Question

A street light is mounted at the top of a 15-ft-tall pole. A man 6 feet tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast (in ft/s) is the tip of his shadow moving when he is 45 feet from the pole

Answer 1

Answer:

25/3 ft/s

Step-by-step explanation:

Height of pole , h=15 ft

Height of man, h'=6 ft

Let BD=x

BE=y

DE=BE-BD=y-x

All right triangles are similar

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

Therefore,

$\frac{AB}{CD}=\frac{BE}{DE}$

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

$\frac{5}{2}=\frac{y}{y-x}$

$5y-5x=2y$

$5y-2y=5x$

$3y=5x$

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

Differentiate w.r.t t

$\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}$

We have dx/dt=5ft/s

Using the value

$\frac{dy}{dt}=\frac{5}{3}(5)=\frac{25}{3}ft/s$

Hence, the tip of  his shadow moving  with a speed 25/3 ft/s when he is 45 feet from the pole.

Answer 2

Answer:

The tip pf the shadow is moving with speed 25/3 ft/s.

Step-by-step explanation:

height of pole = 15 ft

height of man = 6 ft

x = 45 ft

According to the diagram, dx/dt = 5 ft/s.

Now

$\frac{y-x}{y}=\frac{6}{15}\\\\15 y - 15 x = 6 y \\\\y = \frac{5}{3} x\\\\\frac{dy}{dt} = \frac{5}{3}\frac{dx}{dt}\\\\\frac{dy}{dt}=\frac{5}{3}\times 5 =\frac{25}{3} ft/s$