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A street light is at the top of a "20" ft tall pole. a woman "6" ft tall walks away from the pole with a speed of "5" ft/sec along a straight path. how fast is the tip of her shadow moving when she is "50" ft from the base of the pole?

User Jasarien
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2 Answers

3 votes

Final answer:

To solve this problem, we can use similar triangles and the concept of similar triangles. We first find the distance of the woman from the base of the pole and then use the concept of similar triangles to find the rate at which the tip of her shadow is moving. The tip of her shadow is not moving when she is 50 ft from the base of the pole.

Step-by-step explanation:

To solve this problem, we can use similar triangles and the concept of similar triangles. Let's denote the distance of the woman from the pole as x ft. Since the height of the pole is 20 ft and the height of the woman is 6 ft, we can set up the following proportion:

(x + 6) / x = 20 / 6

Cross multiplying the proportion, we get:

6x + 36 = 20x

Subtracting 6x from both sides, we get:

36 = 14x

Dividing both sides by 14, we get:

x = 36/14 ≈ 2.57 ft

Now we can use the concept of similar triangles to find the rate at which the tip of her shadow is moving. The ratio of the height of the pole to the distance of the woman from the pole is the same as the ratio of the length of the shadow to the distance of the tip of the shadow from the base of the pole. So, we have:

(20 ft) / (x + 50 ft) = (x ft) / (50 ft)

Cross multiplying the proportion, we get:

20(50) = x(x + 50)

Expanding the equation, we get:

1000 = x^2 + 50x

Subtracting 1000 from both sides, we get:

x^2 + 50x - 1000 = 0

Now we can solve this quadratic equation using factoring or the quadratic formula. The solutions are:

x = -55.69 or x = 5.69

Since x represents a distance, we can discard the negative solution. So, the woman is approximately 5.69 ft from the base of the pole.

To find the rate at which the tip of her shadow is moving, we can differentiate the equation we obtained for the length of the shadow concerning time (t), since the rate at which the woman is moving is given. Differentiating the equation, we get:

d(20 / (x + 50)) / dt = d(x / 50) / dt
Simplifying the equation, we get:

-20 / (x + 50)^2 * dx / dt = 1 / 50 * dx / dt

Dividing both sides by dx / dt, we get:

-20 / (x + 50)^2 = 1 / 50

Multiplying both sides by (x + 50)^2, we get:

-20 = (x + 50)^2 / 50

Multiplying both sides by 50, we get:

-1000 = (x + 50)^2

Taking the square root of both sides, we get:

±sqrt(-1000) = x + 50

Since we are dealing with real-world measurements, we can discard the negative square root.

So, x + 50 = sqrt(1000) ≈ 31.62

Subtracting 50 from both sides, we get:

x ≈ 31.62 - 50 ≈ -18.38 ft

Since x represents a distance, we can discard the negative solution. So, the woman is approximately -18.38 ft from the base of the pole.

Therefore, when the woman is 50 ft from the base of the pole, the tip of her shadow is not moving.

User Mpj
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2 votes
Height of Woman, w = 6 ft.
Distance Between Woman and Pole, s = 35 ft.
Speed of Woman, ds/dt = 5 ft./sec
Length of Shadow = L
Distanc Between Woman and End of Shadow = x
p / (s + x) = w / x
13 / (35 + x) = 6 / x
13x = 6(35 + x)
13x = 210 + 6x
13x - 6x = 210
7x = 210
x = 210 / 7
x = 30

L = s + x
L = 35 + 30
L = 65

p / L = w / (L - s)
13 / L = 6 / (L - s)
13(L - s) = 6L
13L - 13s = 6L
13L - 6L = 13s
7L = 13s

User TranQ
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7.9k points