed simplifies to s(t)=1=1s(t) = \sqrt{1} = 1. The particle travels at a constant speed of 1. The total distance traveled from t=0t=0 to t=5t=5 is ∫051 dt=[t]05=5\int_0^5 1 \, dt = [t]_0^5 = 5.

Question 14

Find the arc length of the curve x=y44+18y2x = \frac{y^4}{4} + \frac{1}{8y^2} from y=1y=1 to y=2y=2.

  1. 33/1633/16
  2. 123/32123/32 (correct answer)
  3. 125/32125/32
  4. 31/1631/16
Explanation: Since xx is given as a function of yy, we integrate with respect to yy. The formula is L=∫cd1+(dx/dy)2 dyL = \int_c^d \sqrt{1 + (dx/dy)^2} \, dy. First, find the derivative: dxdy=y3−28y3=y3−14y3\frac{dx}{dy} = y^3 - \frac{2}{8y^3} = y^3 - \frac{1}{4y^3}. Next, compute 1+(dx/dy)21 + (dx/dy)^2: 1+(y3−14y3)2=1+(y6−12+116y6)=y6+12+116y61 + (y^3 - \frac{1}{4y^3})^2 = 1 + (y^6 - \frac{1}{2} + \frac{1}{16y^6}) = y^6 + \frac{1}{2} + \frac{1}{16y^6}. This is a perfect square: (y3+14y3)2(y^3 + \frac{1}{4y^3})^2. The integral is ∫12(y3+14y3)2 dy=∫12(y3+14y3) dy\int_1^2 \sqrt{(y^3 + \frac{1}{4y^3})^2} \, dy = \int_1^2 (y^3 + \frac{1}{4y^3}) \, dy. Evaluating the integral: [y44−18y2]12=(164−18(4))−(14−18(1))=(4−132)−(28−18)=4−132−18=4−132−432=4−532=128−532=12332[\frac{y^4}{4} - \frac{1}{8y^2}]_1^2 = (\frac{16}{4} - \frac{1}{8(4)}) - (\frac{1}{4} - \frac{1}{8(1)}) = (4 - \frac{1}{32}) - (\frac{2}{8} - \frac{1}{8}) = 4 - \frac{1}{32} - \frac{1}{8} = 4 - \frac{1}{32} - \frac{4}{32} = 4 - \frac{5}{32} = \frac{128-5}{32} = \frac{123}{32}.

Question 15

A wire is shaped like the curve y=cosh⁡(x)y = \cosh(x) from x=−ln⁡(2)x = -\ln(2) to x=ln⁡(2)x = \ln(2). What is the length of the wire?

  1. 32\frac{3}{2} (correct answer)
  2. 54\frac{5}{4}
  3. 34\frac{3}{4}
  4. 52\frac{5}{2}
Explanation: The length of the wire is the arc length of the curve y=cosh⁡(x)y=\cosh(x). The derivative is y′=sinh⁡(x)y' = \sinh(x). The arc length integrand is 1+(y′)2=1+sinh⁡2(x)\sqrt{1+(y')^2} = \sqrt{1+\sinh^2(x)}. Using the identity cosh⁡2(x)−sinh⁡2(x)=1\cosh^2(x) - \sinh^2(x) = 1, we have 1+sinh⁡2(x)=cosh⁡2(x)1+\sinh^2(x) = \cosh^2(x). So the integrand is cosh⁡2(x)=cosh⁡(x)\sqrt{\cosh^2(x)} = \cosh(x) (since cosh⁡(x)\cosh(x) is always positive). The length is L=∫−ln⁡(2)ln⁡(2)cosh⁡(x) dx=[sinh⁡(x)]−ln⁡(2)ln⁡(2)=sinh⁡(ln⁡(2))−sinh⁡(−ln⁡(2))L = \int_{-\ln(2)}^{\ln(2)} \cosh(x) \, dx = [\sinh(x)]_{-\ln(2)}^{\ln(2)} = \sinh(\ln(2)) - \sinh(-\ln(2)). Since sinh⁡(x)\sinh(x) is an odd function, sinh⁡(−ln⁡(2))=−sinh⁡(ln⁡(2))\sinh(-\ln(2)) = -\sinh(\ln(2)), so the length is 2sinh⁡(ln⁡(2))2\sinh(\ln(2)). Using the definition sinh⁡(z)=(ez−e−z)/2\sinh(z) = (e^z - e^{-z})/2, we have sinh⁡(ln⁡(2))=(eln⁡(2)−e−ln⁡(2))/2=(2−1/2)/2=(3/2)/2=3/4\sinh(\ln(2)) = (e^{\ln(2)} - e^{-\ln(2)})/2 = (2 - 1/2)/2 = (3/2)/2 = 3/4. The total length is 2⋅(3/4)=3/22 \cdot (3/4) = 3/2.

