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)
$$F'\left(75\right)=\frac{F\left(90\right)-F\left(60\right)}{90-60}$$
$$=\frac{\left(354.5-324.5\right)}{\left(90-60\right)}$$
$$=\frac{30}{30}$$
$$F'(75)=1 degF/min$$

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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)
$$L\left(75\right)=F\left(75\right)+F'\left(75\right)\left(t-75\right)$$
$$=342.8+1(t-75)$$

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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\)
$$L(t)=342.8+1(t-75)$$
$$L(72)=342.8+1(72-75)$$
$$L(72)=339.8 degF$$

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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 this answer is exactly right because it is very close to $F(75)=342.8$.

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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)
$$L(t)=342.8+1(t-75)$$
$$L(72)=342.8+1(100-75)$$
$$L(72)=367.8 degF$$

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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 believe my estimate is too large because $F(100)$ is too far from the point: $F(75)=342.8$.
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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)

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