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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)$f'(75)=f'(60)-f'(90)$
   $=324.5-354.5/60-90$
   $= 1°f/min.$


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



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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\)$F(72)=342.8+1(72-75)$
$L(72)=342.8+1(72-75)$
$=342.8+1(-3)$
$=343.8-3$
$=340.80°$

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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)Too large because the estimation of F(72)=340.8° is over from the actual value graphically of 339.8.

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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)$L(75)=F'(75)+F(75-a)$
$=342.8+1(x-75)$
$L(100)=342.8+1(100-75)$
$=343.8+25$
$=368.8°$


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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)L(100) is an overestimation because the graph of L(100)concaves down and the line sits above the graph which provides proof of an overestimation.

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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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!![](https://i.imgur.com/NTf3hmK.png)

(g)Around the point (75,342.8) is a good approximation of F(t).Also it is a good approximation around (60,324.5)to (90,354.5).


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