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)

Step 1:
$$\frac{f(b)-
f(a)}{b-a}$$

Step 2:

a. To solve for $F'(75)$, we will be using the central difference formula.
b.$b-a$ in the denominater is: $90-60$
c. $f(b)-f(a)$ is: 354.5 - 324.5
$$The~formula~ will ~like ~this:

$$\frac{354.5 - 324.5}{90-60}$$

Step 3:
Once you have substituted all the variables for the values of $f(a)$, $f(b)$, [into the central difference formula], your values should apear as:
$$\frac{
30}{30}$$
Which is equal to $$1\frac{℉}{min}.$$

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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)
The linear approximation formula is:
$$L(x)=f(a)+f'(a)(x-a)$$
We then substitute in the values in for x
$$L(x)=f(75)+f'(75)(x-75)$$
These Values we get from the chart of values above.
$$L(x)=342.8+1(x-75)$$

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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\)
$$L(x)=f(a)+f'(a)(x-a)$$
$$L(72)=f(75)+f'(75)(x-75)$$
$$L(72)=342.8+1(72-75)$$
$$L(72)=342.8-3$$
$$=339.8 ℉$$

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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 estimated value for the degrees in (c ) it's small(below our tangent line) but it is a very close opproximation.
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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)
$$L(x)=f(a)+f'(a)(x-a)$$
$$L(72)=f(75)+f'(75)(x-75)$$
$$L(72)=342.8+1(100-75)$$
$$L(72)=342.8+25$$
$$=367.8 ℉$$

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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)
If we see the graph I provided below, the estimated value of $F(100)$ is exatly the same as the oproximation. And although the graph seems concave down, the tangent line still passes directly through $f(100)$ and $f(75)$.

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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 L(t) is a good approximation at F(100) and F(75). Where there is great accuracy in the the crossing at those exact points.


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