Problem 11 - The Calculus with Analytic Geometry

Question

Boyle's law for the expansion of gas is \(P V=C\), where \(P\) is the number of pounds per square unit of pressure, \(V\) is the number of cubic units of volume of the gas, and \(C\) is a constant. At a certain instant the pressure is \(3000 \mathrm{lb} / \mathrm{ft}^{2}\), the volume is \(5 \mathrm{ft}^{3}\), and the volume is increasing at the rate of \(3 \mathrm{ft}^{3} / \mathrm{min}\). Find the rate of change of the pressure at this instant.

Step-by-Step Solution

Answer

\(\frac{dP}{dt} = -1800\) lb/ft\(^2\) per minute.

Step 1: Identify Given Values

Identify and list the given information: Pressure (\(P\)) = 3000 lb/ft\textsuperscript{2}, Volume (\(V\)) = 5 ft\textsuperscript{3}, rate of change of volume (\frac{dV}{dt}) = 3 ft\textsuperscript{3}/min.

Step 2: Write Boyle's Law Equation

Write down Boyle's Law equation: \(P \cdot V = C\).

Step 3: Differentiate Both Sides with Respect to Time

Differentiate both sides of the equation with respect to time (t): \(\frac{d}{dt}(P \cdot V) = \frac{d C}{dt}\).

Step 4: Apply the Product Rule

Since \(C\) is a constant, its derivative equals zero. Therefore, applying the product rule, we get: \(P \cdot \frac{dV}{dt} + V \cdot \frac{dP}{dt} = 0\).

Step 5: Solve for \(\frac{dP}{dt}\)

Rearrange the equation to isolate \(\frac{dP}{dt}\): \(\frac{dP}{dt} = -\frac{P \cdot \frac{dV}{dt}}{V}\).

Step 6: Substitute the Known Values

Substitute the given values into the equation: \(\frac{dP}{dt} = -\frac{3000 \cdot 3}{5}\).

Step 7: Calculate the Rate of Change of Pressure

Simplify the equation to find the rate of change of pressure: \(\frac{dP}{dt} = -1800\) lb/ft\(^2\) per minute.

Key Concepts

Boyle's Law

Boyle's Law is a fundamental principle in physical science that describes how the pressure of a gas tends to increase as the volume of the container decreases, provided the temperature remains constant. This is expressed mathematically as \( P \times V = C \), where \( P \) represents pressure, \( V \) represents volume, and \( C \) is a constant for a given amount of gas at a fixed temperature. In simpler terms, if you compress a gas by reducing its volume, its pressure will increase proportionally, and vice versa.

Differentiation

Differentiation is a key operation in calculus that lets us find how a function changes as its input changes. It's like asking, 'How fast is this function moving?' When we differentiate an equation with respect to time, we essentially measure how one quantity is changing over time relative to another. For Boyle's Law, we need to differentiate with respect to time to understand how pressure and volume change over time. This is especially useful when variables depend on time, like when gas expands or contracts.

Rate of Change

The rate of change is a measure of how much a quantity varies over a particular period. Understanding the rate of change helps us predict future behavior or understand the behavior of a system. In our Boyle's Law example, we are concerned with how the pressure changes over time. Given the volume's rate of change, we can infer the pressure’s rate using calculus. Specifically, knowing that volume changes at 3 ft³/min helps us find the pressure's rate of change using the derived formula.

Product Rule

The product rule is a differentiation technique used when dealing with the product of two differentiable functions. It states that the derivative of a product of two functions \( u(t) \times v(t) \) is given by \[ \frac{d}{dt}[u(t) \times v(t)] = u(t) \times \frac{dv(t)}{dt} + v(t) \times \frac{du(t)}{dt}\]. When applied to Boyle's Law (\( P \times V = C \)), we use the product rule to differentiate both sides with respect to time, keeping in mind that the derivative of the constant (\( C \)) is zero. This step-by-step method allows us to relate the rates of change of pressure and volume.

Pressure and Volume Relationship

In Boyle's Law, the pressure and volume of a gas are inversely related; this means as one goes up, the other goes down. By differentiating the equation and using the given values, we understand the mathematical relationship between pressure and volume's rates of change. For instance, substituting known values into the re-arranged equation \( \frac{dP}{dt} = -\frac{P \times \frac{dV}{dt}}{V} \), we calculate that the pressure changes at a rate of -1800 lb/ft² per minute when the volume increases at the rate of 3 ft³/min. This negative sign indicates that pressure decreases as volume increases, conforming to the inverse relationship.