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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)
$F'(75)=\frac{f\left(90\right)-f\left(60\right)}{90-60}$
$=\frac{354.5-324.5}{30}$
$=1$ degrees 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)$L(x)=f(a)+f'(a)(x-a)$
$L(t)=f(75)+f'(75)(t-75)$
$= 342.8+1(t-75)$
$= t+267.8$


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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\)
$F(72)=72+267.8$
$= 339.8$ degrees 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 my estimate is neither too large or too small because 339.8 falls in between 324.5 and 342.8. Also, the value is closer to 342.8 because 72 is closer to 75 than 60. 


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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)
$F(100)= 100+267.8$
$=367.8$ degrees 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 estimate is more off than the previous estimation becase we are estimating a point that is farther away from the point of tangency. Whether the value is too large or too small depends on if the tengent line is above or under the curve on a graph.  

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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)
$L(t)=t+267.8$
<iframe src="https://www.desmos.com/calculator/pr2arub9oo?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>

L(t) is a good approximation of F(t) when the values reach 75, but they are still good at 65 and 90.

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