tan T = h/60
d tan T /dt = dT/dt sec^2 T = (1/60)dh/dt
dT/dt = (1/60) cos^2 T dh/dt
tan T = 25/60
T = 22.6 deg
so cos^2 T = .852
dT/dt = (1/60)(.852) (8)
in radians !
A balloon rises at the rate of 8 feet per second from a point on the ground 60 feet from an observer. Find the rate of change of the angle of elevation when the balloon is 25 feet above the ground.
I have d(theta)/dt=(1/60)cos^2(theta)(8)
How do I find theta?
2 answers
Let the height of the ballon be h ft
then tanØ = h/60
h = 60tanØ
dh/dt = 60 sec^2 Ø dØ/dt
we know dh/dt = 8
when h = 25
tanØ = 25/60 = 5/12
construct a right-angled triangles with opposite side 5 and adjacent side 12.
By Pythagoras the hypotenuse is 13
so without even finding the actual angle we see that
cosØ = 12/13
cos^2 Ø = 144/169
Picking up where you left off
dØ/dt = (8/60)(144/169)
= 96/845 rad/s or
= appr .1136 rad/s
you were so close
then tanØ = h/60
h = 60tanØ
dh/dt = 60 sec^2 Ø dØ/dt
we know dh/dt = 8
when h = 25
tanØ = 25/60 = 5/12
construct a right-angled triangles with opposite side 5 and adjacent side 12.
By Pythagoras the hypotenuse is 13
so without even finding the actual angle we see that
cosØ = 12/13
cos^2 Ø = 144/169
Picking up where you left off
dØ/dt = (8/60)(144/169)
= 96/845 rad/s or
= appr .1136 rad/s
you were so close
1 answer