Arc Length — Question 3

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Question 3

A bead moves along x=3cos⁡t,y=3sin⁡t,0≤t≤2π3.x=3\cos t,\qquad y=3\sin t,\qquad0\le t\le\frac{2\pi}{3}. Find the distance traveled.

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Original worksheet page 1: question and worked solution for 2-1-003
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Question 3 – Solution

Step 1: Differentiate both coordinates. dxdt=−3sin⁡t,dydt=3cos⁡t.\frac{dx}{dt}=-3\sin t,\qquad\frac{dy}{dt}=3\cos t. Step 2: Compute the speed. (dxdt)2+(dydt)2=9sin⁡2t+9cos⁡2t=9=3.\begin{align*} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} &=\sqrt{9\sin^2t+9\cos^2t}\\ &=\sqrt9=3. \end{align*} Step 3: Integrate the speed. L=∫02π/33dt=[3t]02π/3=2π.\begin{align*} L&=\int_0^{2\pi/3}3\,dt\\ &=[3t]_0^{2\pi/3}=2\pi. \end{align*} L=2π\boxed{L=2\pi}

Original worksheet page 2: question and worked solution for 2-1-003

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