Question

1. A drum of oil in the shape of a cone is leaking oil at a constant rate of . The base radius of the drum is 7 cm and the height of the drum is 15 cm.
1.1 At what rate is the depth of oil in the drum changing when the depth of oil is 5 cm?

Answer 1

Solution. Denote the base radius of the drum by $r$, the height of the drum by $h_0$, and the rate of leaking oil by $l$. Let the variable $V$ denote the volume of oil in the drum, and let the variable $h$ denote the depth of oil in the drum. Then

$$\frac{dV}{dt} = S \frac{dh}{dt}$$

where

• $-\frac{dV}{dt} = l$ (if the rate of leaking oil is positive, the volume of oil is decreasing, this explains the minus sign),
• $-\frac{dh}{dt}$ is the rate of decrease of the depth of oil (if the rate of decrease of the depth of oil is positive, then the depth of oil is decreasing, this explains the minus sign),
• $S$ is the area of the top surface of the oil in the drum, and this surface is the horizontal cross section of the drum at the height $h$.

Assuming that the cone stands on its base, not on its tip, this horizontal cross section of the drum is a disk of radius

$$r \frac{h_0-h}{h_0} = r \left(1 -\frac{h}{h_0}\right),$$

hence its area

$$S = \pi \left(r \left(1 -\frac{h}{h_0}\right)\right)^2.$$

When the depth of oil $h = 5\ \mathrm{cm}$,

$$S = \pi \left(7\ \mathrm{cm}\times \left(1 -\frac{5\ \mathrm{cm}}{15\ \mathrm{cm}}\right)\right)^2 = 68.417\ \mathrm{cm}^2.$$

Then

$$-\frac{dh}{dt} = \frac{l}{S} = \frac{l}{68.417\ \mathrm{cm}^2}.$$

Answer. The depth of oil in the drum is decreasing at the rate

$$\frac{l}{68.417\ \mathrm{cm}^2}$$

where $l$ is the rate of leaking oil.