volumetank=PI*2^2*h
dv/dt=4PI* dh/dt
dh/dt=1/2PI m^3/min
Water is flowing into a vertical cylindrical tank of diameter 4 m at the rate of 2 m3/min. Find the rate at which the depth of the water is rising. (Round your answer to three decimal places.)
2 answers
the area of the water's surface is 4π m^2
since volume = area * thickness, the thickness is changing by
2/(4π) = 1/(2π) m/min
since volume = area * thickness, the thickness is changing by
2/(4π) = 1/(2π) m/min