Question

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?

Answer 1

\[dh \div dt=75\times dN \div dt +100 dE \div dt\]
where \[dN \div dt =V \cos 45\]
\[dE/dt=-V \cos 45\]
find \[dh \div dt\].
now, rate is the height above sea level changing with respect to distance in the direction the car is driving= rate is the height above sea level changing with respect to distance in the direction the car is driving=\[V -dh \div dt\] find it.....

Answer 2

one correction
\[\frac{dN}{dt} = V \sin 45\]
course since angle is 45 , doesn't really matter :|
for changing rate to respect to distance, let m represent miles driven
\[\frac{dh}{dm} = \frac{dh}{dt}*\frac{dt}{dm}\]
where V = dm/dt
put it all together
\[\frac{dh}{dm} = \frac{1}{V}(75V \sin 45 -100V \cos 45) = 75 \sin45 -100 \cos45 = \frac{-25\sqrt{2}}{2}\]