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)\sim\frac{f(60)-f(90)}{60-90}\sim\frac{324.5-354.5}{-30}\sim1Deg.F/minute\\]

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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)\\[L(t)=F(t)+F'(t)(t-a)\\]
\\[L(75)=F(75)+F'(75)(x-75)\\]
\\[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\)\\[L(t)=342.8+1(x-75)\\]
\\[L(72)=342.8+1(72-75)\\]
\\[L(72)=342.8+1(-3)\\]
\\[L(72)=339.8Deg. F\\]
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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 believe my estimate is exactly right because it matches the incerment that the temperature is rising for the given time.

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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+1(100-75)\\]
\\[L(100)=342.8+1(25)\\]
\\[L(100)=367.8Deg. 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) Like the previous example, I believe this to be an over estimate because it continues to increase and begins to straighten out which doesn't follow the curve of the original graph.

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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) On the graph they are rising at a steady interval.On the other hand the approximation for L(100)is too large for the graph. This would lead me to believe that the approximation for L(100)is an over approximation.

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