Question 16

A curve is defined by x=sin⁡3(t)x = \sin^3(t) and y=cos⁡3(t)y = \cos^3(t) for 0≤t≤π/20 \le t \le \pi/2. What is the length of the curve?

  1. 1
  2. 3/4
  3. 2
  4. 3/2 (correct answer)
Explanation: When you encounter a parametric curve like this, you need the arc length formula for parametric equations: L=∫ab(dxdt)2+(dydt)2 dtL = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt First, find the derivatives. For x=sin⁡3(t)x = \sin^3(t): dxdt=3sin⁡2(t)cos⁡(t)\frac{dx}{dt} = 3\sin^2(t)\cos(t). For y=cos⁡3(t)y = \cos^3(t): dydt=−3cos⁡2(t)sin⁡(t)\frac{dy}{dt} = -3\cos^2(t)\sin(t). Now calculate the expression under the square root: (dxdt)2+(dydt)2=9sin⁡4(t)cos⁡2(t)+9cos⁡4(t)sin⁡2(t)=9sin⁡2(t)cos⁡2(t)[sin⁡2(t)+cos⁡2(t)]=9sin⁡2(t)cos⁡2(t)\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 9\sin^4(t)\cos^2(t) + 9\cos^4(t)\sin^2(t) = 9\sin^2(t)\cos^2(t)[\sin^2(t) + \cos^2(t)] = 9\sin^2(t)\cos^2(t) Taking the square root: 9sin⁡2(t)cos⁡2(t)=3sin⁡(t)cos⁡(t)\sqrt{9\sin^2(t)\cos^2(t)} = 3\sin(t)\cos(t) (since both sine and cosine are positive on [0,π/2][0, \pi/2]). The arc length becomes: L=∫0π/23sin⁡(t)cos⁡(t) dt=32∫0π/2sin⁡(2t) dt=32[−12cos⁡(2t)]0π/2=34[1−(−1)]=32L = \int_0^{\pi/2} 3\sin(t)\cos(t) \, dt = \frac{3}{2}\int_0^{\pi/2} \sin(2t) \, dt = \frac{3}{2}\left[-\frac{1}{2}\cos(2t)\right]_0^{\pi/2} = \frac{3}{4}[1-(-1)] = \frac{3}{2} The answer is D) 3/2. Choice A) 1 likely comes from forgetting the factor of 3 in the derivatives. Choice B) 3/4 results from missing the final factor of 2 when evaluating the definite integral. Choice C) 2 might come from computational errors in the trigonometric integration. Strategy tip: Always factor out constants early in parametric arc length problems, and remember that sin⁡(t)cos⁡(t)=12sin⁡(2t)\sin(t)\cos(t) = \frac{1}{2}\sin(2t) is a key identity for these integrals.

Question 17

The distance a particle travels along the curve y=f(x)y=f(x) from x=ax=a to x=bx=b is given by ∫ab1+sec⁡4(x) dx\int_a^b \sqrt{1+\sec^4(x)} \, dx. What could be f′(x)f'(x)?

