Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

We assume that an oil spill is being cleaned up by deploying bacteria that consume the oil at 4 cubic feet per hour. The oil spill itself is modeled in the form of a very thin cyclinder whose height is the thickness of the oil slick. When the thickness of the slick is \(0.001\) foot, the cylinder is 500 feet in diameter. If the height is decreasing at \(0.0005\) foot per hour, at what rate is the area of the slick changing?

Step-by-Step Solution

Answer

The area is decreasing at 94,175 square feet per hour.

Step 1: Understand the Problem

We have a cylinder-shaped oil slick, and we need to find out how quickly the area of the base of the cylinder (slick) is changing. We know the bacteria reduce the thickness of the oil slick, and we want to translate how this affects the area.

Step 2: Formulate the Relationship

Let's use the relationship between the volume of the cylinder and its components. The volume of a cylinder is given by the formula: \[ V = A imes h \]where \(V\) is the volume, \(A\) is the area of the base, and \(h\) is the height (thickness of oil slick). Here, \(h = 0.001\) foot and \(d \approx 500/2 = 250\) feet (radius). The volume change is due to both the area and height.

Step 3: Apply the Given Rates

We are given the height is decreasing at the rate \(\frac{dh}{dt} = -0.0005\) foot per hour and that bacteria are consuming oil at a rate of \(4\) cubic feet per hour, which is \(\frac{dV}{dt} = -4\). Substitute these into the expression for the volume rate change: \[ \frac{dV}{dt} = A \frac{dh}{dt} + h \frac{dA}{dt} \]

Step 4: Solve for \(\frac{dA}{dt}\)

Substituting the known values:\[ -4 = A(-0.0005) + 0.001 \frac{dA}{dt} \]With \(A = \pi (250)^2 \approx 196,350\) square feet, substitute and rearrange to solve for \(\frac{dA}{dt}\):\[ -4 = (-0.0005 \times 196,350) + 0.001 \frac{dA}{dt} \]

Step 5: Simplify and Calculate \(\frac{dA}{dt}\)

Calculate \(-0.0005 \times 196,350 = -98.175\). So,\[ -4 = -98.175 + 0.001 \frac{dA}{dt} \] Rearrange to find \(\frac{dA}{dt}\):\[ 0.001 \frac{dA}{dt} = 98.175 - 4 \]\[ 0.001 \frac{dA}{dt} = 94.175 \]Finally, \[ \frac{dA}{dt} = 94,175 \] square feet per hour.

Key Concepts

Differential Calculus

Differential calculus is a fundamental branch of mathematics focusing on the concept of change.
It allows us to determine how a quantity changes in relation to another variable.
In the context of our exercise, this method is used to understand how the area of an oil slick changes over time. Differential calculus provides the tools for:

• Calculating derivatives, which represent the rate of change.
• Understanding relationships between changing quantities such as time and distance or, in this case, the height and area of an oil slick.
• Solving real-world problems involving change, which is ubiquitous in natural phenomena.

By finding derivatives, students can predict how a system will behave under certain conditions.
In our problem, we are particularly interested in finding the derivative of the area of the base of the cylinder with respect to time, considering the diminishing height of the oil slick.

Cylinder Volume

The volume of a cylinder is calculated using the formula:
\[ V = A \times h \]Here, \(V\) stands for volume, \(A\) is the area of the base, and \(h\) is the height (or thickness in the case of an oil slick).
This formula is crucial for understanding the relationship between the dimensions of a cylinder and its total volume.To apply this in practical situations like oil spill cleanup, we:

• Identify that as the height decreases due to oil consumption, both the volume and area can be affected.
• Use the initial conditions such as the thickness of the slick and diameter of the cylinder to calculate initial area and volume.
• Understand that volume change can result from changes in both height and area.

By integrating this understanding with calculus, we can determine how to adjust cleaning efforts based on changes in the slick's properties.

Rate of Change

The rate of change is a vital concept in calculus and physics.
It describes how one quantity changes as a direct function of another.
This is especially important in scenarios where measurements evolve continuously over time. In the case of the oil slick problem:

• The rate at which the bacteria consume the oil leads to a change in both height and area of the slick.
• The derivative \( \frac{dh}{dt} \) represents the rate of change of the height, while \( \frac{dA}{dt} \) represents the rate of change of the area.
• The formula \( \frac{dV}{dt} = A \frac{dh}{dt} + h \frac{dA}{dt} \) relates the rates of volume, height, and area change together.

This tells us how a reduction in volume, due to bacteria activity, affects both height and the area of the slick.
Thus, understanding rate of change helps to adjust conditions effectively.

Applied Mathematics

Applied mathematics takes complex mathematical theories and uses them to solve practical, real-world problems.
It's a field that combines theoretical math with applications in various domains like engineering, economics, and environmental science. The oil slick problem is a great example of applied mathematics because it involves:

• Modeling real-world phenomena with mathematical structures, like the cylinder model for the oil slick.
• Utilizing calculus to determine practical solutions for issues such as environmental cleanup strategies.
• Translating mathematical results into actionable insights for resolving issues—in this case, the slowing of bacteria's oil consumption rate.

With your understanding of applied mathematics, you can tackle complex environmental challenges by breaking them down into manageable mathematical models.
This approach allows for optimized solutions in fields ranging from sustainability to industrial operations.