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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)
$(F'(75))\sim\frac{\left(354.5-324.5\right)}{90-60}$
$(F'(75))\sim1$ 

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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)
$L(t)=f(75)+f'(75)(x-75)$
$L(t)=342.8+1(x-75)$

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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\)
$L(72)=342.8+1(72-75)$
$L(72)=339.8$ 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 from c is too large because the tangent line is passing above the graph at $F(72)$ since the local linearization was taken from a higher point on a graph that's shape is concave down increasing.


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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)
$L(100)=342.8+1(100-75)$
$L(100)=367.8$ 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)
The estimate of (e) is too small because the local linearization was taken from a much lower value, so the tangent line is likely still higher than the value of $F(100)$

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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)
The line L(t) is close to the value of F(90), but then at L(100) it is exactly the same point as F(100) beacause we have an accurate extrapolation estimate of how many degrees Fahrenheit the potatoes are at 100 minutes.
<iframe src="https://www.desmos.com/calculator/lqfbk6yg32?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>



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