# Problem involving gradients and unit vectors.

• Jul 6th 2010, 05:10 PM
Mattpd
Problem involving gradients and unit vectors.
Find the unit vector in the direction in which \$\displaystyle f(x, y, z) = x2 + 2xyz - yz2\$ increases most rapidly at P(1, 1, 2) and give the rate of change of f in that direction.

I am having trouble setting up the problem. Is the gradient given as P(1, 1, 2) or does it need to be calculated? and the rate of change is the magnitude of the gradient right? But how do I find the unit vector? I know that the dot product of the gradient and the unit vector = the directional derivative.. Does this have anything to do with the solution?
• Jul 6th 2010, 07:02 PM
mr fantastic
Quote:

Originally Posted by Mattpd
Find the unit vector in the direction in which \$\displaystyle f(x, y, z) = x2 + 2xyz - yz2\$ increases most rapidly at P(1, 1, 2) and give the rate of change of f in that direction.

I am having trouble setting up the problem. Is the gradient given as P(1, 1, 2) or does it need to be calculated? and the rate of change is the magnitude of the gradient right? But how do I find the unit vector? I know that the dot product of the gradient and the unit vector = the directional derivative.. Does this have anything to do with the solution?

\$\displaystyle \nabla f\$ evaluated at (1, 1, 2) is a vector in the direction of maximum rate of change of f at the point (1, 1, 2). To get a unit vector, divide the vector by its magnitude. All these things are surely in your classnotes and textbook.
• Jul 8th 2010, 04:58 AM
HallsofIvy
No, the gradient is NOT "given as P(1, 1, 2)". P(1, 1, 2) is a [b]point[/b[]. You are asked to find the gradient at the point (1, 1, 2). The directional derivative has nothing to do with the solution. The "direction in which f increases most rapidly" is the direction in which the gradient points. Find the gradient vector, determine its length to find "the rate of change of f in that direction" and then divide the gradient by its length to find the unit vector in that direction.