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


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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)$F'(75)[t-75]=L(t)-F(75)$ 
$L(t-75)=L(t)-342.8$
$L(t)=t+267$


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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\) $F(72)$
$y=72+207.8=339.8$

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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) By looking at the table we can see that $F(t)$ is decreasing when F is changing t. $F'(t)$ should be negative because $F(t)$ is concaving down. So I believe that the estimate is too large.  

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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) $F'('90)=\frac{F(75)-F(90)}{75-90}=\frac{342.8-354.5}{-15}$
$F(90)=0.78$
$y=L(t)$at points near 90
$L(t)=0.78t+284.3$
F(100) ~ L(100)=0.78*100+284.3
F(100) ~ 362.3
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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) It is very close to the actual value 
$\frac{362.3-359.77}{359.77}$*100%= 0.7% 
The graph that passes through is $f(t)=374.944-304.952*e$^-.03001t

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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://www.desmos.com/calculator/mh6a3quwgl
L(t) is good  because the value of F is close to t=75 and t=90


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