Problem 30 - Differential Equations and Linear Algebra

Question

According to Newton's law of cooling (see Section 1.1), the temperature of an object at time \(t\) is governed by the differential equation $$ \frac{d T}{d t}=-k\left(T-T_{m}\right) $$ where \(T_{m}\) is the temperature of the surrounding medium, and \(k\) is a constant. Consider the case when \(T_{m}=70\) and \(k=1 / 80 .\) Sketch the corresponding slope field and some representative solution curves. What happens to the temperature of the object as \(t \rightarrow \infty .\) Note that this result is independent of the initial temperature of the object.

Step-by-Step Solution

Answer

To sketch the slope field, plug in the given values into the differential equation: \(\frac{dT}{dt} = -\frac{1}{80}(T - 70)\). Then, compute the slopes of the tangent line segments at various points on the \(T-t\) plane. Draw the tangent segments with the calculated slopes to form the slope field. Next, draw some representative solution curves following the path formed by the tangent segments. Observing the behavior of the solution curves as \(t \rightarrow \infty\), the temperature of the object approaches the equilibrium temperature \(T_m = 70^\circ\).

Step 1: Write down the given differential equation and constants

To represent Newton's law of cooling, we have the following differential equation: \[ \frac{dT}{dt} = -k(T - T_m) \] With given values for the temperature of the surrounding medium and the constant \(k\), \(T_m = 70\) and \(k = \frac{1}{80}\).

Step 2: Plug in the given values into the differential equation

Substitute the given values of \(T_m\) and \(k\) into the equation: \[ \frac{dT}{dt} = -\frac{1}{80}(T - 70) \]

Step 3: Sketch the slope field

To sketch the slope field, we need to compute the slopes of the tangent line segments at various points on the \(T-t\) plane using the equation. Pick points on the plane, then substitute their coordinates into the equation to find the slopes of the tangent segments at these points. For example, for point \((t, T) = (1, 69)\), the slope will be: \[ \frac{dT}{dt}(1,69) = -\frac{1}{80}(69 - 70) = \frac{1}{80} \] Do this for several points and draw the tangent segments at each point according to the calculated slopes. The collection of these segments will form the slope field.

Step 4: Draw some representative solution curves

Using the slope field, you can draw some possible solution curves that follow the path formed by the tangent line segments. These curves will represent the temperature changes over time for different initial temperatures.

Step 5: Determine the temperature as time approaches infinity

Observe the behavior of the solution curves as time goes on. As \(t \rightarrow \infty\), the solution curves will approach the equilibrium temperature (which, in this case, is the temperature of the surrounding medium, \(T_m\)). Thus, as \(t \rightarrow \infty\), the temperature of the object will approach \(T_m = 70^\circ\).

Key Concepts

Understanding Differential Equations

Differential equations are a type of equation that relate a function to its derivatives, expressing how the rate of change of a quantity is related to the quantity itself. In the context of Newton's law of cooling, the differential equation takes the form \( \frac{dT}{dt} = -k(T - T_{m}) \), where \( T \) is the temperature of the object, \( t \) is time, \( k \) is a positive constant representing the cooling rate, and \( T_{m} \) is the equilibrium temperature, in this case, the surrounding medium's temperature.
In the solution to the exercise, we see how values for \( k \) and \( T_{m} \) are substituted into the differential equation to describe the specific scenario of cooling. This relationship tells us that the rate at which the temperature of the object changes is proportional to the difference between the object's current temperature and \( T_{m} \)—the larger this difference, the faster the change in temperature. Key concepts to grasp here include: understanding the meaning of each term in the equation, recognizing that differential equations can model physical processes, and knowing how varying parameters affect the system's behavior.

Visualizing Change with Slope Fields

A slope field, also known as a direction field, is a visual representation of a differential equation at a collection of points in the plane. For the given differential equation, the slope field helps us to see how the temperature of an object changes over time at various initial temperatures.To construct one, you calculate the slope given by the differential equation at several points in the \(T-t\) plane, and then draw small line segments with these slopes at the respective points. The step-by-step solution shows how to do this with the point \( (1, 69) \) resulting in a slope of \( \frac{1}{80} \). The collection of these line segments across the plane creates the slope field, and we can then sketch representative solution curves that align with these segments. These curves visually represent possible temperature changes over time for different initial conditions.Visual learners often benefit greatly from slope fields as they show the gradient of potential solutions across the range of initial values. Moreover, observing the orientation and length of these line segments may give clues as to the behavior of solutions, such as their stability and whether they converge to an equilibrium.

Approaching Equilibrium Temperature

Equilibrium temperature in the context of Newton’s law of cooling is the temperature at which the rate of heat loss of the object is equal to the rate of heat gain from the surroundings, resulting in no net change in the object's temperature over time. In the given exercise, \( T_{m} = 70 \) degrees, which is the constant temperature of the surrounding medium.As demonstrated in the final step of the solutions, over time, regardless of the initial temperature of the object, the temperature will stabilize towards this equilibrium temperature. This is due to the nature of the differential equation, where the rate of temperature change diminishes as the difference between \( T \) and \( T_{m} \) becomes smaller. Ultimately, as \( t \) approaches infinity, we expect the temperature \( T \) to approach \( T_{m} \)—implying that the object will become indistinguishable from the surrounding medium in terms of temperature.Understanding equilibrium temperature is crucial as it is a point of stability and often the long-term behavior of a variety of dynamic systems; in engineering, climate science, and even economics, the concept applies to systems seeking a state of balance.