# Path of steepest descent

• October 13th 2009, 03:30 AM
AKTilted
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
• October 13th 2009, 04:43 AM
chisigma
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

$\chi$ $\sigma$
• October 13th 2009, 04:49 AM
AKTilted
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.
• October 13th 2009, 04:52 AM
HallsofIvy
Quote:

Originally Posted by AKTilted
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!
• October 13th 2009, 03:15 PM
B_Miner
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:

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