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Thread: 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...

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

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

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

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

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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 $\displaystyle \nabla z= 2x\vec{i}+ 6y\vec{j}$ so the path of steepest descent always has tangent vector $\displaystyle -\nabla z= -2x\vec{i}- 6y\vec{j}$. If we write the path in terms of the parameter t, the the tangent vector is $\displaystyle \frac{dx}{dt}\vec{i}+ \frac{dy}{dt}\vec{j}$ so we must have $\displaystyle \frac{dx}{dt}= -2x$ and $\displaystyle \frac{dy}{dt}= -6y$.

    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 $\displaystyle z= x^2+ 3y^2$

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

    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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