# Rearranging a formula, transpose for A2 - I'm lost

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.

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It might be useful to see what the answer in the book is. From the answers given below, there are clearly equivalent forms - which might not immediately look the same. – Mark Bennet Mar 29 '12 at 9:30
updated, thanks! – bot_bot Mar 29 '12 at 9:38
So can you see how the answers given can be put in the same form as the answer you require - even though they all look different? – Mark Bennet Mar 29 '12 at 11:00
I'm afraid I can't. I must be missing some fundamental concept. – bot_bot Mar 29 '12 at 11:59
Well you need to get everything under the same square root sign - which means squaring everything outside the square root as you move it inside, and then you need to clear fractions in the numerator and denominator of the main fraction: multiply top and bottom by $q^2$ or $A_1^2$. – Mark Bennet Mar 29 '12 at 12:17

$$q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1}$$

$$q^2=(A_1)^2\frac{2gh}{(\frac{A_1}{A_2})^2-1}$$

$$(\frac{A_1}{A_2})^2-1=(A_1)^2\frac{2gh}{q^2}$$

$$(\frac{A_1}{A_2})^2=(A_1)^2\frac{2gh}{q^2}+1$$

$$\frac{A_1}{A_2}=\sqrt{(A_1)^2\frac{2gh}{q^2}+1}$$

$$A_2=\frac{A_1}{\sqrt{(A_1)^2\frac{2gh}{q^2}+1}}$$

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Thank you for your help – bot_bot Mar 30 '12 at 8:40

$$\frac{q^2}{A_1^2} = \frac{2gh}{\left(\frac{A_1}{A_2}\right)^2-1}$$ $$\Leftrightarrow \frac{q^2}{A_1^2}\left(\left(\frac{A_1}{A_2}\right)^2-1\right) = 2gh$$ $$\Leftrightarrow \frac{q^2}{A_2^2}-\frac{q^2}{A_1^2} = 2gh$$ $$\Leftrightarrow \frac{q^2}{A_2^2}= 2gh+\frac{q^2}{A_1^2}$$ $$\Leftrightarrow \sqrt{\frac{q^2}{2gh+\frac{q^2}{A_1^2}}}= \pm A_2$$

Maybe not the fastest way, but step by step how I did it. Of course this answer can be brought into several equivalent forms.

multiplying the fraction inside the square root with $\frac{1/q^2}{1/q^2}$ gives $$\sqrt{\frac{1}{\frac{2gh}{q^2}+\frac{1}{A_1^2}}}= \pm A_2$$ now with $\frac{A_1^2}{A_1^2}$ to get $$\sqrt{\frac{A_1^2}{\frac{2ghA_1^2}{q^2}+1}}= \pm A_2$$ same expansion with $A_1^2$ starting from my first result gives $$\frac{A_1q}{\sqrt{2ghA_1^2+q^2}}= \sqrt{\frac{A_1^2q^2}{2ghA_1^2+q^2}}= \pm A_2$$ You can keep going as long as you want.... (incidently your closest solution that you have posted does not work due to a wrong invertion as already mentioned in the comments)

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Thank you for your help – bot_bot Mar 30 '12 at 8:40
@mal: no problem. I've modified my solution to show the equivalence of some solutions. Yours is not among them because your "invert again" step was wrong (should have resulted in $\frac{A_2^2}{A_1^2}=\frac{1}{\frac{2ghA_1^2}{q^2}+1}$ ). – example Mar 30 '12 at 9:44

$$\begin{eqnarray*} q &=& A_1\sqrt{\frac{2gh}{(\frac{A_1}{A_2})^2-1} }&\biggr| : A_1, (\;\;)^2\\ \left(\frac{q}{A_1}\right)^2 &=& \frac{2gh}{(\frac{A_1}{A_2})^2-1}&\biggr| (\;\;)^{-1},\cdot 2gh,+1 \\ 2gh\left(\frac{A_1}{q}\right)^2+1 &=& (\frac{A_1}{A_2})^2&\biggr| (\;\;)^{-1/2},\cdot A_1\\ \pm\frac{A_1}{\sqrt{2gh\left(\frac{A_1}{q}\right)^2+1}} &=& A_2\\ \pm\sqrt{\frac{A_1^2q^2}{2ghA_1^2+q^2}} &=& \\ \end{eqnarray*}$$

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thank you for your help – bot_bot Mar 30 '12 at 8:40