Question

My Notes A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

Answer 1

Answer: 6.67ft/s

Step-by-step explanation: PLEASE SEE PICTURE ATTACHED, IT IS A DIAGRAM THAT HELPS YOU TO ANALYSE THE QUESTION.

STEP 1: DEFINE ALL VARIABLES

The man's shadow distance from the pole (Y)

The man's height=6ft

The height of the pole= 15ft

The man's shadow length (X-Y)

The man's speed = 4ft/s

STEP 2: FIND THE MAN'S SHADOW DISTANCE FROM THE POLE (Y)

((X-Y)÷Y) = 6ft÷15ft

Open up bracket and cross multiply

15(X-Y)=6Y

15X-15Y=6Y

collecting like terms together

15X=9Y

Y= 15X/9

Y= 5X/3

STEP 3: FIND THE SPEED OF THE MAN'S SHADOW

Dy/dt = 4ft/s

Therefore

Dy/dx= (4ft/s) × (5X/3)

20ft/3s = 6.67ft/s

The speed of the shadow a cross the pole is 6.67ft/s

Answer 2

6.667 ft/s

Explanation:

Given

Height of street light = 15ft

Height of man = 6ft

Speed away from pole = 4ft/s

Let x represents the distance between the man and the pole, and Let y represents the distance between the tip of the man's shadow to the pole.

This forms a similar triangle (see attachment below).

From similar triangles, we have

(y - x)/6 = y/15 ----- Solve equation

15(y - x) = 6 * y

15y - 15x = 6y ---- Collect Like Terms

15y - 6y = 15x

9y = 15x --- divide through by 9

y = 15x/9

y = 5x/3 ---- differentiate with respect to time, t

dy/dt = 5/3 dx/dt

dx/dt represents rate of change of distance per time = 4ft/s

while dy/dt represents rate of movement of his shadow tips

dy/dt = 5/3 * 4

dy/dt = 20/3 = 6.667 ft/s