Investigating Ratios of Areas and Volumes

Investigating Ratios of Areas and Volumes In this portfolio, I allow be investigating the proportionalitys of the areas and lots formed from a fold in the form y = xn between cardinal arbitrary parameters x = a and x = b, such that a < b. This lead be through by using integ symmetryn to contract the area under the arch or volume of revolution around an axis. The two areas that go away be compared testament be denominate A and B (see figure A). In redact to prove or disprove my presupposes, several different set for n will be employ, including irproportionnal, authoritative rime (? , v2).In addition, the set for a and b will be altered to different value to prove or disprove my conjectures. In order to aid in the calculation, a TI-84 Plus calculator will be used, and Microsoft Excel and WolframAlpha (http//www. wolframalpha. com/) will be used to create and dis ferment charts. Figure A 1. In the starting signal problem, neighbourhood B is the area under the cu rve y = x2 and is move by x = 0, x = 1, and the x-axis. Region A is the kingdom 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 some(prenominal) voices, which is seen below.For neighborhood A, I integrated in relation to y, while for neck of the woods B, I integrated in relation to x. Therefore, the two formulas that I used were y = x2 and x = vy, or x = y1/2. The ratio of surface area A to region B was 21. Next, I calculated the ratio for other portions of the example y = xn where n ? ?+ between x = 0 and x = 1. The first value of n that I tried was 3. Because the formula is y = x3, the rearward of that is x = y1/3. In this lawsuit, 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 value of n and their corresponding ratios of areas, I came up with my first conjecture guesswork 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 be 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 too 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 sincere 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 unfeigned numbers n fr om 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, precisely 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 to a fault 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 obstinate to use different set for the x-parameters, from x = 1 to x = 2. utilise the general formula y = xn, I used 2 for the n value. Again, the ratio was live to the n value.After testing the conjecture treble generation with different parameters, I decided to update my conjecture to hypothesise my findings. The n value did not necessarily pass on to be an integer using fractions such as ? and irrational numbers such as ? did not affect the outcom e. Regardless of the value for n, as long as it was positive, the ratio was always equal to n. In addition, the parameters did not have an termination 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 value from the general case in order to prove than any values a and b will work. So, instead of special(prenominal) 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 equivalent to the ratio of the two regions. . The next part of the portfo lio was to determine the ratio of the volumes of revolution of regions A and B when revolve around the x-axis and the y-axis. First, I fit(p) 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 burn down 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 whole 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 vol ume 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 unrivalled 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 pattern of the ratio be two times the value of n was valid.For this one, I chose n = ? with the parameters be 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 h ave established that they do not play a role in the ratio only the value for n seems to have an effect. Conjecture 3 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 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 exampl es, one beingness 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 significant 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 r egion 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 in 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 prohibit 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 region s 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 positive real numbers.