Showing the solution is a circle: Why is this true?

**karatekid** · April 28th 2012, 12:10 PM

Hi guys,

I've been trying this all day, hopefully I'm just being stupid. The full problem is to show that two simultaneous differential equations in x(t) and y(t) have a solution that is a circle, which I believe is the circle ${(x-\mu)}^2 + y^2 = a$ where mu is given and a is a constant depending on initial conditions. I've reduced the problem to showing $x\dot{y} - y\dot{x} - \mu \dot{y} = const.$

I can't show this, so if anyone could help with that I'd be really grateful.

I'm also confused as to where I'm going wrong here:
If $x = a\cos{\omega t} + \mu$ and $y = a\sin{\omega t}$ then the constant in the last expression above is zero (I hope that's not my mistake!). But then we can rearrange this (when x'(t) is non-zero and x isn't mu) to get $\dot{y}/\dot{x} = y/(x-\mu)$ which is false in most cases. Why can't I divide through?

Thanks!!

**Prove It** · April 28th 2012, 08:04 PM

Originally Posted by karatekid

Hi guys,

I've been trying this all day, hopefully I'm just being stupid. The full problem is to show that two simultaneous differential equations in x(t) and y(t) have a solution that is a circle, which I believe is the circle ${(x-\mu)}^2 + y^2 = a$ where mu is given and a is a constant depending on initial conditions. I've reduced the problem to showing $x\dot{y} - y\dot{x} - \mu \dot{y} = const.$

I can't show this, so if anyone could help with that I'd be really grateful.

I'm also confused as to where I'm going wrong here:
If $x = a\cos{\omega t} + \mu$ and $y = a\sin{\omega t}$ then the constant in the last expression above is zero (I hope that's not my mistake!). But then we can rearrange this (when x'(t) is non-zero and x isn't mu) to get $\dot{y}/\dot{x} = y/(x-\mu)$ which is false in most cases. Why can't I divide through?

Thanks!!

My psychic powers are a little off today, so I have no idea if what you have reduced the original problem to is correct. Please state the original problem and what working you have done so far for the first question.

**karatekid** · April 29th 2012, 03:32 AM

Yeah, sorry. That's the last time I ask a question after midnight.

The original equations are
$\ddot{x} - 2\dot{y} &= \frac{(x-\mu)}{2{r_2}^3} \\ \ddot{y} + 2\dot{x} &= \frac{y}{2{r_2}^3}$

Where ${r_2}^2 = \left( x - \mu \right)}^{2} + y^2$ and in this particular case $\mu = 0.5$ . To try and solve these, I divided the second by the first (or if x = mu, the first by the second) and cross multiplied to get
$x \ddot{y} + 2x\dot{x} - 2\mu\dot{x} - \mu \ddot{y} = y \ddot{x} - 2y\dot{y}$

and we can rearrange and integrate this to get
$x^2 + y^2 - 2\mu x = y\dot{x} - x\dot{y} + \mu \dot{y} + const.$

which is of the form ${(x-\mu)}^2 + y^2 = const.$ if and only if $y\dot{x} - x\dot{y} + \mu \dot{y} = const.$

The context is a spaceship orbiting a large planet. I just don't get why this last equation is true (which it must be, since the circle solves the original problem). Sorry for my vagueness before!

**karatekid** · April 29th 2012, 03:36 AM

Originally Posted by karatekid

I'm also confused as to where I'm going wrong here:
If $x = a\cos{\omega t} + \mu$ and $y = a\sin{\omega t}$ then the constant in the last expression above is zero (I hope that's not my mistake!). But then we can rearrange this (when x'(t) is non-zero and x isn't mu) to get $\dot{y}/\dot{x} = y/(x-\mu)$ which is false in most cases. Why can't I divide through?

I've just worked this last bit out. Late night syndrome obviously, since I was claiming sin^2 + cos^2 = 0 not 1....

I'm still unsure of the first part though.

Thread: Showing the solution is a circle: Why is this true?

LinkBack

Thread Tools

Display

Showing the solution is a circle: Why is this true?

Re: Showing the solution is a circle: Why is this true?

Re: Showing the solution is a circle: Why is this true?

Re: Showing the solution is a circle: Why is this true?

Similar Math Help Forum Discussions

Showing a curve is part of a circle

Showing how pi/2+theta = cos using unit circle.

Showing Transformation is True

showing differential equation solution

Showing a solution for system of equations

Search Tags