Bob B. answered • 04/22/20

Tutor for Algebra, Calculus, Physics, and Electrical Engineering

Hi Connor!

This is a fun little related rates question. The equation for the volume of a cone is V=(1/3)πr^{2}h. Given that dV/dt=20cm^{3}/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)πr^{2}h = (1/3)π(h/2)^{2}h =(1/12)πh^{3}

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(3h^{2}) dh/dt = (1/4)πh^{2} dh/dt

Solving for dh/dt, we have

dh/dt = 4/(πh^{2}) 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 cm^{3}/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 cm^{3}/s) = 4/(π(40000cm^{2}) (20 cm^{3}/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