# Homework Help: Gradient Vector Problem

1. Jan 13, 2013

### bmb2009

1. The problem statement, all variables and given/known data
For a hill the elevation in meters is given by z=10 + .5x +.25y + .5xy - .25x^2 -.5y^2, where x is the distance east and y is the distance north of the origin.

a.) How steep is the hill at x=y=1 i.e. what is the angle between a vector perpendicular to the hill and the z axis?

b.)In which compass direction is the slope at x=y=1 steepest? Indicate whether the angle you provide is the angle measured in the standard way in the counter-clockwise direction from the x axis (east) or whether it is the compass azimuth.

2. Relevant equations

3. The attempt at a solution
a.) So I got the gradient to be ∇f(x,y,z)= .5i - .25j...simply by the definition of the gradient. So i figured to figure the angle between the gradient and the z axis i could use the formula cosθ= (A dotted into B)/(lAl*lBl) but wouldn't be the z axis be the unit vector in the k hat direction making the cosθ=0...which can't be right. Any help on what to do? Also, I was given an equation that says ∇f(x,y,z) perpendicular to a 3 dimensional surface f(x,y,z) = constant. But I don't know how to use that equation to generate an answer.

b.)The gradient gives you the direction of the quickest altitude ascension (at least for this problem) so how do I use what I calculated for the gradient to answer this part?

Thanks for any help!!

2. Jan 13, 2013

### haruspex

Think of the altitude as a potential in 2 dimensions, just x and y. So ∇ is ?

3. Jan 13, 2013

### rude man

grad z gives you the x-y direction of steepest ascent, but now you have to ask yourself: when going in that direction, how much does dz change when you go a distance ds in the aforederived direction.

Think about how the x and y components of the gradient at your point (1,1) tell you what dz/ds would be.

P.S. the angle the problem asks for happens to be (90 deg - the steepness angle). Also, once you did (a) you already have (b). In fact, (b) should come first ... at least the way I look at it. Probably you were not supposed to solve for grad z for part (a) ... I don't know.

4. Jan 13, 2013

### Staff: Mentor

Dear Rude Man,

Very nice job of doping it out. That's exactly the way I would have done it, including doing part (b) first.

Chet

5. Jan 13, 2013

Thanks Chet.