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Math Help - calc gradient question

  1. #1
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    calc gradient question

    Given f(x,y,z)=x^(2)y^(3)z^(6), in what direction is f(x,y,z) increasing the most rapidly at the point P(1,-1,1)? What is its rate of increase in that direction
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  2. #2
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    Quote Originally Posted by bobby77 View Post
    Given f(x,y,z)=x^(2)y^(3)z^(6), in what direction is f(x,y,z) increasing the most rapidly at the point P(1,-1,1)? What is its rate of increase in that direction
    <br />
\nabla f=[2\,x\,y^3\,z^6,\ 3\,x^2\,y^2\,z^6,\ 6\,x^2\,y^3\,z^5]<br />

    so at the point in question:

    <br />
\nabla f=[-2,\ 3,\ -6]<br />
,

    so the unit vector in the direction that f is increasing most rapidly is:

    \frac{\nabla f}{|\nabla f|}=[-2/7,\ 3/7,\ -6/7],

    and the rate of increase in this direction is |\nabla f|=7

    RonL
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  3. #3
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    Quote Originally Posted by bobby77 View Post
    Given f(x,y,z)=x^(2)y^(3)z^(6), in what direction is f(x,y,z) increasing the most rapidly at the point P(1,-1,1)? What is its rate of increase in that direction
    That is the gradient at the point.

    \nabla f = 2xy^3z^6\bold{i}+3x^2y^2z^6\bold{j}+6x^2y^3z^5\bol  d{k}
    Substitute, in the values of x,y,z to get the optimal vector it increases by.

    2(1)(-1)^3(1)^6\bold{i}+3(1)^2(-1)^2(1)^6\bold{j}+6(1)^2(-1)^3(1)^5\bold{k}=-2\bold{i}+3\bold{j}-6\bold{k}
    The norm of this vector is \sqrt{2^2+3^2+6^2}=7.
    Thus, the unit vector is,
    \bold{u}=-\frac{2}{7}\bold{i}+\frac{3}{7}\bold{j}-\frac{6}{7}\bold{k}
    This is the direction of maximal increase.
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