Problem 23 - 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. a. Draw a graph representative of the temperature of the egg \(t\) minutes after the hen leaves the nest. Mathematical Model 5.5.6 Egg cooling. During any short time interval while the egg is uncovered, the change in egg temperature is proportional to the length of the time interval and proportional to the difference between the egg temperature and the air temperature. b. Let \(T(t)\) denote the egg temperature \(t\) minutes after the hen leaves the nest. Consider a short time interval, \([t, t+\Delta t],\) and write an equation for the change in temperature of the egg during the time interval \([t, t+\Delta t]\). c. Argue that as \(\Delta t\) approaches zero, the terms of your previous equation get close to the terms of $$T^{\prime}(t)=-k(T(t)-17)$$ d. Assume \(\mathrm{T}(0)=37\) and find an equation for \(T(t)\). e. Suppose it is known that eight minutes after the hen leaves the nest the egg temperature is \(35^{\circ} \mathrm{C}\). What is \(k ?\) f. Based on that value of \(k,\) if the coldest temperature the embryo can tolerate is \(32^{\circ} \mathrm{C},\) when must the hen return to the nest?

Step-by-Step Solution

Answer

Use differential equations to find that the hen should return after around 21.14 minutes to maintain the egg above 32°C.

Step 1: Set Up the Initial Graph

The graph should start at the point corresponding to the initial temperature of the egg, which is \( T(0) = 37^{\circ} \mathrm{C} \). As \( t \) increases, we expect the temperature \( T(t) \) to gradually decrease, approaching the air temperature of \( 17^{\circ} \mathrm{C} \). This represents an exponential decay process. Thus, the graph should start at \( 37^{\circ} \mathrm{C} \) and slope downwards asymptotically toward \( 17^{\circ} \mathrm{C} \).

Step 2: Derive the Temperature Change Equation

For a short time interval \([t, t+\Delta t]\), the change in temperature \( \Delta T \) can be expressed as \( \Delta T = -k(T(t) - 17) \Delta t \). This reflects the proportionality of temperature change to both the temperature difference \( T(t) - 17 \) and the time interval \( \Delta t \).

Step 3: Approach to Zero Interval

As \( \Delta t \) approaches zero, the equation simplifies to the differential form by considering the instantaneous change, hence we get \( T'(t) = -k(T(t) - 17) \). This reflects the continuous nature of the cooling process and the relation between rate of change of temperature and the temperature difference.

Step 4: Solve the Differential Equation

To find \( T(t) \), solve \( T'(t) = -k(T(t) - 17) \). This is a separable differential equation. Integrating both sides, we derive the solution as \( T(t) = Ce^{-kt} + 17 \). Given \( T(0) = 37 \), we find \( C + 17 = 37 \) which gives \( C = 20 \). Thus, \( T(t) = 20e^{-kt} + 17 \).

Step 5: Determine the Constant k

We're given that \( T(8) = 35 \). Substitute into the equation: \( 35 = 20e^{-8k} + 17 \). Solving for \( e^{-8k} \) gives us \( e^{-8k} = \frac{18}{20} = 0.9 \). Taking the natural logarithm, we calculate \( k = -\frac{\ln(0.9)}{8} \).

Step 6: Find Time for Temperature to Drop to Critical Level

We need to know when \( T(t) = 32 \). Use \( 32 = 20e^{-kt} + 17 \) to solve for \( t \). This leads to \( 20e^{-kt} = 15 \) or \( e^{-kt} = 0.75 \). Next, solve for \( t \): \( t = -\frac{\ln(0.75)}{k} \). Substitute the value of \( k \) found in Step 5 to calculate the exact \( t \).

Key Concepts

Exponential Decay

In many natural processes, you'll find patterns that resemble exponential decay. This occurs when a quantity decreases at a rate proportional to its current value. In the context of our egg cooling example, the egg's temperature diminishes over time as it cools towards the surrounding air temperature of \(17^{\circ} \mathrm{C}\). The concept of exponential decay here states that the rate of change of the egg's temperature is proportional to the difference between its temperature and the ambient air temperature.

Mathematically, this is expressed using the equation \(T'(t) = -k(T(t) - 17)\). The negative sign indicates that the temperature is reducing. The term \(k\) is a constant, representing how quickly the process occurs. Thus, on a graph, this would appear as a curve starting at \(37^{\circ} \mathrm{C}\) that gradually approaches \(17^{\circ} \mathrm{C}\). This slope downward is characteristic of exponential decay.

It's important to note that exponential decay is seen frequently in real life, not only in physics but also in other fields, such as biology, chemistry, and finance. Understanding this concept can be incredibly useful when learning about phenomena like radioactive decay, population declines, or even discount rates in economics.

Newton's Law of Cooling

Newton's Law of Cooling provides another fascinating angle on how objects exchange heat with their environment. It states that the rate of change of the temperature of an object is proportional to the difference in temperature between the object and its surroundings. This principle is beautifully illustrated in the cooling egg example from the exercise.

In practical terms, Newton's Law of Cooling implies that if an object like an egg is placed in a cooler environment, it will release heat. The greater the temperature difference, the faster this process occurs. Conversely, as the egg's temperature nears that of the air, the rate of cooling diminishes. The formula \(T'(t) = -k(T(t)-17)\) captures this relationship, where \(k\) is a positive constant. The law provides a reliable model for a wide range of applications — cooling of hot beverages, air-conditioning systems, and even forensic science.

It's essential to set up the differential equation first to understand how the temperature changes over time. Solving this equation, as shown in the solution steps, gives the predictive insight we need for determining how long the object will continue to cool.

Separable Differential Equations

Separable differential equations are a fundamental tool in solving many real-world problems. They are differential equations in which we can separate variables, meaning that all terms involving the dependent variable (like temperature \(T\)) can be placed on one side of the equation, while those involving the independent variable (like time \(t\)) go on the other.

In our exercise, the equation \(T'(t) = -k(T(t) - 17)\) is separable. By moving all \(T\)-related terms to one side and all \(t\)-related terms to the other, we can integrate both sides to find the solution for \(T(t)\). This leads us to the equation \(T(t) = Ce^{-kt} + 17\), which describes the egg's temperature over time. Here, \(C\) is determined from initial conditions, specifically \(T(0) = 37^{\circ}\), allowing us to solve for \(C = 20\).

This approach is particularly powerful because it applies to many types of problems, whether they involve cooling objects, population dynamics, or chemical reactions. Understanding how to manipulate and solve separable differential equations opens the door to modeling many dynamic systems.