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Calculte growth is some direction question

  1. Aug 12, 2009 #1
    i know that the maximal growth is the gradiend of a function.

    the formula is:

    [tex]

    \triangledown f=f_x'\vec{i}+f_y'\vec{j}+f_z'\vec{k}

    [/tex]



    so when i am given a function

    [tex]f(x,y,2x^2+y^2)=3x-5y[/tex]

    i am was told that it growth on point M(1,2,6) in direction
    [tex]\hat{a}=(\frac{1}{3},\frac{2}{3},\frac{2}{3})[/tex] 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

    [tex]

    (grad f(1,2,6)\dot \hat{a}=1

    [/tex]



    ????
     
  2. jcsd
  3. Aug 12, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi khdani! :smile:

    (have a grad: ∇ and a curly d: ∂ and try using the X2 tag just above the Reply box :wink:)
    That's right :smile:

    or better still, f(x,y,z) = g(x,y) where g = 3x - 5y.

    Now z = 2x2 + y2,

    so, using the chain rule, df/dx = … ?, df/dy = … ?, df/dz = … ? :wink:
     
  4. Aug 12, 2009 #3
    "so, using the chain rule, df/dx = … ?, df/dy = … ?, df/dz = … "

    what???

    you say i would substitute z with[tex] 2x^2+y^2[/tex]
    so i get
    f(x,y,z)=3x-5y
    so [tex]f'_x=3[/tex]
    [tex]f'_y=-5[/tex]
    [tex]f'_z=0[/tex]
     
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