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 $\frac{dh}{dt}=500$:

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

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):

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

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

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