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) $f'(75)=f'(60)-f'(90)$ $=324.5-354.5/60-90$ $= 1°f/min.$ :::info (b) Find the local linearization $y = L(t)$ to the function $y = F(t)$ at the point where $a = 75$. ::: (b) $L(x)=F(a)+F'(a)(x-a)$ $L(75)=F(75)+F'(75)(x-75)$ $=342.8+1(x-75)$ :::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\) $F(72)=342.8+1(72-75)$ $L(72)=342.8+1(72-75)$ $=342.8+1(-3)$ $=343.8-3$ $=340.80°$ :::info (d) Do you think your estimate in (c) is too large, too small, or exactly right? Why? ::: (d)Too large because the estimation of F(72)=340.8° is over from the actual value graphically of 339.8. :::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) $L(75)=F'(75)+F(75-a)$ $=342.8+1(x-75)$ $L(100)=342.8+1(100-75)$ $=343.8+25$ $=368.8°$ :::info (f) Do you think your estimate in (e) is too large, too small, or exactly right? Why? ::: (f)L(100) is an overestimation because the graph of L(100)concaves down and the line sits above the graph which provides proof of an overestimation. :::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)![](https://i.imgur.com/zUVpsdk.png) l(100) is a close estimate however anything after l(100) would not be because the graph curves away from the line after l(100). --- 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.