Solve the general solution of 1- calculus

This is a differential application problem (rate of change with time)

Let any moment t be the water depth h and the plane radius r.

The volume of water at t: V =( 1/3)πr? h

According to similar triangles, cone height/water depth = cone mouth radius/water surface radius.

8/h = 4/r，h = 2r

∴V = ( 1/3)πr？ h =( 1/3)π(h/2)？ h

= ( 1/ 12)πh？

dV/dt = ( 1/4)πh？ dh/dt

dh/dt = 4(dV/dt)/(πh？ )

= 4*4/(25π)

= 16/25π

≈ 0.204m/min

Solution 2- Elementary Algebra Solution

The rising speed of water surface at a certain moment = the volume increase of water at this time/the water surface area at this time.

= 4/(π r 2) = 4/(π 2.5 2) =16/25 π (m/min)

[Note: According to similarity ratio: cone height/water depth = cone mouth radius/water surface radius, r=2.5m)