Related Rates Question

Question

Water is leaking out of an inverted conical tank at a rate of 139000 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 150 meters and the diameter at the top is 35 meters. If the water level is rising at a rate of 270 centimeters per minute when the height of the water is 40 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Answer 1

Hi John!

The first step of this problem is to unpack the text description into a cartoon sketch, a picture. We have a tank that's shaped like an upside-down funnel with water entering the top and leaking out the bottom. The current water level inside the tank is given, and the rate at which this level is currently rising is also given. We need to set up our coordinates so that both water height measured vertically from the tip of the cone (pointed down) is 40 meters at time 0, and the first derivative with respect to time is 270 centimeters per minute.

Mathematically, we have two initial conditions at time t = 0:

h(t = 0) = 40

dh/dt(t = 0) = 0.270

Notice how I converted from centimeters to meters; we can leave the time in minutes for now.

Next step, since we're dealing with a conical tank, is to figure out how much water is currently inside the tank. We're going to need the volume of a cone with a height of 40 meters and diameter...wait, we don't have that number yet! But we can get it by applying similar triangles to the large tank dimensions:

D / 40 = 35 / 150

which leads to

D = 9.333

and this seems reasonable, right? More importantly, since the conical tank is rigid, this means we can always apply this ratio to convert between the water height h(t) and diameter D(t). We have linked them because the angle of the tank is constant with respect to the water level or the flat ground.

So let's recap: we have water flowing into the tank, water leaking out the bottom, and the water level inside is rising. What would be super helpful now is the volume of a cone:

V = pi r^2 h / 3

or in terms of diameter D:

V = pi D^2 h / 6

Cool. We basically have everything we need now. Let's take the derivative with respect to time, remembering to apply the chain rule of the right hand side:

dV / dt = pi / 6 (2 D) h dD / dt + pi / 6 D^2 dh / dt

The term on the LHS is how much the water cone volume is increasing with respect to time; the first term on the RHS give the volume contribution from the growing diameter while the second term on the RHS gives the volume contribution from the increasing height. But we have most of these variables already. The problem statement provided height h and rise velocity dh / dt; we computed diameter D using similar triangles; all that remains is diameter growth rate dD / dt. But dh / dt is linked to dD / dt (how, exactly? see similar triangles above) and this means the problem is almost complete!

The final step is to account for all rates: what's entering the tank either leaves through the leaky bottom or accumulates in the growing cone. Since we know what's leaving and what's accumulating, what enters is also known.

Hope this helps!

Will A.