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Math Help - Another Related Rate Problem

  1. #1
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    Another Related Rate Problem

    Here is the text of my problem:

    A boy is flying a kite at a height of 150ft. If the kite moves gorizontally away from the boy at 20ft/s, how fast is the string being paid out when the kite is 250ft from him?









    Given:
    • y=150ft
    • x=250ft
    • \frac{dx}{dt} = 20ft/s
    Find:
    • \frac{dz}{dt} when x = 250
    Work:
    • by the theorem of pythagoras:
      • z = 50\sqrt{34}
    • \displaystyle\frac{dz}{dt} = \frac{x*\frac{dx}{dt} + y*\frac{dy}{dt}}{z}

    • \displaystyle\frac{dz}{dt} = \frac{(250*20)+ (150*0)}{50\sqrt{34}}

    • \displaystyle\frac{dz}{dt}=\frac{5000}{50\sqrt{34}  }= \frac{100}{\sqrt{34}}}
    The book goves an answer of 16ft/s. Can someone give me a hint of where I might have gone astray? Thanks!
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by dbakeg00 View Post
    Here is the text of my problem:

    A boy is flying a kite at a height of 150ft. If the kite moves gorizontally away from the boy at 20ft/s, how fast is the string being paid out when the kite is 250ft from him?



    Given:
    • y=150ft
    • x=250ft
    • \frac{dx}{dt} = 20ft/s
    Find:
    • \frac{dz}{dt} when x = 250
    Work:
    • by the theorem of pythagoras:
      • z = 50\sqrt{34}
    • \displaystyle\frac{dz}{dt} = \frac{x*\frac{dx}{dt} + y*\frac{dy}{dt}}{z}

    • \displaystyle\frac{dz}{dt} = \frac{(250*20)+ (150*0)}{50\sqrt{34}}

    • \displaystyle\frac{dz}{dt}=\frac{5000}{50\sqrt{34}  }= \frac{100}{\sqrt{34}}}
    The book goves an answer of 16ft/s. Can someone give me a hint of where I might have gone astray? Thanks!
    The distance from the person is the hypotenuse not the adjacent side.

    Another Related Rate Problem-capture.jpg

    This should fix your problem.
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  3. #3
    Member
    Joined
    Mar 2010
    Posts
    150
    ok, I see what you mean.

    That gives me this triangle:
    x=200
    y=150
    z=250

    which now gives me:
    \displaystyle\frac{dz}{dt}=\frac{(x*\frac{dx}{dt})  +(y*\frac{dy}{dt})}{z}

    \displaystyle\frac{dz}{dt}=\frac{(200*20)+(150*0)}  {250}

    \displaystyle\frac{dz}{dt}=\frac{4000}{250}=16ft/s

    Thanks for the help, appreciate it very much. These word problems are sometimes difficult for me to visualize if I'm not careful.
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