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Chapter 6: Problem 37

Heat Diffusion Some food is placed in a freezer. After \(t\) hours, the temperature of the food is dropping at the rate of \(r(t)\) degrees Fahrenheit per hour, where \(r(t)=12+4 /(t+3)^{2} .\) (a) Compute the area under the graph of \(y=r(t)\) over the interval \(0 \leq t \leq 2\). (b) What does the area in part (a) represent?

Short Answer

Expert verified
The area is approximately 24.5333. It represents the total temperature decrease over the interval 0 to 2 hours.

Step by step solution

01

- Set up the Integral

We need to compute the area under the graph of the function. This requires us to set up an integral. The integral for the function over the interval \(0 \leq t \leq 2\) is \[\int_{0}^{2} r(t) \, dt = \int_{0}^{2} \left(12 + \frac{4}{{(t+3)}^{2}}\right) dt\].
02

- Integrate the Constant

First, integrate the constant term 12. The integral of a constant \(a\) is \(a t\). So, \[\int_{0}^{2} 12 \, dt = 12t \bigg|_{0}^{2} = 12(2) - 12(0) = 24.\].
03

- Integrate the Rational Function

Next, integrate the term \(\frac{4}{{(t+3)}^{2}}\). Use the substitution \(u = t+3\), leading to \(du = dt\). The integral then becomes \[(4) \int \frac{1}{{u}^{2}} \, du = -4u^{-1}\]. Returning to the original variable, we get \[-\frac{4}{t+3} \bigg|_{0}^{2} = -\frac{4}{5} - ( -\frac{4}{3}) = \frac{4}{3} - \frac{4}{5} = \frac{20 - 12}{15} = \frac{8}{15}\].
04

- Combine Results

Add the results from the two integrals: \[24 + \frac{8}{15} = 24 \frac{8}{15} = 24.5333.\].
05

- Interpret the Result

The area under the graph of \(y = r(t)\) over the interval \(0 \leq t \leq 2\) represents the total change in temperature of the food over that time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

heat diffusion
Heat diffusion is the process by which heat spreads through a material.
Imagine placing hot food in a freezer. The heat from the food starts to disperse into the cooler surrounding air.
This happens at different rates depending on the temperature difference and other factors.
Understanding heat diffusion helps in many fields, like engineering and environmental science.
In this exercise, you learned how heat diffuses over time by observing temperature changes.
integral calculus
Integral calculus is a part of mathematics focused on accumulation of quantities.
To find the total change in temperature, we calculate the area under the function鈥檚 curve.
The integral symbol \(\int \) represents this accumulation.
In our problem, we used the integral \(\int_{0}^{2} \left(12 + \frac{4}{{(t+3)}^{2}}\right) dt\) to accumulate the change in temperature over 2 hours.
We integrated separate parts of the function step-by-step and then added those parts together.
rate of change
The rate of change is how fast a quantity is changing over time.
In the context of this exercise, it鈥檚 how quickly the temperature of the food drops as it cools in the freezer.
The formula given, \(r(t)=12 + \frac{4}{{(t+3)}^{2}} \), tells us the rate at which the temperature drops at any time t.
Understanding the rate of change helps to understand trends and make predictions.
For example, if the rate is higher, the temperature drops faster.
temperature variation
Temperature variation describes how temperatures change over time.
When we put food in a freezer, the temperature changes from room temperature to freezer temperature.
This concept lets us understand how external conditions impact temperature.
In this problem, by integrating the rate of temperature drop, we found the total temperature variation from the initial placement of food into the freezer to 2 hours later.
It represented the accumulated cooling effect over that period.

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