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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) For this problem we would use the set up: 
$f'\left(75\right)=\frac{f\left(75+15\right)-f\left(75-15\right)}{2\cdot15}$ we use the numbers close to 75, in tis case being 15 above and 15 below 75, and with the chart that was given you just enter the information into the formula we just put. So it would be: 
$f'\left(75\right)=\frac{\left(354.5\right)-\left(324.5\right)}{30}$ and then solve by subtracting the numbers on top and with that answer divide it by the bottom number and the answer we get  is 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) To do this problem we would use the template: 
$y-f\left(a\right)=f'\left(a\right)\left(t-a\right)$ 
With this template we can just plug in our given values from the last problem. So:
$y-342.8=1\left(t-75\right)$ then we can add the 342.8 to the other side in order to solve so we get:
$y=342.8+1\left(t-75\right)$
With this we just solve to get:
$y=t+267.8$


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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\) So using the equation from the last problem:
$y=t+267.8$
then we just input the 72 in place of the t, so it would look like this:
$y=72+267.8=339.8 F$ and that would be our final answer.


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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 for c I believe is actually pretty close because it is close in value to the t value 72 is close to which is 75. So I'd say it is pretty close if not exact. I say this because the value is larger than F(60) which would be below it, and smaller than F(75) which is above it. Either way, I believe that this estimate is pretty close if not exactly right. 

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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) Again using our template from question c:
$y=t+267.8$ and again we just plug in the value 100 for the t in our problem. So it would look like: 
$y=100+267.8=367.8$ 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)This estimation would be exactly right because the line of linearization at t=75 touches the curve. And again, looking at the chart, we are given up to 90 and if you were to continue it is trending up like it was in the chart.

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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) When I plotted both F and L, the line L(t) is near t=75 and this L(t) looks to be over the estimate. L(t) is a good approximation, it touches the curve near t=75 but our initial points are on a parablola and the line L(t) continues to trend in a straight line where F (t) is a parabola. 


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