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Chain Rule, Log Differentiation & Rate of Change · Question 125 of 130

How fast is the area of the water surface growing when the water is 4 m deep?

Question 125

MediumOriginal question

Water is poured at a steady 2 m³/min into a tank shaped like a cone, held vertex down. The tank is 10 m deep and the radius of its top is 5 m. When the water is hh m deep, its surface is a circle of radius rr m.

How fast is the area of the water surface growing when the water is 4 m deep?

Options for question 125

Show the answer and explanation
CorrectNot quiteAnswer A

Correct answer: A

Explanation

  1. A=πr2=πh24A = \pi r^{2} = \frac{\pi h^{2}}{4}, so dAdt=πh2dhdt\frac{dA}{dt} = \frac{\pi h}{2}\frac{dh}{dt}.

  2. At h=4h = 4, dhdt=12π\frac{dh}{dt} = \frac{1}{2\pi}: dAdt=2π×12π=1\frac{dA}{dt} = 2\pi \times \frac{1}{2\pi} = 1 m²/min.

Worth noting

4 m²/min takes the surface radius to be hh instead of h2\frac{h}{2}.

Concept
A rate of change through a related quantity
Typical time
120 s
Your time
—
The rule this testsRate of change of quantities
Practise this chapter →

Ten questions from Chain Rule, Log Differentiation & Rate of Change, at your own pace.

Practise this chapter →

Ten questions from Chain Rule, Log Differentiation & Rate of Change, at your own pace.