Results 1 to 7 of 7

Math Help - Partial Derivatives

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    24

    Partial Derivatives

    Let u = x + y and v = x -y so that x = (u + v)/2 and y = (u-v)/2
    Given a twice continuously differentiable functions f(x,y) set
    g(u,v) = f( (u+v)/2, (u-v)/2) so that f(x,y) = g( x+y, x-y)
    1) Relate ∂f/∂x, ∂f/∂y to ∂g/∂u, ∂g/∂v
    2) Relate ∂^2f/∂x^2, ∂^2f/∂y^2 to ∂^2g/∂u^2 , ∂^2g/∂v^2 and ∂^2g/∂u∂v
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Try taking the partial derivatives and see if you notice anything about them.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2010
    Posts
    24
    how do I take those partials if they aren't defined explicitly?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    You can apply the limit definition of partial derivatives.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle \frac{\partial f}{\partial x}(x,y)=\lim_{h\to 0}\frac{f((x+h),y)-f(x,y)}{h}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2010
    Posts
    24
    well the first part I said, using the chain rule
    ∂g/∂u = ∂g/∂x ∂x/∂u + ∂g/∂y ∂y/∂u = (1/2) ∂f/∂x + (1/2) ∂f/∂y
    ∂g/∂v = ∂g/∂x ∂x/∂v + ∂g/∂y ∂y/∂∂v = (1/2) ∂f/∂x - (1/2) ∂f/∂y because f(x,y) = g(u,v)

    for the second part
    ∂^2/∂x^2 = ∂/∂x (∂g/∂u ∂u/∂x + ∂g/∂v ∂v/∂x)
    where should I go from here?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Playthious View Post
    Let u = x + y and v = x -y so that x = (u + v)/2 and y = (u-v)/2
    Given a twice continuously differentiable functions f(x,y) set
    g(u,v) = f( (u+v)/2, (u-v)/2) so that f(x,y) = g( x+y, x-y)
    1) Relate ∂f/∂x, ∂f/∂y to ∂g/∂u, ∂g/∂v
    2) Relate ∂^2f/∂x^2, ∂^2f/∂y^2 to ∂^2g/∂u^2 , ∂^2g/∂v^2 and ∂^2g/∂u∂v
    \displaystyle f(x,y)=g(x+y,x-y)=g(u,v)=f\left(\frac{u+v}{2},\frac{u-v}{2}\right)

    \displaystyle \frac{\partial f}{\partial x}=\frac{\partial g}{\partial u}

    \displaystyle \frac{\partial f}{\partial y}=\frac{\partial g}{\partial v}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Partial Derivatives
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 19th 2011, 02:46 PM
  2. Partial Derivatives
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 24th 2011, 01:33 PM
  3. Partial derivatives
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 10th 2009, 07:38 AM
  4. first partial derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 17th 2009, 02:18 PM
  5. Partial derivatives
    Posted in the Calculus Forum
    Replies: 7
    Last Post: March 6th 2007, 07:22 PM

Search Tags


/mathhelpforum @mathhelpforum