Arc Length with Parametric Equations — Question 4

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Question 4

Problem

Show that x=etcos⁡tx=e^t\cos t, y=etsin⁡ty=e^t\sin t, 0≤t≤ln⁡20\le t\le\ln2, has length 2\sqrt2.

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Original worksheet page 1: question and worked solution for 3-4-004
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Question 4 – Solution

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Solution

  1. Differentiate using the product rule: x′(t)=et(cos⁡t−sin⁡t),y′(t)=et(sin⁡t+cos⁡t).x'(t)=e^t(\cos t-\sin t), \qquad y'(t)=e^t(\sin t+\cos t).

  2. Compute the squared speed: x′2+y′2=e2t[(cost−sint)2+(sint+cost)2]=e2t[2cos⁡2t+2sin⁡2t]=2e2t.\begin{aligned} x'^2+y'^2 &=e^{2t}\left[(\cos t-\sin t)^2+(\sin t+\cos t)^2\right]\\ &=e^{2t}\left[2\cos^2t+2\sin^2t\right]\\ &=2e^{2t}. \end{aligned} Since et>0e^t>0, the speed is 2et\sqrt2e^t.

  3. Set up and evaluate the length: L=∫0ln⁡22etdt=2[et]0ln⁡2=2(2−1)=2.\begin{aligned} L&=\int_0^{\ln2}\sqrt2e^t\,dt\\ &=\sqrt2\left[e^t\right]_0^{\ln2}\\ &=\sqrt2(2-1)=\boxed{\sqrt2}. \end{aligned}

Original worksheet page 2: question and worked solution for 3-4-004

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