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

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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) $F'\left(75\right)\left[t-75\right]=L\left(t\right)-F\left(75\right)$

$L\left(t-75\right)=L\left(t\right)-342.8$

$L\left(t\right)=t+267.8$

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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\) $F(72)$
$y=72+207.8=339.8$

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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 table we can see $F(t)$ is decreasing, when F is changing t is decreasing.$F'(t)$ should be negative, which means $F(t)$ is concave down leading me to believe the estimate is too large.

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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) $F'('90)=\frac{\left(F\left(75\right)-F\left(90\right)\right)}{75-90}=\frac{\left(342.8-354.5\right)}{-15}$
$F\left(90\right)=0.78$
$y=L(t$) at points near 90
$L\left(t\right)=0.78t+284.3$
$F\left(100\right)\sim$ $L\left(100\right)=0.78\cdot100+284.3$
$F\left(100\right)\sim362.3$
By linear estimation $F(100)=362.3$
The actual graph passes through all the points given by $F(t)=374.944-304.952e^{-0.03001t}$
So, $F(100)=359.77$

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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) It's close to the actual value because
relative error$=\frac{\left(362.3-359.77\right)}{359.77}\cdot100=0.7\%\operatorname{}$

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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) This L(f) is an accurate approximation when trying to find the value of f for a value of t which is close to $t=75$ or $t=90$
<iframe src="https://www.desmos.com/calculator/itk8g5riqm?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>
<iframe src="https://www.desmos.com/calculator/wauqpb0jnx?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>

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