Science:Math Exam Resources/Courses/MATH110/April 2016/Question 06 (a)/Solution 1

We refer to the following picture:

We want to compute $\frac{dD}{dt}$. In order to compute this, we need to come up with an equation relating the rate we know, $\frac{dh}{dt}$, to the rate we want to know, $\frac{dD}{dt}$.

By the Pythagorean theorem, we have

$D^2=1000^2+h^2.$

We want the derivative with respect to $t$. Note that we should take the derivative before plugging in the value of $h$ or $D$.

$2D\frac{dD}{dt}=2h\frac{dh}{dt}$

so

$\frac{dD}{dt}=\frac{h}{D}\frac{dh}{dt}.$

We would like to plug in $h=2000$ and $\frac{dh}{dt}=500$. However, we still need to figure out what value to plug in for $D$. Once again, we will use the Pythagorean theorem. We get

$\begin{array}{rl}{D}^2& =1000^2+2000^2\\ D^2& =5000000\\ D& =\sqrt{5000000}\\ D& =1000\sqrt{5}.\end{array}$

So we plug in the values for $D$, $h$, and $\frac{dh}{dt}$, and get

$\begin{array}{rl}\frac{dD}{dt}& =\frac{h}{D}\frac{dh}{dt}\\ \frac{dD}{dt}& =\frac{2000}{1000\sqrt{5}}500\\ \frac{dD}{dt}& =\frac{1000}{\sqrt{5}}\\ \frac{dD}{dt}& =200\sqrt{5}\end{array}$

So we have that the distance is changing at a rate of $200\sqrt{5}\frac{\text{m}}{\text{sec}}$.

Answer: $200\sqrt{5}\frac{\text{m}}{\text{sec}}$