Vector-valued functions, velocity, and acceleration

Understand the idea

A vector-valued function packages several coordinate functions into one position description. Differentiate each component to obtain velocity and differentiate again to obtain acceleration. These derivatives are vectors, while speed is the scalar magnitude of velocity.

The derivative vector points in the instantaneous direction of travel when it is nonzero. Its components describe coordinate rates, not separate distances to add. At a point where velocity vanishes, speed is zero and a direction cannot be obtained by normalizing the zero vector.

Distinguish a position vector from a velocity vector even when they happen to have similar components. Units provide a useful check: position has distance units, velocity has distance per time, and acceleration has distance per time squared.

\mathbf r(t)=\langle x(t),y(t)\rangle,\quad \mathbf v=\mathbf r',\quad \mathbf a=\mathbf r'',\quad |\mathbf v|=\sqrt{(x')^2+(y')^2}

A method you can apply

Differentiate each component using the ordinary derivative rules.
Evaluate velocity and acceleration at the requested parameter value.
Take the magnitude of velocity only when a scalar speed is requested.

Worked example

For $\displaystyle \mathbf r(t)=\langle t^2,e^t\rangle$ , find velocity, acceleration, and speed at $\displaystyle t=0$ .

Componentwise differentiation gives $\displaystyle \mathbf v(t)=\langle2t,e^t\rangle$ .
A second derivative gives $\displaystyle \mathbf a(t)=\langle2,e^t\rangle$ .
At $\displaystyle t=0$ , these are $\displaystyle \langle0,1\rangle$ and $\displaystyle \langle2,1\rangle$ ; the velocity magnitude is $\displaystyle 1$ .

Velocity is

\displaystyle \langle0,1\rangle

, acceleration is

\displaystyle \langle2,1\rangle

, and speed is

\displaystyle 1

.

Try it yourself

For $\displaystyle \mathbf r(t)=\langle\sin t,\cos(2t)\rangle$ , find velocity and acceleration at $\displaystyle t=\pi/2$ .

Show solution and reasoning

The derivatives are $\displaystyle \mathbf v=\langle\cos t,-2\sin(2t)\rangle$ and $\displaystyle \mathbf a=\langle-\sin t,-4\cos(2t)\rangle$ . At $\displaystyle \pi/2$ , velocity is $\displaystyle \langle0,0\rangle$ and acceleration is $\displaystyle \langle-1,4\rangle$ .