- Topic
- Related rates
- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 4
Related rates practice question 20
A camera on a tripod stands 0.4 km from a straight road. A car drives along the road at 1.5 km/min, moving away from the point on the road nearest the camera, and the camera turns to keep the car in frame. How fast is the camera rotating, in radians per minute, at the instant the car is 0.5 km from the camera?
Give the answer in radians per minute.
The camera is a fixed 0.4 km from the road, while the car’s distance along the road grows. Measuring the camera’s angle from the perpendicular to the road, which trigonometric function links those two lengths?
Let be the angle between the camera’s line of sight and the perpendicular to the road, and the car’s distance along the road from that perpendicular. Then .
Differentiate to get . At the given instant the 0.3–0.4–0.5 triangle gives .
A number.
Theory UBC Math 100 §5CLP-1 §3.2Key results
Worked solution
Strategy
The tangent of the angle relates the rotating line of sight to the car’s position. Differentiate that, then use the right triangle at the given instant to evaluate sec²θ.
Working
Differentiate with respect to time.
The car is 0.3 km along the road at this instant (3–4–5 triangle).
Radians per minute.
Answer
The answer is 2.4 radians per minute.
Check
Only the part of the car’s velocity across the line of sight turns the camera: 1.5 × 0.4/0.5 = 1.2 km/min at a distance of 0.5 km, which is 2.4 rad/min.
Where this goes wrong
Using the 0.5 km distance as x. The 0.5 km is the hypotenuse of the triangle; the car’s distance along the road is 0.3 km, and it is cos θ = 0.4/0.5 that enters the calculation.
Leans on
The derivative of tan, from Exponential and trigonometric functions
Saved in this browser only.