- Topic
- Related rates
- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Related rates practice question 15
A crate on a warehouse floor is dragged toward a wall by a rope tied to it at floor level. The rope runs up to a pulley fixed on the wall 3 m above the floor, and a winch hauls the rope in at 1 m/s. At what rate is the crate’s distance from the wall changing at the moment 5 m of rope lies between the pulley and the crate?
Give a signed rate in metres per second.
The pulley, the foot of the wall and the crate make a right triangle. Which of its sides is fixed, which is the rope, and is the rope’s length growing or shrinking?
Let be the crate’s distance from the wall and the length of rope from the pulley to the crate. The 3 m height is fixed, so at every instant.
Hauling the rope in makes shrink, so . Differentiate, then find at the moment .
A signed rate in m/s, exact or to two decimal places.
Theory UBC Math 100 §5CLP-1 §3.2Key results
Worked solution
Strategy
The wall, the floor and the rope form a right triangle whose height is fixed. Write Pythagoras for it, differentiate with respect to time, and substitute the lengths of this instant and the signed rate of the rope.
Working
Pythagoras. The 3 m height of the pulley never changes.
Differentiate with respect to t; the constant 9 gives zero.
The crate’s distance from the wall at this instant.
The rope is shortening, so dL/dt = −1. The crate moves faster than the rope is hauled.
Answer
The answer is −1.25 metres per second: the crate approaches the wall at 1.25 m/s.
Check
After 0.01 s the rope is 4.99 m long and the crate √(4.99² − 9) ≈ 3.98749 m from the wall, 0.0125 m closer, so it approaches at 1.25 m/s.
Where this goes wrong
Assuming the crate moves at the speed of the rope, 1 m/s. Shortening the slanted rope by a little moves the crate further than that along the floor: here dx/dt = (5/4) dL/dt.
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