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Math Help - calculte growth is some direction question..

  1. #1
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    calculte growth is some direction question..

    i know that the maximal growth is the gradiend of a function.
    the formula is:
    <br />
\triangledown f=f_x'\vec{i}+f_y'\vec{j}+f_z'\vec{k}<br />

    so when i am given a function
    f(x,y,2x^2+y^2)=3x-5y
    i am was told that it growth on point M(1,2,6) in direction \hat{a}=(\frac{1}{3},\frac{2}{3},\frac{2}{3}) is 1
    what is the gradient of f (maximal growth).

    i tried to get it like this:
    first of all i need a function which looks like this f(x,y,z)=...
    in order to find the gradient
    i dont know how to do the gradient of a function which i was given.

    if i substiute the point
    f(1,2,6)=3-10=7

    i know i should write
    <br />
(grad f(1,2,6)\dot \hat{a}=1<br />

    ????
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  2. #2
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    f(x,y,2x^2+y^2)=3x-5y
    i dont know how to do a erivative by x
    by y
    if it were
    f(x,y,z)=3x-5y then
    f'x=3 f'y=-5 f'z=0
    how the f(x,y,2x^2+y^2)=3x-5y
    changes??
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  3. #3
    ynj
    ynj is offline
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    Quote Originally Posted by transgalactic View Post
    f(x,y,2x^2+y^2)=3x-5y
    i dont know how to do a erivative by x
    by y
    if it were
    f(x,y,z)=3x-5y then
    f'x=3 f'y=-5 f'z=0
    how the f(x,y,2x^2+y^2)=3x-5y
    changes??
    you may use the Jacobian Matrix
    (fx,fy)=
    (fx,fy,fz)*
    (xx,xy
    yx,yy
    zx,zy)
    =
    (3,-5,0)
    (1,0
    0,1
    4x,2y)
    =(3,-5)
    another easier method is that fx=fx*1+fz*zx=3
    i know it may be very unclear, but i hope it will help.
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  4. #4
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    yakobian i a method of tranforming into a new varibles
    and its only for integral
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  5. #5
    MHF Contributor
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    Quote Originally Posted by transgalactic View Post
    f(x,y,2x^2+y^2)=3x-5y
    i dont know how to do a erivative by x
    by y
    if it were
    f(x,y,z)=3x-5y then
    f'x=3 f'y=-5 f'z=0
    how the f(x,y,2x^2+y^2)=3x-5y
    changes??
    ??
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