An oil-spill occurs off-shore in an area of calm water with no currents - Leaving Cert Mathematics - Question 8 - 2015

The graph should plot Time (minutes) on the x-axis (1 to 6 minutes) against Volume $(10^6 \text{ cm}^3)$ on the y-axis (0 to 24). The graph should show a linear increase, starting at the origin (0,0) and passing through the points (1,4), (2,8), (3,12), (4,16), (5,20), (6,24).

The equation for the volume of oil on the water after $t$ minutes can be expressed as:

$V(t) = 4 \times 10^{6} t$

The volume $V$ of oil in the slick can be calculated using the formula for the volume of a cylinder:

$V = \pi r^2 h$

Given that the thickness $h = 0.1 \text{ cm}$ (1 millimetre), the equation simplifies to:

$V = \pi r^2 (0.1) = 0.1 \pi r^2 \text{ cm}^3$

To find the rate of change of the radius, we can use the relationship:

$\frac{dV}{dt} = 4 \times 10^{6} \text{ cm}^3 \text{ per minute}$

Using the equation from part (i):

$V = \pi r^2 (0.1)$

Differentiating with respect to time:

$\frac{dV}{dt} = 0.2\pi r \frac{dr}{dt}$

At $r = 50 \text{ m} = 5000 \text{ cm}$, we calculate:

$\frac{dV}{dt} = 4 \times 10^{6} = 0.2\pi (5000) \frac{dr}{dt}$

This results in:

$\frac{dr}{dt} = \frac{4 \times 10^{6}}{0.2 \pi (5000)} \approx 1273.3 \text{ cm per minute}$

The area $A$ of the oil slick is given by:

$A = \pi r^2$

Differentiating with respect to time:

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

Substituting in for $r$ and $\frac{dr}{dt}$:

At $r = 5000$ cm and $\frac{dr}{dt} = 1273.3$ cm/min:

$\frac{dA}{dt} = 2\pi (5000) (1273.3) = 4 \times 10^{7} \text{ cm}^2 \text{ per minute}$

The distance to land is 1 km = 1000 m = 100000 cm.

Given that the radius $r$ expands with $\frac{dr}{dt} = 1273.3 \text{ cm/min}$, the time $t$ it takes to reach land can be calculated as:

$r = 100000 ext{ cm}$

$t = \frac{100000 ext{ cm}}{1273.3 \text{ cm/min}} \approx 78.5 \text{ minutes}$

This converts to hours:

$t \approx \frac{78.5}{60} \approx 1.3 \text{ hours}$

Thus, to the nearest hour:

$t \approx 1 \text{ hour}$