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) The central difference equation is:
$f\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
after pluging in the numbers we get
$f\left(x\right)=\frac{f\left(90\right)-f\left(60\right)}{90-60}$
$=\frac{354.5-324.5}{90-60}$ =$1\frac{degree\ F}{\min}$

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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)
$L\left(t\right)=f\left(t\right)+f'\left(t\right)\left(t-a\right)$
$=f\left(75\right)+f'\left(75\right)\left(t-75\right)$
$L\left(t\right)=342.8+1\left(t-75\right)$

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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\)
$L\left(t\right)=342.8+1\left(72-75\right)$
$=339.8$ Degrees 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)
My estimate in question (C) is too small becuase the curve of the tangent line would sit below $F(72)$.

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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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(e) Using local linearization:

$L\left(t\right)=342.8+1\left(100-75\right)$
$=367.8$ $Degrees F$

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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)
My estiamte in question (e) is too large because the curve of the tangent line sits high above $F(100)$. The tangent line has a slope of one and keeps 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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(g)

$L(t)$ is a good approximationg when $F(75)$ because the tangent line is very close to the slope and thats when the function is equal to the derivative on the given point.

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