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.