Math 1526 Summer 2 Session 1 Lab #2 Part #1 Rate of Change This lab will investigate the relationship between the average rate of change, the slope of a secant line, the instantaneous rate change and the slope of a tangent line. Suppose that a rabbit farmer is keeping data on raising rabbits. From his data he was able to develop an equation using the trend line in Excel. This equation, R(t) = 5t 2-2t+2, gives the number of rabbits t years after the start of breeding of one pair. According to this equation he will have R(1) = 5(1) 2-2(1)+2 = 5 rabbits after one year. We will use Excel to project the number of rabbits over a given time period. t-years R 2 1 5 2 18 3 41 4 74 5 117 6 17 7 233 8 36 9 389 1 482 R Number of Rabbits R(t) = 5t^2-2t+2 6 5 4 3 2 1 5 1 15 t-years The farmer now wishes to know the average rate of change in the number of rabbits from t = 3 years to t = 8 years. Remember that the average rate of change is the difference in the number of rabbits divided by the difference in time. Ave Rate = R/ t = ( R 2 -R 1 )/(t 2 -t 1 ) = ( 36-41)/(8-3) = 265/5 = 53. (notice that R is read as delta R and means the difference in R and the meaning of delta t ( t) also means the difference in t.) This means that on the average the number of rabbits increase about 53 rabbits per year in that 5 year span. Did you notice that the formula for the average rate of change is the same as the slope formula for a line that passes through the points (3, 41) and (8, 36)? The value 53 is the slope of the line shown below. secant line through (3,41)&(8,36) 6 5 4 3 2 1-1 2 4 6 8 1 12-2 t Suppose that the farmer wished to know the instantaneous rate of change in the number of rabbits at t=3. If average rate of change and the slope of the secant line are the same 1
Math 1526 Summer 2 Session 1 value, then it seems reasonable to assume that the instantaneous rate of change and the slope of the tangent line touching the graph at the point (3, 41) would be the same value. This line would be a tangent line since it will only touch the graph at one point. See the graph below. Tangent Line to Graph at (3,41) 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 1 11-1 What is the slope of this tangent line? How do you find it? You need two points but you only have one. You only have the point (3, 41). We will try a different approach to finding the slope value of this tangent line. We will look at the slopes of secant lines that are very very close to this tangent line. We will use these slope values to make a very good guess at the real slope value of the tangent line. We will also be finding the instantaneous rate of change in the number of rabbits at t=3. The first secant line slope we will try to find will be over the interval [3,4]. Therefore the points needed to find the slope of the secant line is (3, 41) and (4, 74). Using the formula for slope m = (74-41)/(4-3) = 33. Therefore the slope of the secant line is 33 and the average rate of change in the number rabbits is also 33. We will now try a smaller interval [3, 3.5]. This gives the two points (3, 41) and (3.5, 56.25) The secant line that passes through these two points is closer to the tangent line than the secant line over [3, 4]. The slope turns out to be m = (56.25 41)/(3.5 3) = 3.5. Using Excel we will continue to look at slope values as intervals become shorter and the secant lines move closer to the tangent line. As the intervals become shorter the distance between the end points becomes smaller. The length of this distance approaches zero. (This distance is called t and t ->.) In general, we will use the following notation. The interval is [t, t+ t] and the slope of the secant line is m =( f(t+ t) f(t)) / ( t+ t t) = (f(t+ t) f(t)) / t. And the slope of the Tangent line at t is given by the Limit ( t ->) [(f(t+ t) f(t)) / t ] We will now use Excel to finish our task. We will let h be the same as t. Label the A column with t and t always equal to 3 since that is the point we are interested in. The second column is h, which is the distance between 3 and the other end point of 2
Math 1526 Summer 2 Session 1 the interval. The third column C is labeled t+h which is the other end point of the interval. In the second cell in that column type in the command =A2+B2 and drag. In the next column D type the label f(t). In the second cell of that column type the command =5*A2^2-2*A2+2 and drag. In column E type the label f(t+h) and in the second cell type in the command = 5*C2^2-2*C2+2 and drag. Notice that f(t) and f(t+h) are the number of rabbits at t and t+h. To find the average rate of change in rabbits over the given interval, type the label [f(t+h)-f(t)]/ h into the F column and in the second cell type in the command =(E2- D2)/B2 and drag, This will give you the average rate of change in rabbits or the slope of the secant line. t h t+h f(t) f(t+h) [f(t+h)-f(t)]/h 3 1 4 41 74 33 3.5 3.5 41 56.25 3.5 3.5 3.5 41 42.4125 28.25 3.5 3.5 41 41.1413 28.25 3.5 3.5 41 41.14 28.25 3.5 3.5 41 41.14 28.25 Now you can guess the slope of the tangent line. The slopes of the secant lines are approaching the slope value of 28 as the intervals get shorter and shorter and corresponding secant lines get closer and closer to the tangent line. So a good guess for the slope of the tangent line is 28. In mathematical termsyou would say that the Limit ( t ->) [(f(t+ t) f(t)) / t ] = 28. Therefore the instantaneous rate of change in the number of rabbits per year is 28. In this lab we will investigate and compare the function and it s derivative. Remember that the derivative is the instantaneous rate of change in the y-values for some value of x. It is also the slope of the