A spotlight on the ground shines on a wall 10 m away. A man 2 m tall walks from the spotlight towards the wall at a speed of 1.2 m/s. How fast is his shadow on the wall decreasing when he is 3 m from the wall?
Answers
Answer:
Explanation:
.
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At the specified moment in the problem, the man is standing at point
D
with his head at point
E
.
At that moment, his shadow on the wall is
y
=
B
C
.
The two right triangles
Δ
A
B
C
and
Δ
A
D
E
are similar triangles. As such, their corresponding sides have equal ratios:
A
D
A
B
=
D
E
B
C
8
12
=
2
y
,
∴
y
=
3
meters
If we consider the distance of the man from the building as
x
then the distance from the spotlight to the man is
12
−
x
.
12
−
x
12
=
2
y
1
−
1
12
x
=
2
⋅
1
y
Let's take derivatives of both sides:
−
1
12
d
x
=
−
2
⋅
1
y
2
d
y
Let's divide both sides by
d
t
:
−
1
12
⋅
d
x
d
t
=
−
2
y
2
⋅
d
y
d
t
At the specified moment:
d
x
d
t
=
1.6
m/s
y
=
3
Let's plug them in:
−
1
12
(
1.6
)
=
−
2
9
⋅
d
y
d
t
d
y
d
t
=
1.6
12
2
9
=
1.6
12
⋅
9
2
=
0.6
m/s
Answer:
At the specified moment in the problem, the man is standing at point
D
with his head at point
E
.
At that moment, his shadow on the wall is
y
=
B
C
.
The two right triangles
Δ
A
B
C
and
Δ
A
D
E
are similar triangles. As such, their corresponding sides have equal ratios:
A
D
A
B
=
D
E
B
C
8
12
=
2
y
,
∴
y
=
3
meters
If we consider the distance of the man from the building as
x
then the distance from the spotlight to the man is
12
−
x
.
12
−
x
12
=
2
y
1
−
1
12
x
=
2
⋅
1
y
Let's take derivatives of both sides:
−
1
12
d
x
=
−
2
⋅
1
y
2
d
y
Let's divide both sides by
d
t
:
−
1
12
⋅
d
x
d
t
=
−
2
y
2
⋅
d
y
d
t
At the specified moment:
d
x
d
t
=
1.6
m/s
y
=
3
Let's plug them in:
−
1
12
(
1.6
)
=
−
2
9
⋅
d
y
d
t
d
y
d
t
=
1.6
12
2
9
=
1.6
12
⋅
9
2
=
0.6
m/s
This is the rate at which the length of his shadow is decreasing at the specified moment.
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