Arc Lengths of Curves: Parametric Equations
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AP Calculus BC › Arc Lengths of Curves: Parametric Equations
A drone traces $x(t)=t^2+1$, $y(t)=3t-t^3$ for $0\le t\le1$; which integral gives the arc length?
$\displaystyle \int_{1}^{0}\sqrt{(2t)^2+(3-3t^2)^2},dt$
$\displaystyle \int_{0}^{1}\sqrt{(2t)^2+(3-3t^2)^2},dt$
$\displaystyle \int_{0}^{1}\sqrt{(t^2+1)^2+(3t-t^3)^2},dt$
$\displaystyle \int_{0}^{1}\sqrt{(2t)+(3-3t^2)},dt$
$\displaystyle \int_{0}^{1}\sqrt{(2t)^2+(3t-t^3)^2},dt$
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.
For $x(t)=\sec t$, $y(t)=\tan t$ on $0\le t\le\frac{\pi}{4}$, which integral sets up the arc length?
$\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec t\tan t)+(\sec^2 t)},dt$
$\displaystyle \int_{0}^{\pi/2}\sqrt{(\sec t\tan t)^2+(\sec^2 t)^2},dt$
$\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec t\tan t)^2+(\sec^2 t)^2},dt$
$\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec t)^2+(\tan t)^2},dt$
$\displaystyle \int_{0}^{\pi/4}\sqrt{(\sec^2 t)^2+(\sec t\tan t)^2},dt$
Explanation
This question tests your understanding of finding the arc length of a curve defined by parametric equations. To find the arc length, compute 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.
A particle moves on $x(t) = t^3 - 1$, $y(t) = 4\sqrt{t}$ for $1 \le t \le 9$; which integral gives arc length?
$\displaystyle \int_{1}^{9}\sqrt{(3t^2)^2+\left(4\sqrt{t}\right)^2},dt$
$\displaystyle \int_{1}^{9}\left[(3t^2)^2+\left(\frac{2}{\sqrt{t}}\right)^2\right]dt$
$\displaystyle \int_{1}^{9}\sqrt{(t^3-1)^2+(4\sqrt{t})^2},dt$
$\displaystyle \int_{1}^{9}\sqrt{(3t^2)^2+\left(\frac{2}{\sqrt{t}}\right)^2},dt$
$\displaystyle \int_{0}^{8}\sqrt{(3t^2)^2+\left(\frac{2}{\sqrt{t}}\right)^2},dt$
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 $(dx/dt)^2$ plus $(dy/dt)^2$ dt. Here, differentiate $x(t) = t^3 - 1$ to get $dx/dt = 3t^2$, and $y(t) = 4\sqrt{t}$ to get $dy/dt = 2 / \sqrt{t}$. Plug these into the formula to obtain $\int_{1}^{9} \sqrt{[(3t^2)^2 + (2/\sqrt{t})^2} , dt$, which matches choice A. A tempting distractor is choice C, which incorrectly uses $4\sqrt{t}$ instead of $2 / \sqrt{t}$ for $dy/dt$, leading to an erroneous integrand. Always remember to differentiate the parametric functions to get the speed components before integrating for arc length.
A curve is $x(t)=\ln(t+1)$, $y(t)=\sqrt{t}$ for $0\le t\le4$; select the correct arc length integral.
$\displaystyle \int_{0}^{4}\sqrt{\left(\frac{1}{t+1}\right)^2+\left(\sqrt{t}\right)^2},dt$
$\displaystyle \int_{0}^{4}\sqrt{\left(\ln(t+1)\right)^2+\left(\sqrt{t}\right)^2},dt$
$\displaystyle \int_{1}^{5}\sqrt{\left(\frac{1}{t+1}\right)^2+\left(\frac{1}{2\sqrt{t}}\right)^2},dt$
$\displaystyle \int_{0}^{4}\sqrt{\left(\frac{1}{t+1}\right)^2+\left(\frac{1}{2\sqrt{t}}\right)^2},dt$
$\displaystyle \int_{0}^{4}\sqrt{\left(\frac{1}{t+1}\right)+\left(\frac{1}{2\sqrt{t}}\right)},dt$
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.
For $x(t)=2t$, $y(t)=t^2$ from $t=0$ to $t=3$, which integral computes the arc length?
$\displaystyle \int_{0}^{3}\sqrt{(2)^2+(2t)^2},dt$
$\displaystyle \int_{0}^{6}\sqrt{(2)^2+(2t)^2},dt$
$\displaystyle \int_{0}^{3}\sqrt{(2)^2+(t^2)^2},dt$
$\displaystyle \int_{0}^{3}\sqrt{(2t)^2+(t^2)^2},dt$
$\displaystyle \int_{0}^{3}\left[(2)^2+(2t)^2\right]dt$
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.
A curve is defined by $x(t)=1+t^4$, $y(t)=2t^2$ for $-1\le t\le1$; which integral gives arc length?
$\displaystyle \int_{-1}^{1}\sqrt{(4t^3)+(4t)},dt$
$\displaystyle \int_{-1}^{1}\sqrt{(1+t^4)^2+(2t^2)^2},dt$
$\displaystyle \int_{0}^{1}\sqrt{(4t^3)^2+(4t)^2},dt$
$\displaystyle \int_{-1}^{1}\sqrt{(4t^3)^2+(4t)^2},dt$
$\displaystyle \int_{-1}^{1}\sqrt{(4t^3)^2+(2t^2)^2},dt$
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=-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.
