Re: eb's question from Yahoo! Answers regarding distance, parametric equations and slope
Hello eb,
I would place Hannah initially at the origin (0,0), and Sally at (3,5).
1.) I would choose to represent the positions of Hannah and Sally parametrically. The unit of distance is km and the unit of time is hr. So, we may state:
$$H(t)=\langle 3t,0 \rangle$$
$$S(t)=\langle 2t+3,5 \rangle$$
and so the distance between them at time $t$ is:
$$d(t)=\sqrt{((2t+3)-3t)^2+(5-0)^2}=\sqrt{t^2-6t+34}$$
Now, letting $d(t)=10$ and solving for $0\le t$, we find:
$$10=\sqrt{t^2-6t+34}$$
Square both sides and write in standard form:
$$t^2-6t-66=0$$
Use of the quadratic formula yields one non-negative root:
$$t=3+5\sqrt{3}\approx11.66$$
2.) Initially the slope of the line through their positions is $$\frac{5}{3}$$, and so we want to equate the slope of the line through their positions at time $t$ to the negative reciprocal of this as follows:
$$\frac{5-0}{2t+3-3t}=-\frac{3}{5}$$
$$\frac{5}{t-3}=\frac{3}{5}$$
Cross-multiply:
$$3t-9=25$$
$$3t=34$$
$$t=\frac{34}{3}$$
To eb and any other guests viewing this topic, I invite and encourage you to post other parametric equation/analytic geometry questions in our http://www.mathhelpboards.com/f21/ forum.
Best Regards,
Mark.