1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parametric Derivative Derivation

  1. Nov 20, 2005 #1
    How does one conclude that [tex] \frac{d^{2} y}{dx^{2}} = \frac{dy\'/dt}{dx/dt} [/tex]?

    Thanks
     
  2. jcsd
  3. Nov 20, 2005 #2

    cronxeh

    User Avatar
    Gold Member

  4. Nov 22, 2005 #3
    [tex]\frac{d^2 y}{dx^2} = \frac{d \dot{y'}}{d \dot{x}}[/tex]
    [tex]= \frac{dy'}{dx} = \frac{dy}{dx^2}[/tex]

    Which is clearly incorrect.
     
    Last edited: Nov 22, 2005
  5. Nov 22, 2005 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    And, you will notice that what you wrote was incorrect. What is true is that [tex]\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex] not the second derivative.
    That follows from the chain rule.
     
  6. Nov 22, 2005 #5
    I didn't write
    . I wrote [tex]\frac{d^{2}y}{dx^{2}}= \frac{\frac{dy'}{dt}}{\frac{dx}{dt}}[/tex]
     
  7. Nov 24, 2005 #6
    Yes. Oops.
    [tex]\frac{d^2 y}{dx^2} = \frac{d \dot{y'}}{d \dot{x}}[/tex]
    [tex]= \frac{dy'}{dx} = y''[/tex]
     
  8. Nov 25, 2005 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Then is y' the derivative with respect to x or t? If with respect to t, then your equation is wrong. If with respect to x, then you are back to the first derivative case.
     
  9. Nov 25, 2005 #8
    [tex]y'' = \frac{d^2 y}{dx^2}[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Parametric Derivative Derivation
Loading...