# 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

Feb 11, 2017

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 $\left(x , 2\right)$

Path of Displacement is $\left(a\right) \left(2 , 6\right) \to \left(x , 2\right) \mathmr{and} \left(b\right) \left(x , 2\right) \to \left(8 , - 3\right)$

(a) Displacement$= \sqrt{{\left(x - 2\right)}^{2} + {\left(2 - 6\right)}^{2}}$
Time taken $= \text{Displacement"/"Velocity} = \frac{\sqrt{{\left(x - 2\right)}^{2} + 16}}{1} s$

(b) Displacement$= \sqrt{{\left(8 - x\right)}^{2} + {\left(- 3 - 2\right)}^{2}}$
Time taken $= \frac{\sqrt{{\left(8 - x\right)}^{2} + 25}}{7} s$

Total Time $t = \sqrt{{\left(x - 2\right)}^{2} + 16} + \frac{\sqrt{{\left(8 - x\right)}^{2} + 25}}{7}$
$\implies t = \sqrt{\left({x}^{2} - 4 x + 4 + 16\right)} + \frac{\sqrt{\left(64 - 16 x + {x}^{2}\right) + 25}}{7}$
Differentiating with respect to $x$ and setting it equal to $0$
$\frac{d}{\mathrm{dx}} \left({\left({x}^{2} - 4 x + 20\right)}^{\frac{1}{2}} + \frac{1}{7} {\left({x}^{2} - 16 x + 89\right)}^{\frac{1}{2}}\right) = 0$
$\implies \frac{1}{2} {\left({x}^{2} - 4 x + 20\right)}^{- \frac{1}{2}} \left(2 x - 4\right) +$
$\frac{1}{7} \times \frac{1}{2} {\left({x}^{2} - 16 x + 89\right)}^{- \frac{1}{2}} \left(2 x - 16\right) = 0$
$\implies {\left({x}^{2} - 4 x + 20\right)}^{- \frac{1}{2}} \left(x - 2\right) + \frac{1}{7} {\left({x}^{2} - 16 x + 89\right)}^{- \frac{1}{2}} \left(x - 8\right) = 0$

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 $\left(2.43 , 2\right)$