OBJECTIVES

When you finish this section, you should be able to:

1. Interpret the derivative of a vector function geometrically (p. 767)
2. Find the unit tangent vector and the principal unit normal vector of a smooth curve (p. 768)
3. Find the arc length of a curve traced out by a vector function (p. 770)

RECALL

The terms normal, perpendicular, and orthogonal all mean “meet at right angles.”

NOTE

There is no tangent vector defined if $\mathbf{r}^{\prime} (t_{0})=\mathbf{0}.$

DEFINITION Tangent Vector to a Curve $C$

Suppose $\mathbf{r}=\mathbf{r}(t)$ is a vector function defined on the interval $[ a,b]$ and differentiable on the interval $( a,b) .$ Let $P_{0}$ be the point on the curve $C$ traced out by $\mathbf{r}=\mathbf{r}(t)$ corresponding to $t=t_{0},$ $a<t_{0}<b.$ If $\mathbf{r}^{\prime} (t_{0})\neq \mathbf{0}$, then the vector $\mathbf{r^{\prime} }( t_{0})$ is a tangent vector to the curve at $t_{0}.$ The line containing $P_{0}$ in the direction of $\mathbf{r}^{\prime} (t_{0})$ is the tangent line to the curve traced out by $\mathbf{r}=\mathbf{r}(t)$ at $t_{0}.$

RECALL

The angle between two nonzero vectors $\mathbf{v}$ and $\mathbf{w}$ is determined by $\cos \theta =\dfrac{\mathbf{v}\,{\cdot}\,\mathbf{w}} {\left\Vert \mathbf{v}\right\Vert \left\Vert \mathbf{w}\right\Vert }$, where $0\leq \theta \leq \pi.$

Show that the acute angle between the tangent vector to the helix $$\mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k}\quad 0\,{\leq}\,t\,{\leq}\,2\pi$$

and the direction $\mathbf{k}$ is $\dfrac{\pi }{4}$ radian.

Solution A tangent vector at any point on the helix is given by $$\begin{equation*} \mathbf{r}^{\prime} (t)=-\sin t\mathbf{i}+\cos t\mathbf{j}+\mathbf{k} \end{equation*}$$

Figure 10 $\mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k}, \\ 0\leq t\leq 2\pi$

Then $\left\Vert \mathbf{r}^{\prime} (t)\right\Vert\;= \sqrt{(-\sin t)^2 + \cos^2 t +1} = \sqrt{1+1} = \sqrt{2}.$

The cosine of the acute angle $\theta$ between $\mathbf{r}^{\prime} (t)$ and $\mathbf{k}$ is $$\begin{eqnarray*} &&\cos \theta =\frac{\mathbf{r}^{\prime} (t)\,{\cdot}\, \mathbf{k}}{\left\Vert \mathbf{r} ^{\prime} (t)\right\Vert \left\Vert \mathbf{k}\right\Vert }=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\qquad \color{#0066A7}{\hbox{$\mathbf{r}^{\prime} (t)\,{\cdot}\, \mathbf{k} = 1$}}\\ \end{eqnarray*}$$

So, $\theta =\dfrac{\pi }{4}$ radian. See Figure 10.

NOW WORK

Problems 9 and 15.

DEFINITION Smooth Curve

Let $C$ denote a curve traced out by a vector function $\mathbf{r}=\mathbf{r}(t)$ that is continuous for $a\leq t\leq b.$ If the derivative $\mathbf{r}^{\prime} (t)$ exists and is continuous on the interval $( a,b)$ and if $\mathbf{r}^{\prime} (t)$ is never $\mathbf{0}$ on $( a,b),$ then $C$ is called a smooth curve.

Show that the unit tangent vector $\mathbf{T}(t)$ to the circle of radius $R$ $$\begin{equation*} \mathbf{r}(t)=R\cos t\mathbf{i}+R\;\sin t\mathbf{j}\quad 0\leq t\leq 2\pi \end{equation*}$$

is everywhere orthogonal to $\mathbf{r}(t).$ Graph $\mathbf{r}=\mathbf{r}( t)$ and $\mathbf{T}=\mathbf{T}( t).$

Solution We begin by finding $\mathbf{r}^{\prime} (t)$ and $\left\Vert \mathbf{r}^{\prime} (t)\right\Vert$: $$\begin{eqnarray*} \mathbf{r}^{\prime} (t)& =&\dfrac{d}{dt}( R\cos t) \mathbf{i}+ \dfrac{d}{dt}( R\;\sin t) \mathbf{j}=-R\;\sin t\mathbf{i}+R\cos t\mathbf{j} \\[6pt] \left\Vert \mathbf{r}^{\prime} (t)\right\Vert & =&\sqrt{(-R\;\sin t)^{2}+(R\cos t)^{2}}=R \end{eqnarray*}$$

Figure 11 $\mathbf{r}( t) =R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j}$

Then the unit tangent vector is $$\begin{equation*} \mathbf{T}(t)=\frac{\mathbf{r}^{\prime} (t)}{\left\Vert \mathbf{r}^{\prime} (t)\right\Vert }=\dfrac{-R\;\sin t\mathbf{i}+R\;\cos t\mathbf{j}}{R}=-\!\sin t\mathbf{i}+\cos t\mathbf{j} \end{equation*}$$

