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


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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) $L\left(x\right)=f\left(a\right)+f'\left(a\right)\left(x-a\right)$ 
$=f\left(75\right)+f'\left(75\right)\left(x-75\right)$
$=342.8+1\left(x-75\right)$
$L\left(x\right)=342.8+1\left(x-75\right)$

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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\) $L\left(72\right)=342.8+1\left(72-75\right)$
$=342.8+1\left(-3\right)$
$=342.8-3$
$=339.8$

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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)I think the estimate is right because 339.8 is less than 342.8, the temperature of the potato at 75 minutes. At 72 minutes, it would make sense that the temperature would be less than the temperature at 75 minutes since the potato has been in the oven for a shorter period of time. Therefore, 339.8 degrees farenheit is a good estimate. 

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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) $L\left(100\right)=342.8+1\left(100-75\right)$
$=342.8+1\left(25\right)$
$=342.8+25$
$=367.8$

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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) I think the estimate is right because the temperature of the potato should be more than it's temperature at 90 minutes since it's been in the oven longer. The temperature at 90 minutes is 354.5 degrees Farenheint, and my result at 100 minutes was bigger, at 367.8 degrees Farenheint.  

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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) $L(t)$ is a good approximation of $F(t)$ around point 60 and so on, because that is when I noticed that $L(t)$ begins to touch $F(t)$ on the graph. 


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