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Question involving gradient

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Hello all,
    I encountered this practice problem for my midterm tomorrow involving the gradient operation.
    Let (r, θ) denote the polar coordinates and (x, y) denote the cartesian coordinates of a point P in the plane. A function is defined via f(P)=xsinθ away from the origin, and let f(O)=0 at the origin O.
    Find f(x, y)
    Also, is the inequality [tex]|\nabla f|\leq \sqrt{2}[/tex] true( at all points where the gradient is defined)?


    Not sure at all how to go about starting this one.
     
  2. jcsd
  3. Nov 1, 2011 #2
    Well we have f(P) = xsin(theta). So we have the x in the equation, which is good. But we have a sin(theta) and we have to write that in terms of x and y. Basically we want xsin(theta) to be written in terms of x and y. So how do you do that? I think to answer the gradient question you first have to write the equation for f in terms of x and y, find the gradient, and then work from there.
     
  4. Nov 1, 2011 #3
    Okay.
    [tex]sin\theta=\frac{y}{\sqrt{y^2+x^2}}[/tex]
    Thus, [tex]f(x, y)=\frac{xy}{\sqrt{y^2+x^2}}[/tex]

    Moreoever, [tex]\nabla =[\frac{y}{\sqrt{x^2+y^2}}-\frac{x^2y}{(x^2+y^2)^{3/2}}]i+[\frac{x}{\sqrt{x^2+y^2}}-\frac{y^2x}{(x^2+y^2)^{3/2}}]j[/tex].

    I understand that the inequality is merely the length of the gradient vector. Do I actually have to compute this in order to determine if that inequality is true, or is there an easier way?
     
  5. Nov 2, 2011 #4
    I'm going to bed. I changed my post like 3 times, so obviously I am tired. After computing the length of the gradient vector, how does one determine if it is indeed less than or equal to a particular number for the domain?
     
  6. Nov 2, 2011 #5
    The short answer to your question is no, there's no faster way. I'm tired too, so I'm assuming that you calculated the gradient correctly. You can make the denominators the same and get that the gradient is y^3 i + x^3 j. Find x and y so that the length of the gradient vector is more than sqrt 2.
     
  7. Nov 2, 2011 #6
    Hello again, I tried manipulating the equation but I cannot seem to eliminate x as you have...
     
  8. Nov 2, 2011 #7
    Actually, I'm pretty sure that it simplifies to
    [tex]\frac{y^3}{(x^2+y^2)^{3/2}}[/tex]
     
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