$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# Smooth Vector-Valued Functions

[ "article:topic", "authorname:pseeburger", "epicycloid", "showtoc:no" ]

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Definition: smoothness

Let $$\vecs{r}(t)=f(t) \,\hat{\mathbf{i}}+g(t) \,\hat{\mathbf{j}} + h(t) \,\hat{\mathbf{k}}$$ be the parameterization of a curve that is differentiable on an open interval $$I$$.

Then $$\vecs r(t)$$ is smooth on the open interval $$I$$, if

$$\vecs r'(t) \neq \vecs 0$$, for any value of $$t$$ in the interval $$I$$.

To put this another way, $$\vecs r(t)$$ is smooth on the open interval $$I$$ if:

1. $$\vecs r(t)$$ is defined and continuous on $$I$$, and
2. $$f', \, g'$$ and $$h'$$ are all continuous on $$I$$, that is, $$\vecs r'(t)$$ is defined and continuous on $$I$$, and
3. $$\vecs r'(t) \neq \vecs 0$$ for all values of $$t$$ in the interval $$I$$.

The concept of smoothness is very closely connected to the application of vector-valued functions to motion.  If you have ever ridden rides at an amusement park, you can probably identify some where the ride felt smooth, and others where there are moments during the ride where it was not smooth.  These moments often correspond to points in the motion where the relevant parameterization is not smooth by the definition above.

To identify open intervals in the domain of $$\vecs r$$ on which $$\vecs r$$ is smooth, we usually identify values of $$t$$ for which $$\vecs r$$ is not smooth.

In addition to using the definition above to identify these points, it is often apparent from the graph of the vector-valued function $$\vecs r(t)$$.  A cusp, i.e., a sharp turning point, is a common occurrence at points where a curve is not smooth.  In Figure $$\PageIndex{3}$$ below, you can quickly identify points at which the parameterization fails to be smooth by locating cusps in the graph.

Figure $$\PageIndex{3}$$:  This epicycloid is not smooth at the points between it's petals.

When we consider the motion described by $$\vecs r(t)$$, there are at least two additional events that can occur at points where the parameterization fails to be smooth: a reversal of motion or a momentary pause in the motion.  To summarize these events:

If a vector-valued function $$\vecs r(t)$$ is not smooth at time $$t$$, we will observe that:

1.  There is a cusp at the associated point on the graph of $$\vecs r(t)$$, or
2.  The motion reverses itself at the associated point, causing the motion to travel back along the same path in the opposite direction, or
3.  The motion actually stops and starts up again, with no visual cue, that is, where the curve appears smooth.

Let's consider a couple examples to see how this can look.

Example $$\PageIndex{4}$$: Determining open intervals on which a vector-valued function is smooth

Determine the open intervals on which each of the following vector-valued functions is smooth:

1. $$\vecs r(t) = \left( 2t^3 - 3t^2\right) \, \,\hat{\mathbf{i}} + \left( t^2 - 2t\right) \, \,\hat{\mathbf{j}} + 2\,\,\hat{\mathbf{k}}$$
2. $$\vecs r(t) = t^3 \, \,\hat{\mathbf{i}} + \sqrt{t^2 - 1} \,\hat{\mathbf{j}}$$
3. $$\vecs r(t) = t^4 \, \,\hat{\mathbf{i}} + \ln(t^2 + 1) \,\hat{\mathbf{j}}$$
4. $$\vecs r(t) = \left( 6\cos t - \cos(6t)\right) \, \,\hat{\mathbf{i}} + \left( 6\sin t - \sin(6t) \right) \, \,\hat{\mathbf{j}}$$, for $$0 \le t \le 2\pi$$

Solution

a.  First we need to consider the domain of $$\vecs r$$.  In this case, $$\text{D}_{\vecs r} : (-\infty, \infty)$$.

