A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 2.1 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

Question

anonymous · Answer

I made a  diagram

anonymous · Answer

|dw:1350149914701:dw|

anonymous · Answer

I can't get the right answer however.

anonymous · Answer

I also know $$\frac{ dx }{ dt }=2.1m/s$$ and I have to use the concept of similar triangles.

phi · Answer

I would let x be the distance from the spotlight (that way dx/dt has the correct sign, meaning x gets bigger with increasing time)

You called the height of the shadow y.  I would move the man closer to the wall (dx amount) and note that the shadow gets shorter.  So expect dy/dx to be negative. 

The question is asking for dy/dt,  and they give you dx/dt.  This suggests using
$$ \frac{dy}{dt}= \frac{dx}{dt}\cdot \frac{dy}{dx} $$

We need to find dy/dx.  That means we need an equation the relates x and y.  Similar triangles will work.
|dw:1350153205237:dw|
we see 
xy= 24
differentiate with respect to x and solve for dy/dx
Can you finish?