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Math Help - Application! Sprinklers and water speed.

  1. #1
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    Application! Sprinklers and water speed.

    Here is the exact problem:

    Lawn Sprinkler: A lawn sprinkler is constructed in such a way that \frac{dy}{dt} is constant, where y ranges between 45(degrees) and 135(degrees). The distance the water travels horizontally is:

    x = (V^2 sin(2y))/32

    where v is the speed of the water. Find \frac {dx}{dt} and explain why the water does not water evenly. What part of the lawn receives the most water?

    Here is the image.
    Application! Sprinklers and water speed.-sprinkelrs.jpg

    If anyone could walk me through this, that would be awesome!
    Last edited by TylaYuhh; September 25th 2011 at 12:42 PM. Reason: I can't use LaTex very well...
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  2. #2
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    Re: Application! Sprinklers and water speed.

    Quote Originally Posted by TylaYuhh View Post
    Here is the exact problem:

    Lawn Sprinkler: A lawn sprinkler is constructed in such a way that \frac{dy}{dt} is constant, where y ranges between 45(degrees) and 135(degrees). The distance the water travels horizontally is:

    x = (V^2 sin(2y))/32

    where v is the speed of the water. Find \frac {dx}{dt} and explain why the water does not water evenly. What part of the lawn receives the most water?

    Here is the image.
    Click image for larger version. 

Name:	Sprinkelrs.JPG 
Views:	14 
Size:	10.5 KB 
ID:	22409

    If anyone could walk me through this, that would be awesome!
    start by taking this derivative w/r to time ...

    \frac{v^2}{32} \cdot \frac{d}{dt}[\sin(2y)]
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  3. #3
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    Re: Application! Sprinklers and water speed.

    Thanks I am going to give that a shot. I'll let you know my results!
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    Re: Application! Sprinklers and water speed.

    Quote Originally Posted by skeeter View Post
    start by taking this derivative w/r to time ...

    \frac{v^2}{32} \cdot \frac{d}{dt}[\sin(2y)]
    w/r = water/rate?
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  5. #5
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    Re: Application! Sprinklers and water speed.

    Quote Originally Posted by TylaYuhh View Post
    w/r = water/rate?
    with respect to
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    Re: Application! Sprinklers and water speed.

    Quote Originally Posted by skeeter View Post
    with respect to
    Oh okay.
    I think I did this right:

    x' = (32(2v sin(2y) + 2v^2 cos(2y))) / 32^2
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    Re: Application! Sprinklers and water speed.

    Quote Originally Posted by TylaYuhh View Post
    Oh okay.
    I think I did this right:

    x' = (32(2v sin(2y) + 2v^2 cos(2y))) / 32^2
    v is a constant
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  8. #8
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    Re: Application! Sprinklers and water speed.

    Quote Originally Posted by skeeter View Post
    v is a constant

    Oops! I hope I got it right this time.

    x' = 32(2v^2 cos(2y)) / 32^2

    I think I messed up here too though. If it is dx/dy, that means I've got to use implicit differentiation on the sin(2y)?
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  9. #9
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    Re: Application! Sprinklers and water speed.

    x = \frac{v^2}{32} \cdot \sin(2y)

    \frac{dx}{dt} = \frac{v^2}{16} \cdot \cos(2y) \cdot \frac{dy}{dt}


    \frac{v^2}{16} and \frac{dy}{dt} are both constant ... what happens to \frac{dx}{dt} as the angle y of the sprinkler varies?
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  10. #10
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    Re: Application! Sprinklers and water speed.

    Quote Originally Posted by skeeter View Post
    x = \frac{v^2}{32} \cdot \sin(2y)

    \frac{dx}{dt} = \frac{v^2}{16} \cdot \cos(2y) \cdot \frac{dy}{dt}


    \frac{v^2}{16} and \frac{dy}{dt} are both constant ... what happens to \frac{dx}{dt} as the angle y of the sprinkler varies?
    The water is spread out further?

    Sorry, I'm just having trouble with this one D: ...Thanks for helping me on this.
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    Re: Application! Sprinklers and water speed.

    The fact that, as skeeter says, \frac{v^2}{16} and \frac{dy}{dt} are both constant, tells you that the entire C= \frac{v^2}{16}\frac{dy}{dt} is a constant and so \frac{dx}{dt}= C cos(2y). That is not a constant because y is not a constant and so cos(2y) is not a constant.
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  12. #12
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    Re: Application! Sprinklers and water speed.

    Quote Originally Posted by HallsofIvy View Post
    The fact that, as skeeter says, \frac{v^2}{16} and \frac{dy}{dt} are both constant, tells you that the entire C= \frac{v^2}{16}\frac{dy}{dt} is a constant and so \frac{dx}{dt}= C cos(2y). That is not a constant because y is not a constant and so cos(2y) is not a constant.

    Thank you for the input. It helped =D
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