Question

A cylindrical water container has developed a leak at the bottom. The water is leaking at the rate of 5 cm$^3$/s from the leak. If the radius of the container is 15 cm, find the rate at which the height of water is decreasing inside the container, when the height of water is 2 meters.

Hint:

When solving related rates problems, express the quantity of interest (height in this case) in terms of the given quantities (volume and radius). Use the chain rule to differentiate and find the rate at which the quantity is changing.

Answer 1

We are given that the volume of water leaking is $ \frac{dV}{dt} = -5 \, \text{cm}^3/\text{s} $. The negative sign indicates that the volume is decreasing. The container is cylindrical, so the volume of water in the container at any time is given by the formula: \[ V = \pi r^2 h \] where $r$ is the radius of the base and $h$ is the height of the water. Step 1: Differentiate the volume formula with respect to time $t$: \[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \] We know that the radius $r = 15$ cm, and the height of the water is given as $h = 200$ cm (since the height of water is 2 meters, which equals 200 cm). Step 2: Substitute the given values into the equation: \[ -5 = \pi (15)^2 \frac{dh}{dt} \] \[ -5 = \pi \times 225 \times \frac{dh}{dt} \] \[ -5 = 225\pi \frac{dh}{dt} \] Step 3: Solve for $\frac{dh}{dt}$: \[ \frac{dh}{dt} = \frac{-5}{225 \pi} \] \[ \frac{dh}{dt} = \frac{-1}{45 \pi} \] Step 4: Approximate the rate: \[ \frac{dh}{dt} \approx \frac{-1}{141.37} \approx -0.0071 \, \text{cm/s} \] Thus, the rate at which the height of water is decreasing is approximately $-0.0071 \, \text{cm/s}$.