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) = $\frac{F'(90)-F'(60)}{90 - 60}$ = $\frac{354.5 - 324.5}{30}$ = $\frac{30}{30}$ = 1 $^{\circ}$F/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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**(b)**
L(t) = f(t) + f'(t)(x - t)
L'(t) = f(75) + f'(75)(x - 75)
L'(75) = 342.5 + 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) = 342.5 + 1(72 - 75)
L'(72) = 339.8 $^{\circ}$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)**
The estimate appears accurate. According to the data chart, at 60 minutes the temperature would be at 324.5 $^{\circ}$F, and at 75 minutes the temperature would reach 342.8 $^{\circ}$F, making my solution point of 72 minutes at a temperature of 339.8 $^{\circ}$F fall between the points given in the data set.

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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 $^{\circ}$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)**
Based on the graph below, my estimate looks to be accurate.

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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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