Problem 66 - College Algebra and Calculus: An Applied Approach

Question

The temperatures \(T\left({ }^{\circ} \mathrm{F}\right)\) at which water boils at selected pressures \(p\) (pounds per square inch) can be modeled by \(T=87.97+34.96 \ln p+7.91 \sqrt{p}\) Find the rate of change of the temperature when the pressure is 60 pounds per square inch.

Step-by-Step Solution

Answer

The rate of change of the temperature when the pressure is 60 pounds per square inch is approximately 1.093186 F/psi.

Step 1: Identifying the function to be differentiated

The function given in the problem is \(T=87.97+34.96 \ln p+7.91 \sqrt{p}\) and the rate of change of temperature with respect to pressure needs to be found, so the derivative of \(T\) with respect to \(p\) is required.

Step 2: Differentiating the function

The derivative of \(T\) with respect to \(p\), denoted as \(dT/dp\) or \(T'\), can be obtained using the rules of differentiation. For a function in the form of \(f(x) = ax^n\), its derivative with respect to \(x\) is \(f'(x) = a*n*x^{(n-1)}\). Additionally, the derivative of the natural logarithm function, \(f(x) = \ln{x}\), is \(f'(x) = 1/x\). So using the chain rule for derivatives, the derivative of \(87.97\) (a constant) with respect to \(p\) is 0, the derivative of \(34.96 \ln p\) is \(34.96/p\), and the derivative of \(7.91 \sqrt{p}\) or \(7.91*p^{0.5}\) is \((7.91*0.5)*p^{-0.5}\). Therefore, \(dT/dp = 0 + 34.96/p + (7.91*0.5)*p^{-0.5}\). Simplifying this, get \(dT/dp = 34.96/p + 3.955/p^{0.5}\) or \(T' = 34.96/p + 3.955/\sqrt{p}\).

Step 3: Evaluating the derivative at p = 60 psi

Substitute \(p = 60\) into \(T' = 34.96/p + 3.955/\sqrt{p}\) to find the rate of change of the temperature at this pressure. So, \(T'(60) = 34.96/60 + 3.955/\sqrt{60} = 0.58266 + 0.510526 = 1.093186\)

Key Concepts

Derivative

Understanding derivatives is like understanding the language of change. In mathematics, derivatives show us how a function changes as its input changes. With this tool, we explore how variables depend on each other. Think of it as the speedometer of your car, telling you how fast your speed is changing at any moment.
In calculus, when we differentiate a function, we find its derivative. This helps us understand how the function's values change with its input variables. For example, in our exercise, we find the derivative of the temperature function, denoted as \( \frac{dT}{dp} \), to analyze how temperature changes with pressure.

• A constant's derivative is zero because constants don't change.
• The derivative of \( \log{x} \) is \( \frac{1}{x} \).
• The derivative of \( x^{n} \) is \( n \cdot x^{n-1} \).

These rules simplify the process, making derivatives easier to understand and calculate.

Temperature

Temperature is a measure of how hot or cold something is. Scientists use temperature to understand various phenomena because it affects physical conditions like pressure and volume. In this exercise, temperature is influenced by pressure, and we explore this through a mathematical model.
In our model, temperature is calculated using a formula that incorporates the natural logarithm and square root of pressure. By analyzing this relationship, we can better grasp how temperature reacts to changes in external conditions like pressure.
Understanding temperature's dynamics is crucial in fields such as meteorology, engineering, and chemistry. By seeing how temperature changes, we can foresee weather patterns, design better engines, and optimize chemical reactions.

Pressure

Pressure is the force exerted over an area and is a key factor in changing temperatures, especially in gases. Pressure and temperature are intertwined in many scientific and everyday scenarios. For instance, when a gas is compressed, its pressure increases, and typically so does its temperature.
The function \( T = 87.97 + 34.96 \ln p + 7.91 \sqrt{p} \) models how temperature varies with pressure in a scientific context. The derivative \( \frac{dT}{dp} \) tells us how the temperature changes with a change in pressure.
Pressure matters greatly in industries like aviation, weather forecasting, and even cooking. Understanding how pressure interacts with temperature lets us control and use these forces effectively in real-world applications.

Natural Logarithm

Natural logarithms, represented as \( \ln \), are a special kind of logarithm with the base \( e \), where \( e \approx 2.71828 \). They appear in many natural growth processes, financial calculations, and scientific models because they naturally describe exponential growth and decay.
In our exercise, \( \ln p \) helps model the relationship between pressure and temperature. The term \( 34.96 \ln p \) in the temperature function shows how pressure logarithmically scales with temperature.
Logarithms simplify complex multiplicative relationships into manageable additive ones. They are widely used in various scientific fields to ease calculations and reveal underlying patterns in data.