A street light is at the top of a 12.0 ft. tall pole.

Question

A street light is at the top of a 12.0 ft. tall pole. A man 6.0 ft tall walks away from the pole with a speed of 3.5 feet/sec along a straight path. How fast is the tip of his shadow moving when he is 31 feet from the pole?

Your answer:______ ft/sec

Answer 1

Create depiction with 12 ft as light pole on left side of right triangle and man somewhere in middle.

Label distance from light pole to man as y, and distance from man to tip of triangle as x. So base of right triangle is x+y and height is 12. Man in the middle somewhere.

By similar triangles 6/x = 12/(x+y) you can determine that irrespective of the values of x,y, x = 6 for the relationship to remain.

Therefore x^2 + 36 = r^2. Differentiating both sides with respect to T and applying the chain rule results in:

2x * (dx/dt) = 2r * (dr/dt). Simplify by dividing both sides by 2 or

X*dx/dt = r*dr/dt.

You can determine R by Pythagorean Theorem to be 31.575 ft when x = 31 feet

Therefore 31/3.5 = 31.575 (dr/dt)

Dr/dt = 0.28 ft/sec