# Math Help - rates of change

1. ## rates of change

The time t, in seconds, taken by an object dropped from a height of s metres to reach the ground is given by the formula t = squareroot(s/5). Determine the rate of change in time with respect to height when the object is 125m above the ground.

so i set up the solution and everything, but the squareroots are driving me mad! please help me solve this?

thank you.

2. ## Rates of change

Hello checkmarks
Originally Posted by checkmarks
The time t, in seconds, taken by an object dropped from a height of s metres to reach the ground is given by the formula t = squareroot(s/5). Determine the rate of change in time with respect to height when the object is 125m above the ground.

so i set up the solution and everything, but the squareroots are driving me mad! please help me solve this?

thank you.
$t = \sqrt{\frac{s}{5}}$

$=\frac{\sqrt{s}}{\sqrt{5}}$

$= \frac{s^{\frac{1}{2}}}{\sqrt{5}}$

$\Rightarrow \frac{dt}{ds} = \frac{\tfrac{1}{2}s^{-\frac{1}{2}}}{\sqrt{5}}$

$=\frac{1}{2\sqrt{5s}}$

When $s = 125$, $\frac{dt}{ds}=\frac{1}{2\sqrt{5\times 125}}$

$=\frac{1}{50}$