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A potato is placed in an oven, and the potato's temperature \(F\) (in degrees Fahrenheit) at various points in time is taken and recorded in the following table. Time \(t\) is measured in minutes. $$\begin{array}{ll} \hline t & F(t) \\ \hline 0 & 70 \\ \hline 15 & 180.5 \\ \hline 30 & 251 \\ \hline 45 & 296 \\ \hline 60 & 324.5 \\ \hline 75 & 342.8 \\ \hline 90 & 354.5 \\ \hline \end{array}$$ a. Use a central difference to estimate \(F^{\prime}(60)\). Use this estimate as needed in subsequent questions. b. Find the local linearization \(y=L(t)\) to the function \(y=F(t)\) at the point where \(a=60\). c. Determine an estimate for \(F(63)\) by employing the local linearization. d. Do you think your estimate in (c) is too large or too small? Why?

Step by step solution

01

Compute the central difference for F'(60)

The central difference formula is \[ F'(t) \approx \frac{F(t+h) - F(t-h)}{2h} \] For t = 60 and h = 15 minutes, use the values from the table: \[ F'(60) \approx \frac{F(75) - F(45)}{2 \times 15} \] Substitute the given values: \[ F'(60) \approx \frac{342.8 - 296}{30} = \frac{46.8}{30} = 1.56 \] Thus, the estimate for \( F'(60) \) is 1.56.

02

Find the local linearization L(t) at t = 60

The local linearization formula is \[ L(t) = F(a) + F'(a)(t - a) \] For our problem, \( a = 60 \), \( F(a) = 324.5 \), and \( F'(a) = 1.56 \). Substituting these values, we get: \[ L(t) = 324.5 + 1.56(t - 60) \]

03

Estimate F(63) using the local linearization

Using the local linearization from the previous step, set t = 63: \[ F(63) \approx L(63) = 324.5 + 1.56(63 - 60) \] This simplifies to: \[ F(63) \approx 324.5 + 1.56 \times 3 = 324.5 + 4.68 = 329.18 \]

04

Determine if the estimate is too large or too small

Considering that the temperature increases at a decreasing rate, the linear approximation might slightly overestimate. Since the rate of temperature increase tends to slow down over time, our linear model might project a slightly higher value than the actual.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Difference Formula

To estimate the rate of change of the potato's temperature at a particular time, we can use the central difference formula. This method provides a numerical approximation of the derivative, which represents the rate of change. Specifically, the central difference formula is:
\[ F'(t) \approx \frac{F(t+h) - F(t-h)}{2h} \]
For our exercise, we use t = 60 minutes and h = 15 minutes. Plugging in the numbers from the table:
\[ F'(60) \approx \frac{F(75) - F(45)}{ 2 \times 15} \approx \frac{342.8 - 296}{30} = \frac{46.8}{30} = 1.56 \]
The central difference provides an estimation of the slope of the temperature graph at t = 60, which tells us how quickly the temperature is changing at that moment.

Linearization

Linearization is a method used to approximate a function near a given point using a linear function. This is particularly useful for making quick estimations. Mathematically, the local linearization of a function at a point a is given by:
\[ L(t) = F(a) + F'(a)(t - a) \]
Let's apply this to our problem using a = 60, F(a) = 324.5, and F'(a) = 1.56:
\[ L(t) = 324.5 + 1.56(t - 60) \]
This formula gives us a line that is tangent to the temperature curve at t = 60 and can be used to approximate the temperature for times close to this point. It converts the potentially complex behavior of the function into a simple linear form, making it easier to work with.

Temperature Rate of Change

Understanding the rate of change of temperature is crucial in many real-world applications, from cooking to climate science. In our specific problem, we estimated the rate of change of the potato’s temperature at t = 60 minutes as 1.56 degrees Fahrenheit per minute. This estimation used the central difference formula:
\[ F'(60) \approx 1.56 \]
This number tells us how fast the potato's temperature is rising at the 60-minute mark. Knowing this rate helps in predicting future temperatures and understanding whether the potato is heating up quickly or gradually slowing down.

Estimation Accuracy

When we make estimations, such as predicting the temperature at t = 63 using the linearization formula, it's crucial to consider the accuracy of our estimates. Linearization offers a good approximation for values close to the point of tangency but can lead to errors if used too far from this point. By setting t = 63 in our linearization formula:
\[ F(63) \approx L(63) = 324.5 + 1.56 \times 3 = 329.18 \]
However, since the temperature increase tends to slow down over time, our linearization might overestimate the true temperature. Therefore, it’s essential to understand the context and the behavior of the actual function to make more accurate predictions.