Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

# Calc hw help?

A street light is mounted at the top of a 15-ft-tall pole. A man 6 feet tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast (in ft/s) is the tip of his shadow moving when he is 30 feet from the pole?

Relevance
• Let the man is x feet away from the pole and the length of the shadow be y, Now [(15-6)/x] = 6/y; 3/x = 2/y; y = 2x/3, dy/dt = (2/3)(dx/dt) = (2/3)*5 = 10/3 ft/s = 3.33 ft/s. His depart from the pole is immaterial for this case.

• Let distance of man’s feet from base of pole be x, so that dx/dt = 5 ft/s

Let distance of shadow tip from base of pole be s, so that shadow length

is x – s, and by similar triangles, s/15 = (s – x)/6

s = (5/3)x

ds/dt = ds/dx * dx/dt = (5/3)*5 = 25/3 ft/s

Note: We do not need “he is 30 feet from the pole” in this example,

because the ratio of speeds only depends on lamp and man's height.

For a 6 ft man with pole height h the ratio of speeds is h/(h – 6)

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• s = distance from pole to tip of shadow.

x = distance from pole to man.

Similar triangles dictate:

15/s = 6/(s-x), or

s = (5/3)x.

ds/dt = (ds/dx)*(dx/dt)

(5/3)(5 ft/s).

The tip of the shadow moves at 8.33 ft/s no matter how far he is from the pole.