To determine whether $\mathbf{r}( t)$ is orthogonal to $\mathbf{T}( t),$ we find their dot product. $$\begin{equation*} \mathbf{T}(t)\,{\cdot}\, \mathbf{r}(t)=( -\!\sin t\mathbf{i}+\cos t\mathbf{j}) \,{\cdot}\, ( R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j}) =-R\;\sin t\cos t+R\;\sin t\cos t=0 \end{equation*}$$

for all $t.$ That is, $\mathbf{T}(t)$ is everywhere orthogonal to $\mathbf{r}(t)$, as shown in Figure 11.

DEFINITION Principal Unit Normal Vector

For a smooth curve $C$ traced out by a vector function $\mathbf{r}=\mathbf{r}(t),$ $a\leq t\leq b,$ that is twice differentiable for $a<t<b,$ the principal unit normal vector $\mathbf{N}(t)$ to $C$ at $t$ is $$\begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\mathbf{N}(t)=\dfrac{\mathbf{T}^{\prime} (t)}{\left\Vert \mathbf{T}^{\prime} (t)\right\Vert }} \end{equation*}$$

Find the principal unit normal vector $\mathbf{N}(t)$ to the circle $\mathbf{r}( t) =R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j}$, $0\leq t\leq 2\pi .$ (Refer to Example 2.) Graph $\mathbf{r}=\mathbf{r}( t),$ and show a unit tangent vector and a unit normal vector.

Solution From Example 2, the unit tangent vector $\mathbf{T}(t)$ is $$\begin{equation*} \mathbf{T}(t)=-\!\sin t\mathbf{i}+\cos t\mathbf{j} \end{equation*}$$

Figure 13 $\mathbf{r}( t) =R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j,} \\ 0\leq t\leq 2\pi$

Since $\mathbf{T}^{\prime} ( t) =-\cos t\mathbf{i}-\sin t\mathbf{j},$ the principal unit normal vector $\mathbf{N}(t)$ is $$\begin{equation*} \mathbf{N}(t)=\frac{\mathbf{T}^{\prime} (t)}{\left\Vert \mathbf{T}^{\prime} (t)\right\Vert }=\frac{-\!\cos t\mathbf{i}-\sin t\mathbf{j}}{\sqrt{(-\!\cos t)^{2}+(-\!\sin t)^{2}}}=-\cos t\mathbf{i}-\sin t\mathbf{j} \end{equation*}$$

Since $\mathbf{r}( t) = R ( \cos t\mathbf{i}+\sin t\mathbf{j}),$ the vector $\mathbf{N}( t)$ is parallel to $-\mathbf{r}(t)$. That is, $\mathbf{N}(t)$ is a unit vector opposite in direction to the vector $\mathbf{r}(t)$, so $\mathbf{N}$ is directed toward the center of the circle, as shown in Figure 13.

NOW WORK

Problem 29.

Show that the principal unit normal vector $\mathbf{N}(t)$ of the helix $$\begin{equation*} \mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k} \end{equation*}$$

is orthogonal to the $z$-axis.

Solution We begin by finding $\mathbf{r^{\prime} }( t)$ and $\left\Vert \mathbf{r}^{\prime} ( t)\right \Vert.$ $$\begin{equation*} \mathbf{r}^{\prime} (t)=-\!\sin t\mathbf{i}+\cos t\mathbf{j}+\mathbf{k}\qquad \left\Vert \mathbf{r}^{\prime }(t)\right\Vert =\sqrt{ (-\!\sin t)^{2}+(\cos t)^{2}+1}=\sqrt{2} \end{equation*}$$

Then the unit tangent vector is $\mathbf{T}(t)=\dfrac{\mathbf{r}^{\prime }(t)}{||\mathbf{r}^{\prime }(t)||}=\dfrac{-\!\sin t\mathbf{i}+\cos t\mathbf{j}+\mathbf{k}}{\sqrt{2}}.$

Figure 14 $\mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t \mathbf{k}$

Since $$\begin{equation*} \mathbf{T}^{\prime} (t)=\dfrac{-\!\cos t\mathbf{i}-\sin t\mathbf{j}}{\sqrt{2} }\qquad \hbox{and}\qquad \left\Vert \mathbf{T}^{\prime }(t)\right\Vert =\sqrt{ \left( -\dfrac{\cos t}{\sqrt{2}}\right) ^{2}+\left( -\dfrac{\sin t}{\sqrt{2}} \right) ^{2}}=\dfrac{1}{\sqrt{2}} \end{equation*}$$

we have $$\begin{equation*} \mathbf{N}(t)=\dfrac{\mathbf{T}^{\prime} (t)}{\left\Vert \mathbf{T}^{\prime} (t)\right\Vert }=-\!\cos t\mathbf{i}-\sin t\mathbf{j} \end{equation*}$$

Since the direction of the $z$-axis is $\mathbf{k}$, it follows that $\mathbf{N}(t)\,{\cdot}\, \mathbf{k}=0$ for all $t.$ So, $\mathbf{N}(t)$ is always orthogonal to the $z$-axis, as shown in Figure 14.

