1. ## Related rates.

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.

2. 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?

3. I just learned related rates and the only problems we did were much simpler than this one unless I'm over-complicating it. I have tried using similar triangles. I just have 2 y values, and 1 dx/dt. I don't know what to do next.

4. Originally Posted by chickeneaterguy
I just learned related rates and the only problems we did were much simpler than this one unless I'm over-complicating it. I have tried using similar triangles. I just have 2 y values, and 1 dx/dt. I don't know what to do next.
go to the link and scroll down to example 6 ...

Pauls Online Notes : Calculus I - Related Rates

5. Originally Posted by skeeter
go to the link and scroll down to example 6 ...

Pauls Online Notes : Calculus I - Related Rates
Thanks for helping. I still don't understand what to do. The examples on that site specified how far away from the light the guy was. I think I'm just missing something very obvious.

6. Originally Posted by chickeneaterguy
Thanks for helping. I still don't understand what to do. The examples on that site specified how far away from the light the guy was. I think I'm just missing something very obvious.
start by making a sketch of the problem, labeling the variable quantities.

let x = distance the man is from the lightpost

now ... what is the mathematical relationship between s and x ?

7. Originally Posted by skeeter
start by making a sketch of the problem, labeling the variable quantities.

let x = distance the man is from the lightpost

now ... what is the mathematical relationship between s and x ?
16/(s+x) = 6/s?

8. Originally Posted by chickeneaterguy
16/(s+x) = 6/s?
good ... now cross multiply, simplify and take the derivative of the resulting equation w/r to time.

9. 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?

10. Originally Posted by chickeneaterguy
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 the rate that the tip of his shadow is moving? Or did I mix up ds/dt and dx/dt?
yes, you did. also, note that dx/dt = -5 ft/s ... why negative?

ds/dt is the rate that the shadow length changes.

finally, ... what would d(x+s)/dt represent?

11. 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?

12. Originally Posted by chickeneaterguy
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?
part (a) wants to know how fast the tip of the shadow in moving w/r to the ground ... that would be d(x+s)/dt.

ds/dt is how fast the length of the shadow is changing ... you have to remember the shadow itself is moving w/r to the ground.

13. okay, so to get ds/dt...

10s = 6x

10ds/dt = 6dx/dt

10ds/dt = 6(-5ft/sec)

10ds/dt = -30ft/sec

ds/dt = -3ft/sec

Then for ds/dt + dx/dt...

-3ft/sec + -5ft/sec = -8ft/sec = rate that the tip of the shadow is moving?

14. to quote Porky Pig ... "that's all folks."

15. Thanks a lot. I feel stupid since the problem was really easy....I was just a bit confused. I have another problem but I'll give it a go before posting it.