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

MATH110 April 2016

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Question 06 (b)

A television camera is positioned $1000{\text{m}}$ from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Answer the questions below. Make sure your solution includes a sketch that is labeled with the same variables used in your calculation. Your answer should be a numerical value, but you do not need to simplify it. Assume the rocket rises vertically and its speed is $500{\text{m}}/{\text{s}}$ when it has risen $2000{\text{m}}$. (b) If the television camera is always kept aimed at the rocket, how fast is the camera’s angle of elevation changing at that same moment?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Try to find a trigonometric function that relates the angle $\theta$ to the height $h$ in the picture.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 ${\frac {1}{10}}{\frac {\text{radians}}{\text{sec}}}$. $\color {blue}{\frac {d\theta }{dt}}={\frac {1}{10}}{\frac {\text{radians}}{\text{sec}}}$