Sandra Allison

2021-12-30

Find the exact length of the curve.

Vasquez

You need to apply the next formula
$L={\int }_{c}^{d}\sqrt{1+\left(\frac{dx}{dy}{\right)}^{2}}dy$ Where $1\le y\le 2$
First step, to find $\frac{dx}{dy}$
$x=\frac{{y}^{4}}{8}+\frac{1}{4{y}^{2}}=\frac{{y}^{4}}{8}+\frac{{y}^{-2}}{4}$
Differentiate both sides with respect to y, you get
$\frac{d}{dy}\left[x\right]=\frac{d}{dy}\left[\frac{{y}^{4}}{8}+\frac{{y}^{-2}}{4}\right]$
$\frac{dx}{dy}=\frac{4}{8}{y}^{3}-\frac{2}{4}{y}^{-3}$
Simplify
$\frac{dx}{dy}=\frac{{y}^{3}}{2}-\frac{{y}^{-3}}{2}$
Now calculate $\left(\frac{dx}{dy}{\right)}^{2}$, so
$\left(\frac{dx}{dy}{\right)}^{2}=\left(\frac{{y}^{3}}{2}-\frac{{y}^{-3}}{2}{\right)}^{2}=\frac{{y}^{6}}{4}-\frac{2}{4}+\frac{{y}^{-6}}{4}$
Simplify
$\left(\frac{dx}{dy}{\right)}^{2}=\frac{{y}^{6}}{4}-\frac{1}{2}+\frac{{y}^{-6}}{4}$
Now calculate $\left(\frac{dx}{dy}{\right)}^{2}+1$, adding 1 to both sides
$\left(\frac{dx}{dy}{\right)}^{2}+1=\frac{{y}^{6}}{4}-\frac{1}{2}+\frac{{y}^{-6}}{4}+1$
You can see that$\frac{{y}^{6}}{4}+\frac{1}{2}+\frac{{y}^{-6}}{4}$, is a perfect square trinomial, so
$\left(\frac{dx}{dy}{\right)}^{2}+1=\left(\frac{{y}^{3}}{2}+\frac{{y}^{-3}}{2}{\right)}^{2}$
Now calculate $\sqrt{\left(\frac{dx}{dy}{\right)}^{2}+1}$
$\sqrt{\left(\frac{dx}{dy}{\right)}^{2}+1}=\sqrt{\left(\frac{{y}^{3}}{2}+\frac{{y}^{-3}}{2}{\right)}^{2}}$
$\sqrt{\left(\frac{dx}{dy}{\right)}^{2}+1}=\frac{{y}^{3}}{2}+\frac{{y}^{-3}}{2}$
Then
$L={\int }_{c}^{d}\sqrt{1+\left(\frac{dx}{dy}{\right)}^{2}}dy={\int }_{1}^{2}\left(\frac{{y}^{3}}{2}+\frac{{y}^{-3}}{2}\right)dt$
Integrate you get:
$L=\left[\frac{{y}^{4}}{2\left(4\right)}+\frac{{y}^{-2}}{2\left(-2\right)}{\right]}_{1}^{2}$
$L=\left[\frac{{y}^{4}}{8}-\frac{{y}^{-2}}{4}{\right]}_{1}^{2}$
Apply the fundamental theorem of calculus
$L=\left(\frac{{2}^{4}}{8}-\frac{{2}^{-2}}{4}\right)-\left(\frac{{1}^{4}}{8}-\frac{{1}^{-2}}{4}\left(=\left(2-\frac{1}{16}\right)-\left(\frac{1}{8}-\frac{1}{4}\right)$
$L=\frac{31}{16}+\frac{1}{8}=\frac{33}{16}$
Answer: $L=\frac{33}{16}$

user_27qwe

karton