Results 1 to 2 of 2

Math Help - Parametric equ's, derivatives, velocity? Confused!

  1. #1
    dub
    dub is offline
    Newbie
    Joined
    Jul 2009
    Posts
    1

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,880
    Thanks
    667
    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
    \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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Very confused...(derivatives)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 13th 2011, 09:17 AM
  2. Calculus Derivatives (confused)
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 18th 2009, 04:31 PM
  3. velocity of parametric function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 12th 2009, 09:51 AM
  4. Parametric question...confused on trig =/
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 2nd 2009, 08:40 AM
  5. Parametric/velocity/distance Q
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 22nd 2007, 05:12 PM

Search Tags


/mathhelpforum @mathhelpforum