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Math Help - rate of change

  1. #1
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    rate of change

    What is the rate of change of the function f(x,y)= ln( square root( x^2 + y^2)) at (0, square root(e)) in direction towards the origin?
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  2. #2
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    Quote Originally Posted by chaddy View Post
    What is the rate of change of the function f(x,y)= ln( square root( x^2 + y^2)) at (0, square root(e)) in direction towards the origin?
    Directional derivative:

    \nabla f \cdot \hat{l} where \hat{l} is a unit vector in the direction from (0, \sqrt{e}) to (0, 0).

    Evaluate this directional derivative at (0, \sqrt{e}).
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  3. #3
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    The unit vector would be (0,1), but I have not been able to find the <br />
\nabla f<br />
    I thought that I would just plug in <br />
(0, \sqrt{e})<br />
and (0,0) into the equation, but that doesn't work because you can't when I plug in (0,0) I get ln(0) which isn't possible.
    Do you know what I am doing wrong?
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  4. #4
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    Quote Originally Posted by chaddy View Post
    The unit vector would be (0,1), Mr F says: Actually it's (0, -1).

    but I have not been able to find the <br />
\nabla f<br />
    I thought that I would just plug in <br />
(0, \sqrt{e})<br />
and (0,0) into the equation, but that doesn't work because you can't when I plug in (0,0) I get ln(0) which isn't possible.
    Do you know what I am doing wrong?
    f = \ln \sqrt{x^2 + y^2} = \ln (x^2 + y^2)^{1/2} = \frac{1}{2} \ln (x^2 + y^2).

    Therefore:

    \frac{\partial f}{\partial x} = \frac{x}{x^2 + y^2}.

    \frac{\partial f}{\partial y} = \frac{y}{x^2 + y^2}.

    Therefore:

    \nabla f = \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j = \frac{1}{x^2 + y^2} \, (x i + y j).

    Evaluate at (0, \sqrt{e}).
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  5. #5
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    So it would wind up being (0, yj/(x^2+y^2)) but what does the j represent? Would the j = sq root of e?
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  6. #6
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    j is the unit vector in the y direction.
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