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Math Help - Directional Derivative

  1. #1
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    Directional Derivative

    Part (c), (d) and (e) of the following problem:





    My attempt (please correct me where I am wrong)

    (c) \hat{u} = \| u \| = \sqrt{1^2 + 1^2} = \sqrt{2} so

    \hat{u} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}

    From part (a) I know that f_x(0,0)=f_y(0,0)=0. Therefore:

    \nabla f(0,0)=f_x(0,0) i + f_y(0,0)j = 0

    Therefore D_u f(0,0)= \nabla f(0,0). \hat{u} = 0

    (d) Letting y=x we will have f(x)=\sqrt[3]{2x^3}. Therefore

    f'(x)= \frac{2^{(1/3)}}{(x^3)^(2/3) x^2}

    \nabla f = f(0)i = 0

    Therefore D_u f(0,0)= \nabla f(0). \hat{u} = 0

    I got 0 again!

    (e) I think maybe the method in part (d) is more efficient but neither work here because in part (b) I showed that the function is not differentiable at (0,0), and if a function is not differentiable at a point then the directional derivative does not exist there for all directions \hat{u}.

    Is this right? I need to submit this tomorrow so I would appreciate any help or suggestions.
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  2. #2
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    Quote Originally Posted by demode View Post
    Part (c), (d) and (e) of the following problem:





    My attempt (please correct me where I am wrong)

    (c) \hat{u} = \| u \| = \sqrt{1^2 + 1^2} = \sqrt{2}
    so

    \hat{u} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}

    From part (a) I know that f_x(0,0)=f_y(0,0)=0. Therefore:

    \nabla f(0,0)=f_x(0,0) i + f_y(0,0)j = 0

    Therefore D_u f(0,0)= \nabla f(0,0). \hat{u} = 0

    (d) Letting y=x we will have f(x)=\sqrt[3]{2x^3}. Therefore

    f'(x)= \frac{2^{(1/3)}}{(x^3)^(2/3) x^2}

    \nabla f = f(0)i = 0

    Therefore D_u f(0,0)= \nabla f(0). \hat{u} = 0
    Then try again! \sqrt[3]{2x^3}= (\sqrt[3]{2})x and its derivative, at x= 0, is NOT 0. You seem to be under the impression that if the numerator of a fraction is 0, then the fraction is equal to 0. That's not true if the denominator is also 0!

    I got 0 again!

    (e) I think maybe the method in part (d) is more efficient but neither work here because in part (b) I showed that the function is not differentiable at (0,0), and if a function is not differentiable at a point then the directional derivative does not exist there for all directions \hat{u}.

    Is this right? I need to submit this tomorrow so I would appreciate any help or suggestions.
    No, if a function is not differentiable at a point then the directional derivative may not exist for some or all directions.
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