Arc Length of Smooth, Planar Curve - AP Calculus BC

1. Two particles move for $0\le t\le1$: $\mathbf r_1=\langle t,t\rangle$, $\mathbf r_2=\langle t,2t\rangle$. Which travels farther?
   Answer: Particle 2, because its speed is larger for all $t$. 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.
2. For $\mathbf{r}(t)=\langle \ln t,\,t^2\rangle$ on $1\le t\le e$, which integral equals the arc length?
   Answer: $\displaystyle \int_1^{e}\sqrt{\left(\frac{1}{t}\right)^2+(2t)^2}\,dt$ 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.
3. A particle's position is $\mathbf r(t)=\langle \sin(2t),\,\cos(2t)\rangle$ for $0\le t\le\pi$. What distance is traveled?
   Answer: $\displaystyle 2\pi$ The position vector r(t) = ⟨sin(2t), cos(2t)⟩ gives us x'(t) = 2cos(2t) and y'(t) = -2sin(2t). The speed is √[(2cos(2t))² + (-2sin(2t))²] = √[4cos²(2t) + 4sin²(2t)] = √[4] = 2. Over the interval [0, π], the distance traveled is ∫₀^π 2 dt = 2π. Choice D shows the squared terms but doesn't simplify using the Pythagorean identity, while choice E incorrectly adds the derivatives linearly. The transferable principle is to recognize when trigonometric identities can simplify the speed before integrating.
4. A drone follows $\mathbf r(t)=\langle e^t,\,e^{-t}\rangle$ for $0\le t\le1$. Which integral gives its distance traveled?
   Answer: $\displaystyle \int_{0}^{1}\sqrt{(e^t)^2+(-e^{-t})^2}\,dt$ For the parametric curve $r(t) = \langle e^t, e^{-t} \rangle$, we calculate distance using derivatives $\frac{dx}{dt} = e^t$ and $\frac{dy}{dt} = -e^{-t}$. The speed formula gives $\sqrt{[(e^t)^2 + (-e^{-t})^2]} = \sqrt{[e^{2t} + e^{-2t}]}$. Choice C omits the negative sign on the y-derivative, but since we square it, $(-e^{-t})^2 = e^{-2t}$, 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 $\frac{dy}{dt}$ matters for direction but not for speed magnitude.
5. A particle has velocity $\mathbf v(t)=\langle 6t,\,-8\rangle$ for $0\le t\le1$; which integral gives distance traveled?​
   Answer: $\displaystyle \int_{0}^{1}\sqrt{(6t)^2+(-8)^2}\,dt$ 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.
6. A particle moves with $x(t)=2\sin t$ and $y(t)=2\cos t$ for $0\le t\le\tfrac{\pi}{2}$. What distance is traveled?
   Answer: $\pi$ This particle moves along a circle of radius 2, with velocity components $\frac{dx}{dt} = 2\cos(t)$ and $\frac{dy}{dt} = -2\sin(t)$. The speed is $\sqrt{(2\cos t)^2 + (-2\sin t)^2} = \sqrt{4\cos^2 t + 4\sin^2 t} = 2\sqrt{\cos^2 t + \sin^2 t} = 2$. With constant speed 2, the distance from t=0 to t=π/2 is $\int_0^{\pi/2} 2 \, dt = 2(\pi/2) = \pi$. Choice B ($2\pi$) 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=2 $, $ \omega=1 $).
7. For $\mathbf r(t)=\langle t,\,t^2\rangle$ on $0\le t\le3$, which integral equals the arc length of the path?
   Answer: $\displaystyle \int_{0}^{3}\sqrt{1+(2t)^2}\,dt$ For the parametric curve r(t) = ⟨t, t²⟩, arc length requires derivatives dx/dt = 1 and dy/dt = 2t. The arc length integral is ∫₀³ √[1² + (2t)²] dt = ∫₀³ √[1 + 4t²] dt. Choice A incorrectly uses √[1 + 2t] instead of √[1 + (2t)²], missing the crucial step of squaring the derivative. Choice C adds without the square root, giving an area-like quantity instead of length. Remember: arc length always uses the formula ∫√[(dx/dt)² + (dy/dt)²] dt, where both derivatives must be squared before adding.
8. Two particles: $\mathbf r_1(t)=\langle t,2t\rangle$, $\mathbf r_2(t)=\langle 2t,t\rangle$ for $0\le t\le1$; compare distances traveled.​
   Answer: They travel the same distance. 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.
9. A robot follows $x(t)=\cos t,\ y(t)=\sin t$ for $0\le t\le \pi$; what distance does it travel?
   Answer: $\displaystyle \pi$ 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=cos t, y=sin t from 0 to π, the speed is 1, so the distance is ∫ from 0 to π of 1 dt = π, matching choice A. A tempting distractor is choice E, 2π, which is the full circumference but ignores the half-interval. Always compute the magnitude inside the integral first to capture the instantaneous speed. This magnitude-first strategy ensures accurate distance for any parametric path.
10. For $x(t)=t$, $y(t)=\ln t$ on $1\le t\le e$, which is the correct arc length integral?​
    Answer: $\displaystyle \int_{1}^{e}\sqrt{1+\left(\frac{1}{t}\right)^2}\,dt$ 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.
11. A particle moves on $\mathbf{r}(t)=\langle t^2,\,t^3\rangle$ for $0\le t\le1$; which integral gives its distance traveled?
