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Differential equations play a crucial role in modeling various physical phenomena, including temperature change in Newton's Law of Cooling. Here, a differential equation describes how the temperature of an object changes over time.

Newton's Law of Cooling can be expressed as: \[ \frac{dT}{dt} = -k(T - T_s) \]
In this equation, \(T\) represents the temperature of the wine, \(T_s\) is the surrounding temperature, and \(k\) is a constant specific to the situation. The term \((T - T_s)\) indicates that the rate of temperature change depends on the difference between the object's temperature and the ambient temperature. This relationship helps us understand how the cooling process evolves over time.

Understanding how temperature changes over time in Newton's Law of Cooling involves acknowledging the interaction between the object and its environment.

Initially, the wine is at room temperature \(70^{\text{°}} \text{F}\). When placed in ice at \(32^{\text{°}} \text{F}\), the wine cools to \(58^{\text{°}} \text{F}\) within 20 minutes. This temperature drop illustrates the key principle of Newton's Law — the rate of temperature change is proportional to the temperature difference.

The progression looks like this:

• Initial temperature: \(70^{\text{°}} \text{F}\)
• Temperature after 20 minutes: \(58^{\text{°}} \text{F}\)
• Surrounding temperature: \(32^{\text{°}} \text{F}\)
• Find when the wine reaches \(50^{\text{°}} \text{F}\)
By modeling these changes with the differential equation, we can predict the future temperature at different time intervals.

Initial conditions are critical for solving differential equations because they provide the necessary starting points. To find the specific solution to Newton's Law of Cooling, we need these initial data points.

For our wine cooling problem, the initial condition is that the wine starts at \(70^{\text{°}} \text{F}\) when \(t = 0\) minutes. Using this initial temperature helps us determine the constant of integration (\(C\)), which is essential for our final temperature function.

By substituting \(T (0) = 70\) into the integrated form, we find \[ C = \text{ln}|70 - 32| = \text{ln} 38.\]These initial conditions ensure that our model accurately reflects the specific scenario we're studying.

Integration is the process used to solve the differential equation in Newton's Law of Cooling.

First, we separate variables: \[ \frac{1}{T - T_s} dT = -k dt. \]

Next, we integrate both sides to get the natural logarithm function: \[ \text{ln}|T - T_s| = -kt + C. \] Using the initial temperature, we solve for the constant \(C\).

Finally, we use another condition — the wine temperature after 20 minutes — to find the constant \(k\). By plugging the values \(T (20) = 58\), we get: \[ k = \frac{\text{ln} 38 - \text{ln} 26}{20}. \]

Once we have \(k\), we can determine the total time needed for the wine to reach \(50^{\text{°}} \text{F}\) through additional integration and solving for \(t\). These steps show how integration helps us derive useful information from the differential equation.