1. ## <unsolved> Gradient vector/optimum path

Hi, i need help in the following question..

Suppose I am on point (3,-8)

i am given F(x,y) = 80 - 2x^2 - y^2,

find the equation of the path i should take (in terms of x and y) if i want to move in the direction of maximum increase of F.

I know that the level curves are ellipses, and was given a hint to solve it by equating v'(u) = gradient vector of F.. where v(u)=(x(u),y(u)) is the vector representing the path.

however im stuck at this point, i wonder if anyone can help me? will very much appreciate it!

2. The rate of change of a scalar function f in the direction of a unit vector $\displaystyle \mathbf{u}$ is given by

$\displaystyle \nabla f \cdot \mathbf{u}$

This is the directional derivative

So, the maximum rate of change is given by

$\displaystyle |\nabla f \cdot \mathbf{u}| = |\nabla f|$

3. Originally Posted by Haven
The rate of change of a scalar function f in the direction of a unit vector $\displaystyle \mathbf{u}$ is given by

$\displaystyle \nabla f \cdot \mathbf{u}$

This is the directional derivative

So, the maximum rate of change is given by

$\displaystyle |\nabla f \cdot \mathbf{u}| = |\nabla f|$

I do know that the maximum change occurs when direction vector u = $\displaystyle \nabla f$ = <-4x, -2y> itself, but i wonder how does it help me to obtain the equation of the path(in terms of x and y) i am suppose to take?

4. anyone has any idea?

5. Remember, that a line can be described by a direction vector (with components a, b, and c) and a point:

$\displaystyle r(t)=<x_0, y_0, z_0> + t<a, b, c>$;

You have a direction vector (it happens to be the gradient), and you have a point (3, -8).

6. Originally Posted by ANDS!
Remember, that a line can be described by a direction vector (with components a, b, and c) and a point:

$\displaystyle r(t)=<x_0, y_0, z_0> + t<a, b, c>$;

You have a direction vector (it happens to be the gradient), and you have a point (3, -8).
i agree this makes sense if i have a constant direction vector, and that the path is a straight line, however i believe the path is not a straight line because the path is a curve that is always tangent to the level curves, which are ellipses.

thus i do not think i am able to use this

7. Originally Posted by navelwort
i agree this makes sense if i have a constant direction vector, and that the path is a straight line, however i believe the path is not a straight line because the path is a curve that is always tangent to the level curves, which are ellipses.

thus i do not think i am able to use this

No, the path of fastest increase is perpendicular to the level curves.

In any case, $\displaystyle \nabla f= -4x\vec{i}- 2y\vec{j}$ so you must have $\displaystyle \frac{dx}{dt}= -4x$, x(0)= 3 and $\displaystyle \frac{dy}{dt}= -2y$, y(0)= -8. Solve those equations to get x and y in terms of the parameter t. You can then, if you wish, eliminate t to get y as a function of x.

8. Originally Posted by HallsofIvy
No, the path of fastest increase is perpendicular to the level curves.

In any case, $\displaystyle \nabla f= -4x\vec{i}- 2y\vec{j}$ so you must have $\displaystyle \frac{dx}{dt}= -4x$, x(0)= 3 and $\displaystyle \frac{dy}{dt}= -2y$, y(0)= -8. Solve those equations to get x and y in terms of the parameter t. You can then, if you wish, eliminate t to get y as a function of x.
opps haha i meant perpendicular, yeah thanks! but just wondering, how do i go about solving those equations?

am i correct to do it like this?

x(t) = -4xt + C, y(t) = -2yt + D

x(0) = 3, => C = 3,
y(0) = -8, => D = -8,

thus x = -4xt + 3, y = -2yt -8

solving t, (x-3)/(-4x) = (y+8)/(-2y)

thus i have -2yx + 6y = -4xy - 32x
=> 3y = xy -16x ?