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helllo

  • 3 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?

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  1. beens
    • 3 years ago
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    where is the light initially?

  2. helllo
    • 3 years ago
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    20 meters from the flagpole! coming from the top

  3. amistre64
    • 3 years ago
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    dL/dt = 5 ....dh/dt = 0 dx/dt =? dx/dt = dx/dL * dx/dt find dx/dL ?? maybe not tho.....

  4. helllo
    • 3 years ago
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  5. beens
    • 3 years ago
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    is there an initial height of the light?

  6. amistre64
    • 3 years ago
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    ahhh...... the shadow moving problem in all the textbooks :)

  7. rejanae13
    • 3 years ago
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    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
    • 3 years ago
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    thats a rough drawing by the way :)

  9. amistre64
    • 3 years ago
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    h/(20+x) = 10/(x) if that helps :)

  10. helllo
    • 3 years ago
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    lol sorry thats what was given to me :p

  11. amistre64
    • 3 years ago
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    y = (200+10x)/x

  12. amistre64
    • 3 years ago
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    dx/dt = x^2(dy/dt)/-200 ?? maybe

  13. amistre64
    • 3 years ago
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    dy/dy = 5

  14. amistre64
    • 3 years ago
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    x=5

  15. amistre64
    • 3 years ago
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    im gonna take a gander and say... 25(5) = 125 125/-200 = ?? -.0625 ?? if I did it right lol

  16. amistre64
    • 3 years ago
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    i think I got the "equation" wrong at the start....

  17. helllo
    • 3 years ago
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    lol thank you! ..but i dont understand how can you have dy/dy? :s

  18. helllo
    • 3 years ago
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    lol so the whole thing is wrong??

  19. amistre64
    • 3 years ago
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    :) fat fingers...little tiny keyboard :)

  20. helllo
    • 3 years ago
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    LMAO! ohhhh i see (:

  21. amistre64
    • 3 years ago
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    it may be right, but I dont trust it yet.... need to find the equation to derive; then the rest is easy.

  22. helllo
    • 3 years ago
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    ya but thats the part i cant get! wahh :(

  23. amistre64
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    this relate y and x..... now implicit it with respect to time right?

  26. helllo
    • 3 years ago
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    yea thats what i gotta do

  27. helllo
    • 3 years ago
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    and thank youuu! for everything so far btw!

  28. amistre64
    • 3 years ago
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    -200(dy/dt) / (y-10)^2 = dx/dt dy/dt = 5 y = 50;

  29. amistre64
    • 3 years ago
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    no prob; I live to math things up lol

  30. amistre64
    • 3 years ago
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    -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
    • 3 years ago
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    hahaha but where did u get negative two hundred?

  32. amistre64
    • 3 years ago
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    ok.... step it out.... 200 x = ------- (y-10) (y-10)(0) - (200)(dy/dt) dx/dt = ---------------------- (y-10)^2

  33. helllo
    • 3 years ago
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    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
    • 3 years ago
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    -200(dy/dt) dx/dt = ------------- (y-10)^2

  35. amistre64
    • 3 years ago
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    when y=50...its right there :)

  36. amistre64
    • 3 years ago
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    dy/dt is given already as the rate of change in "y"... = 5 m/s

  37. amistre64
    • 3 years ago
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    as y rises; x shrinks.... so its (-) direction..

  38. helllo
    • 3 years ago
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    ohhh okay! thank you so so much honestly, it means a lot (: (:

  39. amistre64
    • 3 years ago
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    just let me know if im right :)

  40. helllo
    • 3 years ago
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    i will haha but its the first time im on this site so how can i tell you?

  41. helllo
    • 3 years ago
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    because i have to check with my teacher first

  42. amistre64
    • 3 years ago
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    shout really loud? :) AMISTRE WAS RIGHT!!!.... or if im wrong it would be: AMISTRE WAS WRONG .... ill feel it either way :)

  43. helllo
    • 3 years ago
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    LMAOOOOOOOOO!

  44. amistre64
    • 3 years ago
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    that moving shadow thing is in all the calc textbooks If seen, and it always makes me nervous

  45. helllo
    • 3 years ago
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    lol im sure you got it right though, it seemed to make sense. i found it impossible!

  46. amistre64
    • 3 years ago
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    good luck :)

  47. helllo
    • 3 years ago
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    thank you again!:)

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