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)
$f'\left(75\right)=\frac{f\left(90\right)-f\left(60\right)}{90-60}$
$f'\left(75\right)=\frac{\left(354.5\right)-\left(324.5\right)}{30}$
$=1 f/min$

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

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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\) $f\left(72\right)\sim L\left(72\right)$
$L\left(72\right)=342.8+1\left(72-75\right)$
$L\left(72\right)=342.8+1\left(-3\right)$
$=339.8 F$

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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) The average rate of change for $f(72)$ is the rate of change, it is bigger than the estimate central difference so the estimate in c, is to small.

$AV\left[60,75\right]=\frac{f\left(75\right)-f\left(60\right)}{75-60}$
$=\frac{342.5-324.5}{15}$
$=1.22$

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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)$f\left(100\right)\sim L\left(t\right)$
$L\left(100\right)=342.8+1\left(100-75\right)$
$L\left(100\right)=342.8+1\left(25\right)$
$=367.8 F$

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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)
Based off from the average rate of change from $f(72)$, the same pattern will occur for $f(100)$, so the estimate would still be too small.

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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

The line for $L(t)$ is a good approximation for $f(75)$ but not for $f(72)$ and $f(100)$ because its an underestimate.

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