8. Properties of Curves

b. Limits and Derivatives of Vector Functions

2. Derivatives of Vector Functions

The definition of the derivative of a vector function mimics that for an ordinary function:

The derivative of a vector function \(\vec{f}(t)=\left\langle f_1(t),f_2(t),f_3(t)\right\rangle\) is: \[ \dfrac{d\vec{f}}{dt}=\lim_{h\rightarrow 0}\dfrac{\vec{f}(t+h)-\vec{f}(t)}{h} \] provided the limit exists.

Although this definition is useful and essential for theoretical purposes, the following proposition give a formula which is much better for computational purposes.

The derivative of a vector function \(\vec{f}(t)=\left\langle f_1(t),f_2(t),f_3(t)\right\rangle\) is: \[ \dfrac{d\vec{f}}{dt} =\left\langle \dfrac{df_1}{dt},\dfrac{df_2}{dt},\dfrac{df_3}{dt}\right\rangle \] provided the derivatives exists.
Proof

In other words, to take the derivative of a vector function we simply take the derivative of each component.

We can now define the velocity:

Given a parametric curve \(\vec{r}(t)=\left\langle x(t),y(t),z(t)\right\rangle\), its velocity is its derivative: \[ \vec{v}(t)=\dfrac{d\vec{r}}{dt} =\left\langle \dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt}\right\rangle \]

Find the velocity of the unit circle \(\vec{r}(t)=(\cos t,\sin t)\). Plot the circle and several velocity vectors for \(0 \le t \le \dfrac{3\pi}{2}\).

\( \vec{v}(t)=\dfrac{d\vec{r}}{dt}=\langle -\sin t,\cos t\rangle \) \[\begin{aligned} &\vec{r}(0)=(1,0) &&\vec{v}(0)=\langle 0,1\rangle \\ &\vec{r}\left(\dfrac{\pi}{4}\right) =\left(\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right) &&\vec{v}\left(\dfrac{\pi}{4}\right) =\left\langle \dfrac{-1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right\rangle \\ &\vec{r}\left(\dfrac{\pi}{2}\right)=(0,1) &&\vec{v}\left(\dfrac{\pi}{2}\right)=\langle -1,0\rangle \\ &\vec{r}\left(\dfrac{3\pi}{4}\right) =\left(\dfrac{-1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right)\qquad &&\vec{v}\left(\dfrac{3\pi}{4}\right) =\left\langle \dfrac{-1}{\sqrt{2}},\dfrac{-1}{\sqrt{2}}\right\rangle \\ &\vec{r}(\pi)=(-1,0) &&\vec{v}(\pi)=\langle 0,-1\rangle \\ &\vec{r}\left(\dfrac{5\pi}{4}\right) =\left(\dfrac{-1}{\sqrt{2}},\dfrac{-1}{\sqrt{2}}\right) &&\vec{v}\left(\dfrac{5\pi}{4}\right) =\left\langle \dfrac{1}{\sqrt{2}}, \dfrac{-1}{\sqrt{2}}\right\rangle \\ &\vec{r}\left(\dfrac{3\pi}{2}\right)=(0,-1) &&\vec{v}\left(\dfrac{3\pi}{2}\right) =\left\langle 1,0\right\rangle \\ \end{aligned}\]

In the above graph, notice that the velocity vectors are always tangent to the curve. This is no accident.

The velocity of a parametric curve is always tangent to the curve.

The proof depends heavily on the limit definition of the derivative. It is important to read the proof:

The animation shows a parabola (in blue) along with
    its tangent line at \(\vec{r}(2)\) (in magenta)
    its secant vectors at \(\vec{r}(2)\) (in orange)
    its secant quotient vectors at \(\vec{r}(2)\) (in green).
As \(h\) gets smaller, the directions of the secant vectors and the secant quotient vectors approach the direction of the tangent line. The lengths of the secant vectors approach \(0\) but the lengths of the secant quotient vectors do not. For \(h \gt 1\), the secant quotient is shorter than the secant, but for \(h \lt 1\), the secant quotient is longer.

As a consequence of the proposition, we have:

For a parametric curve, the tangent vector is the velocity vector; they are synonyms.