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Here is the question:

A light is on the ground 20m form a building. A man 2m tall walks directly towards the building at 3m/s. How fast is the length of his shadow on the building changing when he is 8m from the building?

I got ds/dt=20m/s but I am kind of doubting myself since I'm incredibly bad in related rates. Can someone check this over and i'f wrong can please just show me why I'm wrong. Thanks

2007-11-16 08:45:40 · 1 answers · asked by Anonymous in Science & Mathematics Mathematics

1 answers

Did you draw a diagram? See mine: http://i14.tinypic.com/73bqky9.gif

The light is at A. The man is at B, walking toward C. His shadow is CD, whose length is denoted s. AC = 20.

Triangles ACD and ABE are similar. Therefore

CD/BE = AC/AB

so

s/2 = 20/x
s = 40/x

Now differentiate both sides with respect to t, then evaluate at the given values. Note that x is not the distance between the man and the building.

You should get a negative value for ds/dt. What does that mean?

2007-11-16 09:49:59 · answer #1 · answered by Ron W 7 · 0 0

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