- Topic
- Related rates
- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 4
Related rates practice question 21
A camera on the ground is 120 m from a rocket’s launch pad. The rocket rises straight up at 30 m/s, and the camera tilts upward to keep it in view. How fast is the camera’s angle of elevation changing at the instant the rocket is 200 m from the camera?
Give the answer in radians per second.
The camera, the launch pad and the rocket form a right triangle with a fixed 120 m base. Is the 200 m in the question the rocket’s height, or another side?
Let be the rocket’s height and the angle of elevation, so . The 200 m is the hypotenuse of the triangle, not the height.
Differentiate: . Then substitute the height at this instant and .
A fraction, or a decimal to two decimal places.
Theory UBC Math 100 §5CLP-1 §3.2Key results
Worked solution
Strategy
Write the angle directly as an arctan of the changing height over the fixed distance, differentiate with the chain rule, and use Pythagoras to find the height at the given instant.
Working
Pythagoras: the 200 m line of sight is the hypotenuse.
Opposite over adjacent, with the 120 m side fixed; then the derivative of arctan times the inner derivative.
At h = 160, h/120 = 4/3. The angle is increasing, so the rate is positive.
Answer
The answer is 9/100, or 0.09 radians per second.
Check
Across the 200 m line of sight the rocket moves at 30 × 120/200 = 18 m/s, which turns the camera at 18/200 = 0.09 rad/s.
Where this goes wrong
Differentiating arctan(h/120) as if it were h/120, which gives 30/120 = 0.25 rad/s. The factor 1/(1 + u²) is why the camera turns more and more slowly as the rocket climbs.
Leans on
Derivatives of inverse trigonometric functions, from Implicit differentiation and inverse trigonometric functions
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