Given the formula:
$ q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1} $
Transpose for $A_2$
I have tried this problem four times and got a different answer every time, none of which are the answer provided in the book. I would very much appreciate if someone could show me how to do this step by step.
The answer from the book is:
$ A_2=\sqrt\frac{A_1^2q^2}{2A_1^2gh+q^2} $
The closest I can get is the following:
$ q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1} $
$ \frac{q^2}{A_1^2} = \frac{2gh}{(\frac{A_1}{A_2})^2-1} $
Invert: $ \frac{A_1^2}{q^2} = \frac{(\frac{A_1}{A_2})^2-1}{2gh} $ Multiply both sides by $2gh$: $ 2gh\frac{A_1^2}{q^2} = (\frac{A_1}{A_2})^2-1 $
$ \frac{2ghA_1^2}{q^2} = (\frac{A_1}{A_2})^2-1 $ Add 1 to both sides and re-arrange: $ \frac{A_1^2}{A_2^2} = \frac{2ghA_1^2}{q^2} +1 $ Invert again: $ \frac{A_2^2}{A_1^2} = \frac{q^2}{2ghA_1^2} +1 $ Multiply by $A_1^2$: $ A_2^2 = \frac{A_1^2q^2}{2ghA_1^2} +1 $ Get the square root:
$ A_2 = \sqrt{\frac{A_1^2q^2}{2ghA_1^2}+1} $
I cannot see where the $q^2$ on the bottom of the textbook answer comes from.