Calculus ~ Related Rates
posted by Mishaka on .
A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate is the tip of his shadow moving?We've already set this up part of the way. We know that dx/dt = - 2 meters per second, and we're looking for dy/dt.
From what you said, I will assume you defined variables as such....
x is the distance between the man and the streetligh
y is the length of his shadow.
then dx/dt = -2 , find dy/dt
by similar triangles:
2/y = 5/(x+y)
5y = 2x + 2y
3y = 2x
3dy/dt = 2dxdt
dy/dt = 2(-2)/3 = -4/3
so his shadow is decreasing at a rate of 4/3 m/s
(notice , since I said "decreasing", I do not use the negative sign, since the word decreasing implies it)