Derivation of higher deratives of parametric equations

[thomas49th]

Homework Statement

so dy/dx = y dot / x dot

and d^2y/dx^2 = d/dx(y dot/x dot)

can someone please show me the steps to get

d^2y/dx^2 = (x dot y dot dot - y dot x dot dot)/ (x dot)^ 2

I've been trying to get to it for the last half an hour and failing

Thanks

 

[lanedance]

so you have
\frac{dy}{dx} = \frac{y'}}{x'}} = y'(x')^{-1}

now try differntiating using the product & chain rule
(i used dashes instead of dots, but same thing)

 

[thomas49th]

sorry but I don't see where you got x' = y'(x')^-1 from

Thanks
Thomas

 

[Cyosis]

The product rule and chain rule need to be used here. Use the product rule first then use the chainrule. Example \frac{d}{dx} \frac{1}{y}=\frac{d}{dy}\left(\frac{1}{y}\right)\frac{dy}{dx}.

The first step is using the product rule on:

\frac{d}{dx} \frac{\dot{y}}{\dot{x}}.

Can you post the results?

 

[Cilabitaon]

sorry but I don't see where you got x' = y'(x')^-1 from

Thanks
Thomas

That isn't what he meant: he was writing out the equation and it started on a new line.

What he meant was:

\frac{dy}{dx} = \frac{y'}{x'} = y' \cdot (x')^{-1}

and if you ask me you needn't do that step and head straight on to the quotient rule: \frac{vdu - udv}{v^{2}} or \frac{vu'-uv'}{v^{2}} where y' = u \ ; \ x' = v

Just to expand on this:

\frac{du}{dx}=y'' or u'=y''

\frac{dv}{dx}=x'' or v'=x''

both give you the same result(duh) and there is not extra calculation needed.

 

[Cilabitaon]

I hope this has solved your problems.=]