Investigating Ratios of Areas and Volumes

Investigating Ratios of Areas and Volumes In this portfolio, I give be study the symmetrys of the aras and volumes formed from a curve in the form y = xn between ii arbitrary parameters x = a and x = b, such that a b. This forget be d iodine by exploitation integration to find the area under the curve or volume of revolution about(predicate) an axis. The two areas that lead be compared leave behinding be labeled A and B (see presage A). In order to put forward or dis test my meditates, several different determine for n will be apply, including irrational, sincere amount (? , v2).In addition, the values for a and b will be altered to different values to prove or disprove my conjectures. In order to aid in the calculation, a TI-84 Plus calculator will be employ, and Microsoft Excel and WolframAlpha (http//www. wolframalpha. com/) will be employ to create and display graphs. Figure A 1. In the first problem, country B is the area under the curve y = x2 and is boun ded by x = 0, x = 1, and the x-axis. office A is the sphere bounded by the curve, y = 0, y = 1, and the y-axis. In order to find the ratio of the two areas, I first had to calculate the areas of both vicinitys, which is seen below.For region A, I corporate in relation to y, while for region B, I integrated in relation to x. Therefore, the two forms that I engagementd were y = x2 and x = vy, or x = y1/2. The ratio of region A to region B was 21. Next, I calculated the ratio for former(a) get goings of the type y = xn where n ? ?+ between x = 0 and x = 1. The first value of n that I tested was 3. Because the formula is y = x3, the opponent of that is x = y1/3. In this case, the value for n was 3, and the ratio was 31 or 3. I then used 4 for the value of n. In this case, the formula was y = x4 and its inverse was x = y1/4.For the value n = 4, the ratio was 41, or 4. After I analyzed these 3 values of n and their apply ratios of areas, I came up with my first conjecture Conjec ture 1 For all positive integers n, in the form y = xn, where the graph is between x = 0 and x = 1, the ratio of region A to region B is equal to n. In order to test this conjecture further, I used other numbers that were not necessarily integers as n and placed them in the function y = xn. In this case, I used n = ?. The two equations were y = x1/2 and x = y2. For n = ? , the ratio was 12, or ?. I also used ? as a value of n.In this case, the two functions were y = x? and x = y1/?. Again, the value of n was ? , and the ratio was ? 1, or ?. As a result, I concluded that Conjecture 1 was true for all positive in truth numbers n, in the form y = xn, between x = 0 and x = 1. 2. After proving that Conjecture 1 was true, I used other parameters to check if my conjecture was only true for x = 0 to x = 1, or if it could be applied to all possible parameters. First, I tested the formula y = xn for all positive real numbers n from x = 0 to x = 2. My first value for n was 2. The two formulas used were y = x2 and x = y1/2.In this case, the parameters were from x = 0 to x = 2, but the y parameters were from y = 0 to y = 4, because 02 = 0 and 22 = 4. In this case, n was 2, and the ratio was 21, or 2. I also tested a different value for n, 3, with the same x-parameters. The two formulas were y = x3 and x = y1/3. The y-parameters were y = 0 to y = 8. Again, the n value, 3, was the same as the ratio, 31. In order to test the conjecture further, I decided to use different values for the x-parameters, from x = 1 to x = 2. Using the general formula y = xn, I used 2 for the n value. Again, the ratio was equal to the n value.After testing the conjecture dual times with different parameters, I decided to update my conjecture to reflect my findings. The n value did not necessarily agree to be an integer using fractions such as ? and irrational numbers such as ? did not affect the outcome. Regardless of the value for n, as long as it was positive, the ratio was forever equal to n . In addition, the parameters did not have an effect on the ratio it remained equal to the value used for n. Conjecture 2 For all positive real numbers n, in the form y = xn, where the graph is between x = a and x = b and a b, the ratio of region A to region B is equal to n. . In order to prove my second conjecture true, I used values from the general case in order to prove than any values a and b will work. So, instead of specific values, I make the x-parameters from x = a to x = b. By doing this, region A will be the region bounded by y = xn, y = an, y = bn, and the y-axis. Region B is the region enclosed by y = xn, x = a, x = b, and the x-axis. The formulas used were y = xn and x = y1/n. The ratio of region A to region B is n1, or n. This proves my conjecture correct, because the value for n was combining weight to the ratio of the two regions. . The next part of the portfolio was to determine the ratio of the volumes of revolution of regions A and B when rotated around the x-a xis