Find the path of least time given these conditions? Starting point = (2,6) End point = (8,-3) Velocity above the line y=2 is 1 unit/second Velocity below the line y=2 is 7 units/second

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1 Answer
Feb 11, 2017

Answer:

There could be simpler way of doing this but, this is what I followed.

Explanation:

As the intermediate point must lie at the interface #y=2#
Hence the general point must be #(x,2)#

Path of Displacement is #(a)(2,6) to (x,2)and (b)(x,2) to (8,-3)#

(a) Displacement#=sqrt((x-2)^2+(2-6)^2)#
Time taken #="Displacement"/"Velocity"=(sqrt((x-2)^2+16))/1 s#

(b) Displacement#=sqrt((8-x)^2+(-3-2)^2)#
Time taken #=sqrt((8-x)^2+25)/7 s#

Total Time #t=sqrt((x-2)^2+16)+sqrt((8-x)^2+25)/7#
#=>t=sqrt((x^2-4x+4+16))+sqrt((64-16x+x^2)+25)/7#
Differentiating with respect to #x# and setting it equal to #0#
#d/dx((x^2-4x+20)^(1/2)+1/7(x^2-16x+89)^(1/2))=0#
#=>1/2(x^2-4x+20)^(-1/2)(2x-4)+#
#1/7xx1/2(x^2-16x+89)^(-1/2)(2x-16)=0#
#=>(x^2-4x+20)^(-1/2)(x-2)+1/7(x^2-16x+89)^(-1/2)(x-8)=0#

My comp

Plotting this equation with the help of inbuilt graphics utility, gives us
#x=2.43#, rounded to two decimal places, as there is only one point of inflection.
Hence path with least time is through the point #(2.43,2)#