- Level: International Baccalaureate
- Subject: Maths
- Word count: 3311
Math Portfolio Type 1
Extracts from this document...
Introduction
Math Portfolio
Type 1 Task
Investigating areas and volumes
Miras International School
Amrebayeva Nurbala Gr. 12IB
Introduction
In this portfolio, I’m going to investigate ratios of areas and volumes of power functions, which can be generalized as the following:
Y=xn, where n-is the power, n∈R
This function will be graphed between two arbitrary parameters x=a and x=b such that a<b (in other words boundaries or limits). In this investigation, I’m going to use method of math induction, integration using power rule, application of integration (areas under the curve and solids of revolution), also some knowledge about power functions.
Investigation Process
Given the power function y=x2, graph of which is parabola. I need to consider the region formed by this function from x=0 to x=1 and the x-axis, let’s label this area B; and the region from y=0 to y=1 and the y-axis area A. This can be shown on the illustration (Fig.1) below:
Figure 1
To plot this, I used Graph 4-3 Software, which I found very convenient.
So the areas formed and illustrated above (Figure 1), need to be considered in order to find a ratio between them, hence area A: area B. To find that, I need to use integration using power rule. Integration, using power rule is the general integration rule, which is:
where n≠-1
It consists of the following steps:
- Calculate area B, which is -0
- Calculate area of the quadrate (formed by the x and y boundaries of the function), which is
- Calculate area A, which is equal to
area A = (area of the quadrate – area B)
which is SA=
- Step4. Therefore the ratio of the areas, will be
SA:SB=
Middle
0.13
0.87
1: 7
Tab 5
So, the sketches to the functions will be as the followings:
n-rational number (fraction) | |||
n>1 | 0<n<1 | n<0 | |
Tab 6
The general conjecture, for the subset of fractions, will be:
∴ If the power function of the type y=xn, where n – is the power, n∈Q (where Q= and a and b are integers} and has limits such as x=0 and x=1 and x-axis; y=0 and y=1 and the y-axis. The ratios of the areas formed by the function and by the boundaries will be equal to 1:.
To generalize, these examples, I would like to show a graph (Figure 3) Below, this will show, the comparison of the graphs by their powers:
Figure 3
Regardless to the graph and the previous observations for the different subsets of the real numbers, conjecture might be slightly changed. Also note, that I didn’t investigate irrational numbers such as π. So conjecture will be like:
∴If the power function of the type y=xn, where n – is the power,
- where n – is the power, n∈Z+ and has limits such as x=0 and x=1 and x-axis; y=0 and y=1 and the y-axis. The ratios of the areas formed by the function and by the boundaries will be equal to n: 1.
- where n – is the power, n∈Z (where and has limits such as x=0 and x=1 and x-axis; y=0 and y=1 and the y-axis. The ratios of the areas formed by the function and by the boundaries will be equal to 1: |n|.
- where n – is the power, n∈Q (where Q= and a and b are integers} and has limits such as x=0 and x=1 and x-axis; y=0 and y=1 and the y-axis. The ratios of the areas formed by the function and by the boundaries will be equal to 1:.
These conjectures are true for the boundaries, such as x=0 and x=1. I think I need to examine this for the x=0 and x=2; x=1 and x=2. So let’s consider the same function y=x2, but for different limits (Figure 4 and Figure 5)
Figure 4
Figure 5
So, now limits (x=1 and x=2) are changed^{[2]}∗ and the result was done using the same method:
1. SB =
- SA = 8-1-2.33=4.67
- Ratio will be area A : area B=4.67 : 2.33=2 : 1
Or, to calculate the same thing I can use Advanced Calculator 2.0. Computer Software, it’s easy to operate and less time consuming. In order to see a certain pattern, if there is so, let’s take different boundaries, and calculate the area under the curve of the function y=x2 using Advanced CalculatorSoftware:
Boundaries | Area A | Area B | Ratio |
X=0 and x=2 | 4.34 | 2.66 | 2 : 1 |
X=1 and x=2 | 4.34 | 2.66 | 2 : 1 |
X=1 and x=3 | 17.34 | 8.66 | 2 : 1 |
X=2 and x=4 | 37.34 | 18.66 | 2 : 1 |
X=3 and x=6 | 126 | 63.00 | 2 : 1 |
X=7 and x=8 | 112.67 | 56.33 | 2 : 1 |
X=8 and x=10 | 325.34 | 162.66 | 2 : 1 |
Conclusion
∴If the power function of the type y=xn, where n – is the power,
- where n – is the power, n∈Z+ and has limits such as x=0 and x=1 and x-axis; y=0 and y=1 and the y-axis. The ratios of the areas formed by the function and by the boundaries will be equal to n: 1.
- where n – is the power, n∈Z (where and has limits such as x=0 and x=1 and x-axis; y=0 and y=1 and the y-axis. The ratios of the areas formed by the function and by the boundaries will be equal to 1: |n|.
- where n – is the power, n∈Q (where Q= and a and b are integers} and has limits such as x=0 and x=1 and x-axis; y=0 and y=1 and the y-axis. The ratios of the areas formed by the function and by the boundaries will be equal to 1:
∴ If the power functions of the type y=xn, where n – is the power, n∈R and has limits such as x=a and x=b and x-axis. The ratio of the volumes (rotated about x-axis) formed by the function and these boundaries will be equal to 2 n: 1.
∴ If the power functions of the type y=xn, where n – is the power, n∈R and has limits such as y=an and y=bn and y-axis. The ratio of the volumes (rotated about y-axis) formed by the function and these boundaries will be equal to n: 2
[1] - this function is considered with the following boundaries x=0 and x=1 and x-axis; y=0 and y=1 and y-axis
[2]∗- limits, are changed, therefore now area A will be equal to (area of the big rectangular A1B1C1D1 – area of the small rectangular D1B2C2D2 - area B)
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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