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anonymous

  • one year ago

A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out? i got -8/400 but it was wrong

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  1. freckles
    • one year ago
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    |dw:1434325170769:dw| picture looks something like this

  2. freckles
    • one year ago
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    |dw:1434325227788:dw|

  3. freckles
    • one year ago
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    "A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out? i got -8/400 but it was wrong " --- Moves horizontally at a speed of 8ft/s this means x'=8ft/s we are looking for theta (in rad) when y=200ft

  4. freckles
    • one year ago
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    well what is theta when y=200ft

  5. freckles
    • one year ago
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    what trig function can you use here

  6. freckles
    • one year ago
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    |dw:1434325371019:dw|

  7. freckles
    • one year ago
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    hint you have opp and hyp measurements given

  8. anonymous
    • one year ago
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    ok i did sin of 100/200 and got .479

  9. freckles
    • one year ago
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    arcsin(1/2)? which is pi/6

  10. anonymous
    • one year ago
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    why arcsin?

  11. freckles
    • one year ago
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    to solve sin(theta)=1/2 you normally take arcsin( ) of both sides since arcsin(sin(x))=x where x is between -pi/2 and pi/2 |dw:1434325633246:dw| anyways we are going to use x (because we are given x's rate) and theta for sure and I will go ahead and use that constant side \[\tan(\theta)=\frac{100}{x}\] differentiate both side

  12. anonymous
    • one year ago
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    sec^2x=0?

  13. freckles
    • one year ago
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    how did you get that equation?

  14. freckles
    • one year ago
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    sec is never 0

  15. anonymous
    • one year ago
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    im confused...

  16. freckles
    • one year ago
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    Like can you tell me what is confusing you please? Because I don't know how to help if I don't know where to start to help.

  17. anonymous
    • one year ago
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    the derivative of tan=100/x thats where i got confused

  18. freckles
    • one year ago
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    well the equation is actually \[\tan(\theta)=\frac{100}{x} \text{ or you can write it as } \\ \tan(\theta)=100x^{-1}\] so you don't have to use quotient rule if you prefer

  19. freckles
    • one year ago
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    you will need to use chain rule on both sides though

  20. freckles
    • one year ago
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    do you know how to evluate either one of these: \[\frac{d \tan(t)}{dt}=? \\ \frac{d(t^{-1})}{dt}=?\]

  21. freckles
    • one year ago
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    for the first it is easy just to remember derivative of tan(t) w.r.t t is sec^2(t) and the second one I asked you about can be done by power rule have you every used power rule before? I hope so because this question is about application of chain rule

  22. freckles
    • one year ago
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    \[\tan(\theta)=100x^{-1} \\ \text{ differentiate both sides } \\ \sec^2(\theta) \cdot \frac{d \theta}{d t}=100 \cdot \frac{d(x^{-1})}{dt}\] you still to differentiate the right hand side

  23. anonymous
    • one year ago
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    would it be 100/-x^2?

  24. freckles
    • one year ago
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    don't forget you need to write times dx/dt

  25. freckles
    • one year ago
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    \[\sec^2(\theta) \cdot \frac{d \theta}{ d t}=-\frac{100}{x^2} \cdot \frac{dx }{dt}\]

  26. freckles
    • one year ago
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    plug in all the values you found and are given and solve for dtheta/dt

  27. anonymous
    • one year ago
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    would it be sec^2(pi/6)*dtheta/dt=- 100/1ooroot3* 200?

  28. freckles
    • one year ago
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    well the x has a square on it and I thought dx/dt was given as 8ft/sec why you put 200 for dx/dt? dx/dt represents the rate and which the kite moves horizontally

  29. freckles
    • one year ago
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    \[\sec^2(\frac{\pi}{6}) \cdot \frac{ d \theta}{ d t}=\frac{-100}{(100 \sqrt{3})^2} \cdot 8\]

  30. anonymous
    • one year ago
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    i solved the right hand side but then how do i divide by the sec^2 im havinf trouble calculating that?

  31. freckles
    • one year ago
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    1/sec(k)=cos(k)

  32. freckles
    • one year ago
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    \[\frac{d \theta}{dt} =\frac{1}{\sec^2(\frac{\pi}{6})} \frac{-100}{(100 \sqrt{3})^2} \cdot 8 \\ \frac{d \theta}{ dt}=\cos^2(\frac{\pi}{6}) \frac{-100}{(100\sqrt{3})^2} \cdot 8\]

  33. anonymous
    • one year ago
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    okay i got -.o2?

  34. freckles
    • one year ago
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    ok and i have -1/50 those should be equal

  35. freckles
    • one year ago
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    don't forget to put your units

  36. freckles
    • one year ago
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    have fun

  37. anonymous
    • one year ago
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    wait

  38. anonymous
    • one year ago
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    my answer was -8/400 and i got it wrong?

  39. anonymous
    • one year ago
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    or is it not supposed to be negative?

  40. freckles
    • one year ago
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    you can try without did you put the units though ?

  41. anonymous
    • one year ago
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    the units are already tehre rad/s so i just had to type my answer in

  42. freckles
    • one year ago
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    k

  43. freckles
    • one year ago
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    have you got it yet?

  44. anonymous
    • one year ago
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    i only have one more try so im scared to type it and get it wrong

  45. freckles
    • one year ago
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    ok don't know what to tell you we have the answer above says it is decreasing at a rate of 1/50 rad/sec

  46. anonymous
    • one year ago
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    i know this thing is weird -_- but thank you for your help though!!

  47. anonymous
    • one year ago
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    it was supposed to be positive got it right!!!

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