Question: How can I calculate the steepest slope of a hill using its equation?

[Noob.com]

Equation of the height of the hill
z = 2xy - 3x^2 -4y^2 -18x + 28y +12
z: Height of the hill
x: Distance East
y: distance South

Question: In which compass direction is the slope at x = y = 1 steepest?

My question: What does this suppose to mean? The vector that is tangent to that point pointing to the top of the hill? If so how do I do that?

Note: I have already calculated the critical point of the hill and the angle between the normal vector of the hill at x=y=1 and the z-axis.

Thanks!

 

[chrisk]

Take the gradient of the function where the del operator is

\vec{\nabla}=\hat{x}\frac{\partial}{\partial\mbox{x}}+\hat{y}\frac{\partial}{\partial\mbox{y}}+\hat{z}\frac{\partial}{\partial\mbox{z}}

The gradient gives the direction of most rapid change for w = f(x,y,z). In your case, z = f(x,y).

 

[Noob.com]

Gradient of z = (2y - 6x -18 , 2x - 8y +28, 0)

If I put x = y = 1 there, the answer would be (a,b,0) which is a straight line parrallel to z-axis. That doesn't make sense at all!

 

[chrisk]

When x=y=1 you get

\vec{T}=-22\hat{x}+14\hat{y}

where the unit vector x is in the East direction and unit vector y is in the South direction. This gives 22 West by 14 South. Use arctan to find the angle these two components form and this will be the direction.

 

[Noob.com]

Oh so the compass lay on the horizontal direction. I thought it would be something in the z-direction too. Thanks!