  1. sec⁡(x)\sec(x)
  2. sec⁡(x)tan⁡(x)\sec(x)\tan(x)
  3. tan⁡(x)\tan(x)
  4. sec⁡2(x)\sec^2(x) (correct answer)
Explanation: When you encounter an arc length integral, you're looking at the formula ∫ab1+[f′(x)]2 dx\int_a^b \sqrt{1 + [f'(x)]^2} \, dx. This formula gives the distance traveled along a curve from point to point. The key is recognizing that whatever appears under the square root as 1+(something)21 + (\text{something})^2 tells you that "something" equals f′(x)f'(x). Looking at the given integral ∫ab1+sec⁡4(x) dx\int_a^b \sqrt{1+\sec^4(x)} \, dx, you need to identify what derivative, when squared, gives sec⁡4(x)\sec^4(x). Since sec⁡4(x)=(sec⁡2(x))2\sec^4(x) = (\sec^2(x))^2, you have f′(x)=sec⁡2(x)f'(x) = \sec^2(x), which is answer D. Let's check why the other options don't work. For A, if f′(x)=sec⁡(x)f'(x) = \sec(x), then [f′(x)]2=sec⁡2(x)[f'(x)]^2 = \sec^2(x), giving you 1+sec⁡2(x)\sqrt{1 + \sec^2(x)}, not 1+sec⁡4(x)\sqrt{1 + \sec^4(x)}. For B, if f′(x)=sec⁡(x)tan⁡(x)f'(x) = \sec(x)\tan(x), then [f′(x)]2=sec⁡2(x)tan⁡2(x)[f'(x)]^2 = \sec^2(x)\tan^2(x), which doesn't match either. For C, if f′(x)=tan⁡(x)f'(x) = \tan(x), then [f′(x)]2=tan⁡2(x)[f'(x)]^2 = \tan^2(x), again not matching. The strategy here is pattern recognition: whenever you see an arc length integral, immediately identify the expression under the square root and work backwards to find what derivative would produce that form when squared. Remember that arc length problems always follow the template 1+[f′(x)]2\sqrt{1 + [f'(x)]^2}.

Question 18

A particle moves with position (x(t),y(t))(x(t), y(t)) such that dx/dt=1−1/t2dx/dt = 1 - 1/t^2 and dy/dt=2/tdy/dt = 2/t for t≥1t \ge 1. Find the total distance traveled from t=1t=1 to t=e2t=e^2.

  1. e2−e−2e^2 - e^{-2} (correct answer)
  2. e2+e−2e^2 + e^{-2}
  3. e2−e−2−2e^2 - e^{-2} - 2
  4. 2
Explanation: The distance traveled is the integral of the speed. The speed is s(t)=(dx/dt)2+(dy/dt)2s(t) = \sqrt{(dx/dt)^2 + (dy/dt)^2}. We compute: (1−1/t2)2+(2/t)2=1−2/t2+1/t4+4/t2=1+2/t2+1/t4=(1+1/t2)2(1 - 1/t^2)^2 + (2/t)^2 = 1 - 2/t^2 + 1/t^4 + 4/t^2 = 1 + 2/t^2 + 1/t^4 = (1 + 1/t^2)^2. Since t≥1t \ge 1, we have s(t)=1+1/t2s(t) = 1 + 1/t^2. The total distance is D=∫1e2(1+1t2) dt=[t−1t]1e2=(e2−1e2)−(1−1)=e2−e−2D = \int_1^{e^2} (1 + \frac{1}{t^2}) \, dt = [t - \frac{1}{t}]_1^{e^2} = (e^2 - \frac{1}{e^2}) - (1 - 1) = e^2 - e^{-2}.

Question 19

The velocity of a particle moving in the xy-plane is v⃗(t)=⟨sin⁡(t),tcos⁡(t)⟩\vec{v}(t) = \langle \sin(t), t \cos(t) \rangle. Which integral represents the total distance traveled from t=0t=0 to t=πt=\pi?

