# Thread: Solving an equation for the fastest way of traveling (challenging)

1. ## Solving an equation for the fastest way of traveling (challenging)

Hello MHF! I am not sure I posted this question in the right section but this was my best guess.

I will start by saying I'm from Sweden and I apologize for my rather poor English, especially when it comes to math-language.

The problem I have a question about is based on a physics thesis for the way light breaks in water.

It is linked to a picture, which you can study here: Picture

The description is: A boy will run to his girlfriend standing on the other side of a pool. He must therefore run a part on land and also swim one part in water.

1) How will this happen so that the boy will reach his girlfriend as fast as possible? The boy's speed on land is 10 m/s (meters per second) and his speed swimming is 2 m/s.

2) What is the relationship between the angles of the i and b and the velocities v1 and v2 for the period to be as small as possible?

Now, I have solved 2) and I found out that the relationship is:

$\frac{sin(i)}{sin(b)}=\frac{v_{1}}{v_{2}}$

From the calculations to get to that conclusion I know that:
$sin(i)=\frac{x}{\sqrt{x^{2}+100}}$
and
$sin(b)=\frac{x}{\sqrt{100+(20-x)^{2}}}$

x is the horizontal line from the upper left corner of the pool to the point where the boy reaches the pool.

I now have an equation that looks like this to solve 1):
(when $v_{1}=10$ and $v_{2}=2$)

$\frac{\frac{x}{\sqrt{x^{2}+100}}}{\frac{x}{\sqrt{1 00+(20-x)^{2}}}}=\frac{10}{2}$

Is there anyone here on this splendid forum who has clue how to solve this? Or have I done this way to difficult for myself? I feels like there is an easier way of solving the problem but I can't find the way.

I am forever grateful for all the help I can get, even if it's only a mere though/suggestion!

Regards

Liljeros

2. To reach his girl friend as fast as possible, he must take least time to travel the distance between him and his girl friend.
The total time taken by him is given by
T = sqrt(10^2 + x^2)/V1 + sqrt[10^2 + ( 20 - x)^2]/V2.
To make the time least dT/dx must be zero.
0 = x/v1*sqrt((10^2 + x^2) - (20-x)/V2*sqrt[10^2 + ( 20 - x)^2]
Or
sin(i)/v1 = sin(b)/v2.