- Topic
- Related rates
- Difficulty
- Capstone
- Suggested time
- 9 min
- Marks
- 5
Related rates practice question 25
A water trough is 6 m long, and its cross-section is an isosceles trapezoid 1 m wide at the bottom, 2 m wide at the top, and 1 m deep. Water is pumped in at 0.3 m³/min. How fast is the water level rising when the water is 0.5 m deep?
Give the answer in metres per minute.
As the water rises, its surface widens from the 1 m bottom towards the 2 m top. How does the width of the surface depend on the depth ?
Let be the depth. The water surface is wider than the bottom: by similar triangles its width is (it runs from 1 at to 2 at ).
Cross-sectional area of the water is a trapezoid: . Multiply by the length for the volume, differentiate, and substitute .
A fraction, or a decimal to three decimal places.
Theory UBC Math 100 §5CLP-1 §3.2Key results
Worked solution
Strategy
Express the volume of water as a function of depth alone using the geometry of the cross-section, differentiate with respect to time, and solve for dh/dt at the given depth.
Working
Surface width grows linearly from 1 (bottom) to 2 (top) over a depth of 1.
Length times trapezoid area.
Differentiate with respect to time.
At h = 0.5.
Answer
The answer is one thirtieth of a metre per minute, about 0.033.
Check
At 0.5 m deep the water surface is 1.5 m wide and 6 m long, 9 m², and 0.3 m³/min spread over it raises the level 0.3/9 ≈ 0.033 m/min.
Where this goes wrong
Treating the cross-section as a rectangle of fixed width, which gives dh/dt = 0.3/(6 × width). The surface widens as the water rises, and that variable width is the whole difficulty of the problem.
Leans on
Similar triangles, from Functions, graphs and inverses
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