Arc Lengths of Curves: Parametric Equations - AP Calculus BC

1. For $x=\cos(2t)$ and $y=\sin(3t)$ on $0\le t\le\pi$, which integral sets up arc length?
   Answer: $\displaystyle \int_{0}^{\pi}\sqrt{(-2\sin(2t))^2+(3\cos(3t))^2}\,dt$ This problem tests the skill of finding the arc length of a curve defined by parametric equations. Given $x = \cos(2t)$ and $y = \sin(3t)$, find $\frac{dx}{dt} = -2 \sin(2t)$ and $\frac{dy}{dt} = 3 \cos(3t)$. The integrand is $\sqrt{((-2 \sin(2t))^2 + (3 \cos(3t))^2)}$. Integrate from t = 0 to t = $\pi$ to compute the arc length. A tempting distractor like choice E replaces cos(3t) with sin(3t), which is incorrect as it does not use the proper derivative for y. Always remember to use the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$ in the arc length formula and ensure the integration limits correspond to the given parameter range.
2. For $x(t)=\sec t$, $y(t)=\tan t$ on $0\le t\le\frac{\pi}{4}$, which integral sets up the arc length?
   Answer: $\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec t\tan t)^2+(\sec^2 t)^2}\,dt$ 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.
3. For $x=t^2+1$ and $y=3t-t^2$ on $0\le t\le2$, which integral gives the arc length?
   Answer: $\displaystyle \int_{0}^{2}\sqrt{(2t)^2+(3-2t)^2}\,dt$ This problem asks for the arc length of a parametric curve, which requires the formula $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. For $x = t^2 + 1$, we have $\frac{dx}{dt} = 2t$. For $y = 3t - t^2$, we have $\frac{dy}{dt} = 3 - 2t$. Substituting into the arc length formula gives $\int_0^2 \sqrt{(2t)^2 + (3-2t)^2} dt$. 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 compute both derivatives with respect to the parameter, square them, add them, then take the square root.
4. For $x=t^2-2t$ and $y=t^2+2t$, $-1\le t\le 2$, which integral correctly sets up the arc length?
   Answer: $\displaystyle \int_{-1}^{2}\sqrt{(2t-2)^2+(2t+2)^2}\,dt$ This problem applies the parametric arc length formula $\int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. From $x = t^2 - 2t$ and $y = t^2 + 2t$, we find $\frac{dx}{dt} = 2t - 2$ and $\frac{dy}{dt} = 2t + 2$. The arc length integral becomes $\int_{-1}^2 \sqrt{(2t-2)^2 + (2t+2)^2} dt$. Choice B incorrectly uses the original functions $t^2 - 2t$ and $t^2 + 2t$ rather than their derivatives, missing the fundamental concept that arc length depends on the rate of change. Remember: parametric arc length always requires differentiating first, then squaring and adding under the square root.
5. A curve is $x=t-\ln t$, $y=\sqrt{t}$ for $1 \le t \le 4$; which integral gives its arc length?
   Answer: $\displaystyle \int_{1}^{4}\sqrt{\left(1-\frac{1}{t}\right)^2+\left(\frac{1}{2\sqrt{t}}\right)^2}\,dt$ This problem tests the skill of finding the arc length of a curve defined by parametric equations. For $x = t - \ln t$ and $y = \sqrt{t}$, calculate $dx/dt = 1 - \frac{1}{t}$ and $dy/dt = \frac{1}{2\sqrt{t}}$. The integrand is $\sqrt{ \left(1 - \frac{1}{t}\right)^2 + \left( \frac{1}{2\sqrt{t}} \right)^2 }$. Integrate from t = 1 to t = 4 to find the arc length. A tempting distractor like choice C uses $(t - \ln t)^2 + (\sqrt{t})^2$ inside the square root, substituting the functions themselves rather than their derivatives. Always remember to use the derivatives $dx/dt$ and $dy/dt$ in the arc length formula and ensure the integration limits correspond to the given parameter range.
