Problem 46 - University Calculus: Early Transcendentals

Question

An explosion at an oil rig located in gulf waters causes an elliptical oil slick to spread on the surface from the rig. The slick is a constant 9 in. thick. After several days, when the major axis of the slick is 2 mi long and the minor axis is \(3 / 4\) mi wide, it is determined that its length is increasing at the rate of \(30 \mathrm{ft} / \mathrm{hr},\) and its width is increasing at the rate of \(10 \mathrm{ft}\) hr. At what rate (in cubic feet per hour) is oil flowing from the site of the rig at that time?

Step-by-Step Solution

Answer

264,343 cubic feet per hour.

Step 1: Convert units to feet

First, we need to convert the given dimensions from miles to feet since the rates of expansion are given in feet per hour. 1 mile = 5280 feet. Therefore, the initial major axis length: \[2 ext{ miles} = 2 \times 5280 ext{ feet} = 10560 ext{ feet}\]The initial minor axis length:\[\frac{3}{4} ext{ miles} = \frac{3}{4} \times 5280 ext{ feet} = 3960 ext{ feet}\]

Step 2: Calculate initial volume of oil slick

The volume of an elliptical cylinder is given by:\[ V = \pi \times a \times b \times h \]where \(a\) and \(b\) are the semi-major and semi-minor axes, and \(h\) is the thickness. Using the converted full dimensions from Step 1:\[ a = \frac{10560}{2} = 5280 \text{ feet} \]\[ b = \frac{3960}{2} = 1980 \text{ feet} \]\[ h = 9 \text{ inches} = 0.75 \text{ feet} \] (converted from inches to feet)Hence, the initial volume is:\[ V = \pi \times 5280 \times 1980 \times 0.75 \]

Step 3: Differentiate volume with respect to time

To find the rate of change of volume \( \frac{dV}{dt} \), differentiate the volume formula with respect to time \(t\):\[ \frac{dV}{dt} = \pi ( \frac{da}{dt} \times b \times h + a \times \frac{db}{dt} \times h + a \times b \times \frac{dh}{dt} ) \]Since the thickness \( h \) is constant, the term with \( \frac{dh}{dt} \) will be zero.

Step 4: Substitute given rates and calculate

Substitute the given rates and calculated dimensions into the differentiated volume formula. The given rates are:\[\frac{da}{dt} = 30 \text{ ft/hour} \]\[\frac{db}{dt} = 10 \text{ ft/hour} \]\[ \frac{dV}{dt} = \pi (30 \times 1980 \times 0.75 + 5280 \times 10 \times 0.75) \]Compute the rate of change:\[ \frac{dV}{dt} = \pi (44550 + 39600) \]\[ \frac{dV}{dt} = \pi \times 84150 \blockend{inline} \approx 264,343 \text{ cubic feet per hour} \]

Step 5: Final Answer

The oil slick is increasing at approximately 264,343 cubic feet per hour.

Key Concepts

Elliptical Cylinder Volume

Understanding the formula for the volume of an elliptical cylinder is crucial when dealing with shapes like oil slicks on water. Unlike circular cylinders, ellipses have distinct major and minor axes.
To find the volume of an elliptical cylinder, we use the formula: \[ V = \pi \times a \times b \times h \] where:

• \(a\) is the semi-major axis.
• \(b\) is the semi-minor axis.
• \(h\) is the height or thickness of the cylinder.

This formula extends the concept of a circle's area \(\pi r^2\) to an ellipse’s area \(\pi ab\), then multiplies it by the height to find volume, as you would with a cylinder.
In the given problem, we calculated these axes using converted dimensions (5280 feet for semi-major axis and 1980 feet for semi-minor axis) and multiplied by the thickness of 0.75 feet, converted from inches, to find the slick's volume.
This approach helps us create a clear path to determine various parameters of the problem, including volume change over time.

Unit Conversion

Unit conversion is an essential skill in mathematics and physics, ensuring that all metrics align for precise calculations. In the context of the oil slick problem, knowing how to convert between miles, feet, and inches is fundamental.
Miles need to be converted to feet because the rates of change in dimensions are given in feet per hour. Here's the breakdown:

• 1 mile equals 5280 feet. Hence, 2 miles become 10560 feet.
• Similarly, \(\frac{3}{4}\) miles convert to 3960 feet.
• Thickness given in inches (9 inches) is converted to feet by multiplying by \(\frac{1}{12}\), giving us 0.75 feet.

This systematic conversion ensures consistencies in units, allowing seamless integration into further calculations like those required in finding the rate of volume change over time.

Differentiation with Respect to Time

Differentiation with respect to time is a powerful technique in calculus to track how a quantity evolves over time. For related rates problems, it’s how we calculate the rate at which one quantity changes in relation to another.
In the problem of the oil slick, we needed to find \( \frac{dV}{dt} \), the rate of volume change. We began by differentiating the volume formula \( V = \pi a b h \) with respect to time \( t \). The formula for differentiation was:\[ \frac{dV}{dt} = \pi \left( \frac{da}{dt} \times b \times h + a \times \frac{db}{dt} \times h + a \times b \times \frac{dh}{dt} \right) \]Given that the thickness \( h \) is constant, its rate of change \( \frac{dh}{dt} \) becomes zero.
Substituting the known rates of \( \frac{da}{dt} = 30 \text{ ft/hr} \) and \( \frac{db}{dt} = 10 \text{ ft/hr} \), we calculated how fast the volume is changing. This approach gives us insight into dynamic situations, such as the expanding oil slick, by combining algebraic manipulation and calculus principles. Understanding the concept of differentiating with respect to time is immensely useful in physics and engineering applications, where variables frequently change over time.