v = 1/3 pi r^2 h
From the shape of the cone, you know that r = h/6
so, dr/dt = 1/6 dh/dt
dv/dt = pi/3 (2rh dr/dt + r^2 dh/dt)
= pi/3(2(h/6)h(1/6 dh/dt) + (h/6)^2 dh/dt)
= pi/18 h^2 dh/dt
Now just solve for dh/dt when
dv/dt = -1/4
h=4
Water is leaking from the bottom of a tank in the shape of an inverted cone having an altitude of 12 feet and a radius of 2 feet. If the water is leaking at the rate of 0.25 cubic feet per minute, how
fast is the water level decreasing when the water is 4 feet deep?
1 answer
1 answer