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Finding the gradient and maximum gradient at a point P for the surface z=2x^2+3y^2

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data
    For the surface z=2x^2+3y^2, find
    (i) the gradient at the point P (2,1,11) in the direction making an angle a with the x-axis;
    (ii) the maximum gradient at P and the value of a for which it occurs.

    2. Relevant equations
    ma=[itex]\partial[/itex]z/[itex]\partial[/itex]x(cos(a))+[itex]\partial[/itex]z/[itex]\partial[/itex]y(sin(a))

    If dma/da = 0 and d^2ma/da^2<0 the ma, is a maximum for that value of a

    3. The attempt at a solution\partial

    (i) Firstly I calculated:
    [itex]\partial[/itex]z/[itex]\partial[/itex]x = 4x
    [itex]\partial[/itex]z/[itex]\partial[/itex]y = 6y

    Therefore applying equation: ma=[itex]\partial[/itex]z/[itex]\partial[/itex]x(cos(a))+[itex]\partial[/itex]z/[itex]\partial[/itex]y(sin(a)) at P(2,1,11)

    We get:

    ma=8cos(a)+6sin(a)
    (I am quite sure this is correct, I don't have answers so if I am doing something wrong can someone please inform me.

    (ii) This is where I am stuck, I can't quite get the correct working out, can anyone please help out, below is the calculations I attempted:

    If dma/da = 0 and d^2ma/da^2<0 the ma, is a maximum for that value of a.

    dma/da = -8sin(a)+6cos(a)

    dma/da=0 Therefore: -8sin(a) + 6cos(a) = 0 Therefore: 8sin(a)=6cos(a)

    This is what I was up to, got stuck here, not sure if what I done is correct, if it is, how do I proceed from here?

    Thanks in advance
    ------------------------------------------------------

    I have now calculated part (ii), I realised I have a problem in part (i), I need to find a to get the gradient ma, can anybody help me out with this?
     
    Last edited: Jan 17, 2012
  2. jcsd
  3. Jan 17, 2012 #2
    Re: Finding the gradient and maximum gradient at a point P for the surface z=2x^2+3y^

    [itex]\int[/itex]
     
  4. Jan 17, 2012 #3

    ehild

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    Homework Helper
    Gold Member

    Re: Finding the gradient and maximum gradient at a point P for the surface z=2x^2+3y^

    It is correct for the derivative at the given point along the direction defined by the angle α.


    Isolate α: sin(α)/cos(α)=6/8 =tan(α)=0.75 α=?
    Getting α, substitute it back to mα.

    ehild
     
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