The Calculus of Lawn Sprinklers

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In summary, we are given a lawn sprinkler with a constant ## \frac {d \theta}{dt}## where ##\theta## ranges from 45° to 135°. Using the equation ##x = \frac {v^{2}sin2 \theta}{32}##, we can determine the distance the water travels horizontally. The part of the lawn that receives the most water is when ##\frac {dx}{dt} = 0##, which occurs at angles of 45° and 135°. This uneven watering is due to the change in direction of the sprinkler. If the sprinkler rotated continuously in a circle, the watering would be even. By graphing x as a function of θ
  • #1
Cosmophile
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Homework Statement



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!
 
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  • #2
Cosmophile said:

Homework Statement



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:
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.
Cosmophile said:
[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!
 
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  • #3
Mark44 said:
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.

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!
 
  • #4
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.
 
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  • #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.
 
  • #6
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.
 
  • #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!
 

1. How does the calculus of lawn sprinklers work?

The calculus of lawn sprinklers involves using mathematical concepts such as derivatives and integrals to optimize the water coverage and efficiency of a lawn sprinkler system. By using calculus, we can determine the rate at which the sprinkler head rotates and the amount of water it distributes in order to achieve the most uniform and effective watering of a lawn.

2. What is the benefit of using the calculus of lawn sprinklers?

The benefit of using the calculus of lawn sprinklers is that it allows for more precise and efficient watering of a lawn. By using calculations to determine the optimal rate and direction of water distribution, we can minimize water waste and potentially save on water usage costs.

3. Can the calculus of lawn sprinklers be applied to any type of sprinkler system?

Yes, the principles of calculus can be applied to any type of sprinkler system, whether it is a traditional rotating sprinkler or a more advanced drip irrigation system. The main goal is to use mathematical calculations to optimize the distribution of water for the specific type of sprinkler being used.

4. Are there any limitations to using the calculus of lawn sprinklers?

One limitation of using the calculus of lawn sprinklers is that it requires a good understanding of mathematical concepts and equations. It may also be more difficult to implement for larger or more complex lawn areas. Additionally, external factors such as wind and temperature can also affect the effectiveness of the calculations.

5. Is the calculus of lawn sprinklers a proven method for achieving a healthy lawn?

While the calculus of lawn sprinklers has been studied and used by scientists and mathematicians, there is still ongoing research and experimentation in this field. However, using calculus to optimize water distribution has shown promising results in achieving a healthy and well-watered lawn.

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