Question

A person whose height is 6 feet is walking away from the base of a streetlight along a straight path at a rate of 4 feet per second. If the height of a streetlight is 15 feet, what is the rate at which a person’s shadow is lengthening?

Answer 1

Let the distance from the pole be x and the length of the shadow be s. Using the law of similar triangles, we have:
s/6=(x+s)/15
thus
5s=2x+2s
3s=2x

hence
ds/dt=2/3 dx/dt
but
dx/dt=4 ft per second
hence
ds/dt=2/3*4=8/3

Answer 2

The rate at which a person’s shadow is lengthening is

$\rm \dfrac{8}{3}\;ft/sec$

Step-by-step explanation:

Given :

Height of a person = 6 feet

Walking at speed of 4 feet/sec.

Height of a streetlight = 15 feet

Calculation :

Let the distance from the pole be x and the length of the shadow be y.

Using similar triangle property

$\dfrac{y}{6}= \dfrac{x+y }{15}$

$9y=6x$

$y=\dfrac{2x}{3}$   ------ (1)

Differentiate equation (1) with respect to time, we get

$\dfrac{dy}{dt}=\dfrac{2}{3}\dfrac{dx}{dt}$ ----- (2)

Here,

$\rm \dfrac{dx }{dt }= 4 ft/sec$

From equartion (2), we get

$\dfrac{dy}{dt}= \dfrac{8}{3}$

The rate at which a person’s shadow is lengthening is

$\rm \dfrac{8}{3}\;ft/sec$

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https://brainly.com/question/6837143?referrer=searchResults