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) We would first set this function up as: $F'(75)=f(75)-f(45)/(75-45)$
After this we would follow the steps to break it down and solve each part separately.
Next, we would substitute the values and set up the equation as:
$F'(45)=[342.8-296]/(75-45)$
After subtracting the correct numbers, this then becomes:
$F'(75)=52.8/30$
This gives us the final answer of $F'(75)=1$

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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) We need to use the equation form $y-f(a)=f'(a)(t-a)$
We would substitute the values and get $y-342.8=1(t-75)$ and now we need to rearrange the equation to be $y=342.8+(t-75)$ which brings us to our final answer: $y=t+267.8$.

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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\) The estimate for $F(72)$ would be set up as $y=72+267.8$ which added up to $339.8$.

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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 in C is too large because from the table above, you can tell from the table above that $F(t)$ is decreasing which means $F'(t)$ should be negative and concaves down.

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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)The equation of local linearization: $f(t)=t+267.8$ and then is shown as $F(100)=100+267.8=367.8°F$

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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 in E because $t=75$ the line of linearization touches the curve.

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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) It's at/near $t=75$ and with this being said, $L(t)$ is an overestimate.

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