A conical cup of radius 5 cm and height 15 cm is leaking water at the rate of 2 cm^3/min. What rate is the level of water decreasing when the water is 3 cm deep?

1 Answer
May 20, 2015

The "shape" of the water itself will be an upside down cone as well. If hh is the depth/height of the water and rr is the radius of the water at the surface, then the volume of water is V= 1/3 pi r^2 hV=13πr2h.

By using similar triangles (draw a picture), we get h=3rh=3r (since the height of the entire cup is 3 times the radius). Therefore V=1/3 pi r^2\cdot (3r)=pi r^3V=13πr2(3r)=πr3.

Differentiating both sides of this equation with respect to time results in (dV)/dt=3pi r^2 (dr)/dtdVdt=3πr2drdt so that (dr)/dt=\frac{1}{3pi r^2}(dV)/dtdrdt=13πr2dVdt (the Chain Rule was used when differentiating here).

r=1r=1 cm when h=3h=3 cm and (dV)/dt = -2\ (cm^{3})/min.

Hence, at that moment, (dr)/dt = \frac{1}{3pi}\cdot (-2)=-\frac{2}{3pi}\approx -0.2122 so that the water level is going down at a rate of approximately 0.21 cm/min.