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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)  $\frac{f(90)-f(65)}{90-60}$   ~    F'(75)

$\frac{(354.5)-(324.5)}{30}$   ~     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(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\) L(75)= 342.8+ 1(x-75)
L(72)= 342.8+ 1(72-75)
L(72)= 342.8+ 1(-3)
L(72)= 342.8-3
F(72)~ 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 the estimate is too large or an because the graph would be concave down and the line would be above the graph making the approximation an over 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(100)= 342.8+ 1(100-75)
L(100)= 342.8+ 1(25)
L(100)= 342.8+ 25
L(100)~ 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 its dependant if the temperature in the oven is still increasing as in there is no maxium in place or if the oven has been turned off. In this case I am going to assume that the maximum value put in place is over 370 degreesF and state that because of the concanve down graph and guadual increase of temp that is is once again an overestimate because the line of approimation is above the 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)
Using desmos to approximate the valuse of F(t) the graph below indicates the L(t) and f(t). The graph of F(t) shows as if there was a platuea and then a decrease in temperature as if the oven was turned off. The graph of L(T) is an approximation if there was no set maximum temperature. It's hard to tell if the L(t) is a good approximation becase the line of best fit that desmos created does not hit any of the points in question. I feel like the line of L(t) is a good approximation because the points provided at 90 min the temperature still increased from previous time slot 75 indicating that the graph is still concave up and hasn't started the decend yet. 


<iframe src="https://www.desmos.com/calculator/fumh2mnld7?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>



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