A ladder 10 ft long rest against a
vertical wall. If the bottom of the
ladder slides away from the wall at
the rate of 1 ft/sec, how fast is the
top of a ladder standing down the
wall, when the bottom of the ladder
is 6 ft from the wall?
wall
ladder
10
Ground
Answers
Answered by
2
Answer:
Denoting the distance in feet between the wall and the base of the ladder by
x
and the angle in radians between the ladder and the ground by
y
, it is noted
cos
(
y
)
=
x
10
which implies
y
=
arccos
(
x
10
)
Denoting time in seconds by t, it is further noted that
d
y
d
t
=
d
y
d
x
d
x
d
t
(chain rule)
Noting (using standard table of derivatives for convenience)
d
y
d
x
=
−
1
√
1
−
(
0.1
x
)
2
(
0.1
)
(also by chain rule)
that is
d
y
d
x
=
−
0.1
√
1
−
0.01
x
2
It is noted from the question that in this particular system
d
x
d
t
=
0.8
feet per second
So (denoting the derivative as a function of
x
)
d
y
d
t
(
x
)
=
d
y
d
x
d
x
d
t
=
−
0.08
√
1
−
0.01
x
2
So
d
y
d
t
(
8
)
=
d
y
d
x
d
x
d
t
=
−
0.08
√
1
−
0.01
(
64
)
=
−
0.08
√
1
−
0.64
=
−
0.08
√
0.36
−
0.08
0.6
=
−
8
60
=
−
2
15
radians per second
Similar questions