tangent line touching the graph of f (x) at the point (x, f (x) ), The Derivative function f '(x) produces the correct slope value of the tangent line for any chosen x value. Example: Given the parabola f (x) = 3x 2 +12x + 54 what is the slope of the tangent line that touches this parabola at x = 4? To answer the question, take the derivative and evaluate it at 4. [ Derivative Rule: If f(x) = x n then f '(x) = nx n- 1 ] So f '(x) = 6x + 12 and f '(4) = 6(4) + 12 = 36. This tangent line has a positive steep slope of 36, so the tangent line is rising at x = 4 and the parabola is increasing at x = 4. Notice that the graph of the derivative function f '(x) is a line. The graph of the original function f (x) is a parabola. The point representing slope and lying on the graph of the derivative function is (4, f '(4))= (4, 36). The point that lies on the parabola is (4, f (4)) = (4, 15). 3
Math 1526 Summer 2 Session 1 x f (x) f'(x) -1 234-48 -9 189-42 -8 15-36 -7 117-3 -6 9-24 -5 69-18 -4 54-12 -3 45-6 -2 42-1 45 6 54 12 1 69 18 2 9 24 3 117 3 4 15 36 5 189 42 6 234 48 7 285 54 8 342 6 9 45 66 1 474 72 Graph of f(x) and f '(x) 19 17 15 13 11 9 7 5 3 1-12 -1-8 -6-4 -1-2 -3 2 4 6 8 1 12-5 x f (x) f'(x) Looking at the graph above, you will see that f '(x) crosses the x-axis at x = -2. So f '(-2) =. This means that the slope of the tangent line that touches f (x) at x = -2 is zero and the tangent line must be horizontal there. You will also notice that f (x) decreases over the interval [-1, -2] and increases on the interval [2, 1]. What is the slope of the tangent line that touches f (x) at x = -6? Looking at the data the value of f '(-6) = -24. The tangent line touching the parabola at x = -6 has negative slope of 24 and is falling. Therefore f (x) is decreasing. Look at the graph of the parabola and see if you agree. Problems to turn in: 1) Using Example of the rabbit farmer above, find the instantaneous rate of change in rabbits for t = 5. (Redo the table above and make a guess for t = 5) Give your answer in complete sentences. Lab 2 Part #2 In this part we will see how to use Excel to find a "best-fit equation" for your data. We will use the following data as an example. 4
Math 1526 Summer 2 Session 1 Example#1 The national defense spending in the billions for the period from 197 through 1994 is given in the following data already typed into a spread sheet. year billion s 262 2 219 4 185 6 15 8 177.2 1 185 12 214.3 14 214.7 16 276 18 283 2 272 22 25 24 22 Graph this data as you have before using the XY-Scatter Plot. Do not connect data points. 3 25 2 15 1 5 5 1 15 2 25 3 We will now try to produce an equation that will come the close to representing this data. First double click anywhere within the chart. Next click on anyone of the data points. All data points will be highlighted. If you are in Excel 97, click on Chart and then add Trendline. Click on Type and then choose the type of equation that you wish to view over your data. You may wish to try a polynomial of degree (order) 2. Click on OK and see what you have. 5
Math 1526 Summer 2 Session 1 3 25 2 15 1 5 5 1 15 2 25 3 This is not a good fit. Highlight the trendline on your chart, go to Edit-Undo Trendline. That trendline should be gone and we will try again. Highlight the data points again and click on Chart-add Trendline to get the dialogue box and this time we will try polynomial of degree (order) 3. Click on polynomial and then the arrow button to increase the order to 3 3 25 2 15 1 5 5 1 15 2 25 3 This fit is much better. To format your trendline, highlight the trendline and click on Format. You will have a variety of patterns, etc. Now we will develop an equation to "best-fit" our data. Double click on the trendline and click on Chart- add Trendline-Options. Select Display Equation on Chart and click on OK. 3 25 2 15 1 5 y = -.141x 3 + 3.976x 2-37.5x + 269.36 5 1 15 2 25 6
Math 1526 Summer 2 Session 1 You should always label your graph carefully, if you have not already done so. To do this now, click on chart area on the graph and go to Insert and Chart. Redo the chart according to dialogue box. If the equations can not be seen clearly then click on the equation and drag it around until you can see it. You can also increase it s size or make it bold. Note: When you are in the Trendline option box you will notice a box option for a logarithmic function. If that box is blank or it will not work when you click on it, then you have an independent data point of zero included in your data set and log functions are not defined for zero. If you go back and delete that value from your data base then you can make use of the log function options box. Problems to be turned in: The following data sets can possibly be represented by a linear, quadratic, cubic or exponential function. Use Excel to find the best fit function for each data set. Show all work. You answers should be typed neatly on your lab and in complete sentences. Label all graphs clearly and show the equation of the best-fit function found. (Align data when necessary. To align the independent data, set the first value to zero and set all the other values according to its difference from the first vaule.) 1) Laura made and initial deposit of $1 into an account that pays 8% per year compounded yearly: Year 1992 1993 1994 1995 1996 Balance $1. $18. $116.64 $125.97 $136.5 1) Try three different fits for this data: quadratic, cubic and exponential. Show three different graphs with the equations form trendline. Notice that the graphs seem similar. Explain why exponential is the best choice. Align your data before using Excel. 2) Use the exponential equation to estimate the value for 1998. 7