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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)

$$AV[60,90]$$

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

$$f'(75)=\frac{354.5-324.5}{90-60}$$

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

$$f'(75)=1 °F /min$$

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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) 

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




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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\)

$$f(72)≈L(72)$$
$$f(72)≈342.8-1(t-75)$$
$$f(72)≈342.8-1(72-75)$$
$$f(72)≈342.8-1(-3)$$
$$f(72)≈339.8 \ °F $$

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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 the estimate is a little larger than it should be because its is larger than the number we are coparing it to. 

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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)

$$f(100)≈L(100)$$
$$f(100)≈342.8+1(100-75)$$
$$f(100)≈342+1(25)$$
$$f(72)≈367.8 \ °F$$

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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 the answer is too high becuase we are using f(75) and that is too far away from F(100) on the slope line to make an accurate prediction.  



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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/plLajdt.jpg)

$L(t)$ (red graph line) is a good approximation up we reach $f(100)$ the line stays match up with $f(t)$ showing that the approximations is a good approximation until we reach $f(100)$ compaired to $L(100)$



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