Then we need to find $$\vecs r'(t)$$.  Since  $$\vecs r(t) = \left( 2t^3 - 3t^2\right) \, \,\hat{\mathbf{i}} + \left( t^2 - 2t\right) \, \,\hat{\mathbf{j}} + 2\,\,\hat{\mathbf{k}}$$,

$\vecs r'(t) = \left( 6t^2 - 6t\right) \, \,\hat{\mathbf{i}} + \left( 2t - 2\right) \, \,\hat{\mathbf{j}}$

Next we need to consider the domain of  $$\vecs r'$$.  Here,  $$\text{D}_{\vecs r'} : (-\infty, \infty)$$.

Finally we need to determine if there are any values of $$t$$ in $$\text{D}_{\vecs r}$$ where $$\vecs r'(t) = \vecs 0$$.

To do this, we set each component of $$\vecs r'(t)$$ equal to $$0$$ and solve for $$t$$.

\begin{alignat} 6t^2 - 6t = 0 & \quad \text{and} \quad & 2t - 2 = 0 \\ 6t(t -1) = 0 && 2(t - 1) = 0 \\ t = 0 \quad \text{or} \quad t = 1 & & t = 1 \end{alignat}

Since $$t = 1$$ makes both components of $$\vecs r'$$ equal 0, we have $$\vecs r'(1) = \vecs 0$$, so we know that $$\vecs r$$ is not smooth at $$t = 1$$.

Note however that $$\vecs r$$ is smooth at $$t = 0$$, because although one of the components of $$\vecs r'$$ is $$0$$, the other one is not.

Therefore we can conclude that $$\vecs r$$ is smooth on the open intervals:

$(-\infty, 1) \quad \text{and} \quad (1, \infty)$

Note in the figure that there is a cusp in the curve when $$t = 1$$.

b.  The domain of  $$\vecs r$$ is  $$\text{D}_{\vecs r} : (-\infty, -1] \cup [1, \infty)$$, since $$t^2 - 1$$ is less than $$0$$ between $$-1$$ and $$1$$.

Now we calculate $$\vecs r'(t)$$.  Since $$\vecs r(t) = t^3 \, \,\hat{\mathbf{i}} + \sqrt{t^2 - 1} \,\hat{\mathbf{j}}$$, we have

$\vecs r'(t) = 3t^2 \, \,\hat{\mathbf{i}} + \frac{t}{\sqrt{t^2-1}} \, \,\hat{\mathbf{j}}$

Then, since we cannot have zero in the denominator, $$t = -1$$ and $$t = 1$$ are not in the domain of $$\vecs r'$$.  Thus, $$\text{D}_{\vecs r'} : (-\infty, -1) \cup (1, \infty)$$.

Lastly we determine if there are any values of $$t$$ in $$\text{D}_{\vecs r}$$ where $$\vecs r'(t) = \vecs 0$$.

Setting each component of $$\vecs r'(t)$$ equal to $$0$$ and solve for $$t$$ we get:

\begin{alignat} 3t^2 = 0 & \quad \text{and} \quad & \frac{t}{\sqrt{t^2-1}} = 0 \\ t = 0 && t = 0 \end{alignat}

If $$t = 0$$ were in the domain of $$\vecs r'$$, then $$\vecs r$$ would not be smooth there, but since it is not in the domain, we have nothing to remove, and we can conclude that $$\vecs r$$ is smooth on the entire domain of its derivative $$\vecs r'$$, that is,

$\vecs r \, \text{is smooth on the open intervals:} \quad (-\infty, -1) \quad \text{and} \quad (1, \infty).$

Observe the two separate smooth pieces of this function in the graph below.

c.

#### Exercises:

For questions 31 - 39,

a.  Determine any values of $$t$$ at which $$\vecs r$$ is not smooth.

b.  Determine the open intervals on which $$\vecs r$$ is smooth.

c.  Graph the vector-valued function and describe its behavior at the points where it is not smooth.

31)   $$\vecs r(t)= t^3\,\hat{\mathbf{i}}+5t^2 \,\hat{\mathbf{j}}$$

a.  $$\vecs r$$ is not smooth at $$t = 0$$, since $$\vecs r'(0) = \vecs 0$$.
b.  $$\vecs r$$ is smooth on the open intervals $$(-\infty, 0)$$ and $$(0, \infty)$$.
c.  There is a cusp when $$t = 0$$.

