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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)
Using the central difference between 60 min and 90 min, I estimated F'(75) to be 1. The potato increases one degree per minute. 
Work: f'(75)=AV[60,90]=$[f(90)-f(60)]/90-60$
=(354.5-324.5)/90-60
=30/30
=1
=1 deg/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) The Local Linearization formula is L(t)=f(t)+f'(t)(x-t)
=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\) An estimate for F(72) ,or the temperature the potato is at when it has been in the oven for 72 minutes, is 339.8 degrees Farenheight. 
L(72)=f(75)+f'(75)(72-75)
=342.8+1(-3)
=342.8-3
=339.8 deg 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 this estimate is very good however it might be the slightest bit too small. The longer the potato is in the oven, the more time it takes the potato to increase its temperature. The potato is warming up at a faster rate between 60 and 75min than it is from 75 to 90 min. At 72 minutes, the potato may be increasing its temperature more than 1 degree per minute.

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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)=f(75)+f'(75)(100-75)
=342.8+1(25)
=342.8+25
=367.8 degrees F
After being in the oven for 100 minutes the potato is estimated to be around 367.8 deg 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 this estimate is too large because the speed at which the potato gets hotter is slowing down after the 75 minute mark and will be heating up less than one degree per minute.

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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)The line L(t)is a good approximation of F(t) after the 60 minute mark.
![](https://i.imgur.com/Yu3GhSp.png)



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