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) $\frac{\left(F\left(b\right)-F\left(a\right)\right)}{b-a}$ = temperature/minute

F'(75)=$\frac{\left(P\left(90\right)-P\left(60\right)\right)}{90-60}$=1 degree F/ 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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(b)
$L\left(t\right)=f\left(a\right)+f'\left(a\right)\left(x-a\right)$

$L\left(t\right)=f\left(75\right)+f'\left(75\right)\left(x-75\right)$
$L\left(t\right)=342.8+1\left(x-75\right)$

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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\)
F(72)=L(72)
$L\left(t\right)=342.8+1\left(x-75\right)$
$L\left(t\right)=342.8+1\left(72-75\right)$
$L\left(72\right)=339.8$ F

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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) I think that the estimate (339.8 F) is exactly right, because the value of the temperature should be slightly lower than the value of Degrees F after 75 minutes (342.8 F).

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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)
F(100)=L(100)
$L\left(t\right)=342.8+1\left(x-75\right)$
$L\left(t\right)=342.8+1\left(100-75\right)$
$L\left(100\right)=367.8$ F

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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)
I think that this estimate of 367.8 F is exactly right for F(100) because the value should be a little more than the value of F(90) which is 354.5 F.
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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)
<iframe src="https://www.desmos.com/calculator/th7c8o1gjc?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>

L(t) is a good approximation for F(t) after an hour of time has passed.

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