Question

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) \(t=10\).

Short answer

The initial temperature of the food is \(75^{\circ}F\). The rates of change of \(T\) with respect to \(t\) when (a) \(t=1\), (b) \(t=3\), (c) \(t=5\), and (d) \(t=10\) can be found by substituting these values into \(T' (t)\) and simplifying, for each case respectively.

Step-by-step solution

Find Initial Temperature

To find initial temperature, substitute \(t = 0\) into \(T(t)\). Hence, the initial temperature is \(T(0) = 10 \left(\frac{4(0)^{2} + 16*0 + 75}{(0)^{2} + 4*0 + 10}\right) = 75^{\circ}F\).

Find Derivative of \(T(t)\)

We find \(T' (t)\), the derivative of \(T(t)\) with respect to \(t\), using the quotient rule for differentiation. The quotient rule states that \((\frac{f}{g})' = \frac{f'g - g'f}{g^{2}}\), where f and g are functions of t. Thus, we have \(T'(t) = 10 \left(\frac{(8t + 16)(t^{2} + 4t + 10) - (2t + 4)(4t^{2} + 16t + 75)}{(t^{2} + 4t + 10)^{2}}\right)\).

Find Rates of Change

The rates of change of \(T\) with respect to \(t\) when (a) \(t=1\), (b) \(t=3\), (c) \(t=5\), and (d) \(t=10\) respectively are \(T' (1)\), \(T' (3)\), \(T' (5)\) and \(T' (10)\), which can be found by plugging these values into \(T' (t)\).

Simplify

Simplify the expression to find the rates of change for those specific time points.

Key concepts

Initial Temperature Calculation

Understanding the initial temperature of an object is crucial in many scientific and real-world applications, such as food preservation in refrigeration. For our exercise, the model for temperature as a function of time, denoted as T(t), helps us determine how the temperature of food changes over time once placed inside a refrigerator.

The process to find the initial temperature is straightforward. To begin, we consider the time variable, t, to be 0 since we are interested in the temperature at the start, also known as time zero. By substituting t = 0 into the given temperature function, we eliminate terms containing t, simplifying the equation and revealing the initial temperature without any time-dependent changes. Through these steps, we find that the initial temperature of the food is 75 degrees Fahrenheit.

Derivative of Temperature Function

Grasping the concept of derivatives in calculus is essential to understanding rates of change. In the context of our temperature function, the derivative represents how fast the temperature is changing at any given point in time. By finding the derivative T'(t) of our temperature function T(t), we can calculate the precise rate of temperature change with respect to time.

Applying the rules of differentiation allows us to transform our function into one that predicts how rapidly temperature will either rise or fall as time progresses. It involves leveraging known differentiation techniques, including the quotient rule if necessary, to handle complex fractions.

Quotient Rule for Differentiation

The quotient rule is an indispensable tool when it comes to differentiating ratios of functions. It states that the derivative of a function divided by another function, say, f/g, is given by ((f' * g) - (f * g')) / (g^2). Here, f and g are both functions of t, and f' and g' are their respective derivatives.

To use this rule effectively, identify the numerator (top) and the denominator (bottom) of the fraction. Differentiate both separately (as f' and g') and then apply the formula. It's crucial to follow these steps methodically to ensure accuracy - a single sign error or a misstep in calculation can lead to an incorrect rate of change. In our temperature function, the quotient rule helps us derive a formula for T'(t) that is used in subsequent calculations.

Calculating Rates of Change at Specific Points

The practical application of derivatives comes alive when we compute the rate of change at specific moments. By substituting particular values of t into the derivative function T'(t), we determine the temperature change rate at each of these instances. This step is where the concepts of calculus move from the abstract to the concrete, providing real numbers that can inform decisions or provide insights into the problem at hand.

For instance, calculating T'(1), T'(3), T'(5), and T'(10) indicates the temperature rate of change after 1, 3, 5, and 10 hours, respectively. This information could be invaluable for ensuring that the food remains within safe temperature ranges while cooling. It's a process of pinpointing, which not only tests calculus proficiency but also its practical significance.