# Thread: Problems with related rates

1. ## Problems with related rates

It has been a while since I last worked with derivatives so I may be forgetting something basic but here's my problem.

A street light is mounted on top of a 15 ft pole. A man 6 ft tall walks away from the pole at a rate of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

The equation used is $\displaystyle \frac{15}{6}=\frac{x+y}{y}$, where x is the distance from the pole to the man and y is the distance from the man to the tip of the shadow. It makes perfect sense when you picture it as a triangle and make a proportion from the lengths of the legs. Here is where I get stuck. You can multiply each side by y before or after you differentiate, and each way includes different values from the problem. For instance, if you don't multiply each side by y you keep x in the equation, but the left side becomes zero and thus the length of the pole and the man becomes irrelevant. Each way gives me 3.333, but the book gives the answer 8.333

2. Originally Posted by Chokfull
It has been a while since I last worked with derivatives so I may be forgetting something basic but here's my problem.

A street light is mounted on top of a 15 ft pole. A man 6 ft tall walks away from the pole at a rate of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

The equation used is $\displaystyle \frac{15}{6}=\frac{x+y}{y}$, where x is the distance from the pole to the man and y is the distance from the man to the tip of the shadow. It makes perfect sense when you picture it as a triangle and make a proportion from the lengths of the legs. Here is where I get stuck. You can multiply each side by y before or after you differentiate, and each way includes different values from the problem. For instance, if you don't multiply each side by y you keep x in the equation, but the left side becomes zero and thus the length of the pole and the man becomes irrelevant. Each way gives me 3.333, but the book gives the answer 8.333
the rate that the tip of the shadow moves across the ground is $\displaystyle \frac{dx}{dt} + \frac{dy}{dt}$