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 50 ft from the pole

20/3 ft/sec

Step-by-step explanation:

As given the diagram (attached),

x is the distance from the man to the pole, and y is the distance from the tip of the man's shadow to the pole.

I assume that the man and the pole are upright, meaning the 2 triangles are identical.

By similar triangles, we will get

15/y = 6/(y-x)

Triangles are similar if they have the same shape, but can be different sizes. In our case, triangles are in similar shape but in different sizes

15 (y-x) = 6y

15y-15x=6y

9y=15x

y=5/3 x

As we need to find rate of change (How fast), differentiate both sides with respect to  time t. we get

dy/dt = 5/3 dx/dt

As we have given that the man is walking from the with a speed to 4ft/s  

so dx/dt = 4, So, we need to find dy/dt (How fast the tip of shadow moving)

dy/dt = 5/3 *4

dy/dt = 20/3 or 6.666 ft/sec

In this case, the man's distance from the pole doesn't matter, because only his velocity influences how quickly his shadow goes.