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NickR

  • 3 years ago

A kite 50ft above the ground moves horizontally at a speed of 8ft/s. At what rate is the angle between the string and the horizontal decreasing when 300ft of string has been let out?

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  1. Algebraic!
    • 3 years ago
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    |dw:1351143411523:dw|

  2. Algebraic!
    • 3 years ago
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    tan(theta) = 50/x differentiate both sides with respect to time find x when sqrt(x^2+50^2) =300 (*spoiler*) find theta when x= sqrt(300^2 -50^2) plug x, theta and dx/dt into the result of the differentiation to find d(theta)/dt

  3. NickR
    • 3 years ago
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    can you explain the diferrentiatie both sides with respect to time part?

  4. NickR
    • 3 years ago
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    @Algebraic!

  5. Algebraic!
    • 3 years ago
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    I guess, it's just implicit differentiation really...or the chain rule... or whatever you are comfortable thinking of it as... d/dt ( tan ( f(t) ) =

  6. Algebraic!
    • 3 years ago
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    derivative of the 'outside' is...?

  7. NickR
    • 3 years ago
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    sec^2

  8. Algebraic!
    • 3 years ago
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    :)

  9. NickR
    • 3 years ago
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    so it's x = 50/sec^2(theta)?

  10. Algebraic!
    • 3 years ago
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    yep, so sec^2 (f(t) ) * f '(t)

  11. Algebraic!
    • 3 years ago
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    f(t) is theta yeah, I wrote it that way because theta depends on time and I wanted you to see the chain rule

  12. Algebraic!
    • 3 years ago
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    so LHS is sec^2(theta) * d(theta) /dt RHS still needs to be differentiated...

  13. Algebraic!
    • 3 years ago
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    d/dt ( 1/(g(t)) ) =...?

  14. Algebraic!
    • 3 years ago
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    -1/ ..........

  15. NickR
    • 3 years ago
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    what is g(t) in this case?

  16. Algebraic!
    • 3 years ago
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    well, just because I used f(t) for theta... now we're talking about x, which is a different function of time...

  17. NickR
    • 3 years ago
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    ah okay

  18. NickR
    • 3 years ago
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    -1/x^2*dx/dt

  19. Algebraic!
    • 3 years ago
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    great:)

  20. Algebraic!
    • 3 years ago
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    so:\[\frac{ d \theta }{dt } = \frac{ -1 }{x^2 \sec^2\theta }\frac{ d x}{ dt }\]

  21. Algebraic!
    • 3 years ago
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    dx/dt is given.. x and theta are easy to find...

  22. NickR
    • 3 years ago
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    dx/dt = 8

  23. NickR
    • 3 years ago
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    d(theta)/dt = -8/x^2*sec(theta)

  24. Algebraic!
    • 3 years ago
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    x= sqrt(300^2 -50^2) theta = arctan (50/x)

  25. NickR
    • 3 years ago
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    wouldn't x = sqrt(300^2 + 50^2)?

  26. NickR
    • 3 years ago
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    x = 304.138

  27. NickR
    • 3 years ago
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    theta = .1629 rads

  28. NickR
    • 3 years ago
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    @Algebraic!

  29. Algebraic!
    • 3 years ago
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    so x is longer than the hypotenuse?

  30. NickR
    • 3 years ago
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    x = 295.803

  31. Algebraic!
    • 3 years ago
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    k:)

  32. NickR
    • 3 years ago
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    so d(theta)/dt = -8/((295.803)^2*(sec^2(.16744 rads)) which is incorrect.

  33. Algebraic!
    • 3 years ago
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    -8.87E-5?

  34. NickR
    • 3 years ago
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    well you said d(theta)/dt = -1/(x^2*sec^2(theta)) * dx/dt

  35. Algebraic!
    • 3 years ago
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    yes

  36. Algebraic!
    • 3 years ago
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    you didn't get -8.87E-5?

  37. NickR
    • 3 years ago
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    yes I did

  38. Algebraic!
    • 3 years ago
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    and it says it's wrong...

  39. NickR
    • 3 years ago
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    yep

  40. Algebraic!
    • 3 years ago
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    dunno let me look it all over. I don't see any glaring mistakes...

  41. NickR
    • 3 years ago
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    alright

  42. Algebraic!
    • 3 years ago
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    -8.89E-5 rad.s/sec .. best I can do... let's see who else is on, might be able to see if I made a mistake...

  43. RolyPoly
    • 3 years ago
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    May I have the answer to the question?

  44. Algebraic!
    • 3 years ago
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    go for it.

  45. Algebraic!
    • 3 years ago
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    @eSpeX

  46. NickR
    • 3 years ago
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    @RolyPoly id like the answer too!

  47. Algebraic!
    • 3 years ago
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    @callisto is checking it over...

  48. Algebraic!
    • 3 years ago
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    ty @Callisto

  49. Callisto
    • 3 years ago
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    I... am no good at maths... \[tan \theta = \frac{50}{x}\] Differentiate both sides with respect to x. Probably you won't get: \[\frac{ d \theta }{dt } = \frac{ -1 }{x^2 \sec^2\theta }\frac{ d x}{ dt }\]

  50. Callisto
    • 3 years ago
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    My bad, I meant with respect to t

  51. Algebraic!
    • 3 years ago
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    you're saying the differentiation is wrong?

  52. Callisto
    • 3 years ago
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    \[tan \theta = \frac{50}{x}\] Diff. both sides w.r.t. t \[sec^2\theta \frac{d\theta}{dt} = \frac{-50}{x^2} \frac{dx}{dt}\]\[\frac{d\theta}{dt} = \frac{-50}{x^2sec^2\theta } \frac{dx}{dt}\]

  53. Algebraic!
    • 3 years ago
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    arg

  54. Algebraic!
    • 3 years ago
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    forgot the 50 rfl

  55. Algebraic!
    • 3 years ago
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    you so smart cally

  56. Algebraic!
    • 3 years ago
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    :)

  57. NickR
    • 3 years ago
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    -50*-8/((295.803^2)*sec^2(.16744)) = correct answer

  58. NickR
    • 3 years ago
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    thank you @Callisto and @Algebraic!

  59. Algebraic!
    • 3 years ago
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    yes thanks @Callisto !

  60. Callisto
    • 3 years ago
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    \[sec^2\theta=\frac{x^2+50}{x^2}\]You can cancel the x^2 since \[\frac{d\theta}{dt} = \frac{-50}{x^2sec^2\theta } \frac{dx}{dt}=\frac{-50}{x^2(\frac{x^2+50}{x^2}) } \frac{dx}{dt}\]\[=\frac{-50}{(x^2+50) } \frac{dx}{dt}\]And x^2 is easy to find

  61. NickR
    • 3 years ago
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    only took me 4hrs but i got my question worth 1pt right now to get some sleep

  62. Algebraic!
    • 3 years ago
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    heh

  63. NickR
    • 3 years ago
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    you guys have a good night thanks again

  64. Callisto
    • 3 years ago
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    Good night!!~

  65. robtobey
    • 3 years ago
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    A revised Mathematica solution, posted on 25 October 2012, 22:15 California time, is attached.

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