Results 1 to 4 of 4

Math Help - Vector analysis: Gradient of f(M_0)...

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Vector analysis: Gradient of f(M_0)...

    I will appreciate if someone could solve and explain me step by step the following problem:

    Let f:\mathbb{R}^3\to\mathbb{R} be a differentiable in every point and satisfies the following:

    1. f(x,y,2x^2+y^2)=3x-5y for all (x,y) .

    2. The directional derivative of f at point M_0(1,2,6) at direction a=(\frac{1}{3},\frac{2}{3},\frac{2}{3}) equals to 1.

    Compute \nabla f(M_0) .



    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 Also sprach Zarathustra View Post
    I will appreciate if someone could solve and explain me step by step the following problem:

    Let f:\mathbb{R}^3\to\mathbb{R} be a differentiable in every point and satisfies the following:

    1. f(x,y,2x^2+y^2)=3x-5y for all (x,y) .

    2. The directional derivative of f at point M_0(1,2,6) at direction a=(\frac{1}{3},\frac{2}{3},\frac{2}{3}) equals to 1.

    Compute \nabla f(M_0) .



    Thank you.
    The directional derivative in the direction u is given by

    D_u(f)=\nabla f \cdot \frac{\mathbf{u}}{||\mathbf{u}||} using this gives

    \nabla f(1,2,6) \cdot (\mathbf{i}+2\mathbf{j}+2\mathbf{k})=3 =f_{x}(1,2,6)+2f_{y}(1,2,6)+2f_{z}(1,2,6)

    This tells us that f_{z}(1,2,6)=\frac{1}{2}\left[ 3-f_{x}(1,2,6)-2f_{y}(1,2,6) \right]

    Now we have that f(x,y,z(x,y))=f(x,y,2x^2+y^2) \implies z(x,y)=2x^2+y^2

    \nabla f=f_x\mathbf{i}+f_y\mathbf{j}+\left(\frac{\partial f}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial y } \right)\mathbf{k}

    \nabla f=f_x\mathbf{i}+f_y\mathbf{j}+\left( 4xf_z(x,y,z)+2yf_z(x,y,z)  \right)\mathbf{k}

    but we know what f(x,y) is and can take the needed partial derivatives to finish.
    Last edited by TheEmptySet; June 6th 2011 at 12:58 PM. Reason: magnitude of vector incorrect.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by TheEmptySet View Post
    The directional derivative in the direction u is given by

    D_u(f)=\nabla t \cdot \frac{\mathbf{u}}{||\mathbf{u}||} using this gives

    \nabla f(1,2,6) \cdot (\mathbf{i}+2\mathbf{j}+2\mathbf{k})=9 =f_{x}(1,2,6)+2f_{y}(1,2,6)+2f_{z}(1,2,6)

    This tells us that f_{z}(1,2,6)=\frac{1}{2}\left[ 9-f_{x}(1,2,6)-2f_{y}(1,2,6) \right]

    Now we have that f(x,y,z(x,y))=f(x,y,2x^2+y^2) \implies z(x,y)=2x^2+y^2

    \nabla f=f_x\mathbf{i}+f_y\mathbf{j}+\left(\frac{\partial f}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial y } \right)\mathbf{k}

    \nabla f=f_x\mathbf{i}+f_y\mathbf{j}+\left( 4xf_z(x,y,z)+2yf_z(x,y,z)  \right)\mathbf{k}

    but we know what f(x,y) is and can take the needed partial derivatives to finish.
    First, thank you very much! Now, I'm not understand how to arrive to the answer 7i-j-k (the answer appears in a book).

    Thanks again!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Yes I can see why you were confused I wrote the incorrect magnitude for the vector. It should be

    3 =f_{x}(1,2,6)+2f_{y}(1,2,6)+2f_{z}(1,2,6)

    Now we just need to compute two other partial derivatives. Using the chain rule we get that

    \frac{\partial }{\partial x}f(x,y,z(x,y))=f_x+4xf_z=3

    \frac{\partial }{\partial y}f(x,y,z(x,y))=f_y+2yf_z=-5

    3 =f_{x}(1,2,6)+2f_{y}(1,2,6)+2f_{z}(1,2,6)

    This gives a system of 3 equations in three unknowns for the partial derivatives of the function at the point (1,2,6).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Gradient Vector for the Surface f(x,y,z)=0
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 3rd 2009, 12:54 PM
  2. Gradient Vector max/min
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 2nd 2009, 03:54 AM
  3. Gradient vector help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 6th 2009, 04:04 PM
  4. Gradient Vector
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 20th 2009, 05:36 PM
  5. Gradient Vector Field?
    Posted in the Calculus Forum
    Replies: 15
    Last Post: February 23rd 2008, 05:43 PM

Search Tags


/mathhelpforum @mathhelpforum