How Does Walking Direction Affect Your Ascent or Descent on a Quadratic 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.

 

[wais]

(a) If you walk due south, you will start to descend. This can be seen by plugging in the coordinates (50,80) into the equation for z. When x and y decrease, the value of z also decreases.

To find the rate of descent, we can take the partial derivative of z with respect to y and evaluate it at the given point. This gives us:

∂z/∂y = -0.04y

Plugging in y=80, we get a rate of descent of -3.2 meters per meter traveled. This means for every meter you walk south, you will descend 3.2 meters.

(b) If you walk due northwest, you will also start to descend. This can be seen by plugging in negative values for both x and y into the equation for z.

To find the rate of descent, we can take the partial derivative of z with respect to both x and y and evaluate them at the given point. This gives us:

∂z/∂x = -0.02x
∂z/∂y = -0.04y

Plugging in x=-50 and y=-80, we get a rate of descent of -1 meter per meter traveled in the x direction and -3.2 meters per meter traveled in the y direction. This means for every meter you walk northwest, you will descend 1 meter in the x direction and 3.2 meters in the y direction.

(c) The slope is largest in the direction of steepest descent, which is in the direction of the negative gradient vector (-∂z/∂x, -∂z/∂y). To find the rate of ascent in this direction, we can use the magnitude of this gradient vector:

|∇z| = √((-∂z/∂x)^2 + (-∂z/∂y)^2)

Plugging in the values from part (b), we get a rate of ascent of √(1^2 + 3.2^2) = 3.31 meters per meter traveled.

To find the angle above the horizontal at which the path begins, we can use trigonometry. The angle θ can be found by using the tangent function:

tan(θ) = (-∂z/∂y)/(-∂z