An interesting question given by my teacher in class.Two towns are to get their water supply from a river. Both towns are on the same side of the river at distance of 6 km and 18 km respectively from the river bank. If the distance between the points on the river bank nearest to the towns respectively be 10 km, find
(i)Where may a single pumping station be located to require the least amount of pipe?
(ii) How much pipe is needed for the above in (i)?
Any idea on solving?
how can snell's law solve the problem?
well, i think we'll have to draw out the sketch of the towns position and the river..and the pumping station should be at the intersection of the hypothenuses of the right angle triangles. The triangles have their right angles between the side parallel to the river bank and the perpendicular distance of the towns to the river.
any more ideas?
The law of reflection solves the problem because light rays minimise a functional
equivalent to the pipe line length.
A light ray follows a path such that the time of flight is stationary,
as the speed of light is constant in a homogeneous medium this corresponds
to a path who's length is stationary.
In this case it happens to be a minimum.
Thus the pipline will satisfy law of reflection as it is a consequence of the
stationarity of time of flight along the ray.
RonL
Zeez, both the Problem and your idea how to solve it test my understanding of English.
Do I understand what you are saying if I describe the figure according to my understandings?
So there is a vertical line, the river bank. Points A and B, 10 km apart, are on this vertical line.
Perpendicular to line segment AB, from point B, is point C. Point C is 6 km from point B.
Likewise, perpendicular to line segment AB, from point A, is point D. Point D is 18 km from point A.
The pumping station, point P, is on AB.
So the pipe lines are PC and PD.
Yes, your idea is very good.
Total length of pipelines, L = PC +PD
Let P be x km away from A.
So P is (10-x) km away from B.
Then, by Pythagorean Theorem,
L = sqrt[6^2 +(10-x)^2] +sqrt[18^2 +x^2]
Differentiate both sides with respect to x, set dL/dx to zero, and you'd get the least L.
So I solved for the minimum L using the idea that P should be on AB.
I got L = 26 km minimum.
Then, I computed for another idea.
That the pumping station could be at point B.
And so L = BC +CD
L = 6 + sqrt[(18-6)^2 +10^2] = 21.62 km only.
Therefore, I correct my first answer.
Now, for the least pipelines, the pumping station should be on the river bank that is 6 km from the town nearer to the river. The pipeline then goes first to that nearer town, then the pipeline proceeds to the other town. ----------revised answer.