Problem 10 - The Calculus with Analytic Geometry

Question

Boyle's law for the expansion of a gas is \(P V=C\), where \(P\) is the number of pounds per square unit of pressure, \(V\) is the number of cubic units in the volume of the gas, and \(C\) is a constant. Find the instantaneous rate of change of the volume per change of one pound per square unit in the pressure when \(P=4\) and \(V=8\).

Step-by-Step Solution

Answer

\( \frac{dV}{dP} = -2 \)

Step 1: Understand Boyle's Law

Boyle's law is given as \( P V = C \). This means that the product of pressure \( P \) and volume \( V \) is a constant \( C \).

Step 2: Differentiate Boyle's Law

Differentiate both sides of Boyle's law with respect to \( P \): \( \frac{d}{dP}(P V) = \frac{d}{dP}(C) \) Since \( C \) is a constant, its derivative is 0. Using the product rule on the left side, we get: \( V + P \frac{dV}{dP} = 0 \)

Step 3: Solve for \( \frac{dV}{dP} \)

Rearrange the equation to solve for the derivative \( \frac{dV}{dP} \): \( P \frac{dV}{dP} = -V \) \( \frac{dV}{dP} = -\frac{V}{P} \)

Step 4: Substitute the Given Values

Substitute \( P = 4 \) and \( V = 8 \) into the equation: \( \frac{dV}{dP} = -\frac{8}{4} \)

Step 5: Simplify the Expression

Simplify the fraction to get the instantaneous rate of change: \( \frac{dV}{dP} = -2 \)

Key Concepts

Instantaneous Rate of Change

The term 'instantaneous rate of change' refers to the rate at which a quantity changes at a specific instant. It is essentially the derivative of a function at a particular point. In our Boyle's law problem, we are interested in how the volume changes with respect to the pressure at a given instant. When the pressure is 4 units, and the volume is 8 units, we find this rate by differentiating and substituting the values into our derivative formula.

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the derivative of a function. The derivative represents the rate of change of one quantity with respect to another. In our problem, we need to differentiate Boyle's law, which is expressed as \(PV = C\). By applying differentiation, we transform the equation to help us find the rate of change of volume \(V\) with respect to pressure \(P\). Differentiation follows specific rules, such as the product rule, which we use here.

Product Rule

The product rule is a formula used to find the derivative of the product of two functions. If you have two functions, \(u\) and \(v\), their product \(uv\) is differentiated as: \((uv)' = u'v + uv'\). Applying this to Boyle's law \(PV = C\), where \(P\) and \(V\) are functions of pressure, the product rule gives us: \(V + P \frac{dV}{dP} = 0\). This allows us to isolate \( \frac{dV}{dP} \) and find the instantaneous rate of change.

Constant

In mathematics, a constant is a value that does not change. In Boyle's law, \(C\) is a constant, representing the fixed product of pressure and volume. When we differentiate \(PV = C\), the derivative of \(C\) is zero, because it is constant. This simplifies our differentiation process, as constants disappear when differentiated. Therefore, understanding that \(C\) is constant helps streamline our calculation of the instantaneous rate of change.