Posted by **Mishaka** on Thursday, January 5, 2012 at 6:00pm.

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.

- Calculus ~ Related Rates -
**Reiny**, Thursday, January 5, 2012 at 6:52pm
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)

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