Arc Length II

Problem 236: Find the arc length of the graph of the function $f(x)=5\sqrt{x^3}$ from $x=2$ to $x=3$ .

Solution: The arc length is given by

(1) $\begin{equation*} L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \ dx. \end{equation*}$

That is,

(2) $\begin{equation*} f(x) = 5\sqrt{x^3} = 5 (x^3)^{1/2} \implies f'(x) = \frac{15x^2}{2\sqrt{x^3}}. \end{equation*}$

Thus,

(3) $\begin{align*} L & = \int_{2}^{3} \sqrt{1 + [f'(x)]^2} \ dx \\ & = \int_{2}^{3} \sqrt{1 + \left(\frac{15x^2}{2\sqrt{x^3}} \right)^2} \ dx \\ & = \int{2}^{3} \sqrt{1 + \frac{(15x^2)^2}{4x^3}} \ dx \\ & = \int_{2}^{3} \sqrt{\frac{4x^3 + 225x^4}{4x^3}} \ dx \\ & = \int_{2}^{3} \sqrt{4 + \frac{225}{4}x} \ dx \end{align*}$

Using U-substitution to evaluate the integral, we get

(4) $\begin{equation*} L = \int_{2}^{3} \sqrt{4 + \frac{225}{4}x} \ dx \approx 11.88. \end{equation*}$

Therefore, the arc length of the graph between 2 and 3 is approximately

$11.88.$

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