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

To graph the total volume of oil over time, plot the points obtained from the completed table:

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

Draw a straight line connecting these points to show the linear increase of volume over time. Label the x-axis as 'Time (minutes)' and the y-axis as 'Volume (10⁶ cm³)'.

The volume of oil on the water after r minutes can be expressed as:

$V(t) = 4 imes 10^6 t \text{ cm}^3$

where t represents the time in minutes.

The volume of oil in the slick when the radius is r cm can be calculated using the formula for the volume of a cylindrical object:

$V = ext{Area} imes ext{Height}$

Here, area A is given by: $A = \pi r^2$

Since the thickness h is 1 millimetre (or 0.1 cm), we have: $V = \pi r^2 imes 0.1 = 0.1 \pi r^2 \text{ cm}^3$.

To find the rate of increase of the radius, we differentiate the volume equation:

$V = 0.1 \pi r^2$

Differentiating with respect to time:

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

Substituting the known volume rate of change:

$4 \times 10^6 = 0.2 \pi (50) \frac{dr}{dt}$

Now solving for rac{dr}{dt} gives:

$\frac{dr}{dt} = \frac{4 \times 10^6}{0.2 \pi (50)} \approx 1273.3 \text{ cm/min}$.

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 $r = 5000$ cm and the previously found rate rac{dr}{dt}:

$\frac{dA}{dt} = 2 \pi (5000) (1273.3) = 4 \times 10^7 \text{ cm}^2/ ext{min}$.

The distance to land is 1 km, which is 1000 m or 100000 cm. The radius increasing rate gives the speed:

rac{dr}{dt} = 1273.3 ext{ cm/min}

Time taken to reach land can be calculated as:

$t = \frac{100000 ext{ cm}}{1273.3 ext{ cm/min}} \approx 78.5 ext{ min} \approx 1.3 ext{ hours}$

Rounding to the nearest hour, it will take approximately 1 hour for the oil slick to reach land.