Word Problem Involving Tank of Water

Question

A tank of water is in the shape of a cone (assume the “point” of the cone is pointing downwards) and is leaking water at a rate of 20 cm3/sec. The base radius of the tank is 1.5 meter and the height of the tank is 3 meters. When the depth of the water is 2 meters at what rate is the (a) depth changing and (b) the radius of the top of the water changing? Don’t forget units.

Solutions:

a)________________________

b)________________________

How do I solve?

Answer 1

Hi Connor!

This is a fun little related rates question. The equation for the volume of a cone is V=(1/3)πr²h. Given that dV/dt=20cm³/s, we would like to determine what dh/dt and dr/dt are when h=2 m (or 200 cm). This will involve taking the derivative of V with respect to t. The only problem is that both r and h are also functions of t, and taking the derivative of V, in its current form, is a difficult endeavor. What we would like to do is express the V in terms of r, or h, but not both.

That requires us to come up with a between r and t, and I wish I could create a figure. The triangle formed by the cone and the triangle formed by the water levels are similar triangles. That means that the ratio of the cone's height to the cone's radius is the same as the ratio between the water level (h) and the radius at that water level (r). Since the height of the cone (3m) is twice as large as the radius of the cone (1.5 m), it follows that the water level (h) is twice as large as the radius at that level (l), Accordingly, we have h=2r.

With this relation between r and h, we can now express the volume of the cone in terms only one variable, say h. So

V = (1/3)πr²h = (1/3)π(h/2)²h =(1/12)πh³

We can now take the derivative with respect to t recognizing that h is also a function of t. This will give

dV/dt = (1/12)π(3h²) dh/dt = (1/4)πh² dh/dt

Solving for dh/dt, we have

dh/dt = 4/(πh²) dV/dt.

Now, all we have to do is to substitute our values for h and dV/dt. However, we have to be careful because dV/dt is in units of cm³/s and h is in terms of meters. To make the units consistent, I recommend expressing h in cm, which would mean that h = 200 cm. Then, substituting for dV/dt and h, we will have

dh/dt = 4/(π(200cm)²) (20 cm³/s) = 4/(π(40000cm²) (20 cm³/s) = 6.37×10^-4 cm/s

To find dr/dt, all we have to do is use the relationship between r and h that we came up with earlier, h = 2r. Then

dh/dt = 2 dr/dt

or

dr/dt=0.5 dh/dt

Then dr/dt = 0.5 (6.37×10^-4 cm/s) = 3.18×10^-4 cm/s

So for (a) we have dh/dt = 6.37×10^-4 cm/s and for (b) we have dr/dt = 3.18 ×10^-4 cm/s

Hope this works for you. If you need help with anything else, please feel free to contact me.

Bob