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

abidemiokin abidemiokin · Answer 2 · 2022-01-10T13:35:53+01:00

The rate at which her shadow is lengthening is 2.5ft/sec

Triangular altitude theorem:

To get the rate of her shadow lengthening, we will use the similarity theorem of a triangle.

Find the triangle attached. From the triangle;

[tex]\frac{EA}{AC} =\frac{DB}{BC}[/tex]

Substitute the sides as given in the diagram:

[tex]\frac{12}{x+a} =\frac{5}{x}\\12x = 5(x+a)\\ 12x=5x+5a\\7x = 5a[/tex]

Differentiate both sides with respect to t

[tex]7\frac{dx}{dt} = 5\frac{da}{dt} \\7\frac{dx}{dt}=5(3.5)\\7\frac{dx}{dt}=17.5\\\frac{dx}{dt} = 2.5ft/sec[/tex]

The rate at which her shadow is lengthening is 2.5ft/sec

Learn more on similar triangles here: https://brainly.com/question/2644832