A woman 5 ft tall walks at the rate of 3.5 ft/sec away from a streetlight that is 12 ft above the ground. at what rate is the tip of her shadow moving? at what rate is her shadow lengthening? ⇒

Refer to the diagram shown below.

The woman walks at a rate of 3.5 ft/s away from the streetlight. Therefore
[tex] \frac{da}{dt} =3.5 \, ft/s[/tex]

The length of the shadow changes at the rate
[tex] \frac{dx}{dt} [/tex]

Because triangles ACE and BCD are similar  (AAA), therefore
[tex] \frac{AC}{BC}= \frac{AE}{BD} \\\\ \frac{a+x}{x} = \frac{12}{5} [/tex]
12x = 5a + 5x
7x = 5a

Therefore
[tex] \frac{dx}{dt} = \frac{5}{7} \frac{da}{dt} = \frac{5}{7} (3.5 \, ft/s) = 2.5 \, ft/s[/tex]

Answer: 2.5 ft/s