Calc Help Please?

Question

A flask has the shape of a cylinder on top of a truncated cone. The bottom part of the flask is in the shape of a truncated cone with the bottom radius 10 cm, top radius 5 cm, and height 10 cm. Sitting on top of the truncated cone is a tall cylinder whose radius is 5 cm. Liquid is being poured into the flask at a rate of 2 cm^3/s. Note that the volume of a cone whose bottom radius is r and whose vertical height is h is given by V=(1/3)*pi*r^2*h.

Hints:
a) The bottom partial cone has nothing to do with the rate other than that it is already filled.
c) Think of the bottom cone as a full cone with the top taken off. So the volume of the cone: V=(1/3)*pi*r^2*h - (1/3)*PI*[(10 - .5h)^2] * (20 - h)

Questions:

a) When the liquid is 15 cm high in the flask, how fast is the height of the liquid increasing?
b) If the flask is initially empty, how long does it take the liquid to reach a height of 10cm?
c) How fast is the height of the liquid rising when the liquid is 5cm deep?

Answer 1

Let's mark bottom cone base radius R, upper radius of the water r, height of truncated cone H, volume of the truncated cone Vt, volume of cylinder Vc, and variable height h.

a. If the height of the water h > 10 then volume of the water V = Vt + Vc where Vt is constant and Vc(h) = (h-H)pir^2. Therefore dV/dt = dVc/dt = pi*r^2*dh/dt. Since dV/dt = 2 cm^3/s follows that dh/dt = 2/(pi*r^2) = 0.0127 cm/s.

b. Flask_volume = pi*R^2*(2*H)/3 - pi*r^2*H/3 = pi*(H/3)*(2*R^2 - r^2) = 1832.59 cm^3. Divided by rate gives the time to fill the flask = 916.298 sec.

c. Volume of the water within the flask can be calculated as the volume of the big cone minus volume of the big cone not filled with the water. In addition if the water level is h then the radius of the water r = 10 - h/2. Therefore volume of the water V = constant - (1/3)*pi*(20 - h)*(10 - h/2)^2. Then dV/dt = pi*(10 - h/2)^2*dh/dt. From there dh/dt = 0.01132 cm/s.

Answer 2

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