A lawn sprinkler is constructed in such a way that $\dfrac{dθ}{dt}$ is constant, where $θ$ ranges between $45^{\circ}$ and $135^{\circ}$. The distance the water travels horizontally is
$$x=\frac{v^2\sin 2θ}{32}$$
where is the speed of the water. Find $\dfrac{dx}{dt}$ and explain why this lawn sprinkler does not water evenly. What part of the lawn receives the most water?
I differentiated the function $x$ w.r.t. $t$ and I got:
$$\frac{dx}{dt}= \frac{v}{16} \left( \frac{dv}{dt} \, \sin 2θ+v\cos 2θ \, \frac{dθ}{dt} \right)$$
I would say that the lawn sprinkler does not water evenly because the rate of change of the speed of $v$ is not constant. But I'm not sure how to find where the lawn receives the most water.
$\endgroup$ 31 Answer
$\begingroup$Note that without gravity, a circular arc (from $45^{\circ}$ to $135^{\circ}$) receives $Av\Delta t$ amount of water in angle $\Delta \theta$ where $A$ is the cross section of the nozzle. If $\displaystyle \frac{d\theta}{dt}=$ constant, the water is distributed uniformly over the arc.
In this case, there's $Av\Delta t$ amount of water landing on the ground within length $\Delta x$
Assuming uniform $v$, $$\frac{dx}{dt}=\frac{v^{2}}{16} \cos 2\theta \frac{d\theta}{dt}$$
which is $\theta$ dependent and hence not uniform/even.
For maximal the amount of water receiving, $\displaystyle \left| Av\frac{dt}{dx} \right|=\left| \frac{Av}{dx/dt} \right|$ should be the largest.
This achieves when $\cos 2\theta=0$ or $\theta=45^{\circ}$ or $135^{\circ}$.
That is at $\displaystyle x=\pm \frac{v^{2}}{32}$ which is the maximal range of projectile.
See the sprinkle shower with $6^{\circ}$ intervals below:
