- #1

jaderberg

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## Homework Statement

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

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

## Homework Equations

^{}[tex]\frac{dy}{dx}=tan\varphi[/tex]

[tex]\frac{dy}{ds}=sin\varphi[/tex]

[tex]\frac{dx}{ds}=cos\varphi[/tex]

## The Attempt at a Solution

[tex]

s=12(\frac{dy}{ds})^{2}[/tex]

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

which comes out as:[tex] 3y^{\frac{2}{3}}=s[/tex]

now doing the same process for x:

[tex]s=12(1-(\frac{dx}{ds})^{2})[/tex]

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

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

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

so now equating the two equations for s i get

[tex] y^{\frac{2}{3}}-x^{\frac{2}{3}}=4[/tex]

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

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