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)Using the central difference to estimate $F'(75)$

t'(t)=$t(t0+t)-t(t0-t)/20t$

t=15 min and t0=75

$t'(75)=t(75+15)-t(75-15)/2(15)$

=$t(90)-t(60)/30$

=$354.5-324.5/30

t'(75)=1

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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 local linearization at t=a

y-t(a)=t'(a)(t-a)

equation of tangent line or the formula above

a=75

y-342.8=1(t-75)

y=342.8+(t-75)

y=t+267.8

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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\)Using L(t)=t+267.8 from part B to estimate F(72)

F(72)

y=72+267.8

=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)From the data we can see F is decreasing. F is concave down, so that makes tangent lines to F are above function F. That makes F(72) an overestimate

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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 L(t) we estimate F(100)

F(100)=L(100)=0.78(100)+284.3

F(100)=362.3 estimation of F(100) using local linearization at 90

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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 is exactly right because when calculating it, I was able to find the answer and jt fit oerfectky.

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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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