C4 June 2014 Q4: Water is flowing into a vase. When the depth of water is h cm, the volume of water V cm^3 is given by V=4πh(h+4). Water flows into the vase at a constant rate of 80π cm^3/s. Find the rate of change of the depth of water in cm/s, when h=6.

This question wants us to find: dh/dt. We are given: dV/dt=80π and V=4πh(h+4). The equation to use here is: dh/dt = dh/dV x dV/dt. We know dV/dt, but still need to find dh/dV. For this, we can use the reciprocal of dV/dh, which can be found be differentiating the given equation of for V as a function of h. Differentiating we find: dV/dh = 8πh+16π. Therefore: dh/dV = 1/(8πh+16π). Substituting into connected rate of change equation: dh/dt = 1/(8πh+16π) x 80π. This simplifies to: dh/dt = 10/(h+2) At h=6: dh/dt = 10/(6+2) = 1.25 cm.s^-1.