Problem 8 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Suppose the temperature of an object is changing at a rate of \(r(t)=-2 e^{-t}\) degrees Celsius per hour, where \(t\) is given in hours. (a) Is the object heating, or cooling? (b) Between time \(t=0\) and \(t=1\), how much has the temperature changed? (c) Between \(t=1\) and \(t=2\), how much has the temperature changed? (d) If the object was 100 degrees Celsius at time \(t=0\), how hot is it at time \(t=1\) ?

Step-by-Step Solution

Answer

a) The object is cooling. b) The temperature changed by \(2 - 2e^{-1}\) degrees Celsius. c) The temperature changed by \(-2e^{-2} + 2e^{-1}\) degrees Celsius. d) At time \(t=1\), the object is \(100 - (2 - 2e^{-1})\) degrees Celsius hot.

Step 1: Understanding the Problem

The rate of change of temperature \(r(t)=-2 e^{-t}\) indicates the speed and direction at which the temperature of the object changes with respect to time. Since the rate \(r(t)\) is negative, it means the object is cooling down.

Step 2: Calculating the Temperature Change between \(t=0\) and \(t=1\)

To find out the change in temperature between time \(t=0\) and \(t=1\), we apply definite integral over the rate function from 0 to 1. We calculate the integral of \(r(t)=-2 e^{-t}\) from 0 to 1, that is, \(\int_{0}^{1} (-2e^{-t})dt\). This gives \(2 - 2e^{-1}\) degrees Celsius.

Step 3: Calculating the Temperature Change between \(t=1\) and \(t=2\)

The process is the same as in step B to find the change in temperature between \(t=1\) and \(t=2\). The integral to get this change is \(\int_{1}^{2} (-2e^{-t})dt\), which equates to \(-2e^{-2} + 2e^{-1}\) degrees Celsius.

Step 4: Determining the Temperature at Time \(t=1\)

If the object was 100 degrees Celsius at time \(t=0\), then to find its temperature at \(t=1\), we need to subtract the temperature change between \(t=0\) and \(t=1\) from the initial temperature. Doing so gives, \(100 - (2 - 2e^{-1})\) degrees Celsius.

Key Concepts

Rate of Change Calculus

Rate of change calculus, often referred to as differential calculus, is a powerful mathematical tool that measures how a quantity changes over time. It is intrinsic to many fields that involve motion, growth, or decay. In our exercise, the rate of change is represented by the function \( r(t) = -2e^{-t} \) which describes the temperature change of an object over time.

Understanding this rate of change is crucial for predicting future behaviors or states of the system under study. A negative rate of change, as in our function, indicates that the quantity is decreasing—in this case, the object is cooling down. By analyzing this rate of change function, we can extract valuable information about the object's temperature trends and calculate specific temperature changes over intervals of time.

Exponential Decay

Exponential decay refers to a process where a quantity decreases at a rate proportional to its current value. In the context of our exercise, the temperature \( T \) of an object changes over time due to a rate of change that exhibits exponential decay, expressed by the function \( r(t) = -2e^{-t} \).

The exponential function \( e^{-t}\) here suggests that the temperature of the object will decrease rapidly at first and then more slowly over time. This property is characteristic of many natural processes, such as cooling, radioactive decay, and even certain financial investments. Understanding the nature of exponential decay allows scientists and engineers to model and predict the behavior of systems that are cooling or losing value over time in a more effective manner.

Integral Calculus Application

Integral calculus is a branch of mathematics that measures the size, quantity, or total accumulated value of various functions and geometrical shapes. It is essentially the reverse process of differentiation. In our particular example, we apply integral calculus to find the total temperature change of an object over a specific period. This is accomplished by integrating the rate of change function \( r(t) \) over a given interval.

For instance, to calculate the temperature change from \( t = 0 \) to \( t = 1 \) hour, the definite integral of \( r(t) \) is found, symbolically written as \( \int_{0}^{1} (-2e^{-t})dt \). The result of this integral represents the aggregate effect of the cooling process over that hour. Similarly, by integrating over other intervals, we can pinpoint how much the temperature changes over those time frames, thus enabling us to model and understand the thermal dynamics of the object.