Question

A streetlight sits on top of a 20-foot pole. A six-foot tall man walks away from the pole at a rate of 5 ft/sec. At what rate is the man’s shadow growing when the man is 10 feet from the pole?

Answer 1

Answer:

$\frac{25}{3}ft/s$

Step-by-step explanation:

We are given that

Height of pole,h=20 foot

Height of man,h'=6 foot

$\frac{dx}{dt}=5ft/s$

We have to find the rate at which the man's shadow growing when the man is 10 feet from the pole.

Let x be the distance of man from the pole and  y be the distance of tip of man's shadow from the pole.

Distance between man and shadow's tip=y-x

When two triangle are similar then the ratio of their corresponding sides are equal.

$\frac{6}{15}=\frac{y-x}{y}$

$6y=15y-15x$

$15x=15y-6y=9y$

$y=\frac{15}{9}x=\frac{5}{3}x$

Differentiate w.r.t t

$\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}=\frac{5}{3}\times 5=\frac{25}{3}ft/s$

Hence, the man's shadow growing at rate $\frac{25}{3}ft/s$ when the man is 10 feet from the pole.

Answer 2

Answer:

50/7 ft/s

Step-by-step explanation:

Height of pole = 20 ft

height of person = 6 ft

distance of man from pole is x and the length of image is y.

According to the diagram

$\frac{y-x}{y}=\frac{6}{20}$

20 y - 20 x = 6y

y = 20/14 x

y = 10/ 7 x

differentiate with respect to t

dy/dt = 10/7 dx/dt    .... (1)

dx/dt = 5 ft /s

Put in equation (1)

dy/dt = 10 (5) / 7

dy/dt = 50/7 ft/s