Could someone help me with this calculus rate problem?

Question

Water is leaking out of an inverted conical tank at a rate of 10100.0 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 9.0 m and the the diameter at the top is 4.0 m. If the water level is rising at a rate of 26.0 cm/min when the height of the water is 3.5 m, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

 

I've been stuck on this problem for days and don't know what I'm doing wrong. Any help would be appreciated. Thanks in advance!

Answer 1

Volum of cone = V = (1/3)(pi)(r^2)h

Water leaking at rate = dV/dt = -10100cm^3/min

Tank's height = h = 9m = 900cm

Tank's diameter at top = d = 4m = 400cm

Tank's radius at top = r = d/2 = 400/2 = 200cm

Solve for r in terms of h so you only have to deal with one variable in volum equation:

h/r = 900/200

h/r = 9/2

r =(2h)/9

Plug r in Volume equation:

V = (1/3)(pi)(r^2)h

V = (1/3)(pi)((2h/9)^2)h

V = (4/243)(pi)(h^3)

 

Derivative of V with respect to t is:

(dV/dt) = (4/243)(pi)(3h^2)(dh/dt)

(dV/dt) = (4/81)(pi)(h^2)(dh/dt)

dV/dt will have two numbers: one where water is leaking I.e. -10100cm3/min and other is water being pumped. Lets use p for water being pumped.

Water's height is rising at rate = dh/dt = 26cm/min at height = h = 3.5 m = 350cm

 

(dV/dt) = (4/81)(pi)(h^2)(dh/dt)

-10100 + p = (4/81)(pi)(350^2)(26)

p = 10100 + (4/81)(pi)(350^2)(26)

p = 10100 + 494122.104

p = 504222.104 cm^3/min

 

Water is being pumped into the tank at the rate of 504222.104 cubic centimetres per minute.