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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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The central difference equation is 

f(x)=$\frac{f\left(b\right)-f\left(a\right)}{b-a}$

We then plug in and solve

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






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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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L(t)= f(t)+f'(t)(t-a)= f(75)+f'(75)(t-75)
L(t)= 342.8+1(t-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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L(t)= 342.8+1(72-75)= 339.8 DegreesF

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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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This estimate is too small because the tangent line is only an estimate and sits below the curve at F(72).

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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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L(t)= 342.8+1(100-75)= 367.8 DegreesF

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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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This estimate is too large because the point F(100) on the tangent line sits far above the curve. This is because the curve eventually plateaus and the tangent line has a slope of one and is always increasing. 

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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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L(t) is a good approximation at F(75) because the tangent line closely matches or rides the slope at that point. 


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