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title: 'Math 181 Miniproject 4: Linear Approximation and Calculus'

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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) $F'(75) = \frac{\left(f\left(90\right)-f\left(60\right)\right)}{90-30} = 1 degreeF/minute$


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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) $L(t)=f(t)+f'(t)(x-a)$

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

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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\) $L(t)=342.8+1(72-75) = l(t)=342.8-3 = 339.8 degreesF$

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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 beleive my estimate to be not exactly right because the central difference is only an estimatation. I think my estimate may be slightly under because the potato's temperature is on an increasing trend.

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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) $L(t)=342.8+1(100-75) = 342.8 + 25 = 367.8 degreesF$

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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 beleive my extrapolation may still be under because where we last left off, the internal temperature was still increasing.

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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)  ![](https://i.imgur.com/PYiWI4I.png)




The interpolation of (72) matches up very well with the graph of f(x) whereas the extrapolation of (100) is similiar yet slighty bigger which is to be expected since it is a larger number.

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