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  1. #1
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    directional derivative Help

    Consider the function .
    Find the the directional derivative of at the point in the direction given by the angle . Find the unit vector which describes the direction in which is increasing most rapidly at .
    i could get the directional derivative which is -12. but how to get the unit vector out of it?
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  2. #2
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    Quote Originally Posted by zpwnchen View Post
    i could get the directional derivative which is -12. but how to get the unit vector out of it?
    The direction of maximum rate of change is given by \nabla f evaluated at the given point. Divide this vector by it's magnitude to get a unit vector.
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    F_{u} = | \bigtriangledown F(x,y)| . |u| gives the max rate of change.

    so i find the
    <br />
\bigtriangledown F(x,y) = <6x/ \sqrt{36x^2+64y^2} ,8y/ \sqrt{36x^2+64y^2}><br />
    = <12/ \sqrt{36x^2+64y^2}, 8/ \sqrt{36x^2+64y^2}>
    and then take a magnitude of it and gives me 1. so unit vector is going to be "-12" which is not right. :{
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  4. #4
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    Quote Originally Posted by zpwnchen View Post
    F_{u} = | \bigtriangledown F(x,y)| . |u| gives the max rate of change.

    so i find the
    <br />
\bigtriangledown F(x,y) = <6x/ \sqrt{36x^2+64y^2} ,8y/ \sqrt{36x^2+64y^2}><br />
    = <12/ \sqrt{36x^2+64y^2}, 8/ \sqrt{36x^2+64y^2}>
    and then take a magnitude of it and gives me 1. so unit vector is going to be "-12" which is not right. :{
    I don't have the faintest idea what you're doing or why you're doing it.

    The given function is f(x, y) = 3x^2 + 4y^2. I have already told you that \nabla f evaluated at the given point is the direction of maximum rate of change. And I have also told you to divide that vector by it's magnitude to get a unit vector in that direction.

    I cannot see that you have done or attempted to do any of what I said.

    By the way, "-12" is a scalar not a vector.
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  5. #5
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    got it! thank you so much
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