Arc Length — Question 10

Section 9-1 · Calculus pages 1639–1640

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Question 10

A curve satisfies y′(x)=e2x−1,0≤x≤ln⁡3.y'(x)=\sqrt{e^{2x}-1},\qquad0\le x\le\ln3. Find its length. A formula for yy is not needed.

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Question 10 – Solution

Step 1: Substitute the derivative into the arc-length formula. 1+(y′)2=1+(e2x−1)2=1+e2x−1=e2x.\begin{align*} 1+(y')^2 &=1+\left(\sqrt{e^{2x}-1}\right)^2\\ &=1+e^{2x}-1=e^{2x}. \end{align*} Step 2: Simplify the square root. Since ex>0e^x>0, 1+(y′)2=e2x=ex.\sqrt{1+(y')^2}=\sqrt{e^{2x}}=e^x. Step 3: Integrate and evaluate. L=∫0ln⁡3exdx=[ex]0ln⁡3=eln⁡3−e0=3−1=2.\begin{align*} L&=\int_0^{\ln3}e^x\,dx\\ &=[e^x]_0^{\ln3}\\ &=e^{\ln3}-e^0=3-1=2. \end{align*} L=2\boxed{L=2}