Posted by Anonymous on .
Suppose a 6ft. man is 10 ft. away from a 24 ft. tall lamp post. If the person is moving away from the lamp post at at rate of 2 ft/sec, at what rate is the length of his shadow changing?

Calculus 
Reiny,
let the man be x ft from the lamppost.
At that time, let the length of his shadow be y ft
by ratios:
24/(x+y) = 6/y
24y = 6x+6y
18y=6x
3y=x
3dy/dt = dx/dt
but dx/dt = 2
dy/dt = 2/3 ft/s
so his shadow is increasing in length at 2/3 ft/sec,
notice that the end of his shadow would be MOVING at 2 +2/3 or 8/3 ft/se
also notice that the fact that he is 24 feet from the post does not enter the picture at all. 
Calculus 
Anonymous,
Why is it "let the man be x ft from the lamppost. "
Why not 10 feet away?
& Why is it "also notice that the fact that he is 24 feet from the post does not enter the picture at all"
When the lamp post is 24 feet tall? 
Calculus 
Reiny,
I put the wrong number in when I made that last comment. It should have said " .... he is 10 feet from ..."
It has no effect on the solution.
We cannot use the 10 feet in the "general" case, the 10 feet is one specific instant.
in rate of change problems you <bnever use the specific case data until you have differentiated the equation.