Math 181 Miniproject 4: Linear Approximation and Calculus.md --- Math 181 Miniproject 4: Linear Approximation and Calculus === **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**. --- :::info 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. ::: (a)F'(75)=(f(90)-f(60))/90-60 = (354.5-324.5)/30 = 30/30/=1 degrees F/min. :::info (b) Find the local linearization $y = L(t)$ to the function $y = F(t)$ at the point where $a = 75$. ::: (b)F(75)=342. L(t)=f(75)+f'(75)(x-75) y=342.8+1(x-75) :::info (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. ::: (c\) F(72) = L(72) =342.8+1(72-75) =342.8+1(-3) =342.8-3 =339.8 degrees F :::info (d) Do you think your estimate in (c) is too large, too small, or exactly right? Why? ::: (d)I think that the estimate I have calculated for Part C is indeed accurate. This is because my estimate was very close to the actual point, which is a good sign that I was on the right track. If the value you estimate is very different from the known value, you will not come up with a correct estimate. :::info (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. ::: (e)F(100)=L(100) =342.8+1(100-75) =342.8+1(25) =342.8+25 =367.8 degrees F :::info (f) Do you think your estimate in (e) is too large, too small, or exactly right? Why? ::: (f) I think that this estimate is accurate, especially after estimateing the f(100) value. I compared my known values from the data table to the estimated values, and they match what is in the table. Also by using common sense, I can see that the estimates are realistic as well. :::info (g) Plot both $F$ and $L$ and comment on how or when the line $L(t)$ is a good approximation of $F(t)$. ::: (g) Below is the link to what I have plotted using Desmos. F and L were plotted. L(x) is a fair approximation of F(x) at the points where the tangent line intersects the points. As the point values get further away from the tangent line, the estimation becomes less accurate. This increases the margin of error. https://www.desmos.com/calculator/w2sbg3mvw6 --- To submit this assignment click on the Publish button ![Publish button icon](https://i.imgur.com/Qk7vi9V.png). Then copy the url of the final document and submit it in Canvas.