Final answer:
The rate of change of the car's height concerning time, while driving northwest across a sloping plain, is 25v feet per mile per hour, with v representing the car's speed in mph.
Step-by-step explanation:
The question involves calculating the rate of change of the height of a car as it drives across a sloping plain. Given the height function h(x,y) = 5000 + 50x + 25y, where x is the distance in miles north and y is the distance in miles east of the city, we need to express the rate of change of h with respect to time when the car is driving northwest at speed v. The northwest direction implies that the x and y are changing at the same rate but in opposite signs because one coordinate will increase while the other decreases.
To find the rate of change dh/dt, which is the derivative of h with respect to time, we will use the chain rule of differentiation. Calculating the partial derivatives of h will give us dh/dx = 50 and dh/dy = 25. Since the car is moving northwest, the rates of change for x and y with respect to time are equal in magnitude but opposite in sign. Let's denote the rate of change of x with respect to time as dx/dt = v and dy/dt = -v, because the car is driving in a direction that decreases the eastward distance while increasing the northward distance.
Therefore, the rate of change of the height with respect to time is dh/dt = (dh/dx)(dx/dt) + (dh/dy)(dy/dt) = 50v - 25v = 25v. The rate of change of the car's height, in terms of the speed v, is 25v feet per mile per hour.