Answer:
(a) Ascend at 0.8 vertical meter/meter
(b) Descend at -0.2·√2 vertical meter/meter
(c) In the (-0.6, -0.8) direction. The path begins at 45° to the horizontal
Explanation:
The given equation of the shape of the hill is z = 2000 - 0.005·x² - 0.01·y²
The current location = (60, 40, 1966)
The direction of the positive x-axis = east
The direction of the positive y-axis = north
(a) Walking due south = Reducing the y-value 40
From the equation, the elevation varies inversely with the motion towards the north
Therefore, walking south increases the elevation, and we ascend
The rate is given by the partial derivative at in the -j direction, which is 0.02
The rate is therefore 40 × 0.02 = 0.8
(b)The unit vector in the northwest direction u = 1/√2·(-1, 1)
∴ The rate = (-0.01(60), -0.02(40))·u = (-0.6, -0.8)·1/√2·(-1, 1) = -0.2·√2
Therefore we descend
(c) The slope is largest in the grad of the function at the point (60, 40) which is given as follows;
d(2000 - 0.005·x² - 0.01·y²
)/dx, d(2000 - 0.005·x² - 0.01·y²
)/dy = (-0.6, -0.8)
Therefore, the direction is tan⁻¹(-0.8/-0.6) ≈ S 36.87° W
The slope =(√((-0.4)² + (-0.8)²) = 1
Therefore, the angle is 45° to the horizontal.
