Backward design problem for Math 180 class (Calculus I) on Linear Approximations and Differentials Part I - If you were asked to find the following cube roots it would not be difficult: VI =1, V/8 =2, 27 =3, /64 =4, 1000 =10, etc. But what if you were asked to find the cube root of 2 for example (32 =?) , this would not be easy since 2 is not a perfect cube. You would have to find a number that cubes to 2, not an easy task. However, if you look at the graph of f(x) = Vx near the point (2, 2) there is a perfect cube very near at x=1 (1 =1), see the graph below. fix ) 1 H Z 3 Draw a tangent to the curve at the point where x=1 and consider the point on the tangent line at x=2. This point lies directly above the point on the curve for the value V/2 . Although this point does not represent the value we are looking for, it is very close so it could be used to estimate the actual value of V/2 . Try to find the functional value of this point. Part II - Share out your thoughts on this. (Instructor note: Stop break throughs when any student goes too far. These students can then work ahead on their own. Part III - Is this an over estimate or an underestimate?Part IV - Share out your thoughts on this. (Instructor note: Stop break throughs when any student goes too far. These students can then work ahead on their own.) Part V- Now do this again but not for a specific function, do it for a generalized function, call it f. Call the tangent point (a, f(a)), where x is some generalized value a. See the graph below, f (x ) (a, fla)) ( x , f (xx) a Part VI - Share out your thoughts on this. (Instructor note: Stop break throughs when any student goes too far. These students can then work ahead on their own.) Part VII - This approximated value comes from a process called the linearization L(x) of the function at a point *=a. Now repeat this estimation for V2 use the linearization formula we just created to see if we get the same value (estimate). Part VIll - Share out your thoughts on this. (Instructor note: Stop break throughs when any student goes too far. These students can then work ahead on their own.)Part Illl- Now let's look at the differential which accomplishes the same but is more useful. The differential is a function so It can work at any value it whereas the linearization can only he used at a specic point it=a. We now call the tangent point lit. flail] so it can vary, it Its not fitted like with a Ilnearizatlon .The point we want to estimate is then some distance at from the tangent point at ii. see the graph below. Note: do: can also be referred to as tilt in the formula. but our [the change in y for the functionl does not equal dy. So the differential utilizes the slope of the tangent line to determine an estimate for the desired functional yalue. tanscnialnpe = f'ixi= a drrf'iiikg-dr > dy=f'{xi-dx This is the formula for the differential, a; = fix} rd): 2 Now consider a line with a slope of 2. The slope gives us the trajectory of the line at = 2 = T and tells us that for each unit you go to the right you must go 2 units up to get from one point to another on. this line. Draw a line with a slope of 2 in the space below and show both rise and run for that slope using horizontal and vertical lines to illustrate. Cln this same line and using thE same starting point draw the same slope of 2 but using a mm of ill [in = 2 = %] rather than 1 which you just did. Part it Share out your thoughts on this. {Instructor note: Stop hrea it throughs when any student goe 5 too far. These students can then work ahead on their ownl Part X1 - Use the differential formula dy = f'(x). dx wixf'(1) = 2 at some point on the curve, say x=1 to calculate the differential values for the following values of dx, a) dx=1 b) dx=0.5 c) dx=0.1 Now draw each of these (dx versus dy) on the tangent line for the curve shown below, f ( x ) 1 2 Part XII - Share out your thoughts on this. (Instructor note: Stop break throughs when any student goes too far. These students can then work ahead on their own.) Part XIII - Given that the functional value for your tangent point is 3 ( f(1)=3 ) try to determine the y-values for each of the differentials you calculated above, Part XJ. Label these values on the graph of the function shown below. (1, 3) H+ ZPart XIV - Share out your thoughts on this. (Instructor note: Stop break throughs when any student goes too far. These students can then work ahead on their own.) Part XV - This of course was a specific example but we can do the same for the general. In the formula dy = f'(x). cox, f'(x) can be written as a function which can work for any value of x. To see this let's revisit the graph of f(x) = Vx in order to estimate the cube root of two. Like with a linearization we want to choose a point near the point (2, V/2 ) where there is a perfect cube, like at x=1 (Vi =1), which will allow us to easily arrive at a value for '(x) in the formula. You should see that this is the same procedure we used for our linearization above (Part VII), but know that these two ideas are both very useful in their own applications. Draw your result for the differential you obtained and your subsequent estimate for f(2) or the V/2 on the graph below. 2 Part XVI - Share out your thoughts on this. (Instructor note: Stop break throughs when any student goes too far. These students can then work ahead on their own.)