# Thread: finding expression

1. ## finding expression

....S o-----------------------+
.......|Y° *.............(sand)...|
..200|.......*.....................| 200
.......|............*................|
....A.+ - - - - - - - o - - - + B
.......|........a.........X *.......|
..200|........................*..Z°| 200
.......|...(mud).................* |
.......+-----------------------o F
........: - - - - 500m - - - - :

(please ignore the full stops)
details:
if the distance from S to F (hypotenuse) , passing through X =200(secY + sec Z)
and tanY + tanZ=5/2
and the distance from S to F, passing through X in terms of Y=200(secY + 0.5(4sec^2Y - 20tanY+25)^(1/2)

if an object can travel 4m/s through sand and 3m/s through mud, determine the best route it should travel through to get from S to F (hypotenuse), passing through X. and the time taken

2. Originally Posted by asian5
....S o-----------------------+
.......|Y° *.............(sand)...|
..200|.......*.....................| 200
.......|............*................|
....A.+ - - - - - - - o - - - + B
.......|........a.........X *.......|
..200|........................*..Z°| 200
.......|...(mud).................* |
.......+-----------------------o F
........: - - - - 500m - - - - :

(please ignore the full stops)
details:
if the distance from S to F (hypotenuse) , passing through X =200(secY + sec Z)
and tanY + tanZ=5/2
and the distance from S to F, passing through X in terms of Y=200(secY + 0.5(4sec^2Y - 20tanY+25)^(1/2)

if an object can travel 4m/s through sand and 3m/s through mud, determine the best route it should travel through to get from S to F (hypotenuse), passing through X. and the time taken
I think you will find everything you need to do this in the answers to your
earlier question here

RonL

3. Hello, asian5!

You don't need the angles to solve this part of the problem.

Code:
    S o-----------------------+
|   *                   |
200 |       *               | 200
|           *     500-x |
A + - - - - - - - o - - - + B
|       x       P *     |
200 |                   *   | 200
|                     * |
+-----------------------o F
: - - - -  500m - - - - :

If an object can travel 4m/s through sand and 3m/s through mud,
determine the best route it should travel to get from S to F,
passing through P, and the time taken.

Let $\displaystyle AP = x$, then $\displaystyle PB = 500-x$

We have: .distance $\displaystyle SP\;=\;\sqrt{x^2 + 200^2}$ meters.

. . At 4 m/s, it will take: .$\displaystyle \frac{\sqrt{x^2+200^2}}{4}$ seconds.

We have: .distance $\displaystyle PF\;=\;\sqrt{(500-x)^2+200^2}$ meters.

. . At 3 m/s, it will take: .$\displaystyle \frac{\sqrt{(500-x)^2+200^2}}{3}$ seconds.

Hence, the total time is: .$\displaystyle T(x)\;=\;\frac{1}{4}\left[x^2+200^2\right]^{\frac{1}{2}} + \frac{1}{3}\left[(500-x)^2+200^2\right]^{\frac{1}{2}}$

. . . and that is the function you must minimize.

. . . . . . . . . . . Good luck!

4. Soroban, I have already solved this problem (if you see anything wrong with my work in the previous post, please tell me)

I would show you the work, but something's wrong with this post...

5. Hello, Quick!

I think I've straightened out your format . . .

Soroban, I have already solved this problem.
If you see anything wrong with my work in the previous post, please tell me.

and the answer comes out to:

$\displaystyle t=\frac{\sqrt{40000+x^2}}{4.5} +\frac{\sqrt{290000-1000x+x^2}}{2.5}$

which can then be turned into (solving in terms of Y):

$\displaystyle t=\frac{\sqrt{40000+200^2\tan^2Y}}{4.5}$$\displaystyle +\frac{\sqrt{290000-1000(200\tan Y + 200^2\tan^2Y}}{2.5} \displaystyle t=\frac{\sqrt{40000+40000\tan^2Y}}{4.5} +$$\displaystyle \frac{\sqrt{290000-200000\tan Y+ 40000\tan^2Y }}{2.5}$

$\displaystyle t=\frac{\sqrt{40000(1+\tan^2Y)}}{4.5} +\frac{\sqrt{10000(29-20\tan Y + 4\tan^2Y)}}{2.5}$

$\displaystyle t=\frac{200\sqrt{1+\tan^2Y}}{4.5} +\frac{100\sqrt{29-20\tan Y+ 4\tan^2Y }}{2.5}$

I missed your earlier post . . . I will take a look.