ATAR Notes: Forum

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: /0 on November 25, 2009, 06:30:29 pm

Title: A differential equation
Post by: /0 on November 25, 2009, 06:30:29 pm: How do you solve:

$\frac{d^2D}{dt^2} = \frac{-2G}{D^2}$ where G is a constant

I'm not sure if I'm allowed to flip both sides...

lol thanks :p
Title: Re: A differential equation
Post by: humph on November 25, 2009, 11:58:01 pm: By inspection, try considering logarithm answers. And no, you can't flip both sides, you'd have to use techniques beyond Spesh. In fact, this ODE isn't linear, so you can't use most standard techniques. But it's pretty easy to find one family of solutions (though as to whether they'd work depends on your initial values).
Title: Re: A differential equation
Post by: /0 on November 26, 2009, 11:04:55 am: Hmm yeah somehow I can't seem to get them to work though.

The question came from trying to find how long it would take for two 1kg balls 10m apart to meet.

$\frac{d^2x_1}{dt^2} = \frac{G}{D^2}$ , $\frac{d^2x_2}{dt^2} = -\frac{G}{D^2}$

If we let $D = x_2(t)-x_1(t)$

$\frac{d^2D}{dt^2} = -\frac{2G}{D^2}$
Title: Re: A differential equation
Post by: QuantumJG on November 26, 2009, 11:12:16 am: Quote from: /0 on November 26, 2009, 11:04:55 am
Hmm yeah somehow I can't seem to get them to work though.

The question came from trying to find how long it would take for two 1kg balls 10m apart to meet.

$\frac{d^2x_1}{dt^2} = \frac{G}{D^2}$ , $\frac{d^2x_2}{dt^2} = -\frac{G}{D^2}$

If we let $D = x_2(t)-x_1(t)$

$\frac{d^2D}{dt^2} = -\frac{2G}{D^2}$

What subject is this?

I'm guessing this is a question relating to how long it would take under the influence of their gravitation?

All I can think is that if:

$\dfrac{d^{2}r}{dt^{2}}=G\dfrac{m_{1}}{r^{2}}$

$r^{2}dr = Gm_{1}dt$

$\int r^{2}dr = \int Gm_{1}dt$ (r: 10 -> 0 and t: 0 -> t)

$\dfrac{dr}{dt}=-G\dfrac{t}{10} = v$

My logic now is that if the system is held by G/10 J, then its Kinetic energy when r = 0 should be G/10 J (since they have no kinetic energy at r = 10)

Now,

$K = \dfrac{1}{2}*m*v^{2}$

therefore,

$\dfrac{G}{10} = \dfrac{1}{2}*1*G^{2}*\dfrac{t^{2}}{100}$

implying that,

$t = \sqrt{\dfrac{20}{G}} = 547,586 s = 9,126 min = 152 hours \approx 6 days$

Something tells me that my maths probably isn't right (never had to do a question like this before), its a difficult question since your accelleration is changing (depends on r). Interesting question though!
Title: Re: A differential equation
Post by: /0 on November 26, 2009, 11:17:01 am: Not really anything, I just made up the problem (well physics lol)
Title: Re: A differential equation
Post by: QuantumJG on November 26, 2009, 03:00:08 pm: Quote from: Ahmad on November 26, 2009, 03:36:55 pm
It's great that you're thinking about things on your own, highly encourage it! Anyway, numerically solving if they're 10 cm apart then it takes just over 16 minutes and 1 second. For 1 meter apart around 8.44 hours. For 10 meters apart around 11.13 days. Hope I didn't make any mistakes :)

Where can we find those values? More importantly, how did they determine this?
Title: Re: A differential equation
Post by: zzdfa on November 26, 2009, 03:13:53 pm: make a simulation
Title: Re: A differential equation
Post by: Ahmad on November 26, 2009, 03:36:55 pm: It's great that you're thinking about things on your own, highly encourage it! Anyway, numerically solving if they're 10 cm apart then it takes just over 16 minutes and 1 second. For 1 meter apart around 8.44 hours. For 10 meters apart around 11.13 days. Hope I didn't make any mistakes :)
Title: Re: A differential equation
Post by: /0 on November 26, 2009, 03:57:27 pm: LOL sorry I forgot to say what the radiuses of the balls were! I was thinking something like 1cm
(although hmm... I guess you could have point particles after all)
Thanks for trying to work it out Quantum but I don't really understand.. your second order DE seemed to turn into a first order DE

