1 Topic by jsnc (edited by jsnc 2016-07-20 23:46:35)

  • jsnc
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Topic: Polar Equations

Shouldn't the graph of r Sin(t) = Sin (r Cos(t))
r=(Sin (r*Cos (t)))/(Sin (t) )
give me the same thing as y=f(x)=Sin (x) which is the Sine Curve?
y = r Sin(t)
x = r Cos(t) substitute into x,y equation

I tried entering as a polar equation "r(t)=", and then as a relation.
How does one graph the Sine function as a polar equation?\

r(t) - 1+Cos(t) Cardioid graphed just fine.

Why do you indicate the Polar Coordiantes as (θ, r) instead of (r, θ)?

Continuing on, I can graph the relation (x-3)^2 + (y-2)^2=4
A circle with center (3, 2) and radius 2

But after converting this to a Polar Equation 13/(2(3cos(t))+(2Sin(t))) I get a straight line?

Continuing on, I believe that the graph of the equation (4(sin(t))^2+9(cos(t))^2)^.5 should match the graph of the Parametric Equation: 2sin(t), 3cos(t) (I converted the Parametric to the Polar), but it doesn't.
I added my Graph file for both Parametrics and Polar.

Post's attachments

Attachment icon Graphs Polar Equations.grf 3.48 kb, 462 downloads since 2016-07-20 

Attachment icon Parametric Equations.grf 1.25 kb, 467 downloads since 2016-07-20 

Re: Polar Equations

jsnc wrote:

Shouldn't the graph of r Sin(t) = Sin (r Cos(t))
r=(Sin (r*Cos (t)))/(Sin (t) )
give me the same thing as y=f(x)=Sin (x) which is the Sine Curve?

Yes, I believe this is correct.

jsnc wrote:

I tried entering as a polar equation "r(t)=", and then as a relation.
How does one graph the Sine function as a polar equation?\

I don't think that is possible. Unfortunately Graph cannot plot polar functions as relations. I might add support for that one day. And I don't think it is possible to isolate r in the equation, so I think you have to use cartesian coordinates to plot y=sin(x).

jsnc wrote:

Why do you indicate the Polar Coordiantes as (θ, r) instead of (r, θ)?

Where do I do that? I cannot seem to find anywhere I do that, though it sounds like something I might do.

jsnc wrote:

Continuing on, I can graph the relation (x-3)^2 + (y-2)^2=4
A circle with center (3, 2) and radius 2
But after converting this to a Polar Equation 13/(2(3cos(t))+(2Sin(t))) I get a straight line?

Well, it looks like you didn't convert it correctly. The polar coordinates for the center is:
r0=sqrt(13), t0=atan(2/3), a=2

The generic formula for a circle in polar coordinates is:
r(t) = r0*cos(t-t0) + sqrt(a^2-r0^2*sin(t-t0)^2)

I have attached a file that is showing this.

Post's attachments

Attachment icon Polar circle.grf 730 b, 472 downloads since 2016-07-20