Question

Water is pumped into a cylindrical tank, standing vertically, at a decreasing rate given at time minutes r(t) = 130 - 10t ft^3/min for 0≤t≤5.The tank has radius 6 ft and is empty when t=0. Find the depth of the gasoline in the tank at t=3.

Answer 1

dv/dt = 130 - 10 t
so
v = 130 t - 5 t^2 + constant
since v = 0 when t = 0
v = 130 t - 5 t^2
so at t = 3
v = 130(3) - 5(9) = pi r^2 h = pi (36) h
solve for h

Answer 2

rate of height =rateofvolume/area=(130-10t)/(PI*36)

depth= INT rate*dt= (130t-5t^2)/(PI*36) over limits
=(130*3-5*9)/(PI*36) =3.05 ft check all that.