Parametric Equation of an Ellipse

[Buri]

Homework Statement

I have the following:

x(t) = acos(kt) + bsin(kt)
y(t) = aksin(kt) + bkcos(kt)

The Attempt at a Solution

I'd like to show this is an ellipse, by actually explicitly finding the equation, but I honestly I have no clue about how to do this. Wiki gives the equation of an ellipse centered around (0,0) to be

x(t) = acos(t)
y(t) = bsin(t)

Here I could do this, but in my version I can't since it has both cos and sin in each function x and y. How would I go about doing this?

Thanks for the help!

 

[Buri]

Hmm how would I write acos(kt) + bsin(kt) as a cos function and aksin(kt) + bkcos(kt) as a sin function? Maybe this would get me somewhere..

 

[Buri]

So I guess I have:

acos(kt) + bsin(kt) = cos(x)cos(kt) + sin(x)sin(kt)

So in other words:

a = cos(x)
b = sin(x)

Which yields:

b/a = tan(x) or x = arctan(b/a)

So in I have: acos(kt) + bsin(kt) = cos(arctan(b/a) - kt)

And I would get y(t) = ksin(artan(b/a) + kt)

Is this all right? :)

 

[vishal007win]

this is not a case of simple ellipse.. with major and minor axis along x and y direction, instead they are inclined along lines.
work it out by separating cos(kt) and sin(kt) using 2 equations
and then simply cos²(kt) +sin²(kt) =1
all the best

 

[Buri]

And now for the other:

aksin(kt) + bkcos(kt)

I'd like this to be in terms of sine only. So I would like:

k(cos(x)sin(kt) - sin(x)cos(kt)) = ksin(x - kt)

So I'd like:

a = cos(x) and b = sin(x)

So tan(x) = b/a And there again arctan(b/a) = x

So we have:

ksin(x - kt) = ksin(arctan(b/a) - kt)

 

[Buri]

this is not a case of simple ellipse.. with major and minor axis along x and y direction, instead they are inclined along lines.
work it out by separating cos(kt) and sin(kt) using 2 equations
and then simply cos²(kt) +sin²(kt) =1
all the best

What do you mean? Sorry I've never worked with parametric equations before and this is actually for ODE's...

 

[vishal007win]

Homework Statement

I have the following:

x(t) = acos(kt) + bsin(kt)
y(t) = aksin(kt) + bkcos(kt)

manipulating these two equations will lead to
cos(kt) as function of x(t) and y(t) along with some coefficients
and similarly for sin(kt)
then directly use the identity to get a relations between

 

[Buri]

I have this:

cos(kt) = [akx(t) - by(t)]/(a²k - b²k)
sin(kt) = [bkx(t) - ay(t)]/(b²k - a²k)

This is going to get ugly eh?

 

[Buri]

manipulating these two equations will lead to
cos(kt) as function of x(t) and y(t) along with some coefficients
and similarly for sin(kt)
then directly use the identity to get a relations between

What is this supposed to look like in the end? As in something like:

x²/m + y²/n = 1?

It doesn't seem like its going to simplify to something like that.. :(

 

[Buri]

And btw vishal007win I really appreciate your time, I wouldn't be getting this far without your help!

 

[vishal007win]

yes it is..

best i can show you about inclined ellipse...but in the figure , axis are perpendicular, which is not the case here, here the axis are inclined along lines
ak x(t) -b y(t)=0 and bk x(t) -a y(t)=0

 

[Buri]

This is what I have:

( [akx(t) - by(t)]/(a²k - b²k) )² + ( [bkx(t) - ay(t)]/(b²k - a²k) )² = 1

But from here, it just doesn't seem to simply down to something nice...would it be okay to leave it like this?

 

[Buri]

yes it is..

best i can show you about inclined ellipse...but in the figure , axis are perpendicular, which is not the case here, here the axis are inclined along lines
ak x(t) -b y(t)=0 and bk x(t) -a y(t)=0

To be honest with you, I don't know enough about my conic sections anymore to be able to tell whether I'm right or wrong, but the two equations you gave above are in my equations, so maybe I'm right? I haven't done conic sections since high school and that's been over 4 years how...

 

[Buri]

This question actually has to do with a harmonic oscillator question for my ODE's class. I'm told to let A and B be the distances between the origin and the point where the ellipse intersects the positive side of the x-axis and y-axis, respectively. Let a = max(A,B) and b = min(A,B). In other words, a and b are the semimajor and semiminor axis of the ellipse, respectively. How do I actually get these coordinates now?

 

[vishal007win]

so you are dealing with phase potraits ! are you??
if i am getting correctly what you are asking then,
after finding the equation of ellipse , put x = 0 for finding point of intersection with y-axis, and similarly for B put y=0

 

[Buri]

Yes I am actually...I've tried doing this. And the point of it is to show that the eccentricity of all the ellipses in the phase portrat corresponding to the different initial conditions is the same...but it just isn't working :S

 

[Buri]

The actual linear system is
X' = [0 1; -u 0]X

 

[Buri]

Its the most ugliest ellipse equation I've ever seen and nothing seems to be simplifying down..

 

[D H]

Try a change of coordinates.

 

[Buri]

Well to solve the linear system I did use a change of coordinates actually. But its the ellipse that's causing me problems...unless you mean to change coordinates of the ellipse? And that I would have no clue how to do..

 

[D H]

Try to find some angle θ such that upon rotating (x,y) by θ yields a new system, call it (u,v), in which the equation of the ellipse is very simple.

 

[Buri]

I've realized a mistake in my solution (I forgot a negative) and now it actually simplifies down to a really simple ellipse! :) I'd like to thank vishal007win for helping me out D H thanks for your input, I've actually learned a lot by trying to change coordinates for my wrong ellipse. Thanks a lot for the help! I really appreciate both of you guy's time! :)