a person who is 6 feet tall was walking away from a lamp post at the shadow of 40 feet per minute. when the person is 10 feet from the lamp post, his shadow is 20 feet long. find the rate of which the length of the shadow is increasing when he is 30 feet from the lamp post.
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Answer:
Organizing information:
•
dx
dt = 40, when x = 10, s=20
• Goal: Find ds
dt when x = 30.
We set up a ratio of similar triangles.
x + s
h
=
s
6
The height of the pole is a constant. We solve for h by using that when
x = 10, s = 20.
10 + 20
h
=
20
6
6(30) = 20h
h = 180/20 = 9
Now rewrite our orginal ratio equation with the constant height solved for:
x + s
9
=
s
6
6x + 6s = 9s
6x = 3s
Differentiate both sides with respect to t and solve for ds
dt
6
dx
dt = 3
ds
dt
Plug in dx
dt = 40, and solve for ds
dt
ds
dt = 80f t/min
Step-by-step explanation:
Answer:
80 ft/min
Step-by-step explanation:
given:
dx/dt = 40 ft/min
y = 6 ft.
by similar triangle:
L/y = (L+x)/h ----> eq.1
find height of the lamp post, h:
tan∅ = 6/20
∅ = 16.7 deg
tan(16.7) = h/30
h = 9 ft.
subs. value of h to eq.1:
L/6 = (L+x)/9
L = 2x
derive both sides:
dL/dt = 2dx/dt
dL/dt = 2(40)
dL/dt = 80 ft/min