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P5 Intrinsic Coordinates - cartesian equations

This maths work is really getting me down...
Can anyone give me any help on how to change equations in intrinsic coordinates to cartesian equations? I read an earlier post on this, but it wasn't clear about it in general.
Say you have the equation s=acos^2w (where w is psi) and you have to make it into a cartesian equation. How would you do this?
:confused:

Reply 1

BCHL85

Just remember dx/ds = cosw, dy/ds = sinw.
First you differentiate the equation s = f(w).
Then you got dx = cosw.ds = cosw.f'(w)
dy = sinw.ds = sinw.f'(w).
It now becomes a P3 questions :smile:

Reply 2

Da Bachtopus

To find the Cartesian equation of the curve, you need first to find equations for x and y in Psi, i.e. you produce parametric equations.

This means finding and then integrating dy/d(psi) and dx/d(psi), easily done using the Chain Rule (see attachment).

These questions look daunting, but are really just a case of applying the above routine.

psi.jpg8.4KB

Reply 3

lesser weevil

So a general method for converting from a cartesian equation to an intrinsic equation of a function, and vice versa, would be what, exactly?
(Sorry, I'm a bit dense on this subject).
:confused:

Reply 4

m:)ckel

For Intrinsic -> Cartesian:

dy/ds = sinw

Using chain rule: dy/dw = (dy/ds)(ds/dw)

You know dy/ds (above)
You can differentiate s = f(w) [which is the equation given], to get ds/dw.

Now you have dy/dw = g(w) [some function in w]
Integrate with respect to w, to get y = h(w) + c [some function in w, plus a constant]

Then you need to find out what happens to y when w=0 (or similar), to find the constant.

Repeat the process with dx/ds = cosw, and dx/dw = (dx/ds)(ds/dw).

You now have x and y in terms of w. So eliminate w to get the cartesian form.

For Cartesian -> Intrinsic:

You have y = f(x), so differentiate to find dy/dx

Now use s = ∫[1+(dy/dx)²]^1/2dx, from 0 to x.

Sub in dy/dx, and proceed with the integration, to find s = g(x)

Remember that dy/dx = tanw.
Try to manipulate s = g(x) to incorporate dy/dx, and thus incorporate tanw.
[See Example 10, page 104 of the Heinemann book- that's a rather simple one. I'm not too sure if they get any more complicated when doing cartesian -> intrinsic]

Now you have s = h(w)

Reply 5

Feria

Generally, you want to get all the x's and y'x in terms of s's and psi's

using:

dx/ds = cos(psi), dy/ds = sin(psi). dy/dx = tan(psi).

and the Formula for arc length:

s = INT(RT(1+(dy/dx)^2 ) ) dx

to get it into the form s=f(psi)

and the formula for curvature: which is defined as ds/d(psi)

Reply 6

Font

intrinsic coordinates - aghhhhh; much better than geography :biggrin:

- quite fun too :biggrin:

tehre are a few relly nice intrinsic coorinate quetions in the P5 heinnman textbook - anyone worked thrrough them?

Pk

Reply 7

m:)ckel

Not really. Just been trying past papers, mainly. The intrinsic ones in there are pretty doable, once you've got the grip of them...

(I think that's the same for all the topics, i.e. book questions are more challenging)

Reply 8

Font

mockel

Not really. Just been trying past papers, mainly. The intrinsic ones in there are pretty doable, once you've got the grip of them...

only thing i find dodgy in P5 are come of the coordniate geogetry ones - they can be really nasty at times - everything else is piss; and they might give you teh odd dodgy integral if they're feeling nasty!

pk,

the past papers i've done so far (could only find 3!), i've lost a few marks in the coordinate systems bit, and a dumb mistake in an integral i think.

do they have jan P5?'s if so could i have a copy - because i've got no moreP5's to do - except solomans, but looking for actual papers :smile:

Reply 9

m:)ckel

I'm pretty sure they don't have Jan P5's (well, I don't have them anyway).

Yeah, coorinate ones where they make you find the locus of something, are the toughest, I think.
For the trickier ones, I think there's pretty much a set way to approach them, though.