Question 1194694
Here is the man, 6 ft tall, walking towards the light
on a horizontal ground surface, shown when he is at a distance x (in ft) from the light, projecting a shadow with a length s (in ft) on the wall:
{{{drawing(600,300,-3,33,-3,15,
triangle(0,0,18,0,18,6),
red(triangle(0,0,30,0,30,10)),
line(-5,0,35,0),
line(30,0,30,33),
red(circle(0,0,0.3)),
red(circle(0,0,0.2)),
locate(-1,6,red(light)),
red(arrow(0,5,0,0.4)),
rectangle(30,0,29,1),rectangle(18,0,17,1),
locate(8,0.9,x),locate(29.5,5,s),locate(18.2,4,6ft),
locate(30.3,13,wall), locate(20,0,ground),
arrow(15,-1.5,0,-1.5),arrow(15,-1.5,30,-1.5),
locate(14,-1.5,30ft)
)}}}
If we count time from the moment {{{t=0}}} when the man starts walking fron the light,
{{{x=5t}}} with {{{t}}} in seconds.
There are two similar right triangles, so
{{{s/30=6/x}}} --> {{{s=180/x}}}
and if we want {{{s}}} as a function of {{{t}}} 
{{{x=180/5t}}} --> {{{x=36/t}}}
The rate of change of a function with its variable is the derivative of the function, so we could write
{{{ds/dt=-36/t^2}}} and then
calculate the value of {{{ds/dt}}} when{{{5t=15}}} <--> {{{t=3}}} as
{{{-36/3^2=-36/9=highlight(-4)}}} {{{ft/s}}}
which means the shadow's length is decreasing at {{{highlight(4)}}}{{{ft/second}}} at the instant the man is at 15 ft from the light.
 
Alternately, we could write {{{ds/dt=(ds/dx)(dx/dt)}}} and knowing that {{{dx/dt=5}}} as long as the man is walking between light and building
regardless of how we count the time we could calculate {{{ds/dt}}} ,
without caring how long the man has to walk to get 15 ft from the light, we calculate
{{{ds/ds=-180/x^2}}} and for {{{x=15}}} {{{ds/dx=-180/15^2}}} and {{{ds/dt=(-180/15^2)*5=-4}}} to get the same result.