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Garden Sprinkler Mathematical Modelling (Trigonometric Substitutions)?

There are various types of garden sprinkler systems on the market. In our
analysis here we will consider a very basic design, a cylinder with holes
made for water. The cylinder (of negligible size) distributes water to our
rectangular lawn by rotating around its axis at a constant frequency. The
cylinder rotates so that the maximal water angle is 45◦. The angle θ = 0
corresponds to a vertical ...

There are various types of garden sprinkler systems on the market. In our
analysis here we will consider a very basic design, a cylinder with holes
made for water. The cylinder (of negligible size) distributes water to our
rectangular lawn by rotating around its axis at a constant frequency. The
cylinder rotates so that the maximal water angle is 45◦. The angle θ = 0
corresponds to a vertical shoot. We assume no air resistance in our model.

It is clear that the lawn is not watered uniformly. The amount water that
falls on the ground near vertical shoot is less than the amount of water
falling down at the extremes. This is due to the fact that the water moves
quicker on the ground near the vertical shoot. We will now analyse the
water distribution on our rectangular lawn.

(a) Let (v)0 be the (constant) speed of water departure from the cylinder
holes and let g denote the acceleration due to gravity. Show that the
water reach on the ground is given by

(x)0 = [v(0)^2]/g (sin2θ)

where the angle θ is the angle of the holes in the cylinder from the
vertical. Thus the maximal water reach on the ground is given by

x(0)=(v(0)^2]/g

(b) Let f be the constant frequency of the cylinder motion, one cycle
θ ∈ [−π/4, π/4]. Let k ∈ [0, 1] denote the fraction of the ground
distance cover by water starting from zero (vertical) all the way to
the maximal ground reach to x(0). Show that the time t(k) it takes to
cover the ground from zero to kx(0) is given by:

t(k) = [1/(4πf)]sin^-1(k)

in particular, we have t(1)=(1/8f).

Hint set x(0)= 1 and [v(0)^2]/g = 1 for this.

(c) Let pk denote the percentage of total watering for the portion of our
rectangular lawn corresponding to the values [−k, k]. Show that

p(k) = (200/π)sin^-1(k)

Hint: the amount of water in the region [−k, k] is proportional to t(k).

(d) Show that the percentage of watering for the inner half of the lawn
is about 33 % (one third).

(e) Find the variable frequency f = f(θ) so that the lawn gets watered
evenly.

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