Question

Problem 12. A flat, but upward sloped, h has a rise in height of 100 m for every 500 m of displacement to the east and a rise in height of 200 m for every 500 m of displacement to the north. What direction should a car with a fixed speed travel upon this flatl to get the maximum height change with time? If the car has a speed of 5 ms-, what is the time rate of change of its height in this direction?

Answer 1

Consider east direction to be x axis north to be y axis and z axis in vertically upward direction z = z (x, y) dz = partial differential z/ partial differential x dx + partial differential z/ partial differential y dy given partial differential z/ partial differential x = a = 100/500 = 1/5 and partial differential z/ partial differential y = b = 200/500 = 2/5 therefore dz = a dx + b dy implies dz/dt = a dz/dt + b dy/dt equation (1) speed of the car is constant therefore (dx/dt)^2 + (dy/dt)^2 + (dz/dt)62 = v^2 = (5)^2 equation (II) let the car is moving in theta north of exact then dy/dx = v_y/v_x = tan theta equation (III) using (I), (II), (III) we get (dz/dt) = f(theta) {v_z^2 = v^2 (a + b tan theta)^2/(sec^2 theta + (a + b tan theta)^2} for v_z to be maximum d/d theta v_z = 0 will provides us the theta