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)
F'(75) = (354.5-324.5)/(90-60)
F'(75) = 30/30 = 1°F.

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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'(a)(t-a)+F(a)
L(t)= F'(75)(t-75)+F(75)
L(t)= (1)(t-75)+342.8
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)= 72 + 267.8
F(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) This estimate might be a little too large. Mostly because the average rate of change between 75 to 90 can be calculated to be 0.78. The central difference is a little too high compared to the average rate of change.

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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(t)= t + 267.8
F(100)= 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)
The estimate seems about to be right when in comparison to the graph below.

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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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