A cylindrical tank standing upright (with one circular base on the ground) has a radius of 1212 cm for the base. how fast does the water level in the tank drop when the water is being drained at 2323 cm33/sec? note that the volume of a cylinder is v=πr2hv=πr2h where rr is the radius of the base and hh is the height of the cylinder.

Question

Dayanara Beasley

Mathematics

14 July, 22:24

A cylindrical tank standing upright (with one circular base on the ground) has a radius of 1212 cm for the base. how fast does the water level in the tank drop when the water is being drained at 2323 cm33/sec? note that the volume of a cylinder is v=πr2hv=πr2h where rr is the radius of the base and hh is the height of the cylinder.

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Ben Hart · Answer 1

V = pi * r^2*h, r is constant, so that

Change of V in time (dV/dt or V' say) = pi*r^2*v_h, v_h = speed of change of the height.

So, v_h = V' / (pi*r^2) = 2323 / pi / 1212^2 = ... cm/s.

I am not sure you have 1212 or you mean 12.12, and 23.23 cm. Anyway, that is sth easy to do