finding expression

**asian5** · August 5th 2006, 11:41 PM

....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

**CaptainBlack** · August 6th 2006, 01:13 AM

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

**Soroban** · August 6th 2006, 05:36 AM

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 $AP = x$ , then $PB = 500-x$

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

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

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

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

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

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

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

**Quick** · August 6th 2006, 06:34 AM

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...

**Soroban** · August 6th 2006, 07:49 AM

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:

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

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

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

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

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

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

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

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