A trough of water 8 meters long and 5 meters wide has a triangle-shaped base with a height of 2 meters. Water is being pumped in at K meters3 per second. What is the rate of change of the water's depth when the water is 0.12 meters deep? In this related rates problem, it is helpful to sketch a picture of the trough. The volume of any amount of water in the trough is given by V =Bl=(1 /2) bhl since the base is triangle-shaped. The length of the trough is l=8 meters. The base and height vary over time, so we can't just plug in b = 5 and h = 2. Instead, we see that the triangular cross-section for the water in the trough is SIMILAR to the triangular base. So we have b /h=5/2 . Then we have V =(1/2)(5 h /2)h(8)=10 h 2 . Now take the derivative of both sides with respect to t to get dV /dt=20 h dh /dt . The rate of change of the volume is given as K meters3 per second, so we have K=20 h dh /dt . We have now made substitutions for all the rates of change except for the one we are trying to find (i.e. dh/dt ). So we may now plug in the value h=0.12 . Then the rate of change at 0.12 meters deep is dh/dt (h=0.12)=K /0.24 meters per second.