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Math Help - Path of steepest descent

  1. #1
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    Path of steepest descent

    Hi I'm not sure how to get the answer to this question:

    Determine the path of steepest descent along the surface z = x^2 +3y^2 from the point (1,-2,13) to (0,0,0).

    Thank you
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  2. #2
    MHF Contributor chisigma's Avatar
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    The 'steepest descent path' is in the direction of gradient with changed sign. In our case is...

    z=x^{2}+3\cdot y^{2} (1)

    ... so that the components of gradient are...

    \frac{\partial{z}}{\partial{x}}= 2x

    \frac{\partial{z}}{\partial{y}}= 6y (2)

    The 'steepest descent path' in parametric form is the solution of the system...

    \frac{dx}{dt}= -2x

    \frac{dy}{dt}= -6y (3)

    ... with the 'initial conditions' x(0)=1, y(0)=-2...

    Kind regards

    \chi \sigma
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  3. #3
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    Could you please describe this in another way. I'm still not understanding your solution.
    From what I understand the question involves solving a DE of the form f'(t)-k*f(t) = 0 where k is a constant.
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  4. #4
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    Quote Originally Posted by AKTilted View Post
    Hi I'm not sure how to get the answer to this question:

    Determine the path of steepest descent along the surface z = x^2 +3y^2 from the point (1,-2,13) to (0,0,0).

    Thank you
    The "path of steepest descent" is always in the direction opposite the gradient because the gradient always points in the direction of fastest increase. Here the gradient is \nabla z= 2x\vec{i}+ 6y\vec{j} so the path of steepest descent always has tangent vector -\nabla z= -2x\vec{i}- 6y\vec{j}. If we write the path in terms of the parameter t, the the tangent vector is \frac{dx}{dt}\vec{i}+ \frac{dy}{dt}\vec{j} so we must have \frac{dx}{dt}= -2x and \frac{dy}{dt}= -6y.

    Those are easy to solve and will involve one undetermined constant for x and one for y. I puzzled over how to choose coefficients to make this go through two points until I realized that because this is a "path of steepest descent" it must go through (0,0), the lowest point on this paraboloid. When you fall down a mountain, your path is determined by your starting point but you still reach the bottom!

    So just use x(0)= 1 and y(0)= -2 to determine the constants. z, of course, is determined by z= x^2+ 3y^2

    By the way, if you don't like the exponentials you get, you can always write y and z a polynomials in x. Hint: e^{at}= (e^t)^a.

    Blast! Again, Chi Sigma got in ahead of me!
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  5. #5
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    I posted a question about this topic yesterday. I did not get a response, but I think I understand now how it works. What do you guys think: I will use this example.

    You have a function z=x^2+3y^2
    The gradient is <2x, 6y>. The gradient points in the direction of the fastest increase at a point. So, at the starting point <1,-2> the fastest increase is in the direction of <2, -12>.

    The algorithm of steepest descent says to update the point (x,y) as:

    (x,y)-factor*<gradient at x,y>.


    Forget about the "factor" for a second. At first I was confused by why subtract new point (i.e. <gradient at x,y> )
    from old point (i.e. (x,y) above).

    But I think the reason is that to "move in the direction of a vector" from a point means to add to the components of the point the components of the vector.

    For example, think of a a point on the 2 dimensional Cartesian plane at (x,y) = (-3,1). If we move in the direction of the vector <3,2> (with tail at the origin) we move 3 units to the right and 2 units up (so the tip of this vector from (-3,1) is at (0,3). Then, this vector between points (-3,1) and (0,3) is equivalent to the vector we were moving in its direction (i.e. <3,2>). This process was just to add to the point the vector. <-3+3, 1+2> = <0,1>. <3,2> is a position vector and this other one from (-3,1) to (0,3) is a representation.

    BUT since we want to fastest decrease, we subtract.

    This new point (in this example posting) is <1,-2> - <2, -12> = <-1,10>. You then evaluate the gradient at this point and keep going until you reach convergence.

    The "factor" controls how much you move in the direction of the gradient - i.e. i scales the gradient. In practical terms, a small factor (<1) helps the algorithm not get bad results.

    What do you think - have I understood this?

    Thanks
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