A Funnel Has A Circular Top Of Diameter 20cm And A Height Of 30cm. When The Depth Of Liquid In The Funnel Is 12cm, The Liquid Is Dripping From The Funnel At A Rate Of 0.2cm^3/s. At What Rate Is The Depth Of The Liquid In The Funnel Decreasing?


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Oddman Profile
Oddman answered
The diameter of the liquid's surface is a function of the liquid's height. It will be
  d = (20 cm)/(30 cm)*h = (2/3)h
Then, the volume will be
  v = (1/3)*(π/4)d^2*h
  = π/12*(2/3h)^2*h
  = π/27*h^3
The derivative of this expression is
  dv/dt = (π/27)*(3h^2)*dh/dt = π/9*h^2*dh/dt
Solving for dh/dt, we get
  (dv/dt)*9/(π*h^2) = dh/dt
With your numbers, we have
  (0.2 cm^3/s)*9/(π*(12 cm)^2) = dh/dt
  (1.8/(144π)) cm/s = dh/dt
  dh/dt ≈ 0.00398 cm/s

Perhaps the key is to realize that the diameter is a function of height. It is not constant. So, the volume is proportional to the cube of height. The cube introduces a factor of 3 to the derivative.

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