Question

a street light is at the top of a 15 ft tall pole. a woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a straight path. how fast is the tip of her shadow moving when she is 30 ft from the base of the pole?

\\(\frac{\text{ft}}{\text{sec}}\\)

Explanation:

Set up variables and similar triangles

Let \(x\) be the distance from the woman to the base of the pole, and \(s\) be the distance from the tip of her shadow to the base of the pole.
Using similar triangles:

$$ \frac{s}{15} = \frac{s - x}{6} $$

Relate the rates of change

Solve for \(s\) in terms of \(x\):

$$ 6s = 15(s - x) \implies 6s = 15s - 15x \implies 9s = 15x \implies s = \frac{5}{3}x $$

Differentiate both sides with respect to \(t\):

$$ \frac{ds}{dt} = \frac{5}{3}\frac{dx}{dt} $$

Calculate the final rate

Substitute the given rate \(\frac{dx}{dt} = 4\text{ ft/sec}\):

$$ \frac{ds}{dt} = \frac{5}{3}(4) = \frac{20}{3} $$

Answer:

A street light is at the top of a 15 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 30 ft from the base of the pole?

<blank>\(\frac{20}{3}\)</blank> \(\frac{\text{ft}}{\text{sec}}\)