# Can't seem to finish this length of graph problem

• Aug 28th 2006, 03:43 PM
FLTR
Can't seem to finish this length of graph problem
I need help to finish finding the length of the graph of $y=\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}}$ from $(1,-\frac{2}{3}})$ to ${4,\frac{2}{3})$.

So far I have: $\frac{ds}{dx}=\sqrt{1+f'^2} = \sqrt{1+(\frac{1}{2}\sqrt{x} - \frac{1}{2}*\frac{1}{\sqrt{x}})^2dx$ $=\frac{1}{2}\sqrt{\frac{(x+1)^2}{x}}$
• Aug 28th 2006, 04:47 PM
ThePerfectHacker
Quote:

Originally Posted by FLTR
I need help to finish finding the length of the graph of $y=\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}}$ from $(1,-\frac{2}{3}})$ to ${4,\frac{2}{3})$

You have,
$y=\frac{1}{3}x^{3/2}-x^{1/2}$
$y'=\frac{1}{3}\cdot \frac{3}{2}x^{1/2}-\frac{1}{2}x^{-1/2}=\frac{1}{2}x^{1/2}-\frac{1}{2}x^{-1/2}$
Now,
$[y']^2=\left( \frac{1}{2}x^{1/2}-\frac{1}{2}x^{-1/2} \right)^2$
Open,
$\frac{1}{4}x-\frac{2}{1}\cdot \frac{1}{2} \cdot \frac{1}{2}\cdot x^{1/2-1/2}+\frac{1}{4}x^{-1}$
Thus,
$\frac{1}{4}x+\frac{1}{4}x^{-1}-\frac{2}{4}$
The next step is to add one,
$\frac{1}{4}x+\frac{1}{4}x^{-1}+\frac{2}{4}$
Thus,
$\int_1^4 \sqrt{\frac{1}{4}x+\frac{1}{4}x^{-1}+\frac{2}{4}}dx$
$\frac{1}{2}\int_1^4 \sqrt{x+\frac{1}{x}+2} dx$
$\frac{1}{2}\int_1^4 \sqrt{\frac{x^2+2x+1}{x}} dx$
You need to make the square,
$
\frac{1}{2}\int_1^4 \sqrt{\frac{(x+1)^2}{x}dx$

Since $x>0$ you make use the rule,
$\sqrt{a/b}=\sqrt{a}/\sqrt{b}$
Thus,
$\frac{1}{2}\int_1^4 \frac{\sqrt{(x+1)^2}}{\sqrt{x}}dx$= $\frac{1}{2}\int_1^4 \frac{|x+1|}{\sqrt{x}}dx$
But, $x>0\to x+1>0$ thus,
$\frac{1}{2}\int_1^4 \frac{x+1}{\sqrt{x}}dx$
Note that the fraction:
$\frac{x+1}{\sqrt{x}}=\frac{x}{ \sqrt{x} }+\frac{1}{ \sqrt{x} }=x^{1/2}+x^{-1/2}$
Finally,
$\frac{1}{2}\int_1^4 x^{1/2}+x^{-1/2}dx$
• Aug 28th 2006, 10:10 PM
CaptainBlack
Quote:

Originally Posted by FLTR
I need help to finish finding the length of the graph of $y=\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}}$ from $(1,-\frac{2}{3}})$ to ${4,\frac{2}{3})$.

So far I have: $\frac{ds}{dx}=\sqrt{1+f'^2} = \sqrt{1+(\frac{1}{2}\sqrt{x} - \frac{1}{2}*\frac{1}{\sqrt{x}})^2dx$ $=\frac{1}{2}\sqrt{\frac{(x+1)^2}{x}}$

This is good, all you have left to do is compute:

$
s=\int_{x=1}^2 \frac{ds}{dx} dx=\int_1^2\frac{1}{2}\sqrt{\frac{(x+1)^2}{x}} dx
$

RonL