    Answer: $\displaystyle \int_0^1 \sqrt{(2t)^2+(3t^2)^2}\,dt$ 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 or vectorial. For $r(t) = \langle t^2, t^3 \rangle$, the velocity components are $2t$ and $3t^2$, so the integral is $\int_0^1 \sqrt{(2t)^2 + (3t^2)^2} \, dt$, matching choice B. A tempting distractor is choice D, which integrates the components separately and then takes the magnitude, giving the net displacement instead of the path length. Always compute the magnitude inside the integral first to capture the instantaneous speed. This magnitude-first strategy ensures accurate distance for any parametric path.
12. A particle moves on $x(t)=t^2-1$, $y(t)=2t$ for $-1\le t\le2$. Which integral gives its distance?
    Answer: $\displaystyle \int_{-1}^{2}\sqrt{(2t)^2+2^2}\,dt$ Given x(t) = t² - 1 and y(t) = 2t, we need to find the distance traveled from t = -1 to t = 2. Taking derivatives: x'(t) = 2t and y'(t) = 2. The arc length integral is ∫₋₁² √[(2t)² + 2²]dt = ∫₋₁² √[4t² + 4]dt. Choice A incorrectly adds the derivatives (2t + 2) without using the magnitude formula, which would give a signed result rather than total distance. The consistent strategy is to always square the component derivatives, add them, take the square root, then integrate for arc length.
13. For $x(t)=e^t$, $y(t)=e^{-t}$ on $0\le t\le1$, which integral represents the arc length?
    Answer: $\displaystyle \int_{0}^{1}\sqrt{e^{2t}+e^{-2t}}\,dt$ 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.
14. A particle follows $\mathbf{r}(t)=\langle 3\cos t,\,4\sin t\rangle$ for $0\le t\le\frac{\pi}{2}$; which integral gives distance?
    Answer: $\displaystyle \int_0^{\pi/2}\sqrt{9\sin^2 t+16\cos^2 t}\,dt$ 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.
15. A skater moves on $\mathbf{r}(t)=\langle 4t,\,3t\rangle$ for $0\le t\le5$. What total distance is traveled?
    Answer: $25$ The skater moves along the line $r(t) = \langle 4t, 3t \rangle$ with velocity components $\frac{dx}{dt} = 4$ and $\frac{dy}{dt} = 3$. The constant speed is $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$. Over the time interval $[0,5]$, the distance traveled is $\int_0^5 5 \, dt = 5 \times 5 = 25$. Choice D gives just the speed (5), not the total distance. Choice E incorrectly calculates $25\sqrt{7}$, perhaps confusing the speed calculation. For linear motion with constant velocity, distance equals speed times time: here, 5 units/time $\times$ 5 time units = 25 units.
16. A particle moves on $\mathbf{r}(t)=\langle e^t,\,e^{-t}\rangle$ for $-1\le t\le1$; which integral equals its distance traveled?
    Answer: $\displaystyle \int_{-1}^{1}\sqrt{\left(e^t\right)^2+\left(-e^{-t}\right)^2}\,dt$ 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.
17. If $\mathbf r(t)=\langle \sin t,\,\cos t\rangle$ for $0\le t\le \tfrac{\pi}{2}$, what total distance is traveled?
    Answer: $\tfrac{\pi}{2}$ The skill is computing the numerical distance traveled for a quarter-circle parametric path. Distance is the integral of velocity magnitude, not signed displacement, ensuring full arc measurement. For r(t) = $\langle \sin t, \cos t \rangle$, dx/dt = $\cos t$ and dy/dt = $-\sin t$, speed = $1$, so from $0$ to $\tfrac{\pi}{2}$, distance = $\tfrac{\pi}{2}$. Choice A tempts by integrating cos t - sin t without magnitude, yielding a smaller value. Always use magnitude-first: find |v(t)| from derivatives, integrate to get total distance without sign issues.
18. For $\mathbf r(t)=\langle e^t,\,t^3\rangle$ on $0\le t\le 1$, which expression equals the arc length?
    Answer: $\displaystyle \int_{0}^{1}\sqrt{(e^t)^2+(3t^2)^2}\,dt$ This question assesses the ability to compute the arc length of a parametric curve, which represents the total distance traveled by a particle. Distance traveled is found by integrating the speed, the magnitude of the velocity vector $\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$, not the signed displacement which might underestimate by ignoring path curvature. For $\mathbf r(t) = \langle e^t, t^3 \rangle$, dx/dt = $e^t$ and dy/dt = $3t^2$, leading to the integral $\int_0^1 \sqrt{(e^t)^2 + (3t^2)^2} \, dt$. Choice B is a tempting distractor as it uses $(t^3)^2$ instead of $(3t^2)^2$, confusing position with derivative. Always apply the magnitude-first strategy: derive velocity components, square them, sum, take square root for instantaneous speed, then integrate for total distance.
19. A particle travels with $\mathbf r(t)=\langle t,\,t\rangle$ for $0\le t\le3$. What is the distance traveled?
    Answer: $\displaystyle 3\sqrt2$ For the linear parametric curve r(t) = ⟨t, t⟩, we have x'(t) = 1 and y'(t) = 1. The speed is √[1² + 1²] = √2. The distance traveled from t = 0 to t = 3 is ∫₀³ √2 dt = 3√2. Choice D shows (1 + 1) = 2 as the integrand, incorrectly adding the derivatives without using the magnitude formula—this would give 6 instead of 3√2. The key insight is that even for simple linear motion, we must use the speed formula √[(dx/dt)² + (dy/dt)²] to find the correct distance.
20. For $x(t)=t^3$ and $y(t)=t^2$ on $-1\le t\le1$, which expression equals the distance traveled?
    Answer: $\displaystyle \int_{-1}^{1}\sqrt{(3t^2)^2+(2t)^2}\,dt$ 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|.

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