Problem 33 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

At a temperature of \(20^{\circ} \mathrm{C}\), the volume \(V\) (in liters) of \(1.33 \mathrm{~g}\) of \(\mathrm{O}_{2}\) is related to its pressure \(p\) (in atmospheres) by the formula \(V=1 / p\) a. What is the average rate of change of \(V\) with respect to \(p\) as \(p\) increases from \(p=2\) to \(p=3 ?\) b. What is the rate of change of \(V\) with respect to \(p\) when \(p=2 ?\)

Step-by-Step Solution

Answer

a. The average rate of change of \(V\) with respect to \(p\) as \(p\) increases from \(2\) to \(3\) is \(-\frac{1}{6}\) L/atm. b. The rate of change of \(V\) with respect to \(p\) when \(p = 2\) is \(-\frac{1}{4}\) L/atm².

Step 1: Write down the given formula

We are given the formula relating volume \(V\) to pressure \(p\): \[V = \frac{1}{p}\]

Step 2: Calculate average rate of change (Part a)

To find the average rate of change of \(V\) with respect to \(p\) as \(p\) increases from \(2\) to \(3\), we use the average rate of change formula: \[\frac{V(p_2) - V(p_1)}{p_2 - p_1}\] Plug in the given values of pressure, \(p_1 = 2\) and \(p_2 = 3\), into the given formula to find the volume at each pressure: \[V(2) = \frac{1}{2}, \quad V(3) = \frac{1}{3}\] Now, plug these values into the average rate of change formula: \[\frac{\frac{1}{3} - \frac{1}{2}}{3 - 2} = \frac{\frac{-1}{6}}{1} = -\frac{1}{6}\] Thus, the average rate of change of \(V\) with respect to \(p\) as \(p\) increases from \(2\) to \(3\) is \(-\frac{1}{6}\) L/atm.

Step 3: Find the derivative of V with respect to p (Part b)

To find the rate of change of \(V\) with respect to \(p\) when \(p = 2\), we first need to find the derivative of \(V\) with respect to \(p\): \[\frac{dV}{dp}\] Using the given formula, we have: \[V = \frac{1}{p}\] To find the derivative of \(V\) with respect to \(p\), we can rewrite the formula as: \[V = p^{-1}\] Now find the derivative: \[\frac{dV}{dp} = -1 \cdot p^{-2} = -\frac{1}{p^2}\]

Step 4: Evaluate the derivative at p = 2 (Part b)

Now that we have the derivative of \(V\) with respect to \(p\), we can plug in \(p = 2\) to find the rate of change at that pressure: \[\frac{dV}{dp}(2) = -\frac{1}{(2)^2} = -\frac{1}{4}\] Therefore, the rate of change of \(V\) with respect to \(p\) when \(p = 2\) is \(-\frac{1}{4}\) L/atm².

Key Concepts

Average Rate of Change

In calculus, the average rate of change is a concept that measures how much a quantity, such as volume, temperature, or distance, changes on average between two points. It's akin to calculating the slope of the secant line between two points on the graph of a function. When you calculate the average rate of change, you essentially determine the average speed of the quantity over a specific interval.

For instance, the problem given involves the relationship between the volume of oxygen and pressure. To calculate the average rate of change of volume with respect to pressure, we take the volumes at two different pressures and divide their difference by the difference in pressures. This results in a literal interpretation of how much the volume shrinks or expands per unit change in pressure over that interval. Such information helps us understand the behavior of gases in a controlled environment, a staple concept in applied sciences.

Derivative of a Function

The derivative of a function is a cornerstone in calculus, reflecting the rate of change of a function's output with respect to changes in its input instantaneously. In physical terms, it represents how quickly something is moving at a precise moment in time. Derivatives help us understand rates at which changes occur and can predict future behavior under given conditions.

When looking at the function provided, which represents the inverse relationship between volume and pressure, obtaining the derivative involves applying rules of differentiation to the function. The derivative of the volume with respect to pressure, denoted as \(\frac{dV}{dp}\), gives us the rate at which volume changes for a tiny increase or decrease in pressure. This information is essential in fields like engineering and physics, where precise measurements can mean the difference between success and failure in design and implementation of systems.

Calculus in Applied Mathematics

Calculus has a profound impact on applied mathematics, enabling the solving of problems across various disciplines, from economics to engineering. By utilising concepts like the derivative and rate of change, specialists can model and analyse the behavior of systems over time. Applied mathematics often includes maximising or minimising quantities, predicting trends, and understanding complex systems behavior.

In practical terms, when the temperature and mass of oxygen are controlled, calculus helps predict how volume changes with pressure. It helps develop more efficient processes, like optimising the use of gases in industrial applications, which might involve minimizing waste while maximising yield. The problem exercises provided are classic examples of calculus applied in gas laws and thermodynamics—a small window into the larger world of applied mathematic's pivotal role in solving real-world problems.

Inverse Relationships

An inverse relationship is a situation where one variable increases while the other decreases. In the context of the exercise provided, volume and pressure have this kind of relationship, as described by the formula \(V = \frac{1}{p}\). As pressure increases, volume decreases proportionally, and vice versa. Recognizing inverse relationships is important in calculus because it often indicates a specific type of function, one that can be differentiated and integrated to gain insights into the system's behavior.

Understanding the inverse relationships also helps in grasping the fundamentals of how different quantities affect one another, laying the groundwork for more complex mathematical models. These models are then used in creating simulations and predicting outcomes in science, engineering, economics, and beyond.