Problem 58 - Calculus an Applied Approach

Question

Physical Science The temperature \(T\) (in degrees Fahrenheit) of food placed in a refrigerator is modeled by $$T=10\left(\frac{4 t^{2}+16 t+75}{t^{2}+4 t+10}\right)$$ where \(t\) is the time (in hours). What is the initial temperature of the food? Find the rates of change of \(T\) with respect to \(t\) when (a) \(t=1,\) (b) \(t=3,\) (c) \(t=5,\) and (d) \(t=10\).

Step-by-Step Solution

Answer

The initial temperature of the food is 750 degrees Fahrenheit. The rates of change at t=1, t=3, t=5, and t=10 are -48.4, -34.3, -27.2, -21.1 degrees Fahrenheit per hour, respectively.

Step 1: Calculation of Initial Temperature

Substitute \( t = 0 \) into the temperature equation: \( T = 10 * (\frac{4 * 0^{2} + 16 * 0 + 75}{0^{2} + 4 * 0 + 10}) \). This reduces to 750 degrees Fahrenheit.

Step 2: Calculation of Derivative of Temperature Equation

Apply the quotient rule which states that the derivative of \(\(\frac{f}{g}\)\) is \(\(\frac{f'g - fg'}{g^{2}}\)\). The derivative of the temperature equation (\( \frac{dT}{dt} \)) is: - \frac{80*(-16t^{2} + 64t + 225)}{(t^{2} + 4*t + 10)^{2}}

Step 3: Calculate rates of change when t=1

Substitute \( t = 1 \) into the derivative equation, yielding -48.4 degrees Fahrenheit per hour.

Step 4: Calculate rates of change when t=3

Substitute \( t = 3 \) into the derivative equation, yielding -34.3 degrees Fahrenheit per hour.

Step 5: Calculate rates of change when t=5

Substitute \( t = 5 \) into the derivative equation, yielding -27.2 degrees Fahrenheit per hour.

Step 6: Calculate rates of change when t=10

Substitute \( t = 10 \) into the derivative equation, yielding -21.1 degrees Fahrenheit per hour.

Key Concepts

Temperature Modeling

Temperature modeling is the process of using mathematical equations to represent how temperature changes over time. In this exercise, we model the temperature of food cooling in a refrigerator. The temperature, \(T\), depends on the time, \(t\), and is given by the equation \(T = 10 \left(\frac{4t^2 + 16t + 75}{t^2 + 4t + 10}\right)\).

This equation is a rational function, meaning it has both a numerator and a denominator. Such functions are often used to model real-world scenarios where a quantity stabilizes over time.

• The numerator \(4t^2 + 16t + 75\) provides the initial rise and rate of temperature.
• The denominator \(t^2 + 4t + 10\) accounts for the damping effect as the temperature changes.

This setup helps us see how temperature initially spikes and then gradually approaches a steady state as the food cools down.

Derivative Calculation

Calculating the derivative of a function helps us understand how the quantity changes at any point. In calculus, the derivative is a fundamental tool that provides the rate of change of a function with respect to its variable. For this temperature function, the goal is to find \(\frac{dT}{dt}\), which represents how quickly the temperature \(T\) changes concerning time \(t\).

The derivative requires applying calculus techniques such as the quotient rule because our original function is a ratio of two polynomials. This calculation helps us identify specific points where the temperature change is fastest or slowest.

Rate of Change

The rate of change measures how a quantity, like temperature, changes over time. In this exercise, we calculate the rate at various time points to see how quickly the temperature changes as the food cools.

• When \(t=1\), the rate of change is \(-48.4\) degrees Fahrenheit per hour, indicating a steep drop in temperature.
• As \(t\) increases to 3 and 5 hours, the rates are \(-34.3\) and \(-27.2\) respectively, showing a slower decrease in temperature.
• Finally, at \(t=10\), the rate further decreases to \(-21.1\).

This analysis provides a clearer picture of how quickly the temperature stabilizes and reaches the refrigerator's chilled environment. The negative signs indicate the cooling direction.

Quotient Rule

The quotient rule is a method in calculus for finding the derivative of a function that is the ratio of two other functions. Given a function \(\frac{f(t)}{g(t)}\), the quotient rule states that the derivative is:\[ \frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2} \].

This rule helps to solve our exercise because the temperature function is a rational function expressed as the quotient of the polynomials \(4t^2 + 16t + 75\) and \(t^2 + 4t + 10\):

• \(f(t) = 4t^2 + 16t + 75\)
• \(g(t) = t^2 + 4t + 10\)

Following the quotient rule allows us to calculate the derivative \(\frac{dT}{dt}\), thus enabling us to understand how the temperature rate changes over time. Each component's derivative needs careful attention to ensure correct results are used in the final calculation.