Find the length of the curve.
-$$y = \frac { 3 } { 8 } \left( x ^ { 4 / 3 } - 2 x ^ { 2 / 3 } \right) \text { from } x = 1 \text { to } x = 8$$
A) $\frac { 17 } { 2 }$
B) $\frac { 81 } { 8 }$
C) $\frac { 63 } { 8 }$
D) 23

Question

Quizplus · Accepted Answer

We use the formula for arc length, which is $\int_{a}^{b} \sqrt{1+\left[f'(x)ight]^2} dx$. In this case, we have $f(x) = \frac{3}{8}\left(x^{4/3}-2x^{2/3}ight)$, so $f'(x) = \frac{3}{2}x^{1/3}-\frac{3}{4}x^{-1/3}$. Then,
\begin{align*}
\int_{1}^{8} \sqrt{1+\left[f'(x)ight]^2} dx &= \int_{1}^{8} \sqrt{1+\frac{9}{4}x^{2/3}-\frac{9}{4}x^{2/3}+\frac{27}{16}x^{-2/3}} dx \
&= \int_{1}^{8} \sqrt{1+\frac{9}{16}\frac{(2x^{2/3})^2}{(x^{4/3})}} dx \
&= \int_{1}^{8} \sqrt{1+\frac{9}{4}\left(\frac{2}{x}ight)^2} dx.
\end{align*}
Let $u = \frac{2}{x}$. Then $x = \frac{2}{u}$ and $dx = -\frac{2}{u^2} du$, so
\begin{align*}
\int_{1}^{8} \sqrt{1+\frac{9}{4}\left(\frac{2}{x}ight)^2} dx &= \int_{2/8}^{2/1} \sqrt{1+9u^2} \cdot \frac{2}{u^2} du \
&= 2\int_{1/4}^{2} u\sqrt{1+9u^2} du \
&= \frac{2}{27}\int_{1/4}^{2} (1+9u^2)' \cdot \sqrt{1+9u^2} du \
&= \frac{2}{27}\left[\frac{2}{3}(1+9u^2)^{3/2} ight]_{1/4}^{2} \
&= \boxed{\frac{63}{8}}.
\end{align*}

Find the Length of the Curve y=38(x4/3−2x2/3) from x=1 to x=8y = \frac { 3 } { 8 } \left( x ^ { 4 / 3 } - 2 x ^ { 2 / 3 } \right) \text { from } x = 1 \text { to } x = 8y=83​(x4/3−2x2/3) from x=1 to x=8

Find the Length of the Curve $y = \frac { 3 } { 8 } \left( x ^ { 4 / 3 } - 2 x ^ { 2 / 3 } \right) \text { from } x = 1 \text { to } x = 8$