- Topic
- Related rates
- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Related rates practice question 8
A 5 m ladder leans against a vertical wall. The foot of the ladder slides away from the wall at 0.6 m/s. At what rate is the height of the top of the ladder changing when the foot is 3 m from the wall?
Give a signed rate in metres per second, taking upward as positive.
The ladder, the wall and the ground make a right triangle. Which of its sides change as the foot slides, and which stays fixed?
Let be the distance from the wall to the foot and the height of the top. The ladder length is fixed, so at every instant.
Differentiate that relationship with respect to time before substituting .
A decimal to two decimal places, negative if the top is moving downward.
Theory UBC Math 100 §5CLP-1 §3.2Key results
Worked solution
Strategy
Write a relationship between the two changing lengths that holds at every instant, differentiate it with respect to time, and only then substitute the values at the instant of interest.
Working
Pythagoras. The ladder length is constant, which is what makes this equation usable.
Differentiate with respect to time. The constant on the right gives zero.
Find the missing length at this instant, using the original relationship.
Solve for the unknown rate, then substitute.
The negative sign says the top is descending, as it must be while the foot slides away from the wall.
Answer
The answer is −0.45 m/s, that is, descending at 0.45 m/s.
Check
After 0.01 s the foot is at 3.006 m and the top at about 3.99549 m, 0.0045 m lower, so it falls at about 0.45 m/s.
Where this goes wrong
Substituting x = 3 into the Pythagorean relationship before differentiating. That makes x a constant, so dx/dt drops out and 9 + y² = 25 differentiates to 2y·dy/dt = 0, which wrongly says the top is not moving. Substitute only after differentiating.
Leans on
Implicit differentiation, from Implicit differentiation and inverse trigonometric functions
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