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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
up vote 4 down vote accepted

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






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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

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