The v-shaped tank is shown with dimensions L, W, and H. It is partially filled to a level x (where the top surface has a width y). If the rate of filling the tank is 0.002 m³/sec, find the rate at which the depth of the liquid is changing.

To solve this we need to relate the volume of the liquid to the dimensions of the liquid portion:

 .

Since we need to express this in terms of one variable, we need to replace the varying value of y if possible. The triangle end of the tank and the triangle end of the filled portion are similar, so we have

, or

Now we use this to replace y in the volume equation:

Now we can differentiate with respect to time to relate the rates of change of V and x:

.

For W = 0.5m, L = 3.0m, H = 0.6m and x = 0.25m, we use and get

, or