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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) So this would be set up like this: $f'\left(75\right)=\frac{\left[f\left(75\right)-f\left(45\right)\right]}{\left(75-45\right)}$ then following the steps we can break it down and solve each part that is needed.
Next we would do this: $f'\left(75\right)=\frac{\left[342.8-296\right]}{\left(75-45\right)}$ and what we did was we just subsituted the numbers in their corresponding spots. Then what is done is: $f'\left(75\right)=\frac{52.8}{30}$ here what we did was we took the numerator and subtracted it and got 58.2 then took the denominator and subtracted it and then got 30 and doing this math gives us the final answer of $f'\left(75\right)=1$ .


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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) For this we need to use the form $y-f\left(a\right)=f'\left(a\right)\left(t-a\right)$
If we follow that then it would be set up like this: $y-342.8=1\left(t-75\right)$ again just plugging in the numbers into that formula, now doing the math: $y=342.8+\left(t-75\right)$ and our final answer would be: $y=t+267.8$ . 


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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\) $F\left(72\right)$ would be set up like this: $y=72+267.8=339.8$ and that 339.8 would be the final answer for that.

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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) We can see from the table above that f(t) is decreasing and that being said F'(t) should be negative, and is being concave down. So the estimate for C is too large.

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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) equation of local linearization: $f\left(t\right)=t+267.8$ and then is shown like this: $F\left(100\right)=100+267.8=367.8°F$

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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) It is exactly right since at t=75 line of linearization touches 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) It is at or near t=75 and this being said that L(t) is actually an over estimate. 


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