Converting intrinsic equation to cartesian

[jaderberg]

Homework Statement

Intrinsic eqn of a curve is s = 12(sin \varphi)^{2} where s is length of arc from origin and \varphi is angle of tangent at a point with x axis.

Show the cartesian eqn is (8-x)^{\frac{2}{3}}+y^{\frac{2}{3}}=4

Homework Equations

^{}

\frac{dy}{dx}=tan\varphi
\frac{dy}{ds}=sin\varphi
\frac{dx}{ds}=cos\varphi

The Attempt at a Solution

<br /> s=12(\frac{dy}{ds})^{2}

y=\frac{1}{2\sqrt{3}}\int(s^{\frac{1}{2}})ds

which comes out as:3y^{\frac{2}{3}}=s

now doing the same process for x:

s=12(1-(\frac{dx}{ds})^{2})

x=\int(1-\frac{s}{12})^{\frac{1}{2}}ds

x=-8(1-\frac{s}{12})^{\frac{3}{2}}

which comes out as:s=12+3x^{\frac{2}{3}}

so now equating the two equations for s i get
y^{\frac{2}{3}}-x^{\frac{2}{3}}=4

this obviously isn't right so where am i going wrong?!

 

[Tom George]

Seven years too late of course - still worth answering i suppose: I am no expert but I looked through and nothing you appear to have done is wrong. It occurs to me that the two results (correct answer and our answer) are very similar, one is simply shifted to the left of the other. I think you have to remember the constants of integration.

x infact equals -8(1 - s/12)^3/2 + c
so
(c - x) = 8(1 - s/12)^3/2
so
s = 12 - 3(c - x)^2/3
Equating y and x:
4 = (c - x)^2/3 + 3y^2/3(where c appears to be 8). It just so happens that the y constant of integration equals 0.

One intrinsic equation can have more than 1 cartesian equivalent.
Those are my thoughts at least - hope this helps.