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

To represent the total volume of oil on a graph, you would plot the points:

• (1, 4)
• (2, 8)
• (3, 12)
• (4, 16)
• (5, 20)
• (6, 24)

Connect these points with a straight line to indicate a linear increase. The x-axis represents time (minutes), while the y-axis shows the volume of oil (10^6 cm³).

The equation for the volume of oil on the water after $t$ minutes is given by:

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

where $V(t)$ is in cm³ and $t$ is in minutes.

The volume of oil in the slick when the radius is $r$ cm and thickness is 1 mm (or 0.1 cm) is given by:

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

To find the rate of change of the radius, we differentiate:

$\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$

Calculating $\frac{dV}{dr}$: $\frac{dV}{dr} = 0.2 \pi r$

Setting $r = 5000$ cm (which is 50 m): $\frac{dV}{dt} = 4 \times 10^{6}$

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

Calculating gives: $\frac{dr}{dt} \approx 1273.3 \text{ cm/min}$

The area $A$ covered by 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 $\frac{dr}{dt}$ from previous calculations: $\frac{dA}{dt} = 2\pi r (\text{previous rate})$

At $r = 50m (5000 cm)$: $\frac{dA}{dt} = 2\pi (5000)(1273.3) \approx 4 \times 10^{7} \text{ cm}^2/min$

The area covered by the oil slick is given by,

$A = \pi r^2 = \pi (1000)^2 = 10^6 \pi \, \text{cm}^2$

Given the rate of spill: $\text{Rate} = 4 \times 10^{6} \text{ cm}^3/min$

The time $t$ required can be calculated by: $t = \frac{\pi (1000)^2}{4 \times 10^{6}} \approx 785.398 \text{ minutes}$

Converting minutes to hours: $\text{Time} \approx \frac{785.398}{60} \approx 13.09 \text{ hours}$

Thus, to the nearest hour, it will take approximately 13 hours.