Question

a tank shape of inverted cone with diameter 40ft depth of 15 ft being filled at rate of 10cubic ft/min how fast is the depth changing when tank filled of 10 ft?

Answer 1

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Answer 2

Volume cone = 1/3*h*pi*r^2
given is rate volume is changing dV/dt = 10
required is rate depth or height is changing dh/dt
dV/dt = dh/dt * dV/dh
notice how dh will cancel on right side
now to find dV/dh we need V in terms of h
so find relationship between r and h
h of tank = 15, r of tank = 20
r/h = 20/15 --> r = 20h/15
plug this in for r in volume formula
V = 1/3pi*h*(20h/15)^2
= 1/3pi*(16/9)h^3
dV/dh = pi*(16/9)h^2
dV/dt = dh/dt * dV/dh
10=dh/dt * (pi*(16/9)h^2)
dh/dt = 10/(pi*(16/9)h^2)
find dh/dt when h=10
dh/dt = 1/pi*160/9 = .0179