# Simultaneous eqution i think

• March 9th 2013, 05:41 PM
Zashmar
Simultaneous eqution i think
G'day
here is the question ;)
Bobby and Kym are two hikers. Bobby walks at 3klms/h and kym walks at 4klm/h. bobby is exactly klm due west of kym. they start walking immediatley. bobby walks due east and kym walks due south. what is the shortest distance between the walkers?

okay so i thought that maybe i could make the equation
because i used pythagoras
cheers

D2=(5-37)2+(4t)2

<3
• March 10th 2013, 12:05 AM
chiro
Re: Simultaneous eqution i think
Hey Zashmar.

In terms of your distance, it says that bobby is klm due west of kym. If we denote negative values for due west and due south we get a distance equation that is:

D^2 = (x-bobby - x-kym)^2 + (y-bobby - y-kym)^2

Since bobby is walking east and starts at -klm due east (which is equivalent to klm due south), we get bobby-x = -klm + 3*t. Bobby's y co-ordinate doesn't change since we is walking purely in an east direction with no north or south component.

As for kym, she has no east or west component so kym-x = a. Since kym is travelling due south, kyms y component is b - 4*t where b is the initial y position.

So if kym starts at (a,b) then Bobby starts at (a-klm,b) and the distance is over time

D^2 = (a - (a-klm+3t))^2 + (b-4t-b)^2
= (klm-3t)^2 + (4t)^2

Now you can find the turning points that minimize this relationship.
• March 10th 2013, 10:02 AM
Soroban
Re: Simultaneous eqution i think
Hello, Zashmar!

Your work (minus typos) seems to be correct.

Quote:

Bobby and Kym are two hikers. Bobby is exactly 5 km due west of Kym.
They start walking at the same time.
Bobby walks due east at 3km/h and Kym walks due south at 4km/h.
What is the shortest distance between the walkers?

Code:

        3t  P  5-3t     B * - - * - - - - * K               *      |                 *    | 4t               x  *  |                     * |                       * Q
Bobby starts at $B$; Kym starts at $K.$
. . $BK = 5$ km.

In $t$ hours, Bobby walks $3t$ km to point $P.$
. . $PK = 5-3t$

In the same $t$ hours, Kym walks $4t$ km to point $Q.$
Let $x = PQ.$ .Note that $x \ne 0.$

Pythagorus: . $x^2 \;=\;(5-3t)^2 +(4t)^2\quad\Rightarrow\quad x^2 \;=\;25 - 30t + 25t^2$ .[1]

Differentiate: . $2x\frac{dx}{dt} \;=\;-30 + 50t \quad\Rightarrow\quad \frac{dx}{dt} \;=\;\frac{25t - 15}{x}$

Equate to zero: . $\frac{25t-15}{x} \:=\:0 \quad\Rightarrow\quad t \,=\,\tfrac{3}{5}$

Substitute into [1]: . $x^2 \:=\:25 - 30(\tfrac{3}{5}) + 25(\tfrac{3}{5})^2 \quad\Rightarrow\quad x^2 \,=\,16$

Therefore: . $x \:=\:4\text{ km}$