4
$\begingroup$

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.

  • 0
    The (unspecified here) context will determine whether in taking the square root, we should use $\pm$. In general one should, but perhaps the $A_i$ are "naturally" positive.2012-03-30

4 Answers 4

4

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

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

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

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

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

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

3

$\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)

  • 0
    @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 $\frac{A_2^2}{A_1^2}=\frac{1}{\frac{2ghA_1^2}{q^2}+1}$ ).2012-03-30
1

$ \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*} $

0

\begin{align} q &= A_1\sqrt\frac{2gh}{\left(\frac{A_1}{A_2}\right)^2-1} \\ \frac{q^2}{A_1^2} &= \frac{2gh}{\left(\frac{A_1}{A_2}\right)^2-1} \\ \left(\frac{A_1}{A_2}\right)^2-1 &=\frac{2ghA_1^2}{q^2} \\ \frac{A_1}{A_2} &= \sqrt\frac{2ghA_1^2 + q^2}{q^2} \\ A_2 &= A_1\sqrt\frac{q^2}{2ghA_1^2 + q^2} \\ \end{align}