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

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Vector analysis: Gradient of f(M_0)...

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

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

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

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

    Compute $\displaystyle \nabla f(M_0) $.



    Thank you.
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    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 $\displaystyle f:\mathbb{R}^3\to\mathbb{R}$ be a differentiable in every point and satisfies the following:

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

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

    Compute $\displaystyle \nabla f(M_0) $.



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

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

    $\displaystyle \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 $\displaystyle f_{z}(1,2,6)=\frac{1}{2}\left[ 3-f_{x}(1,2,6)-2f_{y}(1,2,6) \right]$

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

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

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

    but we know what $\displaystyle f(x,y)$ is and can take the needed partial derivatives to finish.
    Last edited by TheEmptySet; Jun 6th 2011 at 12:58 PM. Reason: magnitude of vector incorrect.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    The directional derivative in the direction u is given by

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

    $\displaystyle \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 $\displaystyle f_{z}(1,2,6)=\frac{1}{2}\left[ 9-f_{x}(1,2,6)-2f_{y}(1,2,6) \right]$

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

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

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

    but we know what $\displaystyle f(x,y)$ 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!
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  4. #4
    Behold, the power of SARDINES!
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    Yes I can see why you were confused I wrote the incorrect magnitude for the vector. It should be

    $\displaystyle 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

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

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

    $\displaystyle 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).
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