Calculus v=ds/dt making dt the subject

[Starblazer]

velocity is

\(\displaystyle v = \frac{{2\sqrt 3 }}{3}{s^{3/4}}\)

I need to re arrange (equation below) to find t (time)



\(\displaystyle v = \frac{{ds}}{{dt}} \Rightarrow \frac{{2\sqrt 3 }}{3}{s^{3/4}} = \frac{{ds}}{{dt}}\)

In Solution dt is given as

\(\displaystyle dt = \frac{{\sqrt 3 }}{2}{s^{ - 3/4}}ds\)

I do not understand how to get to dt. I know that \(\displaystyle \frac{{ds}}{{dt}}\) can be treated "like a fraction."

Can some please spell out the rearranging process.

Kind Regards

 

[daon2]

Do you know how to solve

\(\displaystyle \dfrac{x}{y} = z\) for \(\displaystyle y\)? If so, is it the exponents that are giving you problems?

 

[Starblazer]

THank you for your help

\(\displaystyle v = \frac{{ds}}{{dt}} \Rightarrow \frac{{2\sqrt 3 }}{3}{s^{3/4}} = \frac{{ds}}{{dt}}\)

\(\displaystyle \frac{{2\sqrt 3 }}{3}{s^{3/4}}dt = ds\)

\(\displaystyle \frac{1}{{\frac{{2\sqrt 3 }}{3}{s^{3/4}}}}ds = dt\)

\(\displaystyle \frac{3}{{2\sqrt 3 }}{s^{ - 3/4}}ds = dt\)

\(\displaystyle \frac{3}{{2\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}{s^{ - 3/4}}ds = dt\)

\(\displaystyle \frac{{3\sqrt 3 }}{6}{s^{ - 3/4}}ds = dt\)

\(\displaystyle \frac{{\sqrt 3 }}{2}{s^{ - 3/4}}ds = dt\) as required

 

[HallsofIvy]

Of course, the original problem was not to "find dt" but to find t. I presume you know that \(\displaystyle \int dt= t\)+ constant, so now you need to integrate both sides of the equation.