- #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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