- Topic
- Related rates
- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Related rates practice question 16
A 3.4 m ladder leans against a vertical wall. Its foot slides away from the wall at 0.45 m/s. At what rate is the angle between the ladder and the ground changing at the moment the foot is 1.6 m from the wall?
Give a signed rate in radians per second.
The ladder’s length is fixed while the foot’s distance from the wall changes. Which trigonometric ratio of the angle with the ground uses exactly those two lengths?
Let be the distance from the wall to the foot and the angle with the ground. The ladder is the hypotenuse, so at every instant.
Differentiate: . At this instant the top is m up the wall.
A signed rate in rad/s, exact or to two decimal places.
Theory UBC Math 100 §5CLP-1 §3.2Key results
Worked solution
Strategy
Relate the angle to the changing length through the trigonometric ratio that uses the fixed hypotenuse, differentiate with respect to time, and read sin θ off the triangle at this instant.
Working
Adjacent over hypotenuse; the length of the ladder is fixed.
Differentiate with respect to t; the chain rule brings in dθ/dt.
The height of the top at this instant, from Pythagoras.
Since 3.4 sin θ = 3. Negative: the ladder is getting flatter.
Answer
The answer is −0.15 radians per second: the angle is shrinking at 0.15 rad/s.
Check
In 0.1 s the foot reaches 1.645 m and the angle arccos(1.645/3.4) falls from about 1.0808 to 1.0658 rad, a drop of 0.015 rad.
Where this goes wrong
Reporting +0.15, either by dropping the sign or by differentiating cos θ as +sin θ. The foot moving out makes the ladder flatter, so the angle with the ground decreases and its rate is negative.
Leans on
Derivatives of sine and cosine, from Exponential and trigonometric functions
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