6. For $x=e^{t}$ and $y=e^{-t}$, $0\le t\le 1$, which integral sets up the arc length?
   Answer: $\displaystyle \int_{0}^{1}\sqrt{(e^{t})^2+(-e^{-t})^2}\,dt$ To find the arc length of this parametric curve, we apply the formula $\int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. Given $x = e^t$ and $y = e^{-t}$, we calculate $\frac{dx}{dt} = e^t$ and $\frac{dy}{dt} = -e^{-t}$. Substituting into the arc length formula yields $\int_0^1 \sqrt{(e^t)^2 + (-e^{-t})^2} dt$. Choice D incorrectly removes the squares from inside the square root, which would give an invalid expression since $e^t - e^{-t}$ could be negative. The essential principle is that arc length measures the magnitude of velocity, requiring both derivatives to be squared before summing.
7. A parametric curve is $x=\sqrt{t}$, $y=\frac{1}{3}t^{3/2}$ for $0\le t\le 9$; which arc length setup is correct?
   Answer: $\displaystyle \int_{0}^{9}\sqrt{\left(\frac{1}{2\sqrt{t}}\right)^2+\left(\frac{1}{2}\sqrt{t}\right)^2}\,dt$ For parametric arc length, we apply $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. With $x = \sqrt{t}$ and $y = \frac{1}{3}t^{3/2}$, the derivatives are $\frac{dx}{dt} = \frac{1}{2\sqrt{t}} \text{ and } \frac{dy}{dt} = \frac{1}{2}t^{1/2} = \frac{1}{2}\sqrt{t}$. This yields $\int_0^9 \sqrt{(\frac{1}{2\sqrt{t}})^2 + (\frac{1}{2}\sqrt{t})^2} dt$. Choice E starts at $t = 1$ instead of $t = 0$, which would miss part of the curve and give an incorrect arc length. Always use the exact parameter bounds given in the problem.
8. A particle moves with $x=\cos t$ and $y=2\sin t$ for $0\le t\le\pi$. Which integral is its path length?
   Answer: $\displaystyle \int_{0}^{\pi}\sqrt{(-\sin t)^2+(2\cos t)^2}\,dt$ This problem requires finding the arc length of a parametric path using the formula $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. Given $x = \cos t$, we compute $\frac{dx}{dt} = -\sin t$. Given $y = 2\sin t$, we compute $\frac{dy}{dt} = 2\cos t$. Plugging these derivatives into the arc length formula yields $\int_0^\pi \sqrt{(-\sin t)^2 + (2\cos t)^2} dt$. Choice B incorrectly uses the original functions instead of their derivatives, which would not measure the instantaneous rate of change needed for arc length. Remember that parametric arc length always requires derivatives of both coordinate functions with respect to the parameter.
9. For $x=t\sqrt{t}$ and $y=\sqrt{t}$ on $1\le t\le4$, which integral sets up the arc length?
   Answer: $\displaystyle \int_{1}^{4}\sqrt{\left(\frac{3}{2}\sqrt{t}\right)^2+\left(\frac{1}{2\sqrt{t}}\right)^2}\,dt$ To find the arc length of this parametric curve, we apply the formula $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. For $x = t\sqrt{t} = t^{3/2}$, we differentiate to get $\frac{dx}{dt} = \frac{3}{2}t^{1/2} = \frac{3}{2}\sqrt{t}$. For $y = \sqrt{t} = t^{1/2}$, we differentiate to get $\frac{dy}{dt} = \frac{1}{2}t^{-1/2} = \frac{1}{2\sqrt{t}}$. Substituting gives $\int_1^4 \sqrt{\left(\frac{3}{2}\sqrt{t}\right)^2 + \left(\frac{1}{2\sqrt{t}}\right)^2} dt$. Choice E incorrectly uses the original functions rather than their derivatives, missing the fundamental requirement that arc length measures instantaneous rate of change. The key to parametric arc length is always differentiating first, then constructing the integral.
10. For $x=t^3$ and $y=\ln(t)$ on $1\le t\le e$, which integral represents the arc length?
    Answer: $\displaystyle \int_{1}^{e}\sqrt{\left(3t^2\right)^2+\left(\frac{1}{t}\right)^2}\,dt$ To find the arc length of this parametric curve, we apply the standard formula $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. Given $x = t^3$ and $y = \ln(t)$, we differentiate to get $\frac{dx}{dt} = 3t^2$ and $\frac{dy}{dt} = \frac{1}{t}$. Substituting these derivatives yields $\int_1^e \sqrt{(3t^2)^2 + (\frac{1}{t})^2} dt$. Choice C incorrectly pairs the derivative of $x$ with the original function $y = \ln t$ instead of its derivative, which would not give the correct arc length. When computing parametric arc lengths, always ensure both components under the square root are derivatives with respect to the parameter.
