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)
$f'(75)≈\frac{(354.5-324.5)}{90-60}=1$

The rate of change at the 75 minute mark is approximately 1 degree Farenheit/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)
Using the formula:

$L(t)=f(a)+f'(a)(x-a)$

$L(t)=342.8+1(x-75)$

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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\)
Using the formula from Part B:

$L(t)=342.8+1(x-75)$

$L(72)=342.8+1[(72)-75]=339.8$

Using this approximation, the temperature of the potato at 72 minutes is approximately 339.8 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)
This is most likely a slight overestimate. If the points in the table are graphed in Desmos, it can be seen that the rate of change starts to decrease with time and the slope calculated at 75 minutes provides a linear function that would therefore have an output value higher than the true output value of the function $F(t)$ (since the function is concave down but the linear approximation continues in the upward direction with the same slope).


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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)
Using the approximation from Part B:

$L(t)=342.8+1(x-75)$

$L(100)=342.8+1(100-75)=367.8$

At the 100 minute mark, the potato is approximated to be 367.8 degrees Farenheit using this extrapolation.

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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)
Estimating the potato to be 367.8 degrees Farenheit at the 100 minute mark is most likely an overestimate of its true temperature. As previously stated, the linear approximation is based off of the formula of a line with the slope of $F(t)$ when t=75 minutes. The linear approximation does not account for the fact that $F(t)$ actually has a slope that decreases over time and is therefore not perfectly linear, like the linear approximation.

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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)
$L(t)$ (when found with $t=75$ minutes) will only be a good approximation of $F(t)$ around input ($x$) values similar to 75. When the input values begin to get "too far" away from $x=75$ in either direction, then the linear approximation becomes too different from the actual output values of $F(t)$.

A Graph of $L(t)$ (red) and Plotting the Measured Points Provided for $F(t)$ (green):

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