A curve is $x(t)=t\cos t$, $y(t)=t\sin t$ for $0\le t\le2$; choose the correct arc length integral.
$$\displaystyle \int_{0}^{2}\sqrt{(t\cos t)^2+(t\sin t)^2},dt$$
$$\displaystyle \int_{0}^{2}\sqrt{(\cos t-t\sin t)+(\sin t+t\cos t)},dt$$
$$\displaystyle \int_{0}^{2\pi}\sqrt{(\cos t-t\sin t)^2+(\sin t+t\cos t)^2},dt$$
$$\displaystyle \int_{0}^{2}\sqrt{(\cos t-t\sin t)^2+(\sin t+t\cos t)^2},dt$$
$$\displaystyle \int_{0}^{2}\left[(\cos t-t\sin t)^2+(\sin t+t\cos t)^2\right]dt$$
Explanation
This question tests your understanding of finding the arc length of a curve defined by parametric equations. To find the arc length, compute $$\int_{0}^{2} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } , dt$$. Here, differentiate x(t) = t cos t using product rule to get $dx/dt = \cos t - t \sin t$, and y(t) = t sin t to get $dy/dt = \sin t + t \cos t$. Plug these into the formula to obtain $$\int_{0}^{2} \sqrt{(\cos t - t \sin t)^2 + (\sin t + t \cos t)^2} , dt$$, 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.
For $x=1+\sin t$ and $y=1-\cos t$ on $0\le t\le \frac{\pi}{2}$, which integral gives arc length?
$\displaystyle \int_{0}^{\pi/2}\sqrt{\left(\sin t\right)^2+\left(-\cos t\right)^2},dt$
$\displaystyle \int_{0}^{\pi/2}\sqrt{1+\left(\frac{\sin t}{\cos t}\right)^2},dt$
$\displaystyle \int_{0}^{\pi/2}\sqrt{\left(1+\sin t\right)^2+\left(1-\cos t\right)^2},dt$
$\displaystyle \int_{0}^{\pi}\sqrt{\left(\cos t\right)^2+\left(\sin t\right)^2},dt$
$\displaystyle \int_{0}^{\pi/2}\sqrt{\left(\cos t\right)^2+\left(\sin t\right)^2},dt$
Explanation
To find the arc length of this parametric curve, we use $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. Given $x = 1 + \sin t$ and $y = 1 - \cos t$, we compute $\frac{dx}{dt} = \cos t$ and $\frac{dy}{dt} = \sin t$. Substituting these derivatives gives $\int_0^{\pi/2} \sqrt{(\cos t)^2 + (\sin t)^2} dt = \int_0^{\pi/2} \sqrt{1} dt = \int_0^{\pi/2} 1 dt$. 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.
For $x=e^{t}$ and $y=e^{-t}$ on $0\le t\le \ln 2$, which integral represents the arc length?
$\displaystyle \int_{0}^{2}\sqrt{\left(e^{t}\right)^2+\left(-e^{-t}\right)^2},dt$
$\displaystyle \int_{0}^{\ln 2}\sqrt{\left(e^{t}\right)^2+\left(e^{-t}\right)^2},dt$
$\displaystyle \int_{0}^{\ln 2}\sqrt{1+\left(\frac{-e^{-t}}{e^{t}}\right)^2},dt$
$\displaystyle \int_{0}^{\ln 2}\sqrt{\left(e^{t}\right)^2+\left(-e^{-t}\right)^2},dt$
$\displaystyle \int_{0}^{\ln 2}\sqrt{\left(e^{t}\right)+\left(-e^{-t}\right)},dt$
Explanation
To find the arc length, we apply $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. Given $x = e^t$ and $y = e^{-t}$, we differentiate to get $\frac{dx}{dt} = e^t$ and $\frac{dy}{dt} = -e^{-t}$. Substituting these derivatives gives $\int_0^{\ln 2} \sqrt{(e^t)^2 + (-e^{-t})^2} dt$. Choice B incorrectly uses $e^{-t}$ instead of $-e^{-t}$ for the derivative of $y$, forgetting the chain rule's negative sign when differentiating $e^{-t}$. Remember that the derivative of $e^{-t}$ is $-e^{-t}$, and this sign matters when squaring in the arc length formula.
A curve is given by $x=\cos t$, $y=\sin(2t)$ for $0\le t\le \pi$; choose the correct arc length setup.
$\displaystyle \int_{0}^{\pi}\sqrt{\left(\cos t\right)^2+\left(\sin(2t)\right)^2},dt$
$\displaystyle \int_{0}^{\pi}\sqrt{1+\left(\frac{2\cos(2t)}{-\sin t}\right)^2},dt$
$\displaystyle \int_{0}^{\pi}\sqrt{\left(-\sin t\right)^2+\left(2\cos(2t)\right)^2},dt$
$\displaystyle \int_{0}^{\pi}\left[\left(-\sin t\right)^2+\left(2\cos(2t)\right)^2\right]dt$
$\displaystyle \int_{0}^{\pi}\sqrt{\left(\sin t\right)^2+\left(\cos(2t)\right)^2},dt$
Explanation
To find the arc length of this parametric curve, we need to apply the formula $L = \int_a^b \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$. For $x = \cos t$ and $y = \sin(2t)$, we calculate $\frac{dx}{dt} = -\sin t$ and $\frac{dy}{dt} = 2\cos(2t)$. Plugging these derivatives into the arc length formula yields $\int_0^\pi \sqrt{(-\sin t)^2 + (2\cos(2t))^2} dt$. Choice B incorrectly uses the original functions instead of their derivatives, which is a common mistake when setting up parametric arc length integrals. The key to parametric arc length is always differentiating first, then applying the Pythagorean sum under the square root.