Question

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?

Answer 1

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

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

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

`r=1` cm when `h=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.