# rates of change

• Feb 5th 2009, 03:53 PM
checkmarks
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.

thank you.
• Feb 5th 2009, 11:03 PM
Rates of change
Hello checkmarks
Quote:

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.

thank you.

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

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

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

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

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

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

$\displaystyle =\frac{1}{50}$