Question on trigo?

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December 24th 2012, 07:34 AM
Tutu

Question on trigo?

Hi I really have no idea which catergory this question falls into, especially so when I don't even know how to start off the question..

If Person A is 20Km north of Person B at 5pm. Person A walks south at a rate of 7Km/h and Person B walks west at a rate of 9Km/h. What time do these two peopole stop approaching each other and instead, drift further apart?

How do I start and what formula should I use? I'm really clueless. Is it under bearings..?

Thank you so much, really!
December 24th 2012, 07:48 AM
skeeter

Re: Question on trigo?

let person B start at the origin, $(0,0)$

person A's position as a function of time, $t$ , in hrs ... $(0, 20-7t)$

person B's position ... $(-9t , 0)$

distance, $r$ ,between the two at any time $t$ in hrs ...

$r = \sqrt{[0 - (-9t)]^2 + [(20-7t) - 0]^2}$

simplify, then determine when $\frac{dr}{dt}$ changes sign from negative to positive.
December 24th 2012, 07:56 AM
MarkFL

Re: Question on trigo?

This is how I would set it up:

Let north be the positive y-axis and west be the positive x-axis. Distances are in km and time is in hrs.

Person A's position at time t is:

$A(t)=(0,20-7t)$

Person B's position at time t is:

$B(t)=(9t,0)$

Let $D(t)$ be the distance separating the two. Hence:

$D^2(t)=(20-7t)^2+(9t)^2=130t^2-280t+400$

We see this is a parabola opening upward, so all you need to do is find the axis of symmetry to find the time when their distance is a minimum.

Recall the axis of symmetry for the parabola $y=ax^2+bx+c$ is $x=-\frac{b}{2a}$ .