Concept preview

What you will be able to do

1. 01Define a vector-valued position function $\vec r(t)=\langle x(t),\,y(t)\rangle$ and differentiate it componentwise to obtain velocity $\vec v=\vec r\,'=\langle x',\,y'\rangle$ and acceleration $\vec a=\vec v\,'=\langle x'',\,y''\rangle$.
2. 02Compute speed as the scalar magnitude $|\vec v|=\sqrt{x'^2+y'^2}$ and clearly distinguish it from the velocity vector $\vec v=\langle x',\,y'\rangle$, which also carries direction.
3. 03Interpret velocity and acceleration geometrically — velocity is tangent to the path, speed is how fast the particle moves along it — and connect this back to the arc-length integrand from 9.3.

Start with the position vector $\vec r(t)=\langle x(t),\,y(t)\rangle$. At each time $t$ it is an arrow from the origin to the particle’s location, and as $t$ varies the arrowhead sweeps out the path. To find how the position is changing, differentiate — and here is the rule that makes everything easy: you differentiate a vector-valued function one component at a time. So the velocity is $\vec v(t)=\vec r\,'(t)=\langle x'(t),\,y'(t)\rangle$. The first component tells you how fast $x$ is changing, the second how fast $y$ is changing; together they form an arrow.