# Intrinsic coordinates.Watch

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#1
I'm on the last P5 exercise which is primarily based on conversion between cartesian and intrinsic equations and finding the radius of curvature. I can do the latter part but i seem to have just a sea of differential relations and no clear strategy for converting from cartesian to intrinsic and vica verca.

For example:
s=12(sinw)^2, s being measured from (-8,0).
(a) Find the radius of curvature at w=pi/6. [Done. Not a problem.]
(b) Show that x^(2/3) + y^(2/3) = 4 is a cartesian equation of the curve. [This being a problem]

Thanks for any help.
0
14 years ago
#2
The two equations relating s and w (which I guess you mean to denote the angle psi) to Cartesian co-ordinates are

ds^2 = dx^2 + dy^2

and

tanw = dy/dx

So if you solve for y you get y = (4-x^(2/3))^(3/2)

Find (ds/dx)^2 = 1 + (dy/dx)^2 from this.

Recalling that ds/dx is positive in this example, take the square root and I got

ds/dx = -2x^(-1/3)

Solving for s I get

s = 12 - 3x^(2/3) [recalling the curve goes through (-8,0)]

You can find 12sin^2w using the identity

sin^2w = tan^2w/(1+tan^2w)

and that should come to the same expression for s.
0
#3
Thanks for the help.

So if i have an intrinsic equation and wish to proof it is a given cartesian equation i should generally start from the cartesian side of things? Is it equally simple to do it the other way around?

So to move from cartesian to intrinsic i need to:
(i) Find s in terms of x, y or t.
(ii) Find dy/dx and set this equal to tanw.
(iii) Replace the s=f(x,y or t) with s=g(w) with a suitable substitution.

sin^2w = tan^2w/(1+tan^2w)
Where does this come from?
0
14 years ago
#4
If they give you the x and y equation I'd work from that - of course, they may not and you'll have to use the formulas the other way around.

The trig relation at the end is a standard trig identity to get sinw (which we want) in terms of tanw (which we know).
0
#5
Thanks again for the help.

Recalling that ds/dx is positive in this example, take the square root and I got

ds/dx = -2x^(-1/3)
Why is it positive, and if it is positive, why are you taking the negative root?

Edit: I also seem to be getting a negative radius of curvature for the ellipse: x^2/a^2 + y^2/b^2 = 1 at a/rt2, b/rt2. The book gives (a^2 + b^2)^1.5 / 2abrt2, where as i get the same expression but with a negative on the bottom.
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