Monday, November 3, 2014

Suppose you have a conical tank that is 24 feet deep and 10 feet across. Water flows in at the rate of 10 cubic feet per minute. How fast is the depth rising when the depth of the tank is 8 feet?

We can use calculus to solve this problem.

The volume of a cone is 1/3(Pi)(Radius squared)(Height)

To solve this, use implicit differentiation. First we need to substitute a value for r in terms of h.  The we know that 5/12 = r/h, therefore r = (5/12)h.

V = (1/3)Pi(5/12 h)^2(h)
    = (1/3)Pi(25/144)h^3

Differentiate to get

V' = (1/3)Pi(75/144)h^2(h')

We know V' = 10, h = 8. Substitute those values into the equation and solve for h'.

10 =  (1/3)Pi(75/144)(8)h'

10 = 4.36h'

h' = 2.29

Therefore, the depth of the tank is rising at the rate of 2.29 cubic feet per minute.

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