Procedure In each case, draw and extend the given series to the fifth generation, then complete the following tasks:


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1 Math IV Nonlinear Algebra 1.2 Growth & Decay Investigation 1.2 B: Nonlinear Growth Introduction The previous investigation introduced you to a pattern of nonlinear growth, as found in the areas of a series of rectangles. You drew a graph of range values, taken from a t table, as well as a graph of the table s first differential. The graph of the differential was shown to contain a function known as the derivative, and you were also introduced to the integral. The integral, finally, was shown to relate to (y) values in the table for a given value of (x). The present investigation extends this work by examining patterns in other figure series. Procedure In each case, draw and extend the given series to the fifth generation, then complete the following tasks: a) Generate a ttable for the areas of the figures in the series, including columns for domain, range, and the first and second differentials. b) Graph (x) against (y), according to the same scales used in the last investigation. Include slope triangles for each segment of the graph.
2 c) Graph (x) against the first differential (m1). Again, use the scales from the last investigation. Include a slope triangle and function for the graph. d) Graph (x) against the second differential (m2). You can extend each series mathematically without actually drawing the figures. Be sure to record all your work, however. Here are the problems, with the dimensions of the first three figures in each series: 1) (1 x 3) (2 x 6) (3 x 9) 2) (2 x 3) (4 x 6) (6 x 9) 3) (3 x 3) ( 4 x 4) (5 x 5) 4) (3 x 6) (4 x 8) ( 5 x 10) 5) (3 x 4) (4 x 5) (5 x 6) 6) (2 x 3) (3 x 4) (4 x 5)
3 Discussion This investigation provides several examples of nonlinear growth patterns and their graphs. In each case, you were asked to draw graphs of the range, and the first and second differentials. You should have found that the graphs of the ranges are curved, and that the graphs of the differentials are linear. Further, you should have noticed that the graphs of first differentials have positive slopes, while graphs of the second differentials have slopes of zero. You ll recall from a previous investigation that the area between the line of a graph and the (x) axis is called the integral. You should also recall that the integral contains some valuable information about the pattern being graphed. Look, for example, at the graph of the first differential in problem 1 of the investigation. You ll find that the graph has a slope of 6 and a (y) intercept of zero. If you consider now the integral for this graph, you ll find that the area of the triangle contained by a vertical line at (x) = 3 can be found by the equation: A = (3) (18)/2 = 27 This is the value of (y) in the ttable where (x) = 3. Similarly, if you look at the integral and imagine a triangle defined by (x) = 4, you ll find its area by: A = (4) (24)/2 = 48 and this is the value for (y) in the table where (x) = 4.
4 The same relationship between the integral and (y) values in a table can be seen in problem 2 of the investigation. Again, examine the integral of this problem s first differential, and imagine a triangle defined by, say, (x) = 4. The triangle will be found to have an area of 96 square units, and 96 is the (y) value in the table where (x) = 4. In sum, the area of the integral, as defined by a given value of (x), will equal the range value for that value of (x) in the table. Graphs of second differentials, meanwhile, contain information about the first differentials. Look at the graph of the second differential in problem 1. The area of its integral, as defined at (x) = 3 is found by the equation: A = (3) (6) = 18 This is the value of the first differential where (x) = 3 in the table. (This may not be apparent unless you examine the table closely. Remember that the first differential does not line up horizontally with domain values in the table. In the case considered here, can you see where the number 18 is hiding?) The relationship between an integral and values in a table, however, is a bit more complex when the integral is not a triangle. This is the case in problems 3 through 6 of the investigation. Look, for example, at the graph of the fist differential in problem 3. Imagine the area of this graph s integral, as defined by (x) = 3. The integral so defined is not strictly in the shape of a triangle, and its area must be calculated in two parts.
5 The parts of the integral are defined by the point where the graph crosses the (m), or vertical, axis. The intersection is at 4, and a horizontal line from this point divides the integral into a rectangle surmounted by a triangle. The rectangle's area can be found by: A = (3) (4) = 12 while the triangle s area is found by: A = (3) (6)/2 = 9 Adding 12 to 9 equals 21, the area of the integral, but this is not the value of (y) where (x) = 3 in the table. However, addition of 4. the value of the (m) intercept, produces a value of 25, which is the correct value for (y) where (x) = 3. Similar relations can be found for the integrals in the other problems, for various values of (x). You might want to verify this by looking for your own examples.
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