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

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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\)
$f(72)=L(72)$

$L(x)=342.8+1(x−75)$
$L(72)=342.8 +1(72-75)$
$= 342.8 +1(-3)$
$=342.8-3$
$=339.8$

$L(72)=339.8°F$
 

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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 feel the vaule I calculated is fairly exact. I feel this because the temperature at 72 minutes isn't greater than 75 minutes. It's just below the degrees for the 75 minutes mark; which I would expect since 72 minutes isn’t considerably far way from 75 minutes. In addition, the temperature isn't  less than the 60 minutes' temperature. 

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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)

$f(100)=L(100)$

$L(x)=342.8+1(x−75)$
$L(100)=342.8 +1(100-75)$
$= 342.8 +1(25)$
$=342.8+25$
$=367.8$

$L(100)=367.8°F$

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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 believe the temperature at 100 minutes is too big. I this this because by looking at the table we can see the rate the temperature is increase is pretty consistent. When we reach at 30 minutes the temperature is raising slowing. We don't t see this at 100 minutes, the increase of the temperature is drastic. In addition by checking the graph, we can see at 100 minutes it was at 352.776 °F. We got 367.8°F, we were over 15.024.
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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)
![](https://i.imgur.com/IXli3jR.png)
I feel that the only time the line L(x) is a good approximation when F(x) is close to our point of reference. After that it would be too far away from our points would be to inaccurate to estimate.



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