Problem 64 - Calculus Single Variable

Question

The initial temperature \(T(0)\) of an object is \(50^{\circ} \mathrm{C}\). If the object is cooling at the rate $$T^{\prime}(t)=-0.2 e^{-0.02 t}$$ when measured in degrees centigrade per second, then what is the limit of its temperature as \(t\) tends to infinity?

Step-by-Step Solution

Answer

The limit of the temperature as \(t\) tends to infinity is \(40^{\circ} \mathrm{C}\).

Step 1: Understand the Problem

We are given the rate of change of temperature of an object as a function of time, and we need to find the limit of the object's temperature as time approaches infinity. The rate of change is given as \( T'(t) = -0.2 e^{-0.02t} \).

Step 2: Integrate the Rate of Change

To find the temperature function \( T(t) \), we need to integrate \( T'(t) \). The integral of \( -0.2 e^{-0.02t} \) with respect to \( t \) is \( T(t) = \int -0.2 e^{-0.02t} \, dt \).

Step 3: Solve the Integral

Compute the integral: \( \int -0.2 e^{-0.02t} \, dt = 10 e^{-0.02t} + C \), where \( C \) is the constant of integration.

Step 4: Use Initial Condition to Find the Constant of Integration

We know \( T(0) = 50 \). Substitute \( t = 0 \) into the solution: \( 10 e^{0} + C = 50 \). This simplifies to \( 10 + C = 50 \), thus \( C = 40 \).

Step 5: Write the Temperature Function

Substituting \( C = 40 \) into the equation for \( T(t) \), we get \( T(t) = 10 e^{-0.02t} + 40 \).

Step 6: Determine the Limit of Temperature as \( t \to \infty \)

As \( t \) approaches infinity, \( e^{-0.02t} \) approaches 0. Therefore, \( T(t) = 10 e^{-0.02t} + 40 \) approaches 40.

Key Concepts

Rate of Change of Temperature

When we talk about the rate of change of the temperature of a cooling object, we are essentially discussing how fast or slow the temperature changes as time progresses.
The formula we are given, \( T'(t) = -0.2 e^{-0.02t} \), represents how the temperature decreases over time. This is measured in degrees centigrade per second and tells us a couple of important things.
- The negative sign indicates that the temperature is decreasing with time.- The presence of the exponential function \( e^{-0.02t} \) implies that the rate of temperature change is not constant.
The rate of change slows down as time increases because the exponential term \( e^{-0.02t} \) itself becomes smaller. This reflects a common real-world phenomenon: as objects cool, their rate of cooling decreases over time.

Integration of Exponential Functions

Integrating an exponential function is a crucial step in finding the temperature function from its rate of change.
To find the temperature function \( T(t) \) from \( T'(t) = -0.2 e^{-0.02t} \), we need to integrate \( T'(t) \) with respect to time \( t \).
When integrating an exponential function of the form \( a e^{bt} \), the result is \( \frac{a}{b} e^{bt} + C \), where \( C \) is a constant of integration:
- Here, \( a = -0.2 \) and \( b = -0.02 \). So the integral is \( 10 e^{-0.02t} + C \).
Integration transforms the rate of change into the actual function, representing temperature in this context.
The constant \( C \) needs to be determined using known conditions, such as initial temperature.

Limit of a Function as Time Approaches Infinity

The concept of the limit of a function as time approaches infinity is important in understanding the long-term behavior of the temperature.
In our problem, we found the temperature function to be \( T(t) = 10 e^{-0.02t} + 40 \).
As time \( t \) becomes very large, the term \( e^{-0.02t} \) converges towards 0.
- With \( e^{-0.02t} \) approaching 0, the equation simplifies to \( T(t) = 10 \times 0 + 40 = 40 \).
Thus, the limit of the temperature as time approaches infinity is 40 degrees centigrade.
Understanding limits helps us determine that after a long time, the object will stabilise at a temperature of 40\(^{\circ}\mathrm{C} \), indicating that it will cease changing beyond this point.