# Thread: Fastest way between two points.

1. ## Fastest way between two points.

I dunno if this is in the right section but anyways here it is:

Runners speed on the path is 3 times the speed he can move in the swamp. What is the fastest route from point 1 to point 2? in other words how many meters does he have to run through swamp and how many on the path for optimal time?
Thanks!

2. ## Re: Fastest way between two points.

There are 3 possible routes:
a) run only on the path- that gives the fastest rate.
b) run on the straight line for 1 to 2 over the swamp- that gives the shortest path.
c) run some distance along the path, then through the swamp.
and we can include (a) and (b) as special cases of (c).

If we let "x" be the distance he runs along the path. That leaves 300- x on that path and 400 as the other leg of a right triangle having his path through the swamp as hypotenuse. By the Pythagorean theorem. the length of that path is $\sqrt{(300- x)^2+ 400^2}$.

Since he runs distance x at speed 3v, his time for that is $\frac{x}{3v}$. Since he runs distance $\sqrt{(300- x)^2+ 400^2}$ at speed v, his time for that is $\frac{\sqrt{(300- x)^2+ 400^2}}{v}$.

You want to find the route that minimizes total time so you want to find x that minimizes
$\frac{x}{3v}+ \frac{\sqrt{(300- x)^2+ 400^2}}{v}$.

3. ## Re: Fastest way between two points.

Hello, Vincent4312!

Runner's speed on the path is 3 times the speed he can move in the swamp.
What is the fastest route from point P to point Q?
In other words, how many meters does he have to run through swamp and how many on the path for optimal time?

Code:
      :  x  R  L-x  : Q
- *-----o-------o
: |    *        |
: |   *         |
W |  *          |
: | *           |
: |*            |
- o-------------*
P : - -  L  - - :
Let $V$ = speed in the swamp (meters per minute).
Let $3V$ = speed on the path.

He runs from $P$ to $R$ through the swamp.
He runs $\sqrt{x^2+W^2}$ meters at $V$ m/min.
This takes: . $\frac{\sqrt{x^2+W^2}}{V}$ minutes.

He run from $R$ to $Q$ on the path.
He runs $L-x$ meters at $3V$ m/min.
This takes: . $\frac{L-x}{3V}$ minutes.

His total time is: $T \;=\;\frac{\sqrt{x^2+W^2}}{V} + \frac{L-x}{3V}$

Differentiate $T \;=\;\frac{1}{V}(x^2+W^2)^{\frac{1}{2}} + \frac{L}{3V} - \frac{1}{3V}x$
. . equate to zero, and solve for $x.$

Got it?

4. ## Re: Fastest way between two points.

Originally Posted by Soroban

Differentiate $T \;=\;\frac{1}{V}(x^2+W^2)^{\frac{1}{2}} + \frac{L}{3V} - \frac{1}{3V}x$
. . equate to zero, and solve for $x.$

Got it?
Yes I understand it now thank you very much . I just have one more problem with this :/ I haven't differentiated anything for a year so could you give me a hand with this?

5. ## Re: Fastest way between two points.

Nevermind I did it!!:-) thank you very much!