32)   $$\vecs r(t) = \langle 3t, 5t^2 - 1\rangle$$

33)   $$\vecs r(t) = \langle t^3 - 3t^2, 7\rangle$$

a.  $$\vecs r$$ is not smooth at $$t = 0$$ and $$t = 2$$, since $$\vecs r'(0) = \vecs 0$$ and $$\vecs r'(2) = \vecs 0$$.
b.  $$\vecs r$$ is smooth on the open intervals $$(-\infty, 0)$$,  $$(0, 2)$$, and $$(2, \infty)$$.
c.  The motion on the curve reverses along the same path at both $$t = 0$$ and $$t = 2$$.

34)    $$\vecs r(t)=\langle 5,\, 2 \sin (t),\, \cos(t)\rangle$$

35)    $$\vecs r(t)=\langle \ln(t^2+4t+5),\left(\frac{t^3}{3} - 4t\right), 5\rangle$$

a.  $$\vecs r$$ is not smooth at $$t = -2$$, since $$\vecs r'(-2) = \vecs 0$$.
Since the domain of $$\vecs r$$ is $$(-\infty, \infty)$$, this is all we have to remove.
b.  $$\vecs r$$ is smooth on the open intervals $$(-\infty, -2)$$ and $$(-2, \infty)$$.
c.  There is a cusp when $$t = -2$$.

36)   $$\vecs r(t)=t^2 \,\hat{\mathbf{i}}+t^3 \,\hat{\mathbf{j}}−5e^{−4t} \,\hat{\mathbf{k}}$$

37)   $$\vecs r(t)=\sqrt{t^3 + 9t^2} \,\hat{\mathbf{i}}+\left(t^2 +12t\right) \,\hat{\mathbf{j}}+7\,\hat{\mathbf{k}}$$

a.  The domain of $$\vecs r$$ is $$[-9, \infty)$$.
And  $$\vecs r$$ is not smooth at $$t = -6$$, since $$\vecs r'(-6) = \vecs 0$$.
The domain of $$\vecs r'$$ is $$(-9, \infty)$$, since $$\vecs r'$$ is undefined at $$t = -9$$.
b.  $$\vecs r$$ is smooth on the open intervals $$(-9, -6)$$ and $$(-6, \infty)$$.
c.  There is a cusp when $$t = -6$$.

38)   $$\vecs r(t) = \left( 5\cos t - \cos(5t)\right) \, \hat{i} + \left( 5\sin t - \sin(5t) \right) \, \hat{j}$$, for $$0 \le t \le 2\pi$$

39)    $$\vecs r(t)= \cos^3 t\,\hat{\mathbf{i}}+\sin t \,\hat{\mathbf{j}}$$, for $$0 \le t \le 2\pi$$

a.  The domain of $$\vecs r$$ is $$(-\infty, \infty)$$.
$$\vecs r'(t) = -3(\cos^2 t)(\sin t)\,\hat{\mathbf{i}}+\cos t \,\hat{\mathbf{j}}$$.  It's domain is also $$(-\infty, \infty)$$.
But note that both components have a factor of $$\cos t$$, so both components will be $$0$$ when $$\cos t = 0$$.
Therefore, $$\vecs r$$ is not smooth at $$t = \frac{\pi}{2}$$ and at $$t = \frac{3\pi}{2}$$, since $$\vecs r'\left( \frac{\pi}{2}\right) = \vecs 0$$ and $$\vecs r'\left( \frac{3\pi}{2}\right) = \vecs 0$$.  Note then that $$\vecs r$$ is not smooth for any odd multiple of $$\frac{\pi}{2}$$, that is for $$t = \frac{(2n + 1)\pi}{2}$$, for any integer value $$n$$.
b.  $$\vecs r$$ is smooth on the open intervals $$(\frac{(2n - 1)\pi}{2}, \frac{(2n + 1)\pi}{2})$$, for any integer value $$n$$.
c.  There is a cusp when  $$t = \frac{(2n + 1)\pi}{2}$$, for any integer value $$n$$.