It rains a lot during the winter in Vancouver. And there are lots of constructions on the UBC campus. Sometimes the construction workers will use traffic cones to warn the drivers that there is a construction ahead. On a rainy day, someone makes fun and puts a traffic cone at the construction near Totem Park residence upside down. So that the rain goes in the cone and start to fill it up. Assume that the bottom of the cone is a circle with a diameter of 50 cm and it has a height of 70 cm. If the height of the water is increasing at a rate of 3 cm/min, what is the rate of the rain filling up the traffic cone as it reaches a height of 40 cm?

Solution:

First of all, we know that the equation of the volume of a cone is V=(π⁄3)r^2(h). We are given that r/h= 25/70, so we can get r= (5/14)h. As we substitute (5/8)h for r and differentiate the function using power rule and chain rule, we can get dV/dt = {(25π)/196}(h^2)(dh/dt). When we plug in 40 cm for h^2 and 3 cm/min for dh/dt, we will get dV/dt = 1923 cm^3/min, which is the rate of the rain filling up the traffic cone.