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) $$F'(75)=354.5-342.8/(90-75)$$
$$F'(75)=0.78$$
F'(75) indicates that there is an increase of 0.78 degrees fahrenheit per minute from 75 minutes after the potato was placed to bake to 90 minutes.

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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)
Following the linear approximation formula and inputting F(75) and F'(75 ) a unique formula for F(t) was created
$$L(t)=F'(a) (t-a)+F(a)$$

$$F(t)= 0.78(t-75)+342.8$$

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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\)$$F(72)= 0.78(72-75)+342.8$$
$$F(72)=340.46$$
This estimated value inicates that at 72 minutes since the initial baking point, the potato is baking at 340.46 degrees farenheit

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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 my estimate is exactly right because compared to the interval [60,75] it falls in between the degrees in fahrenheit of that interval

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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)
$$F(100)= 0.78(100-75)+342.8$$
$$F(100)= 362.3$$

This estimated value indicates that at 100 minutes the potato was baking at 362.3 degrees fahrenheit

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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)
I think my estimate for F(100) using local linearzation is just right since it is a reasonable increase of degrees fahrenheit per minute past 90 minutes of baking.

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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) Observing the graph, I am even more convinced that my estimated values for F(72) and F(100) are right as they fall in place with the values of the initial table

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