Calc Problem [Related Rates]
 

Problem
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A spotlight on the ground shines on a wall 12m away. If a man 2m tall walks from the spotlight toward the building at a speed of 1.6 m/s, how fast is the length of his shadow on the building decreasing when he is 4m from the building?

What I did
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I used related triangles to say:
x/2 = 12/y
xy=24

Taking the derivitive of both sides, I get:
x(dy)+y(dx)=0

Substituting:
3(1.6)+8(dx)=0
*The 8 comes from him being 4m from the wall, which is 12m from the light

4.8+8(dx)=0
8dx=4.8
dx = .6


I know this is wrong, can someone figure this out please? Thanks.

You (almost)get the right answer. Your mistake is in the second to last line. When you move the 4.8 over. That should be negative. 8dx=-4.8 therefore dx=-.6.

er, yeah... that's what i had on paper.

Basically the instructor gave the problem and the answer and we were to show the work. Well I found out today that the answer the instructor gave was wrong and this was correct :/

Thanks vinaur for the reply