# what happens to Gradient Vector at critical points of a multi variable function

• Apr 1st 2010, 06:06 AM
Newtomaths
what happens to Gradient Vector at critical points of a multi variable function
Could any one please explain what happens to the gradient vector at critical point....it becomes a zero/null vector ..right??....But a zero vector should nt have a direction but on the other hand what I understood is at every point on the level curve of function the gradient is orthogonal to the level curve.......Then what happens at the critical point (its just a Null vector , so it simply gets vanished at those points and should nt have any direction)..... This is even more ambiguous if you consider the lagrange multiplier method ..where both the the objective function and constraint function share the same tangent line/ tangent plane and both have the gradient vector perpendicular to their respective function level curves at the crtical points .... thats what a video lecture from MIT says....AM wrong in understanding may be...but would love to get clarified.... So pls reply
• Apr 2nd 2010, 06:58 PM
HallsofIvy
Quote:

Originally Posted by Newtomaths
Could any one please explain what happens to the gradient vector at critical point....it becomes a zero/null vector ..right??....But a zero vector should nt have a direction but on the other hand what I understood is at every point on the level curve of function the gradient is orthogonal to the level curve.......Then what happens at the critical point (its just a Null vector , so it simply gets vanished at those points and should nt have any direction).....

At a critical point, a level curve is a single point or there are two or more level curves passing through the point.

Quote:

This is even more ambiguous if you consider the lagrange multiplier method ..where both the the objective function and constraint function share the same tangent line/ tangent plane and both have the gradient vector perpendicular to their respective function level curves at the crtical points .... thats what a video lecture from MIT says....AM wrong in understanding may be...but would love to get clarified.... So pls reply
Yes, that's true. I'm not sure why you consider that "ambiguous".

Consider the problem of finding a point on the graph of g(x,y)= constant that maximizes f(x,y). The gradient of the function, \$\displaystyle \nabla f\$, points in the direction of greatest increase. If there were no constraint you would simply move in the direction of the gradient to get increasing values of f and eventually get to the maximum point.

But if you are constrained to stay on g(x,y)= constant, what you might do is look at the "projection" of the gradient vector onto the tangent line to the curve and move left or right depending on which way the projection pointed. That would take you at least a little closer- until you reached a point at which the gradient was perpendicular to the curve. Then you could not get closer by moving either "right" or "left". And, of course, since \$\displaystyle \nabla g\$ is perpendicular to g(x,y)= constant, the two gradients must be parallel- \$\displaystyle \nabla f\$ must be a multiple of \$\displaystyle \nabla g\$