# 'Wheel-like' Mathematics (Modulating Trig Functions?)

Tags:
1. May 17, 2015

### Chrono G. Xay

As part of a personal musicology project I found myself with the mathematical model of a geometry which utilizes the equation

a*(a/b)sin(pi*x)

The only problem with this is that I need to take the integral from -1/2 <= x <= 1/2, and according to Wolfram Alpha no such integral exists. I can take the derivative, of course, which is

cos(x)*ln(a)*asin(x)

and I can graph the thing and approximate the area under the curve, but it's incredibly annoying. Is it for situations such as these why Riemann Sums are even taught? (Not bashing on Riemann, it's just incredibly irritating.)

I can approximate the equation itself with

(1/2)[(a-a-1)*sin(x)+(a+a-1)

but it doesn't do it justice. Math people, please rectify this. XD

It's like it has a modulating frequency or something... like a 'chorus' effect for guitar.

Imagine drawing 'a/b' on a circle, including the dividing line between the numerator and denominator, and then slowly rotating it.

The equation came about when I was trying to calculate the length of chord segments when pivoting the initial chord around a point a certain distance from the center of a circle. What it results in is a fraction that changes like a rotating wheel... as though the concept of the equation has a 'visio-mechanical' aspect to it...

Last edited: May 17, 2015
2. May 17, 2015

### Staff: Mentor

While I couldn't integrate it, I found this site that plots it if you replace the a*(a/b) with some number like 3 for 3^sin(pi*x)

https://graphsketch.com

3. May 17, 2015

### Chrono G. Xay

4. May 17, 2015

### Staff: Mentor

This is not an equation. An equation always has = between two expressions.
And this isn't the derivative. I get the derivative as $a\pi \ln(\frac a b)\cos(\pi x) (\frac a b)^{\sin(\pi x)}$. The easiest you to do this is to write $y = a (\frac a b)^{\sin(\pi x)}$ and then take the ln of both sides of this equation. The derivative, with respect to x, of ln(y) is $\frac {y'} y$. The right side is easy to differentiate. After differentiation, multiply both sides by y to get y', the derivative.

5. May 17, 2015

### Chrono G. Xay

Whoops, that was the derivative of asin(x). I had posted this same basic question elsewhere, where the base had been further simplified to just 'a'.

Last edited: May 17, 2015
6. May 17, 2015

### Chrono G. Xay

The original purpose of this pursuit was to calculate the area of a circle using a non-radial(?) point.

7. May 18, 2015

### Staff: Mentor

What is a non-radial point? Any point on a circle is at one end of a line segment from the center to that point; i.e., is on a radius of the circle.

8. May 18, 2015

### Chrono G. Xay

A point that is not at the center, yet inside the circle; i.e., a method that does not use πr2.

9. May 18, 2015

### Chrono G. Xay

Anyway, I think it could be re-written as a wave with an amplitude modulating between a minima (below 'unity'(?) amplitude) of '(1/n)-1' and a maxima (above 'unity'(?) amplitude) of 'n' at intervals of π...

Last edited: May 18, 2015
10. May 19, 2015

### Staff: Mentor

Your description doesn't contain enough details for me to understand what you're trying to do. If you're given a circle of some unknown radius, and a point inside the circle, I don't see how you can find the area inside the circle. BTW, every point inside a circle lies along some radius.

11. May 19, 2015

### Chrono G. Xay

How could one go about calculating the area of a circle without using πr2, but knowing the radius and a point a distance from the center of the circle that lies within the circle?

Get your calculator out, run your graphing program, switch to polar graphing, plug in

"r=sqrt(6^2-2^2)*(sqrt(6^2-2^2)/(6-2))^sin(θ)"

then plug in

"r=12sin(θ)"

and compare the two.

Then switch back to Cartesian graphing and plug the first equation into y=, where θ=x. What I am looking for is:

1. An educated opinion on the behavior of the wave and

2. How to get the same result with a equation that doesn't have a trigonometric function in the exponent, as well as

3. A theory on how the first polar graph might need to be modified in order to transform into the equivalent of 12sin(θ) as 2 approaches 6.

Last edited: May 19, 2015
12. May 19, 2015

### Staff: Mentor

I realize this is obvious, but if you know the radius, that's all you need to calculate the area of the circle. You don't need a point inside the circle.
Where does this formula come from, and what is the significance of 2 and 6 in it? You mentioned that you know a point inside the circle. What are its coordinates?
Where does the 12 come from?
Well, they are either equal or unequal. More interesting to me is why you would want to do this in the first place, given that we know how to calculate the area of a circle, using only its radius (which you state is already known). You can also use integration to calculate the area under, say, the upper half of a circle, provided that you know the equation that represents the circle. Once you know the area of half of the circle, it's a simple matter to get all of the area, just by doubling the value you got for half the area.
What does "2 approaches 6" mean to you? This phrase doesn't mean anything to me, since is constant, and doesn't "approach" anything. It pretty much just sits there on the number line, midway between 1 and 3. You can talk about a variable approaching some number, though.

13. May 21, 2015

### Chrono G. Xay

For my purposes I need to be able to calculate the area of a circle without using πr^2 (Just play along.)

The equation for the length of a chord running through a circle can be expressed as:

c=2*sqrt(r^2-d^2)

Next we can expand it into the sum of the lengths of two individual chord segments:

c=sqrt(r^2-d^2)+sqrt(r^2-d^2)

From there, being two individual chord segments, if we were to pivot a line about a point 'd' from the circle's center, we would notice the individual line segments on either side of the pivot point change in length at a particular rate:

(sqrt(r^2-d^2)/(r-d))^sin(x)

It took a lot of experimentation and meditation on the patterns I was seeing to deduce how I would need the line segments' lengths to change to come up with that equation; to the best of my existing knowledge it's purely of my own creation, and it works.

