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

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

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

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

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

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