Find the length of the curve. $y=\frac{1}{6}\left(x^{2}+4\right)^{3 / 2}, 0 \leq x \leq 3$
A) $6.5000$
B) $7.5000$
C) $8.5000$
D) $5.5000$
E) $4.5000$

Question

Quizplus · Accepted Answer

To find the length of the curve, we use the formula for arc length: $ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $. First, find $\frac{dy}{dx}$ for $y=\frac{1}{6}(x^2+4)^{3/2}$. Using the chain rule, $\frac{dy}{dx} = \frac{1}{6} \cdot \frac{3}{2}(x^2+4)^{1/2} \cdot 2x = \frac{x(x^2+4)^{1/2}}{2}$. Then, $\left(\frac{dy}{dx}\right)^2 = \frac{x^2(x^2+4)}{4}$. The arc length integral becomes $L = \int_{0}^{3} \sqrt{1 + \frac{x^2(x^2+4)}{4}} \, dx$. Simplifying inside the square root gives $\sqrt{\frac{x^4 + 4x^2 + 4}{4}} = \frac{\sqrt{(x^2+2)^2}}{2} = \frac{x^2+2}{2}$. Thus, $L = \int_{0}^{3} \frac{x^2+2}{2} \, dx = \left[\frac{x^3}{6} + x\right]_{0}^{3} = \frac{27}{6} + 3 = 7.5$.

Find the Length of the Curve y=16(x2+4)3/2,0≤x≤3y=\frac{1}{6}\left(x^{2}+4\right)^{3 / 2}, 0 \leq x \leq 3y=61​(x2+4)3/2,0≤x≤3

Find the Length of the Curve $y=\frac{1}{6}\left(x^{2}+4\right)^{3 / 2}, 0 \leq x \leq 3$