Investigating Ratios of Areas and Volumes

Investigating Ratios of Areas and Volumes In this portfolio, I pass on be investigating the balances of the areas and satu proportionalityns makeed from a curve in the fake y = xn mingled with both arbitrary parameters x = a and x = b, much(prenominal) that a b. This leave alone be done by victimization integ dimensionn to find the area under the curve or tawdriness of revolution about an axis. The two areas that go away be compared will be labeled A and B (see figure A). In post to evidence or dis march my ventures, several diverse care fors for n will be employ, including ir proportionalitynal, real bits (? , v2).In addition, the shelters for a and b will be altered to distinct shelters to farm or dis sustain my conjectures. In drift to aid in the calculation, a TI-84 plus calculator will be engage, and Microsoft Excel and WolframAlpha (http//www. wolframalpha. com/) will be use to create and display graphs. Figure A 1. In the low gear problem, theatr ical section B is the area under the curve y = x2 and is bounded by x = 0, x = 1, and the x-axis. Region A is the piece 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 domains, which is seen below.For expanse A, I structured in relation to y, while for office B, I integrated in relation to x. Therefore, the two shapes that I used 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 other berths of the type y = xn where n ? ?+ between x = 0 and x = 1. The first cherish of n that I tested was 3. Because the verbalism is y = x3, the inverse of that is x = y1/3. In this good example, the quantify for n was 3, and the ratio was 31 or 3. I then used 4 for the set of n. In this case, the skeletal systemula was y = x4 and its inverse was x = y1/4.For the value n = 4, the ratio was 41, or 4. after(prenominal) I analyzed these 3 values of n and their jibe ratios of areas, I came up with my first conjecture anticipate 1 For wholly confident(p) 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 impact to n. In order to test this conjecture further, I used other numbers that were not necessarily integers as n and laid them in the escape 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 think 1 was true for every(prenominal) positive real 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 aforementioned(prenominal) 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 jibe to the n value.After testing the conjecture multiple propagation with different parameters, I decided to update my conjecture to reflect my purposes. The n value did not necessarily surrender to be an integer victimization fractions such as ? and i rrational numbers such as ? did not push the out go. 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 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 precise values, I do 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 portfolio was to determine the ratio of the massess of revolution of regions A and B when rotate approximately the x-axis 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 well 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 fork over 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 pattern 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 ground on the 4 examples that I had completed. Because I had used mult iple 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. 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. I n 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 in order to prove than any values a and b will work. This is identical 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. Howeve r, 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 positive real numbers.