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 formula for the central difference for $F'(x)$ is: $F'(x)=\frac{f(x+h)-f(x-h)}{2h}$
$F'(75)=\frac{F(90)-F(60)}{90-60}$
$F'(75)=\frac{354.5-342.8}{30}$
$F'(75)=\frac{11.7}{30}$
$F'(75)= .39$
At 75 minutes the temperature of the potato is increasing by $.39$ $degrees/minute^2$.

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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)The formula for the local linearization is: $y=L(t)= F(a)+F'(a)(t-a)$
$F(75)=342.8$
$F'(75)=.39$

$L(t)=342.8 + .39(t-75)$

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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\)
$L(t)=342.8 + .39(t-75)$
$L(t)=342.8 + .39(72-75)$
$L(t)=342.8 + .39(-3)$
$L(t)=341.63$

At 72 minutes, the potato is estimated to be 341.63 degrees.

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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) I think the estimate of 341.63 degrees for 72 minutes is pretty accuate seeing that at 75 minutes the potato was 342.8 degrees according to the chart.

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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)
$L(100)=342.8 + .39(100-75)$
$L(100)=342.8 + .39(25)$
$L(100)=342.8 + 9.75$
$L(100)=352.55$

According to the local linearization formula, at 100 minutes the potato will be 352.55 degrees.

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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 in e of 352.55 degrees at 100 minutes, is too low considering that at 90 minutes the potato is 354.5 degrees already. The potato would not cool off as it is still cooking.

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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 only accurate very close to the 75 minute mark of $F(t)$ since the local linearization is "local" to 75 minutes. The further from 75 minutes $F(t)$ becomes the more inaccurate the estimate will be.

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