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

We would like to relate the angle $\theta$ to the height $h$. In order to do so, we use trigonometry. By basic trigonometry, we have the equation

$\tan \theta ={\frac {h}{1000}}.$

We will differentiate both sides:

$\sec ^{2}\theta {\frac {d\theta }{dt}}={\frac {1}{1000}}{\frac {dh}{dt}}.$

Now, we solve for ${\frac {d\theta }{dt}}$:

${\frac {d\theta }{dt}}={\frac {1}{1000\sec ^{2}(\theta )}}{\frac {dh}{dt}}.$

We use the fact that $\sec \theta ={\frac {1}{\cos \theta }}$.

${\frac {d\theta }{dt}}={\frac {\cos ^{2}\theta }{1000}}{\frac {dh}{dt}}$

Now, we plug in ${\frac {dh}{dt}}=500$:

${\begin{aligned}{\frac {d\theta }{dt}}&={\frac {500\cos ^{2}\theta }{1000}}\\{\frac {d\theta }{dt}}&={\frac {\cos ^{2}\theta }{2}}\end{aligned}}$

Now, we observe from our picture that $\cos(\theta )={\frac {1000}{D}}$.

${\frac {d\theta }{dt}}={\frac {1000000}{2D^{2}}}$

We plug in $D^{2}=5000000$ (from part a):

${\begin{aligned}{\frac {d\theta }{dt}}&={\frac {1000000}{2\cdot 5000000}}\\&={\frac {1}{10}}\end{aligned}}$

So the rate of change is ${\frac {1}{10}}{\frac {\text{radians}}{\text{sec}}}$.

$\color {blue}{\frac {d\theta }{dt}}={\frac {1}{10}}{\frac {\text{radians}}{\text{sec}}}$