and the y-axis. First, I determined the ratio of the volumes of revolutions when the function is rotated about the x-axis. For the first example, I will integrate from x = 0 to x = 1 with the formula y = x2. In this case, n = 2. When region B is rotated about the x-axis, it can be easily solved with the volume of rotation formula. When region A is rotated about the x-axis, the resulting volume will be bounded by y = 4 and y = x2.The value for n is 2, while the ratio is 41. In this case, I was able to figure out the volume of A by subtracting the volume of B from the cylinder formed when the entire section (A and B) is rotated about the x-axis. For the next example, I integrated the function y = x2 from x = 1 to x = 2. In this case, I would have to calculate region A using a different method. By finding the volume of A rotated around the x-axis, I would also find the volume of the portion shown in figure B labeled Q. This is because region A is bounded by y = 4, y = x2, and y = 1 .Therefore, I would have to then subtract the volume of region Q rotated around the x-axis in order to get the volume of only region A. In this case, the value for n was 2, and the ratio was 41. After this, I decided to try one more example, this time with y = x3 but using the same parameters as the previous problem. So, the value for n is 3 and the parameters are from x = 1 to x = 2. In this case, n was equal to 3, and the ratio was 61. In the next example that I did, I chose a non-integer number for n, to determine whether the current physique of the ratio being two times the value of n was valid.For this one, I chose n = ? with the parameters being from x = 0 to x = 1. In this case, n = ? and the ratio was 2? 1, or 2?. After this, I decided to make a conjecture based on the 4 examples that I had completed. Because I had used multiple variations for the parameters, I have established that they do not play a role in the ratio only the value for n seems to have an effect. Conjectur e 3 For all positive real numbers n, in the form y = xn, where the function is bound from x = a to x = b and a b, the ratio of region A to region B is equal to two times the value of n.In order to prove this conjecture, I used values from the general case in order to prove than any values a and b will work. So, instead of specific values, I made the x-parameters from x = a to x = b. By doing this, region A will be the region bounded by y = xn, y = an, y = bn, and the y-axis. Region B is the region enclosed by y = xn, x = a, x = b, and the x-axis. In this example, n = n and the ratio was equal to 2n1. This proves my conjecture that the ratio is two times the value for n. When the two regions are rotated about the x-axis, the ratio is two times the value for n.However, this does not apply to when they are rotated about the y-axis. In order to test that, I did 3 examples, one being the general equation. The first one I did was for y = x2 from x = 1 to x =2. When finding the volume of revolution in terms of the y-axis, it is important to note that the function must be changed into terms of x. Therefore, the function that I will use is x = y1/2. In addition, the y-parameters are from y = 1 to y = 4, because the x values are from 1 to 2. In this example, n = 2 and the ratio was 11. The next example that I did was a simpler one, but the value for n was not an integer.Instead, I chose ? , and the x-parameters were from x = 0 to x = 1. The formula used was x = y1/?. In this example, the ratio was ? 2, or ? /2. After doing this example, and using prior knowledge of the regions revolved around the x-axis, I was able to come up with a conjecture for the ratio of regions A and B revolving around the y-axis. Conjecture 4 For all positive real numbers n, in the form y = xn, where the function is limited from x = a to x = b and a b, the ratio of region A to region B is equal to one half the value of n.In order to prove this conjecture, I used values from the general case i n order to prove than any values a and b will work. This is similar to what I did to prove Conjecture 3. So, instead of specific values, I made the x-parameters from x = a to x = b. By doing this, region A will be the region bounded by y = xn, y = an, y = bn, and the y-axis. Region B is the region enclosed by y = xn, x = a, x = b, and the x-axis. The ratio that I got at the end was n2, which is n/2. Because the value of n is n, this proves that my conjecture is correct.In conclusion, the ratio of the areas formed by region A and region B is equal to the value of n. n can be any positive real number, when it is in the form y = xn. The parameters for this function are x = a and x = b, where a b. In terms of volumes of revolution, when both regions are revolved around the x-axis, the ratio is two times the value of n, or 2n. However, when both regions A and B are revolved around the y-axis, the ratio is one half the value of n, or n/2. In both situations, n includes the set of all pos itive real numbers.