Sketch a right angles triangle with a base of 400 m, a height of h m , (the balloon), and an angle of Ø opposite the height.
tan Ø = h/400
h = 400 tanØ
dh/dt = 400 sec^2 Ø (dØ/dt)
given: when Ø = 10° or 10(π/180) radians dØ/dt = .2
(you can simply take sec 10° and square it)
dh/dt = 400 sec^2 10° (.2)
= appr 82.5 m/min
check my arithmetic
A hot air balloon is rising vertically upward from the ground. The crew of a boat from a nearby
lake notices this situation and looks upward at an angle of
10 degree
to see the balloon. If the boat is
400meters away from the balloon, and the angle of observation is changing at 0.2 rad/min, how
fast is the balloon rising?
2 answers
f(x)=7
f'(x)=7x^0
=0
f'(x)=7x^0
=0