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Math Help - Check my answer for directional derivative?

  1. #1
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    Question Check my answer for directional derivative?

    Can someone check my answer for this question:
    Find the directional derivative of f(x,y) = x/(x+y) at P(1,0) in the direction of Q (-1,-1).
    Ok so the answer I am getting is 1/sqrt(2) but the answer at the back of my book is 1/sqrt(5). I have rechecked all my workings but still can't get the solution which matches the answer in the book. Can someone tell me if the answer I got is right...thanks
    PS: I am not really sure about the procedure that is needed to be followed for this question because there isn't any example in the book which deals with directional derivatives in the direction of a particular point. Help please.
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  2. #2
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    Quote Originally Posted by skelly83 View Post
    Can someone check my answer for this question:
    Find the directional derivative of f(x,y) = x/(x+y) at P(1,0) in the direction of Q (-1,-1).
    Ok so the answer I am getting is 1/sqrt(2) but the answer at the back of my book is 1/sqrt(5). I have rechecked all my workings but still can't get the solution which matches the answer in the book. Can someone tell me if the answer I got is right...thanks
    PS: I am not really sure about the procedure that is needed to be followed for this question because there isn't any example in the book which deals with directional derivatives in the direction of a particular point. Help please.
    We are moving along the vector \bold{v} = (-2,-1) but we need to make it a unit vector \left( \frac{-2}{\sqrt{5}}, \frac{-1}{\sqrt{5}}\right). Now to find the directional derivative use the fact that D_{\bold{v}}f(1,0) = \nabla f(1,0) \cdot \left( \frac{-2}{\sqrt{5}}, \frac{-1}{\sqrt{5}} \right).
    Can you finish that?
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  3. #3
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    Thumbs up

    Yup I got the answer. Thanks for your help! I had the direction vector wrong. Our lecturer didn't really clear up how the direction vector actually works and I still am not very clear about it . I guess that's where the problem is but thanks for your help
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