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a street light is at the top of a 17ft tall pole. a woman 6ft tall walks away
from the pole with a speed of 8ft/sec along a straight path
how fast is the tip a shadow moving when she is 40 ft from the base of the pole?
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In this problem, there are two right-angled triangles: one has a 17 ft tall pole as vertical leg;
the other has 6 ft tall woman as vertical leg.
The hypotenuse is a straight line from the the pole tip to the woman' shadow tip.
These triangles are similar: they have common acute angle at the tip of the shadow.
From similarity, we can write a proportion
= ,
where L is the current horizontal distance of the woman from the pole and d is the length the shadow.
From this proportion
17d = 6*(L+d). (1)
L and d are functions of time: L = L(t), d = d(t).
Let's differentiate (1) over time t. You will get
17*d'(t) = 6*L'(t) + 6*d'(t). (2)
The derivative L'(t) is the speed of the woman; it is given in the problem, L'(t) = const = 8 ft/s.
So, we substitute 8 ft/s instead of L'(t) into (2). We get then
17*d'(t) = 6*8 + 6*d'(t)
17*d'(t) - 6*d'(t) = 48
11*d'(t) = 48
d'(t) = ft/s.
We get an interesting fact: the length of the shadow in this problem, d(t), is a LINEAR function of time with the constant rate of change.
The initial length of the shadow at t= 0, d(0), is zero; so, we can write d(t) = feet.
The length of the shadow is the linear function d(t) = .
Ok. Let's go further. We are just on the finish line.
In this problem, we need to get the derivative over the time of the sum (L+d), or (L(t) + d(t))' = L'(t) + d'(t).
As we just noticed above, L'(t) = 8 ft/s, the speed of the woman. d'(t) = ; so, the sum is
(L(t) + d(t))' = 8 + = 8 + 4 = 12 ft/s.
ANSWER. The tip of the shadow moves at the constant speed of 12 ft/s.
Solved.
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From this problem/solution, we learned an unexpected fact: the tip of the shadow moves
at the constant speed, which does not depend on the horizontal distance of the person from the pole.