In my earlier example, r=6 and d=2

L=sqrt(6^2-2^2)*[sqrt(6^2-2^2)/(6-2)]^sin(x)

L=sqrt(36-4)*[sqrt(36-4)/4]^sin(x)

L=4*sqrt(2)*[(4*sqrt(2))/4]^sin(x)

L=4*sqrt(2)*sqrt(2)^sin(x)

When x=π/2...

L1=4*sqrt(2)*sqrt(2)^sin(π/2)

L1=4*sqrt(2)*sqrt(2)^1

L1=4*sqrt(2)*sqrt(2)

L1=4*2

L1=8

And when x=-π/2...

L2=4*sqrt(2)*sqrt(2)^sin(-π/2)

L2=4*sqrt(2)*sqrt(2)^-1

L2=4

L1+L2 = 12 = 2r

A circle of radius 6 has a diameter of 12

As d -> r (as "2 approaches 6"), c=0

As d -> 0, c=2r

Last edited: May 21, 2015
14. May 22, 2015

### Chrono G. Xay

When x=π/2,

L1=sqrt(r^2-d^2)*(sqrt(r^2-d^2)/(r-d))^sin(π/2)

L1=sqrt(r^2-d^2)*(sqrt(r^2-d^2)/(r-d))^1

L1=sqrt(r^2-d^2)*sqrt(r^2-d^2)/(r-d)

L1=(r^2-d^2)/(r-d)

L1=((r-d)(r+d))/(r-d)

L1=r+d

When x=0

L=sqrt(r^2-d^2)*(sqrt(r^2-d^2)/(r-d))^sin(0)

L=sqrt(r^2-d^2)*(sqrt(r^2-d^2)/(r-d))^0

L=sqrt(r^2-d^2)*(1)

L=sqrt(r^2-d^2)

When x=-π/2

L2=sqrt(r^2-d^2)*(sqrt(r^2-d^2)/(r-d))^sin(-π/2)

L2=sqrt(r^2-d^2)*(sqrt(r^2-d^2)/(r-d))^-1

L2=sqrt(r^2-d^2)*((r-d)/sqrt(r^2-d^2))

L2=r-d

L1 + L2 = 2r

(r+d)+(r-d) = 2r
r+d+r-d = 2r
r+r+d-d = 2r
2r+0 = 2r
2r = 2r

Thus, sqrt(r^2-d^2)*(((sqrt(r^2-d^2)/(r-d))sin(π/2)+(sqrt(r^2-d^2)/(r-d))sin(-π/2))=2r

15. May 22, 2015

### Staff: Mentor

I'm glad you finally stated what you're doing, which makes things a lot clearer.
I'm not able to look at your work now, but will do so later today or tomorrow.

16. May 23, 2015

### Chrono G. Xay

. . .

17. May 23, 2015

### Chrono G. Xay

Anyway, for my purposes I need to take the integral of (in terms of its form):

a*(a/b)^sin(x)

However, it seems you simply cannot take the integral of a function with such an exponent, so I've been working out an alternative *integratabtle* way to write it using 'n' instead of 'a/b':

I'm not entirely sure what 'q' is, but stopped here for now because it, too, seems to vary periodically (and I was getting frustrated). Another concern of mine has to do with how the earlier equation

y(x)=sqrt(r^2-d^2)*(sqrt(r^2-d^2)/(r-d))^sin(x)

when written as a polar equation (i.e. where y -> r and x -> θ) breaks down more and more as 'd' approaches 'r', which I interpret to mean that something needs to be 'tacked on' so it will or rather could theoretically transition into an equivalent of

r(θ)=2*r*sinθ

18. May 23, 2015

### phion

I would just break it down into quadrants and find the integral for an ellipse with half axes a and b?

19. May 23, 2015

### Staff: Mentor

You show a lot of work on this page, but if you want others to be able to follow it, you need to explain what the symbols represent. I can infer that c is the length of the chord, and r is the radius of the circle, but what is d? This is the first equation in this post, and I'm already lost.
Still no help. A point in the plane has two coordinates. Is d the label for the point or does it represent a number? If it's a number, is it the x-coordinate or y-coordinate or what?

20. May 23, 2015

### Chrono G. Xay

Last edited: May 23, 2015
21. May 23, 2015

### Chrono G. Xay

22. May 25, 2015

### Chrono G. Xay

One observation I made was that if you write the polar equation

r=sqrt(r^2-d^2)*[sqrt(r^2-d^2)/(r-d)]^sin(Θ)

add the expression "2d*sin(Θ)" to it for

r=sqrt(r^2-d^2)*[sqrt(r^2-d^2)/(r-d)]^sin(Θ)+2d*sin(Θ)

and vary 'd' between 'r' and '-r', you have what seems to be the beginnings of an equation that will translate a polar equation along the plane, theoretically without distortion (though *some* distortion is obviously present). I feel rather confident that by reworking it to include 'i' at the proper time variable 'd' could then be taken *beyond* 'r' or '-r'.

(You could, of course, change the direction of translation by simply adding or subtracting from Θ.)

23. May 26, 2015

### Chrono G. Xay

Mark44, the lack of a reply would seem to imply a continued dissatisfaction with my answer to your earlier question. If that is in fact the case, perhaps it is the question itself that needs further explanation. To the best of my understanding I believe I have answered it.

Last edited: May 26, 2015
24. May 26, 2015

### Staff: Mentor

Chrono, I will take a look, but I've been pretty busy this weekend, with guests at the house. I'll try to take another look later tonight or tomorrow.

25. May 27, 2015

### Chrono G. Xay

Oh, ok. Take care.