A tank in the shape of a right circular cone is being filled with water at a rate of 5 ft^3 per minute, but water is also flowing out at a rate of 1 ft^3 per minute. The tank is 60 feet deep and 40 feet across the top.

A) How quickly is the depth of the water changing when the water is 42 ft deep?

B) If the tank is empty, how long will it take for the tank to be 1/2 full?

C) How deep is the water when the tank is 1/2 full? How quickly is the depth of the water changing when the tank is 1/2 full?

I’m really not sure if these are right

A) dv/dt=5-1=4, h=60, r=20

find dh/dt when h=42

h/r=60/20, r=h/3, dr/dt=1dh/3dt

60/20=42/r r=14ft.

V=pr^2h/3 dv/dt=p/3[2rhdr/dt+r^2dh/dt]

4=p/3[2(14)(42)(1dh/3dt)+196dh/dt]

4=p/3[392dh/dt+196dh/dt] = p/3[588dh/dt] = dh/dt=1/49p ft/min.

B) V=pr^2h/3 p(20)^2(60)/3 = 8000p(1/2) = 4000p

dv/dt=4ft/min if 1 min for 4ft, 16 hours 40 min for 4000ft

C) find h when V=4000

3v/pr^2=h h=3(4000)/p(20)^2 = 12000/400p = 30pft^3

find dh/dt when h=30

4=p/3[2(20)(30)(1dh/3dt)+20^2dh/dt]

4=p/3[800dh/dt]

dh/dt=1/66.67p ft/min.