# Parametric equ's, derivatives, velocity? Confused!

• Jul 8th 2009, 12:18 PM
dub
Parametric equ's, derivatives, velocity? Confused!
Im having a little trouble understanding what exactly the derivative of parametric equations gives you, say where x and Y are positions and both a function of time.

I know that regular equations where position is a function of time has a first derivative that gives velocity of position/time. second derivative gives position/time^2.

What exactly does dy/dx give you for parametric? and (d^2y)/(dx^2)?

And how would you go about finding the instantaneous velocity and acceleration of position versus time?

If you can help, thank you , seriously.

Dub
• Jul 8th 2009, 01:37 PM
skeeter
Quote:

Originally Posted by dub
Im having a little trouble understanding what exactly the derivative of parametric equations gives you, say where x and Y are positions and both a function of time.

I know that regular equations where position is a function of time has a first derivative that gives velocity of position/time. second derivative gives position/time^2.

What exactly does dy/dx give you for parametric? and (d^2y)/(dx^2)?

And how would you go about finding the instantaneous velocity and acceleration of position versus time?

If you can help, thank you , seriously.

Dub

$\frac{dy}{dx}$ is the slope of the path of travel at any position $(x(t),y(t))$

$\frac{d^2y}{dx^2}$ is a measure of the concavity of the path of travel at any position $(x(t),y(t))$

velocity vector ...

magnitude (speed) = $\sqrt{x'(t)^2 + y'(t)^2}$

direction, $\theta = \arctan\left(\frac{y'(t)}{x'(t)}\right)$ ... note that you have to pay attention to the signs of $x'(t)$ and $y'(t)$ to determine the correct quadrant for $\theta$

acceleration vector ...

magnitude = $\sqrt{x''(t)^2 + y''(t)^2}$

direction, $\theta = \arctan\left(\frac{y''(t)}{x''(t)}\right)$ ... same note regarding quadrants as for velocity.