  1. ∫0πsin⁡2(t)+t2cos⁡2(t) dt\int_0^{\pi} \sqrt{\sin^2(t) + t^2 \cos^2(t)} \, dt (correct answer)
  2. ∫0π(sin⁡(t)+tcos⁡(t)) dt\int_0^{\pi} (\sin(t) + t \cos(t)) \, dt
  3. ∫0π1+(cos⁡(t)−tsin⁡(t))2 dt\int_0^{\pi} \sqrt{1 + (\cos(t) - t \sin(t))^2} \, dt
  4. ∫0πcos⁡2(t)+(cos⁡(t)−tsin⁡(t))2 dt\int_0^{\pi} \sqrt{\cos^2(t) + (\cos(t) - t \sin(t))^2} \, dt
Explanation: The distance traveled is the integral of the speed. The speed is the magnitude of the velocity vector: s(t)=∣∣v⃗(t)∣∣=(vx(t))2+(vy(t))2s(t) = ||\vec{v}(t)|| = \sqrt{(v_x(t))^2 + (v_y(t))^2}. The components of the velocity are given as vx(t)=sin⁡(t)v_x(t) = \sin(t) and vy(t)=tcos⁡(t)v_y(t) = t \cos(t). Squaring and adding these components gives (sin⁡(t))2+(tcos⁡(t))2=sin⁡2(t)+t2cos⁡2(t)(\sin(t))^2 + (t \cos(t))^2 = \sin^2(t) + t^2 \cos^2(t). The speed is s(t)=sin⁡2(t)+t2cos⁡2(t)s(t) = \sqrt{\sin^2(t) + t^2 \cos^2(t)}. Integrating the speed from t=0t=0 to t=πt=\pi gives the total distance traveled. Therefore, the correct integral is ∫0πsin⁡2(t)+t2cos⁡2(t) dt\int_0^{\pi} \sqrt{\sin^2(t) + t^2 \cos^2(t)} \, dt.

Question 20

Find the length of one arch of the cycloid given by x=a(t−sin⁡t)x = a(t - \sin t) and y=a(1−cos⁡t)y = a(1 - \cos t) for 0≤t≤2π0 \le t \le 2\pi, where a>0a>0.

  1. 4a4a
  2. 6a6a
  3. 8a8a (correct answer)
  4. 2πa2\pi a
Explanation: First, find the derivatives with respect to tt: x′(t)=a(1−cos⁡t)x'(t) = a(1 - \cos t) and y′(t)=a(sin⁡t)y'(t) = a(\sin t). The arc length integrand is (x′)2+(y′)2\sqrt{(x')^2 + (y')^2}. So, (x′)2+(y′)2=a2(1−cos⁡t)2+a2sin⁡2t=a2(1−2cos⁡t+cos⁡2t+sin⁡2t)=a2(2−2cos⁡t)(x')^2 + (y')^2 = a^2(1 - \cos t)^2 + a^2\sin^2 t = a^2(1 - 2\cos t + \cos^2 t + \sin^2 t) = a^2(2 - 2\cos t). Using the half-angle identity 1−cos⁡t=2sin⁡2(t/2)1 - \cos t = 2\sin^2(t/2), this becomes a2(4sin⁡2(t/2))a^2(4\sin^2(t/2)). The integrand is 4a2sin⁡2(t/2)=2a∣sin⁡(t/2)∣\sqrt{4a^2\sin^2(t/2)} = 2a|\sin(t/2)|. For 0≤t≤2π0 \le t \le 2\pi, 0≤t/2≤π0 \le t/2 \le \pi, so sin⁡(t/2)≥0\sin(t/2) \ge 0. The length is L=∫02π2asin⁡(t/2) dt=2a[−2cos⁡(t/2)]02π=−4a[cos⁡(π)−cos⁡(0)]=−4a[−1−1]=8aL = \int_0^{2\pi} 2a\sin(t/2) \, dt = 2a[-2\cos(t/2)]_0^{2\pi} = -4a[\cos(\pi) - \cos(0)] = -4a[-1 - 1] = 8a.