Related rates question

Question

A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. Water is flowing into the tank at a rate of 10 cubic feet per minute. Find the rate of change of the depth of the water when the water is 6 feet deep.

Answer 1

V_cone = (1/3)∏r²h with (dv/dt) = 10 and r = 5 when h = 12 find (dh/dt)

First rewrite the volume equation in terms of h (r/h) = (5/12) so r =(5/12)h

V = (∏/3)[(5/12)h]²•h

V = (∏/3)(25/144)h³

(dv/dt) = 3(∏/3)(25/144)h²(dh/dt)

10 = (25∏/144)(6)²(dh/dt)

10 = (25∏/144)(36)(dh/dt)

10 = (25∏/4)(dh/dt)

10(4/25∏) = (dh/dt)

(8/5∏) = (dh/d)t

Therefore the height of the water in the tank is increasing at a rate of (8/5∏) feet per minute