Results 1 to 2 of 2

Math Help - Partial derivative chain rule

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    5

    Partial derivative chain rule

    Let z = g(x, y) and suppose that x(t) = t^2 + 3t + 2 and y(t) = e^t + sin(3t)

    Find (dz/dt)|t=0

    if (dg/dx)|(1,2) = 6
    (dg/dy)|(1,2) = -2
    (dg/dx)|(2,1) = -3
    (dg/dy)|(2,1) = 8
    (dg/dx)|(0,0) = 0
    (dg/dy)|(0,0) = -4

    Thank You.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by DavidWagner View Post
    Let z = g(x, y) and suppose that x(t) = t^2 + 3t + 2 and y(t) = e^t + sin(3t)

    Find (dz/dt)|t=0

    if (dg/dx)|(1,2) = 6
    (dg/dy)|(1,2) = -2
    (dg/dx)|(2,1) = -3
    (dg/dy)|(2,1) = 8
    (dg/dx)|(0,0) = 0
    (dg/dy)|(0,0) = -4

    Thank You.
    By the Chain rule we get

     \frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t} +\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}=\frac{\partial g}{\partial x}(2t+3)+\frac{\partial g}{\partial y}(e^t+3\cos(3t))

    Now we need to do some calculations

    x(0)=2 and y(0)=1

    so  \frac{\partial g}{\partial x}\bigg |_{(2,1)}=-3
    and
    \frac{\partial g}{\partial y} \bigg |_{(2,1)}=8

    so Finally we get

    \frac{\partial z}{\partial t}=(-3)(2(0)+3)+8(e^0+3\cos(0))=-9+32=23
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Partial Derivative - Chain Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 1st 2011, 07:41 AM
  2. partial derivative and the chain rule
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 8th 2010, 09:22 AM
  3. Chain Rule in Partial Derivative
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 31st 2010, 02:45 PM
  4. Replies: 1
    Last Post: September 25th 2009, 12:35 PM
  5. Partial Derivative and chain rule...
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 3rd 2009, 07:49 AM

Search Tags


/mathhelpforum @mathhelpforum