# Urgent help with rearranging formulae

• March 31st 2008, 01:02 PM
matttbh
Urgent help with rearranging formulae
Hi people!

I have to re-arrange a formula for a current piece of coursework but i'm drawing a complete blank on what to do. I have reason to believe i know what the correct rearrangement is but i simply can't get there! :(

The formula is

s= r - √ (r² - y²)

and i have to rearrange to make y the subject.

I have found what is apparently the correct solution and it does appear to give the correct answer but i have no idea how to get there and i need to show each stage of working !

The formula which gives the right answer is

y = √(2r - s) * √(s)

Thank you to everyone who can help and look forward to becoming a part of your community, Matt
• March 31st 2008, 01:09 PM
topher0805
If I understand correctly, you want to solve for y?

$s = r - \sqrt {r^2 - y^2}$

Add $\sqrt {r^2 - y^2}$ to both sides:

$s + \sqrt {r^2 - y^2} = r$

Subtract s from both sides:

$\sqrt {r^2 - y^2} = r - s$

Square both sides:

$r^2 - y^2 = (r - s)^2$

Subtract r^2 from both sides:

$
-y^2 = (r - s)^2 - r^2$

Multiply both sides by -1:

$y^2 = r^2 - (r - s)^2$

Take the square root of both sides:

$
y = \sqrt {r^2 - (r - s)^2}$
• March 31st 2008, 01:11 PM
Moo
Hello,

$s=r-\sqrt{r^2-y^2}$

Thus,

$\sqrt{r^2-y^2}=r-s$

-> $r^2-y^2=(r-s)^2=r^2-2rs+s^2$

We simplify both sides by r² :

$-y^2=-2rs+s^2=s(-2r+s)$

$y^2=s(2r-s)$

If s, r & y are positive, and supposing 2r-s is positive, we can conclude :

$y=\sqrt{s}*\sqrt{2r-s}$

;)
• March 31st 2008, 01:20 PM
matttbh
thank you very much that was very helpful. I like your name btw moo (Yes)

I have thanked both of you for your contributions, Matt
• March 31st 2008, 01:22 PM
Moo
Why ? :D

I've seen a furtive answer from janvdl if i'm not mistaking, if you really want to thank, you can PM him ! (Rofl)

btw, good luck