Problem 5 - Active Calculus

Question

A cup of coffee has its temperature \(F\) (in degrees Fahrenheit) at time \(t\) given by the function \(F(t)=75+110 e^{-0.05 t}\), where time is measured in minutes. a. Use a central difference with \(h=0.01\) to estimate the value of \(F^{\prime}(10)\). b. What are the units on the value of \(F^{\prime}(10)\) that you computed in (a)? What is the practical meaning of the value of \(F^{\prime}(10) ?\) c. Which do you expect to be greater: \(F^{\prime}(10)\) or \(F^{\prime}(20) ?\) Why? d. Write a sentence that describes the behavior of the function \(y=F^{\prime}(t)\) on the time interval \(0 \leq t \leq 30\). How do you think its graph will look? Why?

Step-by-Step Solution

Answer

a: -2.75 degrees/min. b: Degrees Fahrenheit per minute, indicates cooling rate. c: F'(10) is greater than F'(20). d: The function starts steeply and flattens out.

Step 1: Understand the function

The given function for the temperature of the coffee is \[ F(t) = 75 + 110 e^{-0.05 t} \] where t is time in minutes.

Step 2: Compute the central difference

For part (a), we use the central difference formula with h = 0.01 to estimate F'(10):\[F'(10) \approx \frac{F(10 + h) - F(10 - h)}{2h}\]Evaluate the function at t = 10 + 0.01 and t = 10 - 0.01:\[F(10.01) = 75 + 110 e^{-0.05 \cdot 10.01}\]\[F(9.99) = 75 + 110 e^{-0.05 \cdot 9.99}\]Substitute these values back into the central difference formula to estimate F'(10).

Step 3: Evaluate function values

Let's calculate:\[F(10.01) \approx 75 + 110 e^{-0.5005} \approx 75 + 110 \cdot 0.6065 \approx 141.715\]\[F(9.99) \approx 75 + 110 e^{-0.4995} \approx 75 + 110 \cdot 0.6070 \approx 141.77\]Thus, we have:\[F'(10) \approx \frac{141.715 - 141.77}{0.02} \approx -2.75 \text{ degrees/min}\]

Step 4: Determine units and practical meaning

For part (b), the units of F'(10) are degrees Fahrenheit per minute because we're computing the change in temperature over a small change in time. Practically, F'(10) represents the rate at which the temperature of the coffee is changing at t=10 minutes.

Step 5: Compare derivatives

For part (c), we expect the temperature derivative to be greater (less negative) at t = 10 than at t = 20 because the coffee is cooling down less rapidly as time progresses. Since the exponential decay slows over time, the rate of temperature change decreases in magnitude.

Step 6: Describe the behavior of F'(t)

For part (d), the function y = F'(t) will start with a large negative value, indicating rapid initial cooling, and then approach zero as time increases, indicating slower cooling. The graph would be a curve starting from a steep slope and gently flattening out towards the t-axis.

Key Concepts

Central difference method

The central difference method is a numerical technique used to approximate the derivative of a function. Imagine you want to know how quickly something is changing at a particular moment (like the temperature of the coffee). Instead of using calculus directly, you can estimate this change using values from nearby points.

To find the derivative at time \(t=10\) using the central difference method, we calculate the function's values at \(t=10.01\) and \(t=9.99\):

- First, find \(F(10.01)\): \ F(10.01) = 75 + 110 e^{-0.5005} \ \text{This value is approximately } 141.715

- Then find \(F(9.99)\): \ F(9.99) = 75 + 110 e^{-0.4995} \text{This value is approximately } 141.77

Using these results in our central difference formula, we get:

\ F'(10) \ \text{ \ \approx \frac{F(10.01)-F(10-0.01)}{2h}\ } \ \ \text{Which simplifies to} \ \ \frac{141.715 - 141.77}{0.02} \ \approx -2.75 \ \text{degrees per minute.}

The central difference method is very useful because it uses points on both sides of the desired value, leading to a more accurate estimation.

Temperature function

The temperature function \(F(t) = 75 + 110 e^{-0.05t}\) models the cooling of the coffee over time. Here, \(t\) is time in minutes. Let's break it down:

- **75**: Represents the ambient temperature in degrees Fahrenheit (the room temperature when the coffee is cooling).

- **110**: The difference between the initial coffee temperature and the ambient temperature.

- **\(e^{-0.05t}\)**: This term models exponential decay, which represents how quickly the coffee is losing its heat.

At \(t=0\), the initial temperature is \(75 + 110 = 185^\text{°F}\). As \(t\) increases, the term \(e^{-0.05t}\) decreases because \(e\) raised to a negative exponent becomes smaller over time. This shows the coffee's temperature gradually approaching room temperature. The process of cooling is indicated by how fast this exponent changes.

The units for the derivative \(F'(t)\) are degrees Fahrenheit per minute. For example, at \(t=10\), \(F'(10)=-2.75\text{ degrees/min}\) means the coffee temperature is decreasing by \(2.75\text{ degrees Fahrenheit per minute}\) at that exact moment. Practically, it tells us how quickly the coffee is cooling down.

Exponential decay

Exponential decay describes processes that decrease rapidly at first and then level off over time. This pattern is common in natural settings, such as cooling coffee, radioactive decay, or population declines.

The term \(e^{-0.05t}\) in our temperature function \(F(t)=75+110e^{-0.05t}\) is an exponential decay factor.

Let's explore why this happens:

- **Initial Rapid Decrease**: When coffee is very hot compared to the room temperature, it cools down quickly. The temperature difference is large, making it lose heat rapidly.

- **Slower Cooling Over Time**: As the coffee temperature approaches room temperature, the difference (and hence, the cooling rate) decreases. Thus, the rate of temperature change decreases, and the coffee cools more slowly.

At \(t=0\), \(e^{-0.05 \times 0}=1\), so \(F(0)=185°. As time increases, the exponent \)-0.05t\( makes the \)e^{-0.05t}\( term shrink towards zero, reducing the contribution of the \)110e^{-0.05t}\( term. Since the exponential term gets very small, the temperature asymptotically approaches the room temperature (75°).

Comparing \)F'(10)\( and \)F'(20)\(: Since the exponential decay slows down with time, \)F'(20)\( will be less negative (indicating a slower rate of cooling) compared to \)F'(10). This is why \(F'(10)\) is more negative than \(F'(20)\). The graph of \(F'(t)\) will show a steep decline initially, then gradually level off to approach zero.