Water is flowing into a rightcircular cone at the rate r (volume of water per unit time). The cone has radius a, altitude b and the vertex or "tip" is pointing downwards. Find the rate at which the surface is rising when the depth of the water is y.
Always a good first step to a problem like this, is to draw a diagram, making sure to label any lengths. The unknown in this question, is the rate at which the surface of water, in the cone, is rising. However, this could be better worded (more mathematically worded) as the rate of change of y, since y is the depth of the water. Now as the water goes into the cone, we know that y increases as time goes by. Therefore, the unknown, as a mathematical symbol, is dy/dt. So this is what we want to express in terms of a, b, r and y, since these are the variables we are given in the question (the data). Now, r is a constant, but it's also a rate. More specifically, the rate of change of volume of the water in the cone. Therefore, it can be expressed as dV/dt. Now, we can express V in terms of y, where y is the altitude of a smaller cone within the main cone (diagram). We do not know the radius of this new cone, but we can call it x. So now we have V=(pi)x2y/3. The question now is, is x independent of y, or more simpy, can x be expressed in terms of y. Well, x does increase as y increases, so you'd expect there to be some kind of connection. These two cones are similar, just like similar triangles. Therefore, x : y = a : b, or x/y = a/b. So x = ay/b. Now, if we substitute this into the equation for V, we have:V = (pi)a2y3/3b2.Now, there are two variables here, which change with respect to time, V and y. So if we differentiate this with respect to time, using partial differentiation, we should get a dV/dt term and dy/dt term. Both these terms are good, as one is our unknown, dy/dt, and the other is a constant given to us, dV/dt = r.Differentiate:dV/dt = r = (pi)a2y2(dy/dt)/b2.Rearange to express dy/dt:dy/dt=rb2/(pi)a2y2
