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Math Help - Directional Derivative

  1. #1
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    Directional Derivative

    Ok, so I kept getting an Latex Error:Unknown Error when I tried to enter my stuff in LaTex. Then I scanned my handwritten work and tried to attach it...it just timed out on me everytime. Sorry this is so primitive, but I just typed everything out. I'm looking for someone to check my work and help me with the last part of #20. Here are the two problems:

    19. Suppose that z=e^{xy+x-y}. How fast is z changing when we move away from the origin toward (2,1)?

    20. In problem 19 in what direction should we move away from the origin for z to change most rapidly? What is the maximum rate of change? In what directions is the derivative zero at the origin?

    Here is my work:

    19. z=e^{xy+x-y}
    P=(0,0)
    v=<2,1>
    u={1/sqrt(5)}<2,1>
    DZ/DX=(y+1)*e^{xy+x-y}
    DZ/DY=(x-1)e^{xy+x-y}
    DZ/DX(0,0)=1
    DZ/DY(0,0)=-1
    Gradient of z(0,0)=<1,-1>
    Directional Derivative of z at (0,0)=<1,-1> . (1/sqrt{5})<2,1>=(1/sqrt{5})
    Answer to 19: (1/sqrt{5})

    20. The gradient of z(0,0)=<1,-1> is the direction we should move away from (0,0) for z to change most rapidly.

    The maximum rate is sqrt{2}

    -I'm stuck on the last part of question 20: "In what directions is the derivative zero at the origin?" Can someone help me with this part? Thanks.
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  2. #2
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    Re: Directional Derivative

    I finally clued into the fact that Latex is only working with [tex] tags, so I retyped everything to make it easier.

    19. z=e^{xy+x-y}

    P: (0,0)
    \vec{v}=<2,1>

    \vec{u}=\frac{1}{\sqrt{5}}<2,1>

    \frac{\delta z}{\delta x}=(y+1)*e^{xy+x-y}=1

    \frac{\delta z}{\delta y}=(x-1)*e^{xy+x-y}=-1

    D_{\vec{u}}f(0,0)=<1,-1> \cdot\ \frac{1}{\sqrt{5}}<2,1>

    D_{\vec{u}}f(0,0)=\frac{1}{\sqrt{5}}

    20.

    <1,-1> is the direction we should move away from the origin for z to change most rapidly.

    The max rate of change is \sqrt{2}

    I still have yet to figure out how to solve the last question in problem #20.

    Any help with where I'm stumped or in just reviewing the work I have completed so far would be appreciated. Thanks.
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  3. #3
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    Re: Directional Derivative

    The "directional derivative" in the direction of unit vector, v, is the dot product of grad f and v. In particular, the derivative will be 0 is that dot product is 0 which means v must be perpendicular to grad f. What vectors are perpendicular to <1, -1>?
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  4. #4
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    Re: Directional Derivative

    <-1,-1> & <1,1>
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