Quote from: Ahmad on November 26, 2009, 03:36:55 pm
It's great that you're thinking about things on your own, highly encourage it! Anyway, numerically solving if they're 10 cm apart then it takes just over 16 minutes and 1 second. For 1 meter apart around 8.44 hours. For 10 meters apart around 11.13 days. Hope I didn't make any mistakes :)

Wow that's interesting, I never thought it would happen so fast. Did you use point particles?
(Oh and what program did you use? :P)
Title: Re: A differential equation
Post by: Ahmad on November 26, 2009, 04:12:53 pm: Firstly, I'd like to correct myself, the values I gave were for 10 kg point particles, just to keep things reasonable. How did I do it? I used mathematica. How could you do it if you only had a hand calculator? I've attached an excel file which shows how one might do something like this. I've used a very simple algorithm which is conceptually easy but not as accurate as something like mathematica would use. Even so, it seems to give the answer for 10 kg point particles separated by 10 cm correct to at least 1 decimal place. :)

Back in year 12 I made a program which simulates gravity and shows how planets orbit each other (chaotically at times!) which I'm happy to share if anyone is interested.
Title: Re: A differential equation
Post by: kamil9876 on November 26, 2009, 10:12:07 pm: I remember a similair, but more complicated DE comming from the same situation, except the balls both had a charge q :P
Title: Re: A differential equation
Post by: /0 on November 26, 2009, 10:14:36 pm: Quote from: kamil9876 on November 26, 2009, 10:12:07 pm
I remember a similair, but more complicated DE comming from the same situation, except the balls both had a charge q :P

Do you remember how it was solved? Or was it solved by numerical methods?
Title: Re: A differential equation
Post by: kamil9876 on November 26, 2009, 11:15:43 pm: actually, I made it up myself for the sake of revision, althuogh to be honest i had a slightly different situation: ball with mass m, charge q dropped from height h directly above another ball of same charge q fixed to the ground, (this is happening on Planet earth so i used constant acceleration of g). Heuristically one would imagine that initially it falls, then rises once the coloumb force gets big etc. then coloumb force gets weaker and eventually it turns back around. Using conservation of energy(remember; physics revision this was)(assuming no air resistance), you get that the equation for the values of y at which kinetic energy is zero(turning points) is a quadratic, hence two solutions only. Meaning that it is an osscilatary function with some constant amplitude!, however, trigonometric substitution didn't work for the actual differential equation.
Title: Re: A differential equation
Post by: /0 on November 26, 2009, 11:18:22 pm: lol nice
Coulomb vs. Gravity is like a really long spring in a changing gravitational field
Title: Re: A differential equation
Post by: QuantumJG on November 26, 2009, 11:24:18 pm: Quote from: QuantumJG on November 26, 2009, 11:12:16 am
Quote from: /0 on November 26, 2009, 11:04:55 am
Hmm yeah somehow I can't seem to get them to work though.

The question came from trying to find how long it would take for two 1kg balls 10m apart to meet.

$\frac{d^2x_1}{dt^2} = \frac{G}{D^2}$ , $\frac{d^2x_2}{dt^2} = -\frac{G}{D^2}$

If we let $D = x_2(t)-x_1(t)$

$\frac{d^2D}{dt^2} = -\frac{2G}{D^2}$

What subject is this?

I'm guessing this is a question relating to how long it would take under the influence of their gravitation?

All I can think is that if:

$\dfrac{d^{2}r}{dt^{2}}=G\dfrac{m_{1}}{r^{2}}$

$r^{2}dr = Gm_{1}dt$

$\int r^{2}dr = \int Gm_{1}dt$ (r: 10 -> 0 and t: 0 -> t)

$\dfrac{dr}{dt}=-G\dfrac{t}{10} = v$

My logic now is that if the system is held by G/10 J, then its Kinetic energy when r = 0 should be G/10 J (since they have no kinetic energy at r = 10)

Now,

$K = \dfrac{1}{2}*m*v^{2}$

therefore,

$\dfrac{G}{10} = \dfrac{1}{2}*1*G^{2}*\dfrac{t^{2}}{100}$

implying that,

$t = \sqrt{\dfrac{20}{G}} = 547,586 s = 9,126 min = 152 hours \approx 6 days$

Something tells me that my maths probably isn't right (never had to do a question like this before), its a difficult question since your accelleration is changing (depends on r). Interesting question though!

