- Level: International Baccalaureate
- Subject: Maths
- Word count: 948
Investigating Ratios of Areas and Volumes
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Introduction
Introduction
In this portfolio, I will investigate the ratio of areas formed when y = xn is graphed between two arbitrary parameters x = a and x = b such that a < b.
Ratio of Area
Given the function y = x2, I will first consider the region formed by this function from
x = 0 to x = 1 and the x-axis. The region will be labeled B. The region from y = 0 to y = 1 and the y-axis will be labeled A. A diagram of this is show below:
Finding the ratio of area A: area B:
By integrating this function from x = 0 to x = 1, we can find the area under the curve, area B.
B = = = − 0 =
Since the total area of area A and B is 1, we can simply subtract the area of B from 1 to get area A.
A = 1 − =
Thus the ratio of area A: area B is 2:1.
Middle
1 − =
1:2 = ½
y = x5/3
= = − 0 =
1 − =
5:3 = 5/3
The graphs of these functions are shown below:
One important subset of real numbers has been excluded from further consideration is negative numbers. This is because no corresponding y-value exists at x = 0. There is a vertical asymptote at x = 0 and a horizontal asymptote at y = 0, as they approach infinity. Thus y = x-ncannot be integrated as infinity is not a valid area. The graph below demonstrates them phenomena for
y = x-2:
To further test the conjecture, I will now consider the areas between other x-values. I will examine if the conjecture holds for x = 0 and x = 2, x = 1 and x = 2, and x = 2 and x = 5.
First, I will consider functions of the type y = xn, between x = 0 and x = 2. The conjecture seems to hold for the bounds x = 0 and x = 2.
Function | Area of B (under the curve) | Area of A | Ratio |
y = x0 | = 2 | 0 | 0:2 = 0 |
y = x2 | = = − 0 = | 8 − = | 2 |
y = x2/3 | = = 1.9048… − 0 = 1.9048… | − 1.9038… = 1.2699… | 1.2699…/ 1.9038… = 2/3 |
Area of A and B is now defined as the following:
Area A: y = xn, y = an, y = bn, and the y-axis
Area B: y =x, x = a, y = b, and the x-axis
The graph below is an example of the bounds for y = x2 from x=1 to x=2
Next, I will consider functions of the type y = xn, between x = 1 and x =2. The conjecture seems to hold for the bounds x = 1 and x = 2.
Function | Area of B (under the curve) | Area of A | Ratio |
y = x0 | = 1 | 0 | 0:1 = 0 |
y = x2 | = = − = | 7 − = | 14:7 = 2 |
y = x2/3 | == 1.9048… − = 1.3048… | (−1) − 1.3048…= 0.8699… | 0.869…/ 1.3048… = 2/3 |
Conclusion
Area A: y = xn, y = an, y = bn, and the y-axis
Area B: y =x, x = a, y = b, and the x-axis
Proof:
B = = =
The formula for area A can be defined as:
= = = − =
Thus the ratio of A: B is n:
== n
Ratios of Volume
After considering and proving that the ratio of the area A: area B is simply n, I will now consider a general formula for the ratios of the volumes of revolution generated by the regions A and B when they are rotated about the x-axis and y-axis.
Proof for x-axis conjecture:
I will use the disc method to solve for volumes of revolution: V = , y is the function
y = xn
VB = = = = =
VA = = =
Thus ==
Proof for the y-axis:
VA = ==== =
VB = ===
= =
Conclusion
Thus the ratio of area A: area B is for the x-axis and for the y-axis.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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