# 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 $\displaystyle y=\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}}$ from $\displaystyle (1,-\frac{2}{3}})$ to $\displaystyle {4,\frac{2}{3})$.

So far I have:$\displaystyle \frac{ds}{dx}=\sqrt{1+f'^2} = \sqrt{1+(\frac{1}{2}\sqrt{x} - \frac{1}{2}*\frac{1}{\sqrt{x}})^2dx$$\displaystyle =\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 \displaystyle y=\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}} from \displaystyle (1,-\frac{2}{3}}) to \displaystyle {4,\frac{2}{3}) You have, \displaystyle y=\frac{1}{3}x^{3/2}-x^{1/2} \displaystyle 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, \displaystyle [y']^2=\left( \frac{1}{2}x^{1/2}-\frac{1}{2}x^{-1/2} \right)^2 Open, \displaystyle \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, \displaystyle \frac{1}{4}x+\frac{1}{4}x^{-1}-\frac{2}{4} The next step is to add one, \displaystyle \frac{1}{4}x+\frac{1}{4}x^{-1}+\frac{2}{4} Thus, \displaystyle \int_1^4 \sqrt{\frac{1}{4}x+\frac{1}{4}x^{-1}+\frac{2}{4}}dx \displaystyle \frac{1}{2}\int_1^4 \sqrt{x+\frac{1}{x}+2} dx \displaystyle \frac{1}{2}\int_1^4 \sqrt{\frac{x^2+2x+1}{x}} dx You need to make the square, \displaystyle \frac{1}{2}\int_1^4 \sqrt{\frac{(x+1)^2}{x}dx Since \displaystyle x>0 you make use the rule, \displaystyle \sqrt{a/b}=\sqrt{a}/\sqrt{b} Thus, \displaystyle \frac{1}{2}\int_1^4 \frac{\sqrt{(x+1)^2}}{\sqrt{x}}dx=\displaystyle \frac{1}{2}\int_1^4 \frac{|x+1|}{\sqrt{x}}dx But, \displaystyle x>0\to x+1>0 thus, \displaystyle \frac{1}{2}\int_1^4 \frac{x+1}{\sqrt{x}}dx Note that the fraction: \displaystyle \frac{x+1}{\sqrt{x}}=\frac{x}{ \sqrt{x} }+\frac{1}{ \sqrt{x} }=x^{1/2}+x^{-1/2} Finally, \displaystyle \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 \displaystyle y=\frac{1}{3}x^{\frac{3}{2}}-x^{\frac{1}{2}} from \displaystyle (1,-\frac{2}{3}}) to \displaystyle {4,\frac{2}{3}). So far I have:\displaystyle \frac{ds}{dx}=\sqrt{1+f'^2} = \sqrt{1+(\frac{1}{2}\sqrt{x} - \frac{1}{2}*\frac{1}{\sqrt{x}})^2dx$$\displaystyle =\frac{1}{2}\sqrt{\frac{(x+1)^2}{x}}$

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

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

RonL