Posted by dave on .
a balloon rises at a rate of 3 meters per second from a point on the ground 30 meters from an observer. find the rate of change of the angle of elevation of the balloon from the observer when the balloon is 30 meters above the ground...
so i drew a small picture depicting it but i'm still confused on what to do. am i basically finding the rate of change?? as in the DR/DT? i need someone to explain this to me, pleaseee! thank you.
calculus : related rates -
call the angle of elevation α (alpha)
so what you want to find is dα/dt
from your triangle
tan α = x/30, where x is the height of the balloon
then x = 30 tan α
and dx/dt = 30 (sec^2 α)dα/dt
when x = 30 , α = 45º and sec^2 45º = 2
so 3 = 30(2)dα/dt
dα/dt = 1/10
when the ballooon is 30 m hight, dα/dt = 1/10 radians/sec