calculus: climbing a hill

[tandoorichicken]

Suppose you are climbing a hill whose shape is given by the equation z = 1000 - 0.01x^2 - 0.02y^2, where x, y, and z are measure in meters, and you are standing at a point with coordinates (50,80,847). The positive x-axis points east and the positive y-axis points north.
(a)If you walk due south, will you start to ascend or descend? At what rate?
(b)If you walk due northwest, will you start to ascend or descend? At what rate?
(c)In which direction is the slope largest? What is the rate of ascent in that direction? At what angle above the horizontal does the path in that direction begin?

for (a) and (b), I am pretty sure you ascend, but I don't know how to find the rate. For (c), I am totally lost.

[s_a]

z = 1000 - 0.01x^2 - 0.02y^2
grad(z) = (-0.02x, - 0.04y)

at (50,80,847),
grad(z) = (-0.02*50, - 0.04*80)
= (-1,-3.2) (this vector points in the direction of the steepest uphill slope)

a) The unit vector for "south" is (0,-1), so (-1,-3.2) . (0,-1) = 3.2 > 0. So a walk south from the point (50,80,847) is uphill with gradient 3.2

b) The unit vector for "northwest" is (-1/sqrt(2), 1/sqrt(2)), so (-1,-3.2) . (-1/sqrt(2), 1/sqrt(2)) = 1/sqrt(2) * (1 - 3.2) = -2.2/sqrt(2) = -11*sqrt(2)/10 < 0. So a walk northeast from (50,80,847) is downhill with gradient -11*sqrt(2) / 10.

c) The slope is the largest (i.e. most uphill) in the direction of grad(z), and has gradient |grad(z)|. The direction of steepest ascent at (50,80,847) is in the direction of the vector (-1,-3.2) with gradient sqrt(1^2 + 3.2^2) = 3.35.