A balloon rises at the rate of 8 ft/sec from a point on the ground 60 ft from the observer. How do you find the rate of change of the angle of elevation when the balloon is 25 ft above the ground?
Answers
To solve this related rates (of change) problem:
Let
y
= the height of the balloon and let
θ
= the angle of elevation.
We are told that
d
y
d
t
=
8
ft/sec.
We are asked to find
d
θ
d
t
when
y
=
25
ft.
Draw a right triangle with base = 60 ft (that doesn't change), height
y
and angle opposite height
θ
.
Then
tan
θ
=
y
60
and
y
=
60
tan
θ
.
Differentiating with respect to
t
gives us:
d
d
t
(
y
)
=
d
d
t
(
60
tan
θ
)
.
d
y
d
t
=
60
sec
2
θ
d
θ
d
t
.
We are asked to find
d
θ
d
t
when
y
=
25
.
We have:
8
=
60
sec
2
θ
d
θ
d
t
, so
d
θ
d
t
=
8
60
cos
2
θ
=
2
15
cos
2
θ
.
We need
cos
θ
when
y
=
25
.
With base = 60 and height = 25, we get hypotneuse
c
=
√
60
2
+
25
2
=
√
(
5
⋅
12
)
2
+
(
5
⋅
5
)
2
=
5
√
(
12
)
2
+
(
5
)
2
=
5
⋅
13
=
65
.
So, when
y
=
25
, we have:
cos
θ
=
60
65
=
12
13
.
So
d
θ
d
t
=
2
15
cos
2
θ
=
2
15
(
12
13
)
2
=
96
845
radians / sec
.
(Remember, in order to use
d
d
θ
(
tan
θ
)
=
sec
2
θ
, we must have
θ
either a real number or the radian (not degree) measure of an angle.