OBJECTIVES

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

1. Find the domain of a vector function (p. 758)
2. Graph a vector function (p. 758)
3. Find the limit and determine the continuity of a vector function (p. 761)
4. Find the derivative of a vector function (p. 762)
5. Find the derivative of a vector function using derivative formulas (p. 763)

spanDEFINITIONspan Vector Function

A vector-valued function of a real variable, or simply, a vector function, \(\mathbf{r}=\mathbf{r}(t)\), is a function whose domain is a subset of the set of real numbers and whose range is a set of vectors.

NOTE

As with vectors and scalars, the symbol for a vector function is written using boldface roman type and the symbol for a real function is written using lightface italic type.

Finding the Domain of a Vector Function

If \(\mathbf{r}(t) =\left(2t+2\right) \mathbf{i}+\sqrt{t}\mathbf{j}-\ln \left(t+3\right) \mathbf{k}\), then the components of \(\mathbf{r=r}(t)\) are \begin{equation*} {x(t) =2t+2\qquad y(t) =\sqrt{t}\qquad z(t) =- \ln \left( t+3\right)} \end{equation*}

Since no domain is specified, it is assumed that the domain consists of all real numbers \(t\) for which each of the component functions is defined. Since \(x=x(t)\) is defined for all real numbers, \(y=y(t)\) is defined for all real numbers \(t\geq 0,\) and \(z=z(t)\) is defined for all real numbers \(t\,{>}\!\,-\!3\), the domain of the vector function \(\mathbf{r=r}(t)\) is the intersection of these three sets, namely, the set of nonnegative real numbers, {\(t|t\geq 0\)}.

NOW WORK

Problem 17.

NEED TO REVIEW?

Parametric equations are discussed in Section 9.1, pp. 637-643.

NOTE

From now on we will just say “the curve \(C\) is traced out by the vector function \(\mathbf{r}=\mathbf{r}(t)\)” rather than “traced out by the tip of the vector function.”

Graphing a Vector Function

Figure 4 \(\mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j,}~0\leq t\leq 2\pi\)

Discuss the graph of the vector function \begin{equation*} \mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}\qquad 0\leq t\leq 2\pi \end{equation*}

Solution The components of \(\mathbf{r}=\mathbf{r}(t)\) are the parametric equations \begin{equation*} x=x(t) =\cos t\qquad y=y(t) =\sin t\qquad 0\leq t\leq 2\pi \end{equation*}

Since \(x^{2}+y^{2}=\cos ^{2}t\,{+}\,\sin ^{2}t=1\), the vector function \(\mathbf{r=r}(t)\) traces out a circle with its center at the origin and radius \(1\). See Figure 4. The curve begins at \(\mathbf{r}( 0) =\cos 0\,\mathbf{i}\,{+}\,\sin 0\,\mathbf{j}=\mathbf{i},\) contains the vectors \(\mathbf{r} \!\left( \dfrac{\pi }{2}\right) =\cos \dfrac{\pi }{2}\,\mathbf{i}\,{+}\,\sin \dfrac{\pi }{2}\,\mathbf{j}=\mathbf{j},\) \(\mathbf{r}(\pi) =\cos \pi \mathbf{i}\,{+}\,\sin \pi \mathbf{j}=-\mathbf{i},\) and \(\mathbf{r}\!\left( \dfrac{3\pi }{2}\right) =-\mathbf{j}.\) It ends at \(\mathbf{r}( 2\pi )=\mathbf{i},\) so the positive direction of the circle is counterclockwise.

NOW WORK

Problems 11 and 31.

NEED TO REVIEW?

Parametric equations and symmetric equations of a line are discussed in Section 10.6, pp. 734-736.

Graphing a Vector Function

Graph the curve \(C\) traced out by the vector function \begin{equation*} \mathbf{r}(t)=(2+3t)\mathbf{i}+(3-t)\mathbf{j}+2t\mathbf{k} \end{equation*}

Figure 5 \(\mathbf{r}(t)=(2+3t)\mathbf{i}+(3-t)\mathbf{j}+2t\mathbf{k}\)

Solution The components of \(\mathbf{r=r}(t)\) are the parametric equations \begin{equation*} x(t)=2+3t\qquad y(t)=3-t\qquad z(t)=2t \end{equation*}

These are the parametric equations of a line in space containing the point \((2,3,0)\) (corresponding to \(t=0\)) and in the direction of the vector \(3\mathbf{i}-\mathbf{j}+2\mathbf{k}\). Solving for \(t\), we obtain the symmetric equations of this line \begin{equation*} \frac{x-2}{3}=\frac{y-3}{-1}=\frac{z}{2} \end{equation*}

Since \(\mathbf{r}(0)=2\mathbf{i}\,{+}\,3\mathbf{j}\) and \(\mathbf{r}(1)=5\mathbf{i}\,{+}\,2\mathbf{j}\,{+}\,2\mathbf{k}\), the positive direction of \(C\) is given by \(\mathbf{r}(1)-\mathbf{r}(0)=3\mathbf{i}-\mathbf{j}\,{+}\,2\mathbf{k}\). See Figure 5.

Graphing a Vector Function

Graph the curve \(C\) traced out by the vector function \begin{equation*} r(t)=a\cos t\mathbf{i}+a\sin t\mathbf{j}+t\mathbf{k}\qquad t\geq 0 \end{equation*}

where \(a\) is a positive constant.

Solution The parametric equations of the curve \(C\) are \begin{equation*} x=x(t) =a\cos t\qquad y=y(t) =a\sin t\qquad z=z(t) =t \end{equation*}

Since \(x^{2}+y^{2}=a^{2}\cos ^{2}t+a^{2}\sin ^{2}t=a^{2},\) for any real number \(t\), any point \((x,y,z)\) on the curve \(C\) will lie on the right circular cylinder \(x^{2}+y^{2}=a^{2}\).

If \(t=0\), then \(\mathbf{r}( 0) =a\mathbf{i}\) so the point \((a,0,0)\) is on the curve. As \(t\) increases, the vector \(\mathbf{r}=\mathbf{r}(t)\) starts at \((a,0,0)\) and winds up and around the circular cylinder, one revolution for every increase of \(2\pi\) in \(t\). See Figure 6.

Figure 6 \(r(t)=a\cos t\mathbf{i}\,{+}\,a\sin t\mathbf{j}\,{+}\,t\mathbf{k}, t\geq 0\)

NOW WORK

Problems 37 and 43.

spanDEFINITIONspan Limit of a Vector Function

For a vector function \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\) in the plane \begin{equation*} \lim_{t\rightarrow t_{0}}\mathbf{r}(t)=\Big[ \lim_{t\rightarrow t_{0}}x(t) \Big] \mathbf{i}+\Big[ \lim_{t\rightarrow t_{0}}y(t)\Big] \mathbf{j} \end{equation*}

provided \(\lim\limits_{t\rightarrow t_{0}}x(t)\) and \(\lim\limits_{t\rightarrow t_{0}}y(t)\) both exist.

For a vector function \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+z(t)\mathbf{k}\) in space \begin{equation*} \lim_{t\rightarrