Parametric Derivatives: Understanding Second Derivatives of Parametric Equations

[sooyong94]

Homework Statement

Given a pair of parametric equations,
$x=f(t)$ and $y=g(t)$ ,
The first derivative is given by
$\frac{dy}{dx}=\frac{g'(t)}{f'(t)}$
and the second derivative is actually
$\frac{d}{dt}(\frac{dy}{dx})$
But why we cannot find the second derivative of a parametric equation by doing this?
$\frac{d^2 y}{dx^2}=\frac{g''(t}{f''(t)}$

Homework Equations

Parametric derivatives

The Attempt at a Solution

I know the parametric derivatives can be done by treating
$\frac{dy}{dt}$ and $\frac{dx}{dt}$ as a fraction, but why can't we do that to second derivatives?

 

[vanhees71]

Start with deriving the correct equation
$$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{g'(t)}{f'(t)}.$$
Then you should get an idea, how to get the 2nd derivative.

 

[tiny-tim]

hi sooyong94!

… and the second derivative is actually
$\frac{d}{dt}(\frac{dy}{dx})$

you mean $\frac{d}{dx}\left(\frac{dy}{dx}\right)$

But why we cannot find the second derivative of a parametric equation by doing this?
$\frac{d^2 y}{dx^2}=\frac{g''(t}{f''(t)}$

no, that's dv/du where u(t) = f'(t), v(t) = g'(t)

 

[sooyong94]

hi sooyong94!


you mean $\frac{d}{dx}\left(\frac{dy}{dx}\right)$


no, that's dv/du where u(t) = f'(t), v(t) = g'(t)

I don't get it though... I known I can use quotient rule to find the second derivative...

 

[Curious3141]

I know the parametric derivatives can be done by treating
$\frac{dy}{dt}$ and $\frac{dx}{dt}$ as a fraction, but why can't we do that to second derivatives?

Because you're not really treating them as fractions. You're using Chain Rule. World of difference.

 

[tiny-tim]

Because you're not really treating them as fractions. You're using Chain Rule. World of difference.

yes, chain rule: dy/dx = dy/dt*dt/dx = dy/dt*(1/(dx/dt)) = g'(t)/f'(t) …

nothing to do with quotients ​

 

[sooyong94]

I got it right now... So it's about chain rule, not treating them as fractions.