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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)
$f'(75)=\frac{f(90)-f(60)}{90-60}$
$f'(75)=\frac{354.5-324.5}{30}$
$f'(75)=\frac{30}{30}$ $=1 deg(F)/min$

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

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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\)
$L(72)= 342.8+1(72-75)$
$L(72)= 342.8+1(-3)$
$L(72)= 342.8-3$
$L(72)= 339.8 deg(F)$

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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)
The estimate above is pretty accurate because 72 minutes is close to 75 minutes, which is what the local linearization is based off of. However, the estimate is a little large than the actual and we know this by knowing that 72 minutes falls above the curve by just a little, which means the degrees fahrenheit is an overestimate of the actual value. 

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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)
$L(100)= 342.8+1(100-75)$
$L(100)= 342.8+1(25)$
$L(100)= 367.8deg(F)$
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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)
Our estimation in (e) is too large from what the actual value would be. We know this because the baked potato at 100 minutes is way above the curve at 75 minutes, which makes the estimation less accurate because 100 minutes is far away from the 75 minutes which is what the local linearlization is based off of. From knowing this our estimate is too large, or an overestimate because it falls above the curve.

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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)
<iframe src="https://www.desmos.com/calculator/dkh2mesxkb?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>

In the image above it shows both the local linearization and the table plotted on the graph. Based on the plots we know the L(t) is a good approximation of F(t) at 75 minutes or any value closest to 75 minutes because that's where the local linearization intersects the plot from the table.We know when that happens, it means the values are the same, or the estimate is accurate to the actual amount.

The image below shows both the estimates at 72 minutes and 100 minutes from the previous problems. As you can see, at 72 minutes is fairly accurate, but just a little too small. You can also tell that at 100 minutes the value in degrees(F) is too large because it ends up being above the curve.

<iframe src="https://www.desmos.com/calculator/wiwm0iku4r?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>


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