Problem 19 - Applied Calculus

Question

A bar of metal is cooling from \(1000^{\circ} \mathrm{C}\) to room temperature, \(20^{\circ} \mathrm{C}\). The temperature, \(H,\) of the bar \(t\) minutes after it starts cooling is given, in \(^{\circ} \mathrm{C},\) by $$I I=20+980 e^{-0.1 t}$$ (a) Find the temperature of the bar at the end of one hour. (b) Find the average value of the temperature over the first hour. (c) Is your answer to part (b) greater or smaller than the average of the temperatures at the beginning and the end of the hour? Explain this in terms of the concavity of the graph of \(H\)

Step-by-Step Solution

Answer

(a) Calculate \( H(60) = 20 + 980 e^{-6} \). (b) Find average with \( \frac{1}{60} \int_0^{60} (20 + 980 e^{-0.1t}) dt \). (c) The average from (b) is less than the arithmetic middle due to concavity.

Step 1: Convert Time to Correct Units

1 hour is 60 minutes, thus when evaluating the formula for the temperature at the end of the hour, use \( t = 60 \) minutes.

Step 2: Calculate Temperature at 60 Minutes

Substitute \( t = 60 \) into the given formula \( H = 20 + 980 e^{-0.1t} \). \( H = 20 + 980 e^{-0.1 \times 60} = 20 + 980 e^{-6} \). Evaluate this to find the temperature.

Step 3: Find Average Temperature Over One Hour

Use the integral formula for average value: \( \frac{1}{b-a} \int_a^b H(t) \, dt \), where \( a = 0 \), \( b = 60 \). Integrate \( 20 + 980 e^{-0.1t} \) from 0 to 60 and divide by 60.

Step 4: Calculate Definite Integral

The integral \( \int (20 + 980 e^{-0.1t}) \, dt \) is \( 20t - 9800e^{-0.1t} \). Evaluate from 0 to 60: \( [20(60) - 9800e^{-0.1\times60}] - [20(0) - 9800e^0] \).

Step 5: Calculate Average Value Using Definite Integral Result

Compute the definite integral results from Step 4 and divide by 60 to find the average temperature over the first hour.

Step 6: Compare with Arithmetic Average

The arithmetic average of the temperatures at the beginning (1000°C) and end (temperature calculated in Step 2) is \( \frac{1000 + H(60)}{2} \). Compare this with the result of the average temperature calculated in Steps 3-5.

Step 7: Explain in Terms of Concavity

The function \( H(t) \) is exponentially decreasing, which is concave up (the second derivative \( H''(t) > 0 \)). Thus, the integral average will be lower than the arithmetic average because the function decreases at a decreasing rate.

Key Concepts

Temperature Function

In the context of this exercise, the temperature function, denoted as \( H(t) \), models how the temperature of a cooling metal bar changes over time. The given formula for this function is \[ H(t) = 20 + 980 e^{-0.1t} \] where:

• \( t \) is the time in minutes after the cooling process has started.
• The constant 20 represents the room temperature to which the bar is cooling.
• The term \( 980 e^{-0.1t} \) describes the exponential decay of the initial temperature difference of 980°C between the hot metal and the room temperature.

This function captures the essence of how the metal bar's temperature exponentially decreases from its initial high temperature of \( 1000^{\circ} \mathrm{C} \). As \( t \) increases, the influence of the exponential term diminishes, bringing the temperature closer to 20°C. This characteristic makes it easier to predict the cooling behavior over time.

Average Value of a Function

The average value of a function over a specified interval helps us understand the typical output value of that function within the given range. For functions like the temperature function \( H(t) \), we use the concept of integrations to find this average.
To calculate the average value of \( H(t) \) over the first hour (0 to 60 minutes), we apply the formula:
\[ \frac{1}{b-a} \int_a^b H(t) \, dt \] where \( a = 0 \) and \( b = 60 \).
Integrating the function over this interval gives us:

• \( 20t - 9800 e^{-0.1t} \) evaluated from 0 to 60.

By evaluating this, we determine the integral value over the period and then divide by 60 (the duration) to find the average temperature. This process shows how the integral's output reflects the overall behavior of \( H(t) \) beyond just snapshot values at specific times.

Exponential Decay

Exponential decay is a mathematical concept describing processes that decrease at a rate proportional to their current value. In this exercise, the temperature function exhibits exponential decay, as evidenced by the term \( 980 e^{-0.1t} \).
Some key features of exponential decay include:

• It starts quickly and slows as it progresses, eventually approaching but never quite reaching a limiting value—in this case, 20°C.
• The rate of decrease is proportional to the difference between the current temperature and the ambient temperature.
• Each unit of time sees a consistent proportionate decrease, governed by the decay constant in the exponent, here given as \(-0.1\).

This mathematical feature allows us to better understand real-world phenomena like cooling and radioactive decay, where the change's intensity diminishes over time. In this particular exercise, it influences why the average temperature over an hour can differ from the simple mean of the starting and ending temperatures, as exponential decay shifts the behavior of the function nonlinearly.