## helllo 4 years ago PLEASE HELP! PROBLEM ON RELATED RATES :) A ball is located at point Q on top of a flagpole 10m above the ground. a light is being raised vertically at the rate of 5 m/s casting the shadow of Q on the ground. how fast is the shadow moving along the ground when the light is 50m high?

1. beens

where is the light initially?

2. helllo

20 meters from the flagpole! coming from the top

3. amistre64

dL/dt = 5 ....dh/dt = 0 dx/dt =? dx/dt = dx/dL * dx/dt find dx/dL ?? maybe not tho.....

4. helllo

5. beens

is there an initial height of the light?

6. amistre64

ahhh...... the shadow moving problem in all the textbooks :)

7. rejanae13

well you need to cross multiply and divide. so it would be 20m/50m by 5m/x.... that would be 50x by 100... divide 50 by 100 to get x alone and you get x=2

8. amistre64

thats a rough drawing by the way :)

9. amistre64

h/(20+x) = 10/(x) if that helps :)

10. helllo

lol sorry thats what was given to me :p

11. amistre64

y = (200+10x)/x

12. amistre64

dx/dt = x^2(dy/dt)/-200 ?? maybe

13. amistre64

dy/dy = 5

14. amistre64

x=5

15. amistre64

im gonna take a gander and say... 25(5) = 125 125/-200 = ?? -.0625 ?? if I did it right lol

16. amistre64

i think I got the "equation" wrong at the start....

17. helllo

lol thank you! ..but i dont understand how can you have dy/dy? :s

18. helllo

lol so the whole thing is wrong??

19. amistre64

:) fat fingers...little tiny keyboard :)

20. helllo

LMAO! ohhhh i see (:

21. amistre64

it may be right, but I dont trust it yet.... need to find the equation to derive; then the rest is easy.

22. helllo

ya but thats the part i cant get! wahh :(

23. amistre64

I am thinking it is the relation of congruent triangles; "y" being the height of the lamp: x = length of shadow. 10/x = y/(20+x) is what I come up with; there is no question about the distance of the light to the tip of the shadow...

24. amistre64

10(20+x) = y(x) (200+10x)/x = y 200/x + 10 = y 200/x = y-10 200/(y-10) = x ??? sounds good

25. amistre64

this relate y and x..... now implicit it with respect to time right?

26. helllo

yea thats what i gotta do

27. helllo

and thank youuu! for everything so far btw!

28. amistre64

-200(dy/dt) / (y-10)^2 = dx/dt dy/dt = 5 y = 50;

29. amistre64

no prob; I live to math things up lol

30. amistre64

-0.625 is what i get, and I am almost sure its right... but i get that nagging feeling that I am just being stupid :)

31. helllo

hahaha but where did u get negative two hundred?

32. amistre64

ok.... step it out.... 200 x = ------- (y-10) (y-10)(0) - (200)(dy/dt) dx/dt = ---------------------- (y-10)^2

33. helllo

lol im on the phone with my friend trying to figure this out and im reading her what u wrote and we both agreed your funny, just thought id let you know :p but yea we dont understand also where you got the 50 from? (: care to explain?

34. amistre64

-200(dy/dt) dx/dt = ------------- (y-10)^2

35. amistre64

when y=50...its right there :)

36. amistre64

dy/dt is given already as the rate of change in "y"... = 5 m/s

37. amistre64

as y rises; x shrinks.... so its (-) direction..

38. helllo

ohhh okay! thank you so so much honestly, it means a lot (: (:

39. amistre64

just let me know if im right :)

40. helllo

i will haha but its the first time im on this site so how can i tell you?

41. helllo

because i have to check with my teacher first

42. amistre64

shout really loud? :) AMISTRE WAS RIGHT!!!.... or if im wrong it would be: AMISTRE WAS WRONG .... ill feel it either way :)

43. helllo

LMAOOOOOOOOO!

44. amistre64

that moving shadow thing is in all the calc textbooks If seen, and it always makes me nervous

45. helllo

lol im sure you got it right though, it seemed to make sense. i found it impossible!

46. amistre64

good luck :)

47. helllo

thank you again!:)