Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

An egg is covered by a hen and is at \(37^{\circ} \mathrm{C}\). The hen leaves the nest and the egg is exposed to \(17^{\circ} \mathrm{C}\) air. After 20 minutes the egg is at \(34^{\circ} \mathrm{C}\). Draw a graph representative of the temperature of the egg \(t\) minutes after the hen leaves the nest. Mathematical Model. During any short time interval while the egg is uncovered, the decrease in egg temperature is proportional to the difference between the egg temperature and the air temperature. a. Introduce notation and write a dynamic equation representative of the mathematical model. b. Write a solution equation for your dynamic equation. c. Your dynamic equation should have one parameter. Use the data of the problem to estimate the parameter.

Step-by-Step Solution

Answer

The egg's temperature model is given by \( T(t) = 17 + 20e^{-0.00747t} \).

Step 1: Introduce Notation

Let's define the necessary variables for our model. Let \( T(t) \) be the temperature of the egg at time \( t \) in minutes. The initial temperature of the egg is \( T(0) = 37^{\circ} \mathrm{C} \). The air temperature is \( T_a = 17^{\circ} \mathrm{C} \). After 20 minutes, the temperature of the egg is \( T(20) = 34^{\circ} \mathrm{C} \).

Step 2: Write a Dynamic Equation

According to Newton's Law of Cooling, the rate of change of the temperature of the egg is proportional to the difference between the temperature of the egg and the surrounding air temperature. This can be modeled by the differential equation: \[ \frac{dT}{dt} = -k(T(t) - T_a) \] where \( k \) is a proportionality constant.

Step 3: Solve the Differential Equation

To solve the differential equation, separate the variables and integrate: \[ \int \frac{1}{T(t) - T_a} \, dT = -k \int dt \] \[ \ln|T(t) - T_a| = -kt + C \] Solving for \( T(t) \), we get: \[ T(t) = T_a + Ce^{-kt} \] where \( C \) is a constant determined by initial conditions.

Step 4: Use Initial Conditions to Determine Constants

We use the initial condition \( T(0) = 37^{\circ} \mathrm{C} \) to find \( C \): \[ 37 = 17 + Ce^{0} \] \[ C = 20 \] Therefore, \( T(t) = 17 + 20e^{-kt} \).

Step 5: Estimate Parameter Using Given Data

Use the data \( T(20) = 34^{\circ} \mathrm{C} \) to estimate \( k \): \[ 34 = 17 + 20e^{-20k} \] \[ 17 = 20e^{-20k} \] \[ e^{-20k} = \frac{17}{20} \] \[ -20k = \ln \left( \frac{17}{20} \right) \] \[ k = -\frac{1}{20} \ln \left( \frac{17}{20} \right) \] Calculating \( k \) gives an approximate value of \( k \approx 0.00747 \).

Step 6: Final Solution Equation

The solution to the dynamic equation with the estimated parameter is: \[ T(t) = 17 + 20e^{-0.00747t} \] This represents the temperature of the egg at any time \( t \) minutes after the hen leaves the nest.

Key Concepts

Mathematical Modeling

Mathematical modeling is an essential process utilized in various fields of science and engineering to understand real-world phenomena. It involves creating a mathematical representation or equation of a real-world situation. In the context of Newton's Law of Cooling, mathematical modeling helps us understand how the temperature of an object changes over time when it is exposed to a surrounding environment.

Newton's Law of Cooling assumes that the rate at which an object's temperature changes is proportional to the difference between its temperature and the ambient temperature. This idea can be represented in a model using differential equations. By introducing initial conditions, such as initial temperatures and times, the model can predict future temperature changes.

• Real-world phenomena are translated into mathematical language.
• Models help in predicting future states of the system.
• They help identify key variables and their interrelations.

Mathematical models are valuable because they can be adjusted or calibrated using actual data, enabling accurate prediction of outcomes and aiding decision-making in practical applications. They serve as a bridge between theory and experiment.

Differential Equations

Differential equations play a central role in mathematical modeling, especially in situations like temperature change under Newton’s Law of Cooling. A differential equation is a mathematical equation that involves functions and their derivatives, describing how a rate of change in one variable is related to other variables.

In the egg cooling scenario, the differential equation helps us understand how the egg's temperature changes with time. The equation derived for this scenario is:
\[ \frac{dT}{dt} = -k(T(t) - T_a) \] where:

• \( \frac{dT}{dt} \) represents the rate of change of temperature over time.
• \( T(t) \) is the temperature of the egg at time \( t \).
• \( T_a \) is the constant ambient temperature, which is \( 17^{\circ} \mathrm{C} \) in our example.
• \( k \) is a proportionality constant that quantifies how quickly the egg comes to the ambient temperature.

Solving this differential equation requires integration and the use of initial conditions to find specific solutions. The process involves separating variables, integrating, and using known conditions like \( T(0) = 37^{\circ} \mathrm{C} \) to solve for constants. Understanding differential equations allows us to predict changes and create reliable models for many physical systems.

Temperature Change

Temperature change, especially in the context of Newton's Law of Cooling, can be fascinating to study since it helps us understand heat transfer between an object and its environment. According to the law, the temperature change of an object is directly linked to the initial temperature differences and how well or quickly heat is exchanged.

For the egg losing heat to the air, we modeled the temperature change as:
\[ T(t) = T_a + Ce^{-kt} \] where:

• \( T_a \) is the ambient temperature, which is \( 17^{\circ} \mathrm{C} \).
• \( C \) is a constant derived from the initial temperature condition.
• \( e^{-kt} \) represents the exponential decay of the temperature difference over time.

The exponential factor \( e^{-kt} \) indicates that temperature change follows a clear exponential trend. This trend reflects that heat exchange decreases as the temperature difference narrows. Initially, when the temperature difference is greatest, the rate of heat exchange is fastest, meaning the egg cools quickly at first. As time passes and the temperature difference decreases, the rate of cooling slows down.

Understanding these dynamics helps in many practical applications, like cooking or environmental science, where predicting temperature changes is crucial.