A street light is at the top of a 15 foot tall pole. A 6 foot tall woman 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 50 feet from the base of the pole?

1 Answer
May 10, 2018

$d ' \left({t}_{0}\right) = \frac{20}{3} = 6 , \overline{6}$ ft/s

Explanation:

Using Thales Proportionality theorem for the triangles $A \hat{O} B$, $A \hat{Z} H$

The triangles are similar because they have $\hat{O} = 90$°, $\hat{Z} = 90$° and $B \hat{A} O$ in common.

We have $\frac{A Z}{A O} = \frac{H Z}{O B}$ $\iff$

ω/(ω+x)=6/15 $\iff$

15ω=6(ω+x) $\iff$

15ω=6ω+6x $\iff$

9ω=6x $\iff$

3ω=2x $\iff$

ω=(2x)/3

Let $O A = d$ then

d=ω+x=x+(2x)/3=(5x)/3

• $d \left(t\right) = \frac{5 x \left(t\right)}{3}$

• $d ' \left(t\right) = \frac{5 x ' \left(t\right)}{3}$

For $t = {t}_{0}$ , $x ' \left({t}_{0}\right) = 4$ ft/s

Therefore, $d ' \left({t}_{0}\right) = \frac{5 x ' \left({t}_{0}\right)}{3}$ $\iff$

$d ' \left({t}_{0}\right) = \frac{20}{3} = 6 , \overline{6}$ ft/s