Question

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? ⇒

Answer 1

Refer to the diagram shown below.

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

The length of the shadow changes at the rate
$\frac{dx}{dt}$

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

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

Answer: 2.5 ft/s

Answer 2

Victor views his switch to mathematics as a very overwhelming experience, he tends to compare it to hard equations of physics and chemical formula .Victor has always struggled with STEM relevant subjects and to take up Mathematics has been one of the biggest challenges of his college course.