Question

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

Answer 1

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`

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)`