Math Portfolio Type 1

Tab 3

The table above represents the values of the areas and their ratios (SA: SB). In order to see it clearly, I can show you a chart with functions presented there.

Tab 4

I would like to pay attention, on that I’m ignoring the negative sign, when I’m doing a ratio; therefore I would take it as an absolute value - . Therefore my conjecture for the set of negative integers will be as the following:

∴ If the power function of the type y=xn, 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|.

Next subset of rational numbers will be fractions, so will consider following functions represented on the following table:

     

Tab 5

So, the sketches to the functions will be as the followings:

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∗ and the result was done using the same method:

 1.         SB =

2. SA = 8-1-2.33=4.67
3. 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 Calculator Software:

Tab. 3.0.

From the table above, I can say that, the conjecture is hold for different limits (positive values of x are only taken). So, limits don’t affect the entire conjecture, there is no pattern seen. However, I would like to mention that with change of the domain (x-values, limits) range (y-value) changes respectively. In order to find area using other way (not manual one) I used Advanced Calculator 2.0. Computer Software.  

Area under the curve is an application of integration, so there should be some general formulae for it, so let’s consider a power function of the type y=xn where n ∈R from x=a to x=b such that a<b and for the regions defined below:

Area A: y=xn, y=an, y=bn and the y-axis

Area B: y=xn, x=a, x=b and the x-axis

For better understanding, I will use the following diagram (Figure 6):

Figure 6

On the diagram there is a graph of power function y=xn, with the boundaries x=a and x=b; y=an and y=bn. And I need to find a general formula for the ratio of the areas A and B.

Now, I will try to derive a general formula for the area B:

        (i)

Then area A, but I also would like to explain generally, how I’m going to find it; first of all, I need to consider area of the small rectangular, lets say area C; then the whole area A will be equal to (area D-area C- area B).

(ii)

Because S=ab, where a and b-sides of the rectangular

Therefore just substitute the values of these areas into formula (ii)

And, final result, the ratio of the areas:

    SA : SB =n: 1

I can prove that using method of math induction:

Let P (n) be the proposition that SA : SB=n: 1

1.        Test for n=1

SB=

SA=

SA : SB=

2.        Let P (n) be true for n=k, i.e.,

SB=

SA=

SA : SB=

3. Test for n=k+1

SB=

SA=

SA : SB=

Thus, if the proposition is true for n=k then its true for n=k+1 as proved. As it true for n=1, then it must be true for n=1+1(n=2). As it true for n=2 then it must hold for n=2+1(n=3) and so on for the set of real numbers.

That is, by the principal of mathematical induction P (n) is true.  

So, my general conjecture for the area will be as the following:

∴ 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; and y=xn, x=a, x=b and the x-axis. The ratio of the areas formed by the function and these boundaries will be equal to n: 1.

Also, I need to derive general formulae for the volumes of revolution generated by the region A and B when they are each rotated about

1. the x-axis
2. the y-axis

1. For the volumes of revolution, those are rotating around the x-axis, the following will be true:

                 (iii) where x=a and x=b are boundaries, such as f(x) its a given function, particularly in that case it’s a power function f(x)=xn

This can be seen on the Figure 6, which is plotted below

Figure 6

This graph shows two volumes (A and B); region, which is colored in blue, is VB and region, colored in green is VA. I need to find a ratio of those. So, I will try to derive a general formula for volumes of revolution rotated around x-axis.

According, to the formulae (iii), volume B will be equal to:

     (iv)

In order to determine VA; I need to consider volume of the cylinder, because when it’s rotating around x-axis, it creates a cylinder. The volume of the cylinder (ABCD) equals

VC= πR2H, where R is bn and height is b

And also, I need to consider V0 that is another small cylinder (A1B1C1D1) which creates something like a hole inside the rotating figure, which is

V0=πR2H, where R is an and height is a

So, therefore VA=VC-VB-V0

(v)

So, the general formula for this is:

VA : VB=2n: 1

I can prove this formula, using method of math induction:

Let P (n) be the proposition that VA : VB=2n: 1

1.        Test for n=1

VB=

VA=

VA : VB=

2.        Let P (n) be true for n=k, i.e.,

VB=

VA=

VA : VB=

3.        Test for n=k+1

VB=

VA=

VA : VB=

Thus, if the proposition is true for n=k then its true for n=k+1 as proved. As it true for n=1, then it must be true for n=1+1(n=2). As it true for n=2 then it must hold for n=2+1(n=3) and so on for the set of real numbers.

That is, by the principal of mathematical induction P (n) is true.  

Hence my conjecture for the solids of revolution around x-axis will be as the following: ∴ 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 formed by the function and these boundaries will be equal to 2|n|: 1.

In order to see, I would like to present some examples, which were found using Advanced Calculator 2.0. Software  

 

Note, that the boundary is from x=1 to x=2; and it is rotated around x-axis. So, the results were as the following:

Tab. 3.0.

This table only proves that the conjecture works, for n∈Z, Q, and R.

b) For the volumes, which are rotating around y-axis, the following will be true:

 V= π ∫ [f-1(y)] 2dy, where y=an and y=bn are the boundaries of the given function y=f(x)

Fig. 7.0.

On the figure above, there is a rotation around y-axis. Let say that the area shaded in blue is a VB and just white region is VA. As far as I need to integrate an inverse of the given function, I would like to show the process itself; because it might cause a mistake.

1.        Given, f (x) = xn

2.        Need to find, f -1(x); which is y=xn

x=

I will try to derive a formula for solids of revolution around y-axis.

VA= (vi)

In order to find VB; I need to consider volume of the cylinder created by

VC= πR2H, where R is b and height is bn

And V0=πR2H, where R is a and height is an

So, therefore, VB=VC-VA-V0 where VC is a volume of the big cylinder (ABCD) and V0 is a volume of the small cylinder inside (A1B1C1D1)

VB=(vii)

Then, I plug in the derived formulas (vi) and (vii) the ratio of the volumes will be:

VA : VB =

I can prove this formula, using method of math induction (as it was shown previously) Therefore, I can state my conjecture as the following:

∴ 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 formed by the function and these boundaries will be equal to n: 2.

Also, I can show some examples, using Advanced Calculator Software, basically same functions, but rotating around y-axis now. That’s what I got:

Tab 4This table, also proves that conjecture works, I decided to calculate that, just in order to make sure. It’s kind of a method for proving the conjecture, by collecting the relevant data.

Conclusion

In this investigation, I was considering power functions of the type y=xn, where n∈R. I had to find, ratios of the areas and volumes, with given boundaries. My limitations are basically, that the boundaries are positive, so a and b >0. The conjectures were proved, by using the method of math induction. I got the following results:

∴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

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