# Two people walk from P (1, 0) to the point Q (-1, 0) along the paths r_1 (t) = < cos t, sin t >...

## Question:

Two people walk from {eq}P (1,\ 0) {/eq} to the point {eq}Q (-1,\ 0) {/eq} along the paths {eq}r_1 (t) = \langle \cos t,\ \sin t \rangle {/eq} and {eq}r_2 (t) = \langle -t,\ -5 (t^2 - 1) \rangle {/eq} respectively. Who arrives at {eq}Q {/eq} first and why? Graph each curve and indicate the time when they reach point {eq}Q {/eq}.

## Time Dependent Position

The coordinates of the position of a particle can be expressed as function of the time parameter. With that, it can be found the time at which the particle reaches to a given point. Time is not necessarily constrained to the positive values. Negative values means time before time parameter starts to be measured.

## Answer and Explanation:

We must find which times correspond to each point for each path. In the first path, we have:

{eq}\begin{align} r_1(t)=\left(\cos t,\sin t\right)\\ P(1,0)=\left(\cos t,\sin t\right)\\ t_{1,P}=0\\ Q(-1,0)=\left(\cos t,\sin t\right)\\ t_{1,Q}=\pi\\ \end{align} {/eq}

and for the second path:

{eq}\begin{align} r_2(t)=\left(-t,-5(t^2-1)\right)\\ P(1,0)=\left(-t,-5(t^2-1)\right)\\ t_{1,P}=-1\\ P(-1,0)=\left(-t,-5(t^2-1)\right)\\ t_{1,Q}=1\\ \end{align} {/eq}

Then the person moving along the second path arrives first at point Q (at t=1).

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#### Learn more about this topic:

from Precalculus: High School

Chapter 24 / Lesson 3