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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)

$f'\left(75\right)\ =\ \frac{f\left(90\right)-f\left(60\right)}{30}$

$=f'\left(75\right)\ =\ \frac{30}{30}$

$=1$.

At the 75 min mark, the potato is heating up at a rate of 1 degF/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) 


$L\left(x\right)=\ f\left(a\right)+f'\left(a\right)\left(x-a\right)$

$L(75)= f(75)+f'(75)(x-75)$

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



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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\) 

$F(72) = L(72)$

$= 342.8+(1)(72-75)$

$= 342.8-3$

$= 339.8$ degrees. 
    



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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 the estimate for F(72) is over. Because of the concavity (concave down) of the graph, the estimate of the slope is always over the curve, making F(72) an ovrestimate 

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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(100) = L(100)

= 342.8 + 1(100-75)

= 367.5 degrees

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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)

The estimate for F(100) is to large because our tangent line apporiximation always stays above the curve of F. 

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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)

Zooming into this image, the tangent line is most accurate at F(72). Thus, x-values that are close to 75 will have a similar slope and be more accurate.The farther away the x-valuse are from 75, they farther the slope L gets from F. Pretty har to see from this image, but I think the tagent line begins to ben noticlby different around x = 70 and x= 80. 

![](https://i.imgur.com/fXTe1QK.png)
 




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