User:Webbam/sandbox

I'll be as explicit as possible.

First, square roots (or any powers) are distributive over multiplication, so, for example

$$ \sqrt{AB} = \sqrt{A}\sqrt{B}$$

and $$\sqrt{\frac{A}{B}}$$ can be written in all of the following ways

$$\sqrt{\frac{A}{B}} = \sqrt{A\frac{1}{B}} = \frac{\sqrt{A}}{\sqrt{B}} = \sqrt{A}\sqrt{\frac{1}{B}} = \sqrt{A}\frac{1}{\sqrt{B}}$$

Being comfortable with these rules makes it easier to see what's going on with your equations. Starting with:

$$2C \sqrt{\frac{k_ak_mk_gk_p}{\tau_m}} = \frac{1}{\tau_m}$$

A slight rewrite to make the next step clearer:

$$2C \sqrt{k_ak_mk_gk_p}\frac{1}{\sqrt{\tau_m}} = \frac{1}{\tau_m}$$

Divide/multiply each side by various terms to get $$C$$ on its own:

$$C = \frac{1}{2}\frac{1}{\sqrt{k_ak_mk_gk_p}}\sqrt{\tau_m}\frac{1}{\tau_m}$$

The next bit that might be confusing: what's $$ \sqrt{\tau_m}\frac{1}{\tau_m}$$? Well,

$$\frac{\sqrt{A}}{A} = \frac{\sqrt{A}}{(\sqrt{A})^2} = \frac{\sqrt{A}}{\sqrt{A}\sqrt{A}} = \frac{1}{\sqrt{A}}$$

so

$$C = \frac{1}{2}\frac{1}{\sqrt{k_ak_mk_gk_p}}\frac{1}{\sqrt{\tau_m}}$$

The rest of the minor rearrangements/tidying up comes from the rules of multiplying fractions (since you can do multiplications and divisions in any order, you can collect the terms of the numerator together and the same for the denominator, e.g., $$\frac{A}{B}\frac{C}{D} = \frac{AC}{BD}$$), and again, the fact that taking the square root distributes over multiplication, so

$$C = \frac{1}{2}\frac{1}{\sqrt{k_ak_mk_gk_p}\sqrt{\tau_m}}$$

$$C = \frac{1}{2}\frac{1}{\sqrt{k_ak_mk_gk_p\tau_m}}$$

Now, being less explicit, these are the thought processes and steps I would go through:

$$2C \sqrt{\frac{k_ak_mk_gk_p}{\tau_m}} = \frac{1}{\tau_m}$$

"Divide and multiply each side..."

$$C = \frac{1}{2}\sqrt{\frac{\tau_m}{k_ak_mk_gk_p}}\frac{1}{\tau_m}$$

"Ahh, $$\sqrt{\tau_m}/\tau_m = 1/\sqrt{\tau_m}$$"

$$C = \frac{1}{2}\frac{1}{\sqrt{k_ak_mk_gk_p}}\frac{1}{\sqrt{\tau_m}}$$

"Tidy up"

$$C = \frac{1}{2}\sqrt{\frac{1}{k_ak_mk_gk_p\tau_m}}$$