To find the rate at which oil is flowing from the site of the rig, we'll calculate the rate of change of the volume of the oil slick.

Step 1: Express the Volume of the Oil Slick

The oil slick is in the shape of an elliptical cylinder with a constant thickness. The volume $V$ of the slick can be given by:

$V = \text{Area of the ellipse} \times \text{Thickness}$

The area $A$ of an ellipse is given by:

$A = \pi \times \frac{a}{2} \times \frac{b}{2}$

where $a$ is the length of the major axis and $b$ is the length of the minor axis.

Therefore, the volume $V$ of the oil slick is:

$V = \pi \times \frac{a}{2} \times \frac{b}{2} \times \text{Thickness}$

Since the thickness is 7 inches, we need to convert it to feet:

$\text{Thickness} = \frac{7}{12} \text{ feet}$

So,

$V = \pi \times \frac{a}{2} \times \frac{b}{2} \times \frac{7}{12}$

Simplifying, we get:

$V = \frac{\pi \times a \times b \times 7}{48} \text{ cubic feet}$

Step 2: Differentiate the Volume with Respect to Time

To find the rate at which the volume is changing, we differentiate the volume $V$ with respect to time $t$:

$\frac{dV}{dt} = \frac{\pi \times 7}{48} \times \left(\frac{d(a \times b)}{dt}\right)$

Using the product rule for differentiation:

$\frac{dV}{dt} = \frac{\pi \times 7}{48} \times \left(a \frac{db}{dt} + b \frac{da}{dt}\right)$

Step 3: Substitute the Given Values

We are given:

• $a = 8$ miles $= 8 \times 5280$ feet
• $b = 0.2$ miles $= 0.2 \times 5280$ feet
• $\frac{da}{dt} = 40$ feet per hour
• $\frac{db}{dt} = 4$ feet per hour

First, calculate the dimensions in feet:

$a = 8 \times 5280 = 42240 \text{ feet}$ $b = 0.2 \times 5280 = 1056 \text{ feet}$

Now, substitute the values into the derivative:

$\frac{dV}{dt} = \frac{\pi \times 7}{48} \times \left(42240 \times 4 + 1056 \times 40\right)$

Calculate each term:

$42240 \times 4 = 168960$ $1056 \times 40 = 42240$

Add these results:

$168960 + 42240 = 211200$

Now, plug this into the derivative:

$\frac{dV}{dt} = \frac{7\pi}{48} \times 211200$

Finally, simplify:

$\frac{dV}{dt} = \frac{7 \times 211200 \times \pi}{48} \approx 96434.3 \times \pi \text{ cubic feet per hour}$

So the rate at which oil is flowing from the site of the rig is approximately:

$\frac{dV}{dt} \approx 302998 \text{ cubic feet per hour}$

Final Answer:

Oil is flowing at approximately 302,998 cubic feet per hour.

Would you like more details or have any questions? Here are some related questions to consider:

1. How would the rate of oil flow change if the thickness of the slick were doubled?
2. What is the effect of increasing the rate of expansion of the minor axis?
3. If the thickness of the oil slick decreases over time, how would that impact the volume rate of change?
4. What happens to the oil flow rate if both the length and width expansion rates are doubled?
5. How can this model be adjusted for non-elliptical shapes?

Tip: Remember to always convert all units to the same system before performing calculations, especially when dealing with physical quantities like volume and rate of change.