Have you tried nothing at all? Look at the hint! Draw a picture showing the street light, the man, and the tip of his shadow. Draw the line from the street light to the tip of the shadow. Do you not see the similar triangles refered to?
A man 6 feet tall walks at the rate of 5 ft/sec toward a streetlight that is 16 feet above the ground. (a) At what rate is the tip of his shadow moving? [Hint: Use similar triangles] (b) At what rate is the length of his shadow changing? [Hint: Use similar triangles] I have no idea how to figure this one out.
go to the link and scroll down to example 6 ...
Pauls Online Notes : Calculus I - Related Rates
so first cross multiply: 16s = 6s + 6x --> 10s = 6x. Next Take d/dt: 10 ds/dt = 6 dx/dt since the man walks at a rate of 5ft/sec, that's ds/dt? 10(5ft/sec) = 6 dx/dt --> 50ft/sec = 6 dx/dt --> (50/6)ft/sec = dx/dt --> (25/3)ft/sec = dx/dt --> 8.333 ft/sec = dx/dt or did I mix up dx/dt with ds/dt?
I had it as negative but erased it because I'm an idiot
It's negative because we drew it with the pole being on the left of the man, so the man is moving towards the pole/light.
I'm not sure what d(x+s)/dt would represent...but since x+s is the distance from the pole to the tip of the man's shadow, I guess it would represent that rate?
Isn't ds/dt all I need for part a?