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Math Help - Rate of change (partial differentiation)

  1. #1
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    Rate of change (partial differentiation)

    Greetings,

    Rate of change (partial differentiation)-scan_marts-28-2014-2-28-27-961-pm.pngI am almost done with this exercise, but I don't get a sqrt(5) in the denominator in my final answer, thus the correct answer is: 5/sqrt(5). How come I don't get a correct answer? Is the formula used incorrect?
    Last edited by kaemper; March 28th 2014 at 07:27 AM.
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  2. #2
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    Re: Rate of change (partial differentiation)

    When you calculate a directional derivative, you need to do the inner product with a unit vector. The unit vector in the direction \vec{i}+2\vec{j} is \frac{1}{\sqrt{5}}\vec{i}+\frac{2}{\sqrt{5}}\vec{j  }. So when you take the dot product, you should get \frac{4}{\sqrt{5}}. I assume your answer 5/sqrt(5) is a typo.

    - Hollywood
    Thanks from kaemper and HallsofIvy
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  3. #3
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    Re: Rate of change (partial differentiation)

    Yes, it is certainly a mistake.
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    Re: Rate of change (partial differentiation)

    You also have the gradient itself wrong. The gradient of x^2y at (-1, -1) is NOT 2\vec{i}+ \vec{j}.
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  5. #5
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    Re: Rate of change (partial differentiation)

    It's not? I calculated \nabla(x^2y) = \vec{i}\frac{\partial}{\partial{x}}(x^2y) + \vec{j}\frac{\partial}{\partial{y}}(x^2y) = 2xy\vec{i}+x^2\vec{j}, and substituting x=y=-1 gives 2\vec{i}+\vec{j}.

    What am I missing?

    - Hollywood
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    Re: Rate of change (partial differentiation)

    Quote Originally Posted by HallsofIvy View Post
    The gradient of x^2y at (-1, -1) is NOT 2\vec{i}+ \vec{j}.
    Why is it so, HallsofIvy. The two of us cannot understand that.

    Respectfully,
    Kaemper
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    Re: Rate of change (partial differentiation)

    I agree with previous (we say about unnormed vector)
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    Re: Rate of change (partial differentiation)

    Quote Originally Posted by Cartesius24 View Post
    I agree with previous (we say about unnormed vector)
    Do you agree with HallsofIvy?
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    Re: Rate of change (partial differentiation)

    With You
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  10. #10
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    Re: Rate of change (partial differentiation)

    Quote Originally Posted by hollywood View Post
    It's not? I calculated \nabla(x^2y) = \vec{i}\frac{\partial}{\partial{x}}(x^2y) + \vec{j}\frac{\partial}{\partial{y}}(x^2y) = 2xy\vec{i}+x^2\vec{j}, and substituting x=y=-1 gives 2\vec{i}+\vec{j}.

    What am I missing?

    - Hollywood
    Okay, I was wrong. For some reason I didn't realize that we were multiplying two negatives!
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