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Math Help - Directional Derivative and Gradient Help

  1. #1
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    Directional Derivative and Gradient Help

    I am having some difficulty with the following problems:
    1. For the function
    f(x,y)=\left\{\begin{array}{cc}\frac{x|y|}{\sqrt(x  ^2+y^2)},&\mbox{ if }<br />
(x,y) != (0,0)\\0, & \mbox{ if } (x,y) = (0,0)\end{array}\right.

    use definition  D_vf(a) = \lim_{h \to 0}\frac{f(a + hv) - f(a)}{h} to determine for which unit vectors v = vi + wj the direction derivative  D_vf(0,0) exists.

    I am able to get the function down to the following:
    \lim_{h \to 0}(\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)} - 0)/h

    Simplifying I get:
    \lim_{h \to 0}(\frac{h^2v|w|}{\sqrt(v^2+w^2)}
    This is where I get lost. The answer is supposed to be v|w|. The limit I have goes to 0.
    ------------------------
    Problem 2
    The depth of a lake is given by f(x,y) = 400 - 3x^2y^2 find the direction a swimmer should swim so that there is no change in the depth. The swimmer begins at (1,-2).

    This is part 2 of the problem. Part one has me find the direction the depth increases most rapidly and I got \frac{1}{\sqrt5}(-2i + j)

    To be honest I am not quite sure where to begin. Would I set the gradient equal to zero at that point and go from there?

    Thanks for any and all help.
    Last edited by Alterah; October 11th 2009 at 11:28 AM.
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  2. #2
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    Quote Originally Posted by Alterah View Post
    I am having some difficulty with the following problems:
    1. For the function
    f(x,y)=\left\{\begin{array}{cc}\frac{x|y|}{\sqrt(x  ^2+y^2)},&\mbox{ if }<br />
(x,y) != (0,0)\\0, & \mbox{ if } (x,y) = (0,0)\end{array}\right.

    use definition  D_vf(a) = \lim_{h \to 0}\frac{f(a + hv) - f(a)}{h} to determine for which unit vectors v = vi + wj the direction derivative [tex] D_vf(0,0) exists.

    I am able to get the function down to the following:
    \lim_{h \to 0}(\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)} - 0)/h

    Simplifying I get:
    \lim_{h \to 0}\frac{h^2v|w|}{\sqrt(v^2+w^2)}
    This is where I get lost. The answer is supposed to be v|w|. The limit I have goes to 0.
    ------------------------
    Problem 2
    The depth of a lake is given by f(x,y) = 400 - 3x^2y^2 find the direction a swimmer should swim so that there is no change in the depth. The swimmer begins at (1,-2).

    This is part 2 of the problem. Part one has me find the direction the depth increases most rapidly and I got \frac{1}{\sqrt5}(-2i + j)

    To be honest I am not quite sure where to begin. Would I set the gradient equal to zero at that point and go from there?

    Thanks for any and all help.
    In the first part, you dropped an h so you should have

    \lim_{h \to 0}\frac{h^2v|w|}{h^2\sqrt{v^2+w^2}} = v|w| since u^2+v^2 = 1 (it's a unit vector)

    Second part. \nabla f = <-6x y^2, - 6x^2 y> which at your point is \frac{<-2,1>}{\sqrt{5}}. If the direction you move in is {\bf u} = <u,v>, then D_u f = \nabla f \cdot {\bf u} = \frac{<-2,1>}{\sqrt{5}} \cdot <u,v> and if you want this to be zero then -2u + v = 0. So pick a u and v that satisfies this then make it a unit vector.
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  3. #3
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    For the first problem, in the first problem on this part:

    \lim_{h \to 0}(\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)} - 0)/h<br />

    The square root symbol is over the entire line. So, if we factor out  h^2 then we get \sqrt(h^2)\sqrt(v^2+w^2) in that line. Then shouldn't \sqrt(h^2) = |h|? We can then move the h in the very bottom up to the top. I suppose I am not seeing where I am losing an h
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  4. #4
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    Quote Originally Posted by Alterah View Post
    For the first problem, in the first problem on this part:

    \lim_{h \to 0}(\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)} - 0)/\color{red}{h}<br />

    The square root symbol is over the entire line. So, if we factor out  h^2 then we get \sqrt(h^2)\sqrt(v^2+w^2) in that line. Then shouldn't \sqrt(h^2) = |h|? We can then move the h in the very bottom up to the top. I suppose I am not seeing where I am losing an h
    See above in red.
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  5. #5
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    Quote Originally Posted by Danny View Post
    See above in red.
    I suppose I am figuring that I didn't lose that h because we can move it to the numerator. Below is all the steps I am going through.

    \lim_{h \to 0}(\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)} - 0)/h

    is the same as:
    \lim_{h \to 0}\frac{\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)}- 0} {h}


     \lim_{h \to 0} \frac{\frac{hv|h||w|}{|h|\sqrt{v^2+w^2}}}{h}

    Becomes:

     \lim_{h \to 0} \frac{h^2v|h||w|}{|h|\sqrt{v^2+w^2}}

     \lim_{h \to 0} \frac{h^2v|w|}{\sqrt{v^2+w^2}}

    But, I am obviously forgetting something or doing something wrong.
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  6. #6
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    Quote Originally Posted by Alterah View Post
    I suppose I am figuring that I didn't lose that h because we can move it to the numerator. Below is all the steps I am going through.

    \lim_{h \to 0}(\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)} - 0)/h

    is the same as:
    \lim_{h \to 0}\frac{\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)}- 0} {h}


     \lim_{h \to 0} \frac{\frac{hv|h||w|}{|h|\sqrt{v^2+w^2}}}{h}

    Becomes: It's this step

     \lim_{h \to 0} \frac{h^2v|h||w|}{|h|\sqrt{v^2+w^2}}

     \lim_{h \to 0} \frac{h^2v|w|}{\sqrt{v^2+w^2}}

    But, I am obviously forgetting something or doing something wrong.
    Above in red. Note that \frac{\frac{1}{a}}{b} = \frac{1}{ab}.
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  7. #7
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    Quote Originally Posted by Danny View Post
    Above in red. Note that \frac{\frac{1}{a}}{b} = \frac{1}{ab}.
    Thanks. Yeah, I am finally seeing where I went wrong.
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