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(1 point) a street light is at the top of a 16.5 ft. tall pole. a man 5.5 ft tall walks away from the pole with a speed of 7.0 feet/sec along a straight path. how fast is the tip of his shadow moving when he is 33 feet from the pole? your answer: ft/sec hint: draw a picture and use similar triangles.

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Using similar triangles, the tip of the man’s shadow is moving at a rate of 10.5 ft/sec when he is 33 feet from the pole.

Similar triangles are triangles with the same shape but differences in size.

Similar triangles have congruent corresponding angles and equal proportional corresponding sides.

The height of the street light at the top of a tall pole = 16.5 ft.

The height of the man who walks away from the pole = 5.5 ft

The man's walking speed = 7.0 feet/second

The last distance of the man from the pole = 33 feet

Equation:

Let the distance between the man and the pole = x

Let the length of the man’s shadow = y

Therefore,

y / (16.5 ft) = (y + 5.5 ft) / x

Simplifying this equation:

y = (16.5 ft * 5.5 ft) / (x - 5.5 ft)

Differentiating both sides concerning time:

dy / dt = (16.5 ft * 5.5 ft) / (x - 5.5 ft)^2 * dx / dt

When the man is 33 feet from the pole, we have:

x = 33 ft

dx / dt = 7.0 ft/sec

Substituting these values into the equation above:

dy / dt = (16.5 ft * 5.5 ft) / (27.5 ft)^2 * 7.0 ft/sec

Simplifying the above expression:

dy / dt = 10.5 ft/sec

Thus, the tip of the man’s shadow is moving at a rate of 10.5 ft/sec when he is 33 feet from the pole.

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