Transposing of formulae

mikewhant · #1 $Old$ October 26th, 2009, 11:09 AM

Hi there, thanks in advance for any help

Make A the subject of the formula:

Q = C.A square root (2g.h / (1-A^2/x^2))

I have got this far but unsure what to do with the fraction of a fraction... the (x^2) bit.

Q^2 / C^2.A^2 = (2g.h / (1-A^2/x^2))

Thank you again for any assitance

Mike

HallsofIvy · #2 $Old$ October 26th, 2009, 01:07 PM

Quote:

Originally Posted by mikewhant $View Post$

Hi there, thanks in advance for any help

Make A the subject of the formula:

Q = C.A square root (2g.h / (1-A^2/x^2))

I have got this far but unsure what to do with the fraction of a fraction... the (x^2) bit.

Q^2 / C^2.A^2 = (2g.h / (1-A^2/x^2))

Thank you again for any assitance

Mike

I would first isolate that square root:
[LaTeX Error: Syntax error]
and then square both sides.

[math]\frac{Q^2}{C^2A^2}= \frac{2gh}{1- \frac{A^2}{x^2}}

Multiply both numerator and denominator, on the right, by $x^2$
$\frac{Q^2}{C^2A^2}= \frac{2ghx^2}{x^2- A^2}$

Get rid of the fractions by multiplying both sides by the denominators, $C^2A^2$ and $x^2- A^2$

$Q^2(x^2- A^2)= 2ghx^2C^2A^2$
$Q^2x^2- Q^2A^2= 2ghx^2C^2A^2$

Add $Q^2A^2$ to both sides

$Q^2x^2= 2ghx^2C^2A^2+ Q^2A^2$

Factor [math]A^2[/quote] out of the right side

$Q^2x^2= (2gx^2C^2+ Q^2)A^2$

Divide both sides by $2gx^2C^2+ Q^2$

$\frac{Q^2x^2}{2gx^2C^2+ Q^2}= A^2$

Finally, take the square root of both sides

$A= \sqrt{\frac{Q^2x^2}{2gx^2C^2+ Q^2}}$

mikewhant · #3 $Old$ October 26th, 2009, 03:00 PM

Thanks alot, great help, i just needed a worked example, now I can do them!

Cheers