# Thread: Calculate the length of the curve

1. ## Calculate the length of the curve

x12 Calculate the length of the curve
between x=0 och x=1.

I think i managed to solve until

2. Originally Posted by sf1903
x12 Calculate the length of the curve
between x=0 och x=1.

I think i managed to solve until

$y=\frac{x^{3\slash 2}}{3}-x^{1\slash 2}\Longrightarrow y'=\frac{x^{1\slash 2}}{2}-\frac{1}{2x^{1\slash 2}}$

$1+(y')^2=1+\frac{1}{4}x-\frac{1}{2}+\frac{1}{4x}=\frac{1}{4}\left(x+2+\fra c{1}{x}\right)=\frac{1}{4}\left(x+\frac{1}{x}\righ t)^2$

So the function in the integral is pretty simple: ${1}{2}\left(x+\frac{1}{x}\right)$ ...unfortunately this improper integral diverges at zero since $\ln x\xrightarrow[x\to 0]{}-\infty$ ...

Tonio

3. Originally Posted by tonio
$y=\frac{x^{3\slash 2}}{3}-x^{1\slash 2}\Longrightarrow y'=\frac{x^{1\slash 2}}{2}-\frac{1}{2x^{1\slash 2}}$

$1+(y')^2=1+\frac{1}{4}x-\frac{1}{2}+\frac{1}{4x}=\frac{1}{4}\left(x+2+\fra c{1}{x}\right)=\frac{1}{4}\left(x+\frac{1}{x}\righ t)^2$
No, tonio, that last term is
$\frac{1}{4}\left(x^{1/2}+ \frac{1}{x^{1/2}}\right)^2$.

The integral becomes $\frac{1}{2}\int_0^1 \left(x^{1/2}+ x^{-1/2}\right)dx$ which, while an improper integral, does have a finite value.

So the function in the integral is pretty simple: ${1}{2}\left(x+\frac{1}{x}\right)$ ...unfortunately this improper integral diverges at zero since $\ln x\xrightarrow[x\to 0]{}-\infty$ ...

Tonio

4. Originally Posted by HallsofIvy
No, tonio, that last term is
$\frac{1}{4}\left(x^{1/2}+ \frac{1}{x^{1/2}}\right)^2$.

The integral becomes $\frac{1}{2}\int_0^1 \left(x^{1/2}+ x^{-1/2}\right)dx$ which, while an improper integral, does have a finite value.
This is how far I have come, am I correct?

5. Originally Posted by HallsofIvy
No, tonio, that last term is
$\frac{1}{4}\left(x^{1/2}+ \frac{1}{x^{1/2}}\right)^2$.

The integral becomes $\frac{1}{2}\int_0^1 \left(x^{1/2}+ x^{-1/2}\right)dx$ which, while an improper integral, does have a finite value.

You are, of course, right. I copied, pretty stupidly, from the line before the last one instead of he alst one...and forgot.
Thanx....oh, and then the integral converges (it still is an improper integral)

Tonio