11. A particle travels on $x=\cos t$, $y=\sin t$, $0\le t\le \pi$; which integral gives the path length?
    Answer: $\displaystyle \int_{0}^{\pi}\sqrt{(-\sin t)^2+(\cos t)^2}\,dt$ This problem requires finding the arc length of a parametric curve using the formula $\int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. For $x = \cos t$ and $y = \sin t$, we compute $\frac{dx}{dt} = -\sin t$ and $\frac{dy}{dt} = \cos t$. The arc length integral becomes $\int_0^\pi \sqrt{(-\sin t)^2 + (\cos t)^2} dt$, which simplifies to $\int_0^\pi \sqrt{\sin^2 t + \cos^2 t} dt = \int_0^\pi 1 dt = \pi$. Choice A incorrectly uses the original functions instead of their derivatives, missing the fundamental requirement of the arc length formula. Remember that parametric arc length always uses the derivatives of the position functions, not the functions themselves.
12. A curve is defined by $x(t)=1+t^4$, $y(t)=2t^2$ for $-1\le t\le1$; which integral gives arc length?
    Answer: $\displaystyle \int_{-1}^{1}\sqrt{(4t^3)^2+(4t)^2}\,dt$ 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.
13. For $x=2t-\sin t$ and $y=1-\cos t$ on $0\le t\le2\pi$, which integral gives arc length?
    Answer: $\displaystyle \int_{0}^{2\pi}\sqrt{(2-\cos t)^2+(\sin t)^2}\,dt$ This problem tests the skill of finding the arc length of a curve defined by parametric equations. For x = 2t - sin t and y = 1 - cos t, find dx/dt = 2 - cos t and dy/dt = sin t. The integrand is √((2 - cos t)² + (sin t)²). Integrate this from t = 0 to t = 2π to compute the arc length. A tempting distractor like choice C replaces (sin t)² with (1 - cos t)², which is incorrect as it does not use the proper derivative for y. Always remember to use the derivatives dx/dt and dy/dt in the arc length formula and ensure the integration limits correspond to the given parameter range.
14. A cyclist's path is $x=2\cos t$, $y=3\sin t$ for $0\le t\le \frac{\pi}{2}$; which arc length integral is correct?
    Answer: $\displaystyle \int_{0}^{\pi/2}\sqrt{(-2\sin t)^2+(3\cos t)^2}\,dt$ To find arc length of the parametric curve, we use $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. For $x = 2\cos t$ and $y = 3\sin t$, the derivatives are $\frac{dx}{dt} = -2\sin t$ and $\frac{dy}{dt} = 3\cos t$. This gives the integral $\int_0^{\pi/2} \sqrt{(-2\sin t)^2 + (3\cos t)^2} dt$. Choice B incorrectly computes the derivatives as $2\cos t$ and $3\sin t$, forgetting the chain rule that introduces negative signs and swaps trig functions. When differentiating trigonometric parametric equations, carefully apply derivative rules to avoid sign errors.
15. For $x=\ln t$ and $y=t^2$, $1\le t\le e$, which integral gives the curve's arc length?
    Answer: $\displaystyle \int_{1}^{e}\sqrt{\left(\frac1t\right)^2+(2t)^2}\,dt$ This problem applies the parametric arc length formula $\int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. Given $x = \ln t$ and $y = t^2$, we find $\frac{dx}{dt} = \frac{1}{t}$ and $\frac{dy}{dt} = 2t$. The arc length integral is $\int_1^e \sqrt{(\frac{1}{t})^2 + (2t)^2} dt$. Choice B incorrectly uses the original functions $\ln t$ and $t^2$ rather than their derivatives, missing the essential concept that arc length measures the rate of change of position. Remember: always differentiate the parametric equations before setting up the arc length integral.
