# Thread: solving for a variable

1. ## solving for a variable

Can the following be solved for b in terms of x?

$x=\frac{1}{4}\left(\sinh^{-1}(2b)+2b\sqrt{1+4b^2}\right)$

I post in this forum because the subject equation comes from evaluating the (arc length) integral

$x=\int_{0}^{b}\sqrt{1+4x^2}\,dx$,

and also because sinh(x) is a very calculus-y function.

Thanks,
Dfrtbx

2. Originally Posted by Dfrtbx
Can the following be solved for b in terms of x?

$x=\frac{1}{4}\left(\sinh^{-1}(2b)+2b\sqrt{1+4b^2}\right)$

I post in this forum because the subject equation comes from evaluating the (arc length) integral

$x=\int_{0}^{b}\sqrt{1+4x^2}\,dx$,

and also because sinh(x) is a very calculus-y function.

Thanks,
Dfrtbx
Please post the entire original question.

3. Originally Posted by mr fantastic
Please post the entire original question.
That is the entire original question. It didn't come out of a textbook, if that's what you're asking. Here's a graph of $y_1=x^2$ superimposed on $y_2=\frac{1}{4}\left(\sinh^{-1}(2x)+2x\sqrt{1+4x^2}\right)$:

http://i51.tinypic.com/avk3v7.jpg

Gold is the former, green the latter. Essentially what I'm trying to do is find an inverse function for $y_2$.

Maxima (my computer algebra system) is unable to solve the original equation $x=\frac{1}{4}\left(\sinh^{-1}(2b)+2b\sqrt{1+4b^2}\right)$ for b in terms of x.

EDIT: sorry... I sorta implied it in my original post, but I guess I should be more explicit. $y_2=\frac{1}{4}\left(\sinh^{-1}(2x)+2x\sqrt{1+4x^2}\right)$ is the arc length of the function $y_1=x^2$ from 0 to some x.

4. Originally Posted by Dfrtbx
Can the following be solved for b in terms of x?

$x=\frac{1}{4}\left(\sinh^{-1}(2b)+2b\sqrt{1+4b^2}\right)$

[snip]
No.