Date: 10/13/2003 at 13:49:13 From: Chris Subject: Mathematical Methods A man 6 feet tall walks at 4 miles an hour directly away from a lampost 18 feet tall. Why does his shadow lengthen at a constant rate? What is this rate and how can I work it out?
Date: 10/15/2003 at 11:54:52 From: Doctor Luis Subject: Re: Mathematical Methods Hi Chris, If you let h be the height of the man, L be the height of the lamppost, x be the distance of the man from the lamppost, and y be the length of the shadow, then you can see from the following diagram that the proportion L/(x+y) = h/y holds, from similar triangles. So we have Ly = h(x + y), or (L - h)y = hx. Since x and y are related linearly, then so are dx/dt and dy/dt. This means that if dx/dt is constant, then dy/dt is also constant. You can use the equation between x and y to find the relationship between dx/dt and dy/dt. I hope this helped! Let us know if you have any more questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/
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