Math 181 Miniproject 4: Linear Approximation and Calculus.md
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Math 181 Miniproject 4: Linear Approximation and Calculus
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**Overview:** In this miniproject you will put the idea of the *local linearization* of a function to build linear approximations to complex functions and then make *interpolations* and *extrapolations* using them.

**Prerequisites:** Sections 1.8 in *Active Calculus*, which focuses on this topic. **Completion of Miniprojects 1 and 2 is recommended before doing this miniproject**.

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1\. 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. The time $t$ is measured in minutes.
| $t$ | 0 | 15 | 30 | 45 | 60 | 75 | 90 |
|----- |---- |------- |----- |----- |------- |------- |------- |
| $F$ | 70 | 180.5 | 251 | 296 | 324.5 | 342.8 | 354.5 |

(a) Use a central difference to estimate $F'(75)$. Use this estimate as needed in subsequent questions in this problem.
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(a) Using the central difference to estimate F'(75) would look like this.
$F'(75)=\frac{\left(f\left(90\right)-f\left(60\right)\right)}{30}$
$F'(75)=\frac{354.5-324.5}{30}$
Which makes $F'(75)$=$\frac{30}{30}$=$1$.
This means that the slope in tempereature is 1 per minute.

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(b) Find the local linearization $y = L(t)$ to the function $y = F(t)$ at the point where $a = 75$.
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(b) The local linearization of $y=L(t)$ where $a=75$ would be $L(75)=f(75)-f'(75)(x-75)$
$L(75)=342.8+1(x-75)$

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(c\) Determine an estimate for $F(72)$ by employing the local linearization. Terminology: This estimate is called an *interpolation* because we are estimating a value that lies within a data set, between two known data points.
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(c\) By employing the previously found local linearization the result would be
$L(72)=342.8+1(72-75)$
$L(72)=339.8$ ℉

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(d) Do you think your estimate in (c) is too large, too small, or exactly right? Why?
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(d) By examining the given graph this estimate looks just right because $f(75)=342.8$ and when finding f(72) the result is in between the given results (60 and 75) on the chart.

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(e) Use your local linearization to estimate $F(100)$. Terminology: This estimate is called an *extrapolation* because we are estimating a value that lies outside the range of values of a data set.
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(e) Using the local linearization from step b).
$L(100)=342.8-1(100-75)$
$L(100)=367.8$ ℉

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(f) Do you think your estimate in (e) is too large, too small, or exactly right? Why?
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(f) This estimate is right or close to the resulting temperature. $L(100)$ would have to be larger because the chart shows a rise in temperature.

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(g) Plot both $F$ and $L$ and comment on how or when the line $L(t)$ is a good approximation of $F(t)$.
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(g) When L(t) is a good approximation of F(t) is L(72) because it matces perfectly with the graph of f(x). Though L(100) was close L(72) was more accurate.

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