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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)
$F'\left(75\right)=\frac{F\left(90\right)-F\left(60\right)}{90-60}$
$=\frac{354.5-324.5}{30}$
$=1$


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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)
$L(t)=f(a)+f'(a)(x-a)$
$L(75)=f(75)+f'(75)(x-75)$
$=342.8+1(x-75)$


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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\)
$L(72)=342.8+1((72)-75)$
$=339.8$

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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)
Exactly right. Because the value is accurately proportionate between the $F$ values of $60$ and $75$

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

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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)
Too large. Because according to the graph, the longer the potato is in the oven, the less the temperature increases.

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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/vvOmkd1.png)

$L(t)$ is a good approximation only when the value is between 60min. and 90min.



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