NOW WORK

Problem 37.

THEOREM Arc Length of a Vector Function

If a smooth curve $C$ is traced out by the vector function $\mathbf{r}=\mathbf{r}(t),$ $a\leq t\leq b$, the arc length $s$ along $C$ from $t=a$ to $t=b$ is $$\begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {s=\int_{a}^{b}\left\Vert \mathbf{r}^{\prime} (t)\right\Vert dt} \end{equation*}$$

NEED TO REVIEW?

The arc length of a smooth curve is discussed in Section 9.2, pp. 651-653.

Find the arc length of:

(a) The circle $\mathbf{r}(t)=R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j}$ from $t=0$ to $t=2\pi$

(b) The circular helix $\mathbf{r}(t)=R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j}+t\mathbf{k}$ from $t=0$ to $t=2\pi$

Figure 15 $\mathbf{r}(t)=R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j}, \\ 0\leq t\leq 2\pi$

Solution (a) We begin by finding $\mathbf{r^{\prime} }( t)$ and $\left\Vert \mathbf{r^{\prime} }( t) \right\Vert$: $$\mathbf{r}^{\prime} (t)=-R\;\sin t\mathbf{i}+R\;\cos t\mathbf{j}\qquad \left\Vert \mathbf{r}^{\prime} (t)\right\Vert =\sqrt{(-R\;\sin t)^{2}+(R\;\cos t)^{2}}=R$$

Now we use the formula for arc length. $$\begin{equation*} s=\int_{a}^{b}\left\Vert \mathbf{r}^{\prime} (t)\right\Vert dt=\int_{0}^{2\pi }Rdt=\big[ R\hbox{ }t\big] _{0}^{2\pi }=2\pi R \end{equation*}$$

which is the familiar formula for the circumference of a circle. See Figure 15.

(b) We begin by finding $\mathbf{r^{\prime} }( t)$ and $\left\Vert \mathbf{r^{\prime} }( t) \right\Vert.$ $$\begin{eqnarray*} \mathbf{r}^{\prime} (t) &=&-R\;\sin t\mathbf{i}\,{+}\,R\;\cos t\mathbf{j}+\mathbf{k},\\[6pt] \left\Vert \mathbf{r}^{\prime} (t)\right\Vert &=& \sqrt{\!(-R\;\sin t)^{2}\,{+}\,(R\;\cos t)^{2}\,{+}\,1^{2}}\,{=}\,\sqrt{R^{2}\,{+}\,1} \end{eqnarray*}$$

Figure 16 $\mathbf{r}(t)=R\;\cos t\mathbf{i}+R\;\sin t\mathbf{j}+t\mathbf{k,} \\ 0\leq t\leq 2\pi$

Then $$\begin{equation*} ~s=\int_{a}^{b}\left\Vert \mathbf{r}^{\prime} ( t) \right\Vert dt=\int_{0}^{2\pi }\sqrt{R^{2}+1}~dt=\left[ \sqrt{R^{2}+1}\,t\right] _{0}^{2\pi }=2\pi \sqrt{R^{2}+1} \end{equation*}$$

See Figure 16.

NOW WORK

Problem 51.

Figure 17 $\mathbf{r}( t) =2\cos t\mathbf{i}+3\sin t\mathbf{j,} \\ 0\leq t\leq \dfrac{\pi }{2}$

Use technology to find the arc length $s$ of the ellipse shown in Figure 17 traced out by the vector function $\mathbf{r}( t) =2\cos t\mathbf{i}+3\sin t\mathbf{j}$ from $t=0$ to $t=\dfrac{\pi }{2}.$

Solution We begin by finding $\mathbf{r^{\prime} }( t)$ and $\left\Vert \mathbf{r^{\prime} }( t) \right\Vert .$ $$\mathbf{r}^{\prime} (t)=-2\sin t\mathbf{i}+3\cos t\mathbf{j}\qquad \left\Vert \mathbf{r}^{\prime} (t)\right\Vert =\sqrt{4\sin ^{2}t+9\cos ^{2}t}$$

Now we use the formula for arc length. $$\begin{equation*} s=\int_{a}^{b}\left\Vert \mathbf{r}^{\prime} (t)\right\Vert dt=\int_{0}^{\pi /2}\sqrt{4\sin ^{2}t+9\cos ^{2}t}\,dt \end{equation*}$$

This is an integral that has no antiderivative in terms of elementary functions. To obtain a numerical approximation to the arc length, we use a CAS or a graphing utility.

When entering the integral in WolframAlpha, we find $$\begin{equation*} \int_{0}^{\pi /2}\sqrt{4\sin ^{2}t+9\cos ^{2}t}\,dt\approx 3.96636 \end{equation*}$$

Figure 18

If we use a TI-84 to find the value of this integral, we obtain the same result. The screen shot is given in Figure 18.

NOW WORK

Problem 55.