Before doing this mathematical mess, I should have read what humph said:

Quote
In fact, this ODE isn't linear, so you can't use most standard techniques.

i.e. My first year uni maths is pretty useless.

So basically what I did was all wrong.
Title: Re: A differential equation
Post by: QuantumJG on November 26, 2009, 11:27:47 pm: how do you run that excel program?
Title: Re: A differential equation
Post by: kamil9876 on November 26, 2009, 11:32:17 pm: The perils of Liebniz notation.
Title: Re: A differential equation
Post by: Ahmad on November 26, 2009, 11:43:26 pm: Highlight cells A3 to D3, right click and copy them. Then highlight the first 4 cells of the next 100 rows, right click and paste. Then you'll probably get the idea and paste into the next 60,000 rows or whatever :P
Title: Re: A differential equation
Post by: humph on November 27, 2009, 02:03:04 am: Good old non-linear ODEs. They come up all the time in electromag stuff. In first year I remember deriving an ODE to find the acceleration of a metal pellet out of a Gauss gun:
$\frac{d^2 x}{dt^2} = \frac{\tau n^2 I^2 \mu_0}{ms^2} \left(\frac{s/2 + x}{\sqrt{(s/2 + x)^2 + r^2}} - \frac{s/2 - x}{\sqrt{(s/2 - x)^2 + r^2}}\right) \left(\frac{1}{\left((s/2 + x)^2 + r^2\right)^{3/2}} - \frac{1}{\left((s/2 - x)^2 + r^2\right)^{3/2}}\right)$
where
$x$ is position of the projectile inside the Gauss gun,
$t$ is time,
$m$ is the mass of the projectile,
$\tau$ is its volume,
$s$ is the length of the coil of the Gauss gun,
$I$ is the current running through the coil,
$r$ is the radius of the coil,
$n \phantom{}$ is the number of turns of the coil, and
$\mu_0 = 4\pi \times 10^{-7}$ is a constant.

Needless to say, even trying to solve this ODE numerically was a bit beyond us :P
Title: Re: A differential equation
Post by: /0 on November 27, 2009, 02:24:06 am: Quote from: Ahmad on November 26, 2009, 11:43:26 pm
Highlight cells A3 to D3, right click and copy them. Then highlight the first 4 cells of the next 100 rows, right click and paste. Then you'll probably get the idea and paste into the next 60,000 rows or whatever :P

lol excel is throwing a tantrum about all that data
Title: Re: A differential equation
Post by: humph on November 27, 2009, 03:43:39 am: Quote from: kamil9876 on November 26, 2009, 11:32:17 pm
The perils of Liebniz notation.
It's why you should always just write $f''(x) = -\frac{2G}{f(x)}$ . Though it does make solving separable equations seem less elementary, though all you really have to do is remember change of variables rule when integrating.
Title: Re: A differential equation
Post by: QuantumJG on November 27, 2009, 12:25:27 pm: Quote from: humph on November 27, 2009, 02:03:04 am
Good old non-linear ODEs. They come up all the time in electromag stuff. In first year I remember deriving an ODE to find the acceleration of a metal pellet out of a Gauss gun:
$\frac{d^2 x}{dt^2} = \frac{\tau n^2 I^2 \mu_0}{ms^2} \left(\frac{s/2 + x}{\sqrt{(s/2 + x)^2 + r^2}} - \frac{s/2 - x}{\sqrt{(s/2 - x)^2 + r^2}}\right) \left(\frac{1}{\left((s/2 + x)^2 + r^2\right)^{3/2}} - \frac{1}{\left((s/2 - x)^2 + r^2\right)^{3/2}}\right)$
where
$x$ is position of the projectile inside the Gauss gun,
$t$ is time,
$m$ is the mass of the projectile,
$\tau$ is its volume,
$s$ is the length of the coil of the Gauss gun,
$I$ is the current running through the coil,
$r$ is the radius of the coil,
$n \phantom{}$ is the number of turns of the coil, and
$\mu_0 = 4\pi \times 10^{-7}$ is a constant.

Needless to say, even trying to solve this ODE numerically was a bit beyond us :P

That is an intense second order ODE!
Title: Re: A differential equation
Post by: zzdfa on November 27, 2009, 11:13:15 pm: coincidence o.o

http://forums.xkcd.com/viewtopic.php?f=17&t=50471
Title: Re: A differential equation
Post by: kamil9876 on November 27, 2009, 11:16:25 pm: great minds think alike :P