What is the shortest possible pipeline to minimize the amount of time the gas will travel? - See Details below.
Engineers want to design an auxiliary pipeline from a new natural gas source 3 km from the main pipeline, to a power plant located 10 km down the main pipeline. If the oil can flow 10 km/hr along the auxiliary pipeline and 15 km/hr along the main pipeline, where should the engineers join the auxiliary pipeline to the main pipeline to get the gas to P as soon as possible?
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Let the distance for the connection be x down the main pipe line. Therefore the distance form the connection to the power plant is (10-x).
The distance from the new well to the connection is `sqrt(3^2+x^2) = sqrt(9+x^2)`
the time required to flow is t, then
`t = (10-x)/15 + sqrt(9+x^2)/10`
`(dt)/(dx) = (-1)/15 + 1/10*1/2*2x(9+x^2)^(-1/2)`
`(dt)/(dx) = (-1)/15 + x/(10sqrt(9+x^2))`
For critcal points, dt/dx = 0
`(-1)/15 + x/(10sqrt(9+x^2)) = 0`
`x/(10sqrt(9+x^2)) = 1/15`
`3x = 2sqrt(9+x^2)`
by taking the square of each side,
`9x^2 = 4(9+x^2)`
`5x^2 = 36` this gives `x = 6/sqrt(5)`
To check whether this is a minimum, we have to check for the sign of the second derivative, `(d^2t)/(dx^2)`
`(d^2t)/(dx^2) = 1/10[(sqrt(9+x^2)-x*1/2*2x*(9+x^2)^((-1)/2))/(9+x^2)]`
`(d^2t)/(dx^2) = 1/10[(sqrt(9+x^2)-x^2/sqrt(9+x^2))/(9+x^2)]`
`(d^2t)/(dx^2) = 1/10[((9+x^2)-x^2)/(9+x^2)^(3/2)]`
`(d^2t)/(dx^2) = 9/(10(9+x^2)^(3/2))`
This is positive for any value there for at `x =6/sqrt(5)` the second derivative is positive and we have a minimum value there for the function t.
Therefore to minimize the time the engineers have to fix the connection `6/sqrt(5)` distance down the main pipeline.
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