Problem 18 - 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 $$ H=20+980 e^{-0.1 t} $$ (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) 22.43°C; (b) 182.92°C; (c) Less due to concavity of the cooling graph.

Step 1: Calculate temperature at one-hour mark

Given that the temperature at time \( t \) is described by the equation \( H = 20 + 980e^{-0.1t} \), we substitute \( t = 60 \) minutes into the formula to find the temperature at the end of one hour.\[H = 20 + 980e^{-0.1 \times 60}\]Calculating the exponent first:\[-0.1 \times 60 = -6\]Now, calculate \( e^{-6} \) using a calculator and substitute it back:\[H \approx 20 + 980 \times 0.002478752\]\[H \approx 20 + 2.429\]\[H \approx 22.429 \text{ degrees Celsius}\]So, the temperature at the end of one hour is approximately \( 22.43^{\circ} \mathrm{C} \).

Step 2: Find the average temperature over the first hour

To find the average temperature over the first hour, we integrate the temperature function from \( t = 0 \) to \( t = 60 \) and divide by the total time interval (which is 60 minutes).\[\text{Average Temperature} = \frac{1}{60} \int_0^{60} \left( 20 + 980e^{-0.1t} \right) \ dt\]Separate the integral and solve:\[ \int_0^{60} 20 \, dt = 20t \Bigg|_0^{60} = 20 \times 60 = 1200 \]\[\int_0^{60} 980e^{-0.1t} \, dt = -9800 e^{-0.1t} \Bigg|_0^{60} \]\[= -9800(e^{-6} - e^{0}) = -9800(0.002478752 - 1)\]\[= 9800 \times (0.997521248) \approx 9775 \]Combine results:\[\text{Total integral} = 1200 + 9775 = 10975\]\[\text{Average Temperature} = \frac{10975}{60} \approx 182.92 \]The average temperature over the first hour is approximately \( 182.92^{\circ} \mathrm{C} \).

Step 3: Compare average to midpoint temperature

Calculate the midpoint temperature by averaging the initial and final temperatures: The initial temperature (at \( t = 0 \)) is \( 1000^{\circ} \mathrm{C} \).The final temperature (at \( t = 60 \)) is approximately \( 22.43^{\circ} \mathrm{C} \).\[ \text{Midpoint Temperature} = \frac{1000 + 22.43}{2} \approx 511.215 \]Since the average temperature over the first hour (\( 182.92^{\circ} \mathrm{C} \)) is less than the midpoint temperature (\( 511.215^{\circ} \mathrm{C} \)), we conclude that the function \( H(t) \) must be concave down. This is expected because the cooling rate decreases over time, implying that \( H(t) \) is decreasing concavely.

Key Concepts

Exponential Decay

When a bar of metal cools, its temperature decreases over time, which can often be modeled using the concept of exponential decay. In this situation, the temperature function given is \[H = 20 + 980 e^{-0.1t}\]Exponential decay is characterized by a rapid decrease initially, which slows down as time progresses. This is due to the presence of a negative exponent, \(-0.1t\), in the exponential function.

• The base of the exponential, \( e \), is a mathematical constant approximately equal to 2.71828.
• The exponent \(-0.1t\) decreases over time, leading to a decay effect.

Initially, the decrease in temperature is steep when the metal is hottest. As it approaches room temperature (20°C), the rate of decay lessens. This phenomenon reflects the nature of exponential decay: fast declines that taper to a more gradual slope as equilibrium is approached.

Average Temperature Calculation

To find the average temperature over a time period, integration is used. For the metal bar cooling scenario, the average temperature was calculated over the first hour (0 to 60 minutes) with the following steps:

• Use the formula for the average value of a function: \[\text{Average Temperature } = \frac{1}{60} \int_0^{60} \left(20 + 980e^{-0.1t}\right) dt\]
• Calculate the individual integrals for each term. Constant integration yields a linear result, while exponential integration involves calculating changes over the interval and adjusting with any constant factors.
• Combine both results to find a total integral, then divide by the time period (60 minutes in this case). The result gives us the average temperature.

This method provides an important understanding of thermal processes over time, capturing the entire cooling experience rather than snapshots at specific moments.

Concavity in Graphs

Concavity gives insights into the behavior of graphs, showing how a function curves. For the function \( H(t) = 20 + 980e^{-0.1t} \), analyzing concavity helps us understand how the cooling rate changes over time.If a graph is *concave down*, the curve bows downwards like a frown. This behavior indicates a decreasing rate of change. In this case, the temperature decreases quickly at first, then more slowly.

• Mathematically, a function is concave down if its second derivative is negative.
• Evaluating rates at the start and end of the hour showed more rapid decreases at first, aligning with the expected concave-down behavior of this cooling process.

The average temperature being lower than the midpoint temperature (between starting and ending temperatures) confirms this theory. The graph dips below the straight-line average, showing the slowing rate inherent in concave-down patterns.