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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) The central difference formula is 
\\[f'(x)=\frac{f(x+h)-f(x-h)}{2(h)}\\]\\[f'(75)=\frac{f(75+15)-f(75-15)}{2(15)}\\] \\[f'(75)=\frac{f(90)-f(60)}{30}\\] \\[f'(75)=\frac{354.5-329.5}{30}\\]\\[f'(75)=1\\] The average rate of change in 75 minutes is 1 in degress Fahrenheit. 

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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) The local linerization formula is 
\\[y-f(a)=f'(a)(t-a)\\]\\[y-342.8=1(t-75)\\]\\[y=342.8+(t-75)\\]\\[y=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\) The local linerization we just found is \\[L(t)=t+267.8\\]
$F(72)$ gets plugged into the eqaution\\[L(72)=72+267.8\\]\\[L(72)=339.8 in degrees Fahrenheit\\]
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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) The estimate in (c) is exactly right. The average rate of change is 1 in degrees Fahrenheit and when I subtract the value for $f(75)$ and $f(72)$, I get 1 in degrees Fahrenheit exaclty. 
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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) The eqaution that was just found for local inerization is 
\\[L(t)=t+267.8\\]\\[L(t)=100+267.8\\]\\[L(100)=367.8 in degrees Fahrenheit\\]


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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) According the grapah of $F$ and $L$, the estimate in C is excatly right because it the tangent line of $L$ coincides with the graph of $F$.
<iframe src="https://www.desmos.com/calculator/pfu7ajzqsd?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>
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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) Line L(t) is a good aproximation at or near $t=75$ otherwise the value of L(t) is an over estimate. 
<iframe src="https://www.desmos.com/calculator/pfu7ajzqsd?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>


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