An oil-pipe has to connect the oil-well O and the factory F, between which there is a river whose banks are parallel.

Question

@JonathanZ Thanks for going all the way to help me out.

An oil-pipe has to connect the oil-well O and the factory F, between which there is a river whose banks are parallel. The pipe must cross the river perpendicular to the banks. Find the position and nature of the shortest such pipe and justify your answer.

I tried out this problem with some numbers and got the idea of path of minimum pipe length. How can I do it more rigorously?

This is what I got through my intuition. I think blue one is the minimal path.

Answer 1

Since we just want to know where to cross the river wrt to factory I have set the y coordinate of the factory to 1.

$d_1^2=a^2+x^2$

$d_2^2=b^2+(1-x)^2$

$y=d_1+d_2=\sqrt{(a^2+x^2 )}+\sqrt{(b^2+(1-x)^2 )}$

Note that w does not appear here, so it doesn't matter. We can set it to zero and looking at the diagram the answer becomes obvious, but let's just verify that:

We need to find the x that minimises y

$\frac{dy}{dx}=\frac{d(\sqrt{a^2+x^2}+\sqrt{b^2+(1-x)^2})}{dx}=\frac{x}{\sqrt{a^2+x^2}}-\frac{1-x}{\sqrt{b^2+(1-x)^2}}=0$

$\frac{1-x}{\sqrt{b^2+(1-x)^2}}=\frac{x}{\sqrt{a^2+x^2}}$

${1-x}{\sqrt{a^2+x^2}}={x}{\sqrt{b^2+(1-x)^2}}$

$(1-x)^2(a^2+x^2)-x^2(b^2+(1-x)^2)=0$

$(a^2 - b^2 ) x^2- 2 a^2 x + a^2=0$

$x=\frac{a}{a+b}$ or $x=\frac{a}{a-b}$

Only the first one makes sense for a minimum so the other must be a max (check by getting 2nd derivative)

So pick x so that it is in the proportion a/(a+b) of the y distance to the factory, i.e. on a straight line to the factory if the river was not there.

Answer 2

Draw two lines perpendicular to the river, one through $O$, one through $F$.
If they coincide, that's already the path of mininal lenght.
Else, mark points $O'$ and $F'$ on their respective lines, each moved towards the river by an amount equal to the rivers width.
Connect $O$ and $F'$ with a line and mark point $C_1$ where this line crosses the river bank near $O$.
Connect $O'$ and $F$ with a line and mark point $C_2$ where this line crosses the river bank near $F$.

The two segments $OC_1$ and $C_2F$ are parallel, and of combined length $OF'=O'F$, which is the minimal distance in case of a river with zero width; and in case the river does have width, its crossing pipe is always of length equal to the rivers width.

The pipe of minimal length follows the chain $OC_1C_2F$

Answer 3

PROOF

Place the two 'diagonal' pipes end to end. Combined in this way they have to cover a fixed distance perpendicular to the river (say $L$) and a fixed distance parallel to the river (say $M$). The total of their lengths is therefore at least $\sqrt(L^2+M^2)$ and this is achieved only when the two pipes form a straight line.

In your diagram the two diagonal sections must therefore be parallel. The pipe from $A$ must therefore head towards a point a distance $x$ to the left of $C$ and the pipe from $C$ must head towards a point a distance $x$ to the right of $A$.

Answer 4

Cross the river 'first' and then join the two endpoints.

Answer 5

The optimal crossing is solid green, the solid red one is not optimal.

Answer 6

Ignore river width at first by a river width shift and later restore with a shift vector of opposite sense.

S is start point and G is goal or destination.

Shift G left by a length of width w or b of river to new point B.

Join S to B by a straight line representing a taut thread or ray of light that goes in a straight line. SB cuts left bank at H.

Now move B back to G, and H to I across river as horizontal displacement parallel to x-axis same vector of opposite sense.

Required path is along vectors $(a,b,c).$

Lengths found from similar triangles:

$$ \frac{HE}{SE}=\frac{GK}{KS} = \tan \phi,$$

the inclination angle $\phi$ makes to x-axis.