A stone drops into a pond, creating a circular ripple - HSC - SSCE Mathematics Extension 1 - Question 8 - 2017 - Paper 1

To find the rate at which the area is increasing, we start with the formula for the area of a circle:

$A = heta r^2$

where $A$ is the area and $r$ is the radius. We need to find the rate of change of the area with respect to time, which is given by:

rac{dA}{dt} = rac{dA}{dr} \cdot \frac{dr}{dt}

First, we calculate rac{dA}{dr}:

$\frac{dA}{dr} = 2\pi r$

Substituting in the value for $r = 15$ cm:

$\frac{dA}{dr} = 2\pi (15) = 30\pi$

Next, we know from the problem statement that the radius increases at a rate of rac{dr}{dt} = 5 cm s^-1.

Now, substituting the values into the rate of change formula:

$\frac{dA}{dt} = (30\pi) \cdot (5) = 150\pi \, \text{cm}^2 \, \text{s}^{-1}$

Therefore, the area enclosed within the ripple is increasing at a rate of $150\pi \, \text{cm}^2 \, \text{s}^{-1}$ when the radius is 15 cm.