- Topic
- Related rates
- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 4
Related rates practice question 23
A balloon is released from the ground and rises straight up at 2 m/s. At the moment of release, a runner 30 m away runs straight toward the release point at 4 m/s. At what rate is the distance between the runner and the balloon changing 3 seconds after the release?
Give a signed rate in metres per second.
The release point, the runner and the balloon form a right triangle that changes shape. Where is each of them after 3 seconds, and is each leg growing or shrinking?
Let be the runner’s distance from the release point and the balloon’s height, seconds after the release. Then and ; work out both at .
The distance between them satisfies . Differentiate, with and .
A signed rate in m/s, exact or to two decimal places.
Theory UBC Math 100 §5CLP-1 §3.2Key results
Worked solution
Strategy
Find where the runner and the balloon are at t = 3 from their starting positions and speeds, then differentiate the Pythagorean relation for the distance between them and substitute.
Working
Positions 3 seconds after the release.
The release point, the runner and the balloon form a right triangle.
Differentiate with respect to t and divide by 2.
The runner’s approach outweighs the balloon’s climb, so the distance is shrinking.
Answer
The answer is negative root 10, about −3.16 metres per second: the distance is shrinking.
Check
At t = 3.01 the runner is 17.96 m out and the balloon 6.02 m up, so D ≈ 18.9421 m, down about 0.0316 from √360 ≈ 18.9737.
Where this goes wrong
Using the starting positions, 30 m and 0 m, which gives −4. The speeds stay the same but the positions do not, and the rate of change of the distance depends on both; find where each one is at t = 3 first.
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