A conical tank has height 3 m and radius 2 m at the top. Water flows in at a rate of 1.8m^3/min. How fast is the water level rising when it is 2.7 m?

2 answers

What is the radius of the tank at the bottom?

volumetank=PI*rad^2*h
volume of airspace=PI*rad^2*(3-h)^2/3^2 *(3-h)

volumewater=volumetank-volumeairspace
= PI(rad^2*h-rad^2*(3-h)^2/9 * (3-h)
take derivative wrespect to time, you know dVwater/dh=1.8, and h=2.7, solve for dh/dt
r = 2/3 h

v = 1/3 h π r^2 = 1/3 h π 4/9 h^2 =
... 4/27 π h^3

dv/dt = 4/27 π 3 h^2 dh/dt = 4π/9 h^2 dh/dt

dh/dt = 1.8 / [(4π/9) 2.7^2]