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Thread: Parametric equ's, derivatives, velocity? Confused!

  1. #1
    dub
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    Question 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
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    Quote Originally Posted by dub View Post
    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
    $\displaystyle \frac{dy}{dx}$ is the slope of the path of travel at any position $\displaystyle (x(t),y(t))$

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


    velocity vector ...

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

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


    acceleration vector ...

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

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