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Homework Help: Gradient of a two term equation

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data
    1. for the surface z=9 - x^2 - y^2 find,
    i) the gradient at (1,1,7) in the direction making an angle alpha with the x axis
    ii) the max gradient at the point (1,1,7) and the value of alpha for which it occurs


    2. find the stationary point of z=x^2 +2x +3y^2 -3xy + 5 and determine the nature of this point

    2. Relevant equations

    this is what i need, im not sure of the equation you use to differentiate an equation like this.


    3. The attempt at a solution

    I know you have to partially diff it with respect to x and then with respect to y,
    but I don't know how to combine the 2 together so you can just put the numbers in and find the gradient.
     
  2. jcsd
  3. Aug 27, 2011 #2

    vela

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    Look up the directional derivative in your textbook. That's what you need for (i).
     
  4. Aug 28, 2011 #3
    ah ok I read that and looked through the example but in the example the alpha was given

    so i ended up with du f(x,y) = -2(1)cos(alpha) + -2(1)sin(alpha)

    I'm not sure what to do next, if i put 7 on the left hand side the equation is false.

    Or is this the answer? gradient m = -2cos(alpha) + -2sin(alpha) and i simply leave it at that and go onto part ii?


    EDIT: ok im going to assume that the answer is: m = -2cos(alpha) + -2sin(alpha)

    so now using that i got 2 * SQRT(2) = max gradient
    and alpha = -135 degrees and alpha = 225 degrees
    is this correct?
     
    Last edited: Aug 28, 2011
  5. Aug 28, 2011 #4
    also for the 2nd question I got x=-4 and y=-2 but I don't know what it is asking for when it says determine the nature. does it just mean if it's a minimum or maximum?
     
    Last edited: Aug 28, 2011
  6. Aug 28, 2011 #5

    vela

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    Yes, that's right.
    How did you find these angles? One is correct; the other isn't.
    I think so. Remember it doesn't have to be a minimum or a maximum. It could also be a saddle point, for example.
     
  7. Aug 28, 2011 #6
    i found the angles by solving for alpha 2*2^.5 = -2cos(alpha) - 2sin(alpha)
    this gave alpha=45(8n-3)

    then i just solved for n=0 and n=1
    I'm not sure how many solutions this should have a bit more explanation of how to do the final step would be appreciated.


    from graphing i believe the point to be a minimum right?
     
  8. Aug 28, 2011 #7

    vela

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    Oh, sorry, I completely missed the sign on -135 degrees. That's the same angle as 225 degrees, so obviously, both answers are correct.

    What was your reasoning that the maximum gradient was [itex]2\sqrt{2}[/itex]? It's correct, but I'm wondering how you figured it out.

    In the second problem, the point is a minimum. You should be able to show it analytically by calculating derivatives.
     
  9. Aug 28, 2011 #8
    2 * root(2) was found like this
    m=-2xi -2yj
    substitute 1,1
    m(1,1)=-2i-2j
    then take sqrt of the squares added together
    SQRT((-2)^2 + (-2)^2)
    = 2*sqrt(2)

    but with the angles, when you solve the equation how many answers should I give since obviously the solution to anthing=sin(alpha) has infinite solutions, should I give the 2 angles already stated? more? less? leave the answer as alpha=45(8n-3)?
     
  10. Aug 28, 2011 #9

    vela

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    OK, so you've implicitly used the fact that m=∇f points along the direction of maximal change. That's what I was trying to figure out if you knew.

    You do realize that the angles -135 degrees and 225 degrees specify the same angle and direction, right?
     
  11. Aug 28, 2011 #10
    oh wow i just saw that, i honestly thought they were different directions and there were 2* n number of solutions. Herp derp
    ok so i guess i just submit 225 as the angle and im done, thank you for your time and assistance
     
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