Q.(a) Assume that water issuing from the end of horizontal pipe, 7.5 m above the ground, describes a parabolic path. The vertex of the parabolic path is at the end of the pipe. At a position 2.5 m below the line of the pipe, the flow of water has curved outward 3 m beyond the vertical line through the end of the pipe. How far beyond this vertical line will the water strike the ground ? OR

(a) Sets up the downward-opening parabolic path with vertex at the pipe, fits it using the given point, and evaluates at the ground level; (b) solves the linear ODE $y'+y/x=\sin x$ using the integrating factor $x$ and integration by parts. Both alternatives answered below.

(a) Path of water from the pipe

1. Set up coordinates. Let the vertex of the parabolic path be at the end of the pipe (origin), with $x$ = horizontal distance from the vertical line through the pipe end, and $y$ = vertical distance below the pipe (so $y$ increases downward). The path is a parabola opening downward from the vertex: $x^2=4ay$ for some $a>0$.

2. Use the given data point. At $2.5$ m below the pipe ($y=2.5$), the water has moved $3$ m sideways ($x=3$):

$3^2=4a(2.5)\ \Rightarrow\ 9=10a\ \Rightarrow\ a=0.9$

So the path is $x^2=3.6y$.

3. Ground level. The pipe is $7.5$ m above the ground, so the ground corresponds to $y=7.5$:

$x^2=3.6\times7.5=27\ \Rightarrow\ x=\sqrt{27}=3\sqrt3\approx5.196\ \text{m}$

4. The water strikes the ground about $3\sqrt3\approx5.2$ m beyond the vertical line through the pipe's end.

(b) Solve $\dfrac{dy}{dx}+\dfrac yx=\sin x$

1. Standard linear form. $P(x)=\dfrac1x$, $Q(x)=\sin x$.

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