Jump to content
ALine

question about pi

Recommended Posts

Hello, I wanted to present something so I could see where I went wrong.

So when I was calculating the area of a circle I wanted to get a better understanding of the circumference of a circle so I found the equation

for it online.

( y = sqrt(r^2 - x^2))

now this equation is for a semi-circle, but I can multiply it by 2 in order to get the full circle.

so when I substituted the y term with the circumference / 2 and the x term with the diameter I got the following,

( C / 2 ) = sqrt(r^2 - (D)^2)

from here you can replaced the diameter term with it radius counterpart

(C / 2) = sqrt(r^2 - (2*r)^2)

from this you could, in turn, combine the two r terms into the following

(C/2) = sqrt(r^2-4r^2)

(C/2) = sqrt(-3r^2)

C = 2*sqrt(-3r^2)

from this I fully reduced it to

C = 2*[sqrt(-3)]*r

now what made be ask this question is because the sqrt(-3) term is where pi should be.

I thought I clearly made a mistake, which I most likely still have.

However when I did the same this in order to find the area using integration I got the following ( solving process below )

(A/2) = integral(0->Diameter) of (sqrt(r^2)-D^2) dDiameter

going through this and applying all of the substitutions got me

A = (sqrt(-3)) * r^2

from going through and solving both for the circumference and area gave the same equations as if you were to replace sqrt(-3) with a pi term.

[C = 2 * pi * r] vs [C = 2 * (sqrt(-3)) * r]

[A = pi*r^2] vs [A = (sqrt(-3))*r^2]

I must have done something wrong, I am thinking it may have something to do with the diameter term and maybe I should use this in place of an x-term.

Help would be much appreciated.

Thank you for your time

P.S. Apologies, can't really LATeX well

Share this post


Link to post
Share on other sites
20 minutes ago, ALine said:

So when I was calculating the area of a circle I wanted to get a better understanding of the circumference of a circle so I found the equation

for it online.

( y = sqrt(r^2 - x^2))

That is not the equation for the circumference of a circle.

That will give you the y coordinates for the (two) points on a circle (with radius r) corresponding to the given x value. You could, perhaps, integrate over a range of x values to get the circumference but it seems unnecessarily complicated because:

C = 2 pi r

 

Share this post


Link to post
Share on other sites
27 minutes ago, ALine said:

So when I was calculating the area of a circle I wanted to get a better understanding of the circumference of a circle so I found the equation

for it online.

( y = sqrt(r^2 - x^2))...........................1

now this equation is for a semi-circle, but I can multiply it by 2 in order to get the full circle.

so when I substituted the y term with the circumference / 2 and the x term with the diameter I got the following,

( C / 2 ) = sqrt(r^2 - (D)^2)..................2

What are you trying to do here?

Equation 1 is an unusual way of writing the standard equation of a circle

x2 + y2 = r2   

(note you don't need LaTex to write this, just use the sub and super script icons on the text input editor, labelled   x2 and x2)

your way puts you into immediate trouble since if D is the diameter D = 2r so equation 2 becomes


[math]\frac{C}{2} = \sqrt {{r^2} - {{\left( {2r} \right)}^2}}  = \sqrt { - 3{r^2}} [/math]


So you are trying for the square root of a negative number.

Furthermore if C is the length of the circumference, you can't just replace y with half of it.

Edited by studiot

Share this post


Link to post
Share on other sites
13 minutes ago, Strange said:

That is not the equation for the circumference of a circle.

That will give you the y coordinates for the (two) points on a circle (with radius r) corresponding to the given x value. You could, perhaps, integrate over a range of x values to get the circumference but it seems unnecessarily complicated because:

C = 2 pi r

got it, thank you for the correction strange.

13 minutes ago, studiot said:

What are you trying to do here?

Did not really have a goal in mind, just "playing around with the numbers" I guess.

13 minutes ago, studiot said:

Furthermore if C is the length of the circumference, you can't just replace y with half of it.

thank you for the correction as well studiot.

I feel as if I have an issue with the conversion between physical geometry and x and y relationship chart representations, I think that is the right way to say this.

Need to get better at understanding this.

 

Also, stupid question, so pi is just the relationship between Circumference and Diameter? 

Like if I wanted to know the circumference and am given the diameter then if I multiply this diameter with pi it will give me pi?

And is pi the relationship between diameter and circumference? Like could it be described as being the number which gives a circle its "circle" characteristic or property? if that makes any sense.

p.s. thank you guys for answering my questions

Edited by ALine

Share this post


Link to post
Share on other sites
4 minutes ago, ALine said:

thank you for the correction as well studiot.

I feel as if I have an issue with the conversion between physical geometry and x and y relationship chart representations, I think that is the right way to say this.

Need to get better at understanding this.

Yes possibly.

Strange's comment supplied the formula (I know it has an equals sign) for the circumference of any circle.
That circle may be anywhere, hanging on a tree, spinning on the ice, just chilling in a deck chair.
or even referenced to an x-y coordinate system.

If you choose to reference it to an x-y coordinate system you can properly call the expression with an equals sign an equation.

And you can then do much more with it.

But you have to get the algebra right, so you need to revise your algebra.

Furthermore once you have referenced your circle to a particular x-y system, it can no longer be anywhere.
You have fixed it in space.
Also you have to work with  x and y , you can't just switch them into something else and bring in a freewheeling formula.

 

Share this post


Link to post
Share on other sites
8 minutes ago, studiot said:

You have fixed it in space.

when you say, " fixed it in space." what do you mean by this.

When I hear this it makes me think that someone is analogously "tacking" it to a wall. Or even taking like a snap shot of it and you cannot change anything about it, whereas you can only look at it and then trace over it with an x and y graph.

8 minutes ago, studiot said:

it can no longer be anywhere.

I am also having a problem understanding this concept. Can you please explain.

When I hear this it makes me think that you are analogously "removing it from reality and placing into a mathematical universe where it can be dissected."

 

So like taking a picture of it, and then analyzing and understanding that "image."

Edited by ALine
added a part about an image in the end

Share this post


Link to post
Share on other sites
2 minutes ago, ALine said:

When I hear this it makes me think that someone is analogously "tacking" it to a wall. Or even taking like a snap shot of it and you cannot change anything about it, whereas you can only look at it and then trace over it with an x and y graph.

That's a pretty fair analogy.

As soon as you introduce x and y, they have values, and every pair of values is unique.

 

Share this post


Link to post
Share on other sites
4 minutes ago, studiot said:

As soon as you introduce x and y, they have values, and every pair of values is unique.

So would it still be considered a "circle" at that point? Or would its "circular value" is lost, where it is only like a background image for the different points in space?

Also, if you do not mind me asking, how would this relate the x and y's and functions? Where functions are relating the x points with the y points in space.

Also would the definition of the x and y graph dependent on the "space" you define them in?

Like can there be different "types" of spaces?

Like is x and y considered dimensions of a given space and if so does this mean that you can use other variables like time or velocity as be part of the same "dimensional space."

( when I say dimension I just mean a variable which can have a specific value on a number line)

p.s. sorry if I am speculating to much, I have a bad habit of "guess to far ahead."

Share this post


Link to post
Share on other sites

After you have chosen a specific "xy" Cartesian coordinate system, then the set of (x, y) points satisfying $x^2+ y^2= r^2$ gives a circle with center at the origin of that coordinate system and radius r in the units used for that coordinate system.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.