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Homework Help: The Calculus of Lawn Sprinklers

  1. Apr 22, 2015 #1
    1. The problem statement, all variables and given/known data

    A lawn sprinkler is constructed in such a way that ## \frac {d \theta}{dt}## is constant, where ##\theta## ranges between 45° and 135°. The distance the water travels horizontally is

    [tex] x = \frac {v^{2}sin2 \theta}{32}, 45^{\circ} \leq \theta \leq 135^{\circ} [/tex]

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

    2. The attempt at a solution

    [tex] \frac {dx}{dt} = \frac {d}{dt} [\frac {v^{2}sin2 \theta}{32}] [/tex]
    [tex] =(\frac {v^2}{32}) \frac {d}{dt}[sin2 \theta] [/tex]
    [tex] =\big( \frac {v^2}{16}\big) \frac {d \theta}{dt} [cos2 \theta] [/tex]

    What part of the lawn receives the most water?

    Here, my reasoning is that the part of the lawn that will receive the most water is that which is receiving water when ##\frac {dx}{dt} = 0## because the water will be hit with more water as the sprinkler changes direction (similar to how if you were to track the position of a ball in a parabolic trajectory, the highest density would be seen at the peak of the parabola, which is when ##\frac {dx}{dt} = 0## because the ball is changing directions). So:

    [tex] \frac {dx}{dt} = 0 = cos2 \theta [/tex]
    [tex] arccos 0 = 2 \theta [/tex]
    [tex] \frac {arccos 0}{2} = \theta [/tex]
    [tex] \theta = 45^{\circ}, 135^{\circ} [/tex]

    Is my reasoning correct? As for the first part of the question, "why does this lawn sprinkler not water evenly," my thought is that the watering is uneven due to the change in direction. If the sprinkler instead rotated in a circle with no change in ##\theta##'s direction, the watering would be even, correct?

    Also, I apologize for any sloppy LaTeX. I'm still learning. If it is preferred for the equations to be formatted differently (say, lined up horizontally as opposed to vertically), let me know! Thanks in advance!
    Last edited: Apr 22, 2015
  2. jcsd
  3. Apr 22, 2015 #2


    Staff: Mentor

    I think that a better view of the watering pattern can be realized by looking at the graph of x vs. ##\theta##. Also (minor points), the curve isn't a parabola, and there isn't a ball in this problem -- it's about a sprinkler head.
  4. Apr 22, 2015 #3
    Certainly; I did visualize x vs ##\theta##, but I wanted to return to the common example of a ball in an arc in order to justify/describe my thought process and make sure that my reasoning was solid. Thanks for the input!
  5. Apr 22, 2015 #4


    Staff: Mentor

    My point in looking at the graph of x as a function of θ was that the graph will show you exactly what parts of the lawn get watered.
  6. Apr 22, 2015 #5
    Actually, would you mind helping me figure out how to show x as a function explicitly of theta so I can graph it on Wolfram? I think I'm visualizing x v ##\theta## in my head properly, but would really like to test it.
  7. Apr 22, 2015 #6


    Staff: Mentor

    My thinking is that v, the speed of the water, is constant, so it suffices to look at x = sin(2θ), for θ between 45° and 135°. That graph will give you the shape, but not the actual size, of the portion of lawn being watered.
  8. Apr 22, 2015 #7
    Awesome, that's what I figured. I just wanted to be sure. I put it in wolfram, and it gave a 3D plot. I appreciate it!
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