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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$F'(75)= \frac{F(90)-F(60)}{90-60}$
$F'(75)= \frac{354.5-324.5}{90-60}$
$F'(75)= \frac{30}{30}$
$F'(75)=1$ degF/min.

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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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$L(75)=F(75)+F'(75)(x-75)$
$L(75)=342.8+1(x-75)$

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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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$F(72)$~$L(72)$
$L(72)=342.8+1(72-75)$
$L(72)=342.8+1(-3)$
$L(72)=342.8-3$
$L(72)=339.8$ degF

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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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I believe that the estimate calculated in section (c) to be exactly right. Being that 72 degF is close to 75 degF, we do not expect to observe a huge deviation in calculating the temperature using the local lineraization.

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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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$F(100)$~$L(100)$
$L(100)=342.8+1(100-75)$
$L(100)=342.8+1(25)$
$L(100)=367.8$ degF

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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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The value estimated for $F(100)$ done above using the local linerization is too large. Utilization of the local linearization when $a=75$ yields the accurate estimations when the value is closest to $a$; $a=100$ compared to $a=75$ is a large difference. When we input $a=100$ in the formula for the local linerization for when $a=75$ this will yield an overestimation due to the large difference.

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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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The local linearization when $a=75$ ($L(75)$) is depicted in black; $F(t)$ is depicted in red.

$L(100)$ is not a good approximation for estimating the value for $F(100)$ as shown in the graph.

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