Arc Length — Question 5

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Question 5

Find a>0a>0 so that y=23x3/2,0≤x≤a,y=\frac23x^{3/2},\qquad0\le x\le a, has length 14/314/3.

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Original worksheet page 1: question and worked solution for 2-1-005
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Question 5 – Solution

Step 1: Differentiate. y′=x.y'=\sqrt{x}. Step 2: Write and evaluate the length. L=∫0a1+xdx=[23(1+x)3/2]0a=23((1+a)3/2−1).\begin{align*} L&=\int_0^a\sqrt{1+x}\,dx\\ &=\left[\frac23(1+x)^{3/2}\right]_0^a\\ &=\frac23\left((1+a)^{3/2}-1\right). \end{align*} Step 3: Set the length equal to 14/314/3 and solve. 23((1+a)3/2−1)=143,(1+a)3/2−1=7,(1+a)3/2=8,1+a=82/3=4,a=3.\begin{align*} \frac23\left((1+a)^{3/2}-1\right)&=\frac{14}{3},\\ (1+a)^{3/2}-1&=7,\\ (1+a)^{3/2}&=8,\\ 1+a&=8^{2/3}=4,\\ a&=3. \end{align*} a=3\boxed{a=3}

Original worksheet page 2: question and worked solution for 2-1-005

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