Suppose you are climbing a hill whose shape is given by the equation z = 2000 − 0.005x2 − 0.01y2, where x, y, and z are measured in meters, and you are standing at a point with coordinates (60, 40, 1966). 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? ascend descend Correct: Your answer is correct. At what rate? Correct: Your answer is correct. vertical meters per horizontal meter (b) If you walk northwest, will you start to ascend or descend? ascend descend Correct: Your answer is correct. At what rate? (Round your answer to two decimal places.) Correct: Your answer is correct. vertical meters per horizontal meter (c) In which direction is the slope largest? What is the rate of ascent in that direction? vertical meters per horizontal meter At what angle above the horizontal does the path in that direction begin? (Round your answer to two decimal places.)

(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.

electrical to thermal

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.