I'm given that x=cos3θ and that y=sin3θ

if (d2y/dx2)=[(dy/dθ)/(dx/dθ)]/[dx/dθ] is right, wouldn´t the second derivative of the parametric be:

1/3c3θ ??

I got this by using dy/dθ=3sin2θ,

and dx/dθ=-3cos2θsinθ

any idea what's wrong? or is it right?

## Answers and Replies

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if (d2y/dx2)=[(dy/dθ)/(dx/dθ)]/[dx/dθ] is right
I'm pretty sure this isn't correct. you need to wrap a (d/dθ) around the first term ([(dy/dθ)/(dx/dθ)]), which gives you
((d/dθ)[(dy/dθ)/(dx/dθ)])/[dx/dθ]
=((d/dθ)[(dy/dθ)/(dx/dθ)])/[1/(dx/dθ)]
=[(dx/dθ)*(d^2y/dθ^2)-(d^2x/dθ^2)(dy/dθ)]/[(d^3x/dθ^3)]
or (x' y'' - y' x'')/y'''
where ' means derivative wrt θ

$$\frac {d^2y} {dx^2} = \frac {d \frac {dy}{dx} } {dx} = \frac { d\frac {dy}{dx} } {d\theta} \frac {d\theta} {dx} = \frac { d (\frac {dy}{d\theta} \frac {d\theta} {dx})} {d\theta} \frac {d\theta} {dx} = \frac { d [\frac {dy}{d\theta} / \frac {dx} {d\theta}]} {d\theta} / \frac {dx} {d\theta} = \frac { d \frac {y'} {x'}} {d\theta} / x' = \frac {y''x' - y'x''}{x'^3}$$