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Math Help - Directional Derivative in direction of max change

  1. #1
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    Directional Derivative in direction of max change

    I need help! I am suppose to find the directional derivative of function xy + z^2 at point (2,3,4)

    I know that this means in the direction of the gradient of the function.

    So., using the partial derivatives of xy + z^2 which are y , x, 2z I have y ivectors + x j vectors + 2z k vectors.

    When I come up with my vector in the greatest direction and then the unit vector for this and multiple it times the 3 i + 2 j + 8 k I get a value of 1!!!! I must be doing something wrong but I do not know what!!!! Frostking in a blizzzzzard of confusion.
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  2. #2
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    find the directional derivative of function xy + z^2 at point (2,3,4)
    What direction are you headed?. That is not stated.

    Let's say it is headed in the direction of the vector a=2i+j-2k

    f(x,y,z)=f_{x}=y

    f_{y}=x

    f_{z}=2z

    {\nabla}f(x,y,z)=yi+xj+2zk

    f(2,3,4)=3i+2j+8k

    A unit vector in the direction of a is:

    \frac{u}{||a||}=\frac{2i+j-2k}{\sqrt{9}}=

    =\frac{2}{3}i+\frac{1}{3}j-\frac{2}{3}k

    Therefore, D_{u}(2,3,4)={\nabla}f(2,3,4)\cdot u=3(\frac{2}{3})+2(\frac{1}{3})-8(\frac{2}{3})=\frac{-8}{3}
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  3. #3
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    Directionn derivative in direction of max change

    The direction is that of the maximum change so I thought this was in the direction of the gradient of the function and when I used this I got 1 for an answer. Does this clarify? Thanks for helping!!!!
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  4. #4
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    Quote Originally Posted by Frostking View Post
    The direction is that of the maximum change so I thought this was in the direction of the gradient of the function and when I used this I got 1 for an answer. Does this clarify? Thanks for helping!!!!
    Yes, the direction of maximum change is the direction of the gradient. The derivative in the direction of the gradient is just the length of the gradient vector.

    Since f(x,y,z)= xy+ z^2 \nabla f= y\vec{i}+ x\vec{j}+ 2z\vec{k}. At (2, 3, 4) that is 3\vec{i}+ 2\vec{j}+ 8\vec{k} and its length is \sqrt{9+ 4+ 64}= \sqrt{77}


    If you take the dot product of any vector with a unit vector in the same direction, you get the length of that vector, not 1. I can only conclude that you had done some calculation wrong.
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