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 May 4th, 2009, 02:31 PM #2 Guest   Joined: Posts: n/a Thanks: Re: Related rates #2: Using similar triangles. Let x=distance between man and pole and y=length of his shadow. $\frac{x+y}{18}=\frac{y}{6}$ Solving for y: $y=\frac{1}{2}x$ $\frac{dy}{dt}=\frac{1}{2}\frac{dx}{dt}$ We are told dx/dt=200 $\frac{dy}{dt}=\frac{1}{2}(200)=100$

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# related rates y=x^2 x and y changing at same rate

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