16. A path is $x=e^t$ and $y=e^{-t}$ for $0\le t\le1$. Which integral represents its arc length?
    Answer: $\displaystyle \int_{0}^{1}\sqrt{(e^t)^2+(-e^{-t})^2}\,dt$ This problem requires finding the arc length using the parametric formula $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. Given $x = e^t$, we differentiate to get $\frac{dx}{dt} = e^t$. Given $y = e^{-t}$, we differentiate to get $\frac{dy}{dt} = -e^{-t}$. Substituting these derivatives into the arc length formula gives $\int_0^1 \sqrt{(e^t)^2 + (-e^{-t})^2} dt$. Choice E incorrectly omits the negative sign in the derivative of $e^{-t}$, though this doesn't affect the final answer since we square the term anyway. The crucial concept is that parametric arc length always uses derivatives of the coordinate functions, properly squared and summed.
17. For $x=1+t^2$ and $y=t^3-1$ on $1\le t\le3$, which integral represents the arc length?
    Answer: $\displaystyle \int_{1}^{3}\sqrt{(2t)^2+(3t^2)^2}\,dt$ To find the arc length of this parametric curve, we use the formula $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. For $x = 1 + t^2$, differentiating gives $\frac{dx}{dt} = 2t$. For $y = t^3 - 1$, differentiating gives $\frac{dy}{dt} = 3t^2$. Substituting these derivatives into the arc length formula yields $\int_1^3 \sqrt{(2t)^2 + (3t^2)^2} dt$. Choice E incorrectly uses the original parametric functions $(1+t^2)$ and $(t^3-1)$ 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.
18. A curve is $x=2\sin t$ and $y=3\cos t$ for $0\le t\le \pi$; which integral gives its arc length?
    Answer: $\displaystyle \int_{0}^{\pi}\sqrt{\left(2\cos t\right)^2+\left(-3\sin t\right)^2}\,dt$ This problem asks for the arc length using $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. For $x = 2\sin t$ and $y = 3\cos t$, we find $\frac{dx}{dt} = 2\cos t$ and $\frac{dy}{dt} = -3\sin t$. Plugging these into the formula gives $\int_0^\pi \sqrt{(2\cos t)^2 + (-3\sin t)^2} dt$. Choice D incorrectly removes the square root operation, which would calculate the integral of the squared speed rather than the actual arc length. The arc length formula measures distance along the curve, requiring the square root to convert from squared speed to speed before integrating.
19. For $x=\sqrt{t}$ and $y=t^{3/2}$, $1\le t\le 4$, which integral correctly sets up arc length?
    Answer: $\displaystyle \int_{1}^{4}\sqrt{\left(\frac{1}{2\sqrt{t}}\right)^2+\left(\frac{3}{2}\sqrt{t}\right)^2}\,dt$ This problem requires the parametric arc length formula $\int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. Given $x = \sqrt{t} = t^{1/2}$ and $y = t^{3/2}$, we find $\frac{dx}{dt} = \frac{1}{2}t^{-1/2} = \frac{1}{2\sqrt{t}}$ and $\frac{dy}{dt} = \frac{3}{2}t^{1/2} = \frac{3}{2}\sqrt{t}$. The arc length integral becomes $\int_1^4 \sqrt{(\frac{1}{2\sqrt{t}})^2 + (\frac{3}{2}\sqrt{t})^2} dt$. Choice B incorrectly uses the original functions rather than their derivatives, missing the fundamental requirement of the arc length formula. Always remember to differentiate first when setting up parametric arc length integrals.
20. For $x=\ln(1+t)$ and $y=t^2$ on $0\le t\le3$, which integral sets up the arc length?
    Answer: $\displaystyle \int_{0}^{3}\sqrt{\left(\frac{1}{1+t}\right)^2+(2t)^2}\,dt$ To find the arc length of this parametric curve, we apply the formula $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. For $x = \ln(1+t)$, we use the chain rule to get $\frac{dx}{dt} = \frac{1}{1+t}$. For $y = t^2$, we have $\frac{dy}{dt} = 2t$. Substituting these derivatives gives $\int_0^3 \sqrt{\left(\frac{1}{1+t}\right)^2 + (2t)^2} dt$. Choice E incorrectly uses the original functions $\ln(1+t)$ and $t^2$ 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.

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