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)= F(90)-F(60)/90-60
= (354.5-324.5)/30
=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) F'(75)[t-75]
=L(t)-F(75) L(t-75)
=L(t)-48
L(t)=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(72) = L(72) = 72+267.8
F(72) = 339.8
estimation based on L(t)

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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) By looking at the data we see that F' is decreasing which also means it is concaving down, therefore the tangent lines to F are above the function of F. Which means that F(72) is an **overestimate**

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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) Backwards quotient difference:
F'(90)= (F'(75)-F(90))/75-90
=342.8-354.5/-15
=F'(90)=0.78
y=L(t)
L(t)= F'(90)(t-90)+F(90)
=0/78(t-90)+354.5
L(t)= 0.78t + 284.3

F(100)=L(100)
0.78x100+284.3
F(100)=362.3

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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 think that my estimation is way too large compared to what it should be as F(100) should be around t= 1-15

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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

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