This airplane is flying at 500 km/h in a straight direction at a constant 5000 m.
is the angle of elevation from a fixed point of observation. What is the rate of change of
when it is 30°? Required to answer in DMS/s and round to one decimal place.Solution:
Drew the right triangle as follows: Vertical segment labeled 5000 m = 5 km, assigned variable h
j. At the top, a stick figure of a jet plane. Horizontal segment labeled 5
km, assigned variable b
. Hypotenuse labeled 10 km, assigned variable c
. The 5
and the 10 we get from the 30-60-90 relationship. Angle of elevation is angle
= 30 deg (using degree symbol).
We also know that
= 500 km/h and
= 0. So,
= h/b and
/dt = (b dh/dt - h db/dt)/h2
/dt = (0 - 5*500)/25 = -100. But sec
= 10/5sqrt3, then sec2
= 4/3 and
/dt = -100. Multiplying both sides by 3/4 gives d
/dt = -75; i.e. -75 radians/h.
Using a calculator, I get -75 /h * 1 h / 60 min * 1 min / 60 s = approx -0.0208333333 / s which converts to -1.19+°, or (-1 deg 11 min 37.2 sec) /s