Question : A street light is at the top of a 13 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 50 ft from the base of the pole?
I drew up a triangle with the vertical height as 13 and a horizontal height of 50, but I could not figure out how to proceed after that. Is this a case involving similar triangles and then finding the derivative using implicit differentiation? If so, what equation should I use for the differentiation?
Let $x$ equal the distance the woman is from the pole. Let $y$ represent the distance the woman is to the tip of her shadow. So I can setup the following equation $\frac{13}{x+y}=\frac{6}{y}$. I would make this easy and cross multiply and then differentiate. Your goal is to find $(x+y)'$. By the way you already know $x'=4 \frac{ft}{sec}$