Arc Length — Question 2

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Question 2

For 0≤x≤20\le x\le2, compare C1:y=0,C2:y=23x3/2.C_1:y=0,\qquad C_2:y=\frac23x^{3/2}. Find L2−L1L_2-L_1.

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Original worksheet page 1: question and worked solution for 2-1-002
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Question 2 – Solution

Step 1: Find the length of C1C_1. Since y′=0y'=0, L1=∫021+02dx=∫021dx=2.L_1=\int_0^2\sqrt{1+0^2}\,dx=\int_0^2 1\,dx=2. Step 2: Differentiate C2C_2. y′=23⋅32x1/2=x.y'=\frac23\cdot\frac32x^{1/2}=\sqrt{x}. Step 3: Find the length of C2C_2. L2=∫021+(x)2dx=∫021+xdx=[23(1+x)3/2]02=23(33−1)=23−23.\begin{align*} L_2&=\int_0^2\sqrt{1+(\sqrt{x})^2}\,dx\\ &=\int_0^2\sqrt{1+x}\,dx\\ &=\left[\frac23(1+x)^{3/2}\right]_0^2\\ &=\frac23(3\sqrt3-1)=2\sqrt3-\frac23. \end{align*} Step 4: Subtract. L2−L1=23−23−2=23−83.L_2-L_1=2\sqrt3-\frac23-2=2\sqrt3-\frac83. L2−L1=23−83\boxed{L_2-L_1=2\sqrt3-\frac83}

Original worksheet page 2: question and worked solution for 2-1-002

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