Asked by dave
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.
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.
Answers
Answered by
Reiny
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
and finally
dα/dt = 1/10
when the ballooon is 30 m hight, dα/dt = 1/10 radians/sec
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
and finally
dα/dt = 1/10
when the ballooon is 30 m hight, dα/dt = 1/10 radians/sec
Answered by
john
you made a mistake its 1/20
Answered by
Matthew Martin
Just wondering, since 1/20 radians/sec is the answer, would 9/pi degrees/sec also work because 1/20 * 180/pi = 180/20pi = 9/pi, or should one just keep the answer as 1/20 radians/sec?
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