1. Mar 2, 2009

### cmajor47

1. The problem statement, all variables and given/known data
The elevation of a mountain above sea level at (x,y) is 3000e^$$\frac{-x^2-2y^2}{100}$$ meters. The positive x-axis points east and the positive y-axis points north. A climber is directly above (10,10). If the climber moves northwest, will she ascend or descend and at what slope.

2. Relevant equations
$$\frac{d}{dx}$$eu=eu$$\frac{du}{dx}$$

3. The attempt at a solution
$$\nabla$$f(10,10)=-600e-3i-1200e-3j

I know that the climber will descend but I don't know how to figure out the slope that she will descend at. Can anyone help?

2. Mar 2, 2009

### Dick

To find the slope you want to take the dot product of your gradient with a unit vector pointed in the northwest direction. Can you find one?

3. Mar 2, 2009

### cmajor47

Would the unit vector be 600e-3i-1200e-3j since west is the opposite of east?

4. Mar 2, 2009

### Dick

No. A UNIT vector pointing west would be -i. A unit vector pointing north would be j. Do you see why? Northwest is at a 45 degree angle to both of those. Finding a unit vector pointing NW has nothing to do with your gradient vector. It's a whole different problem.