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A car is driving northwest at v mph across a sloping plain whose height, in feet above sea level, at a point N miles north and E miles east of a city is given by: h(N,E)=4750+75N+100E.
At what rate is the height above sea level changing with respect to distance in the direction the car is driving?

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h(N,E) = 2500 + 100N + 50E 
dh/dN = 100 feet per mile north 
dh/dE = 50 feet per mile east 
sqrt(100^2 + 50^2) = 111.8 feet per mile northeast
facundo3141592

Answer: The car is driving northwest at v mph.

if h(N,E) = 4750 + 75N + 100E

where N is the miles north, and E the miles at east.

But the car is moving northwest, so we should replace E for -W, then

h(N,W) = 4750 + 75N -100W

Now this equation says that for each mile to the north, the height increases by 75 miles, and for each mile to west, the height decreases by 100 miles.

So if the car is moving exactly at northwest, for each mile it does to north, also does one mile to west, This means that for each mile to north (or west), the height decreases by 25 miles.

then if the car does V miles per hour, then the velocity pointing north and west are the same, and are Vₙ = V/[tex]\sqrt{2}[/tex]

if the car does X =Vₙ*1h miles in one hour, then the Height will decrease by X*25 per hour

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