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

The rate of water leakage is given as $ \frac{dV}{dt} = -5 \, \text{cm}^3/\text{s} $, signifying a decrease in volume. The container is cylindrical, with its water volume described by $V = \pi r^2 h$, where $r$ is the radius and $h$ is the water height. Step 1: Differentiate the volume formula with respect to time: $ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} $. Given $r = 15$ cm and $h = 200$ cm. Step 2: Substitute known values: $ -5 = \pi (15)^2 \frac{dh}{dt} $, simplifying to $ -5 = 225\pi \frac{dh}{dt} $. Step 3: Solve for $ \frac{dh}{dt} $: $ \frac{dh}{dt} = \frac{-5}{225 \pi} $, which reduces to $ \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} $. The water height is decreasing at an approximate rate of $-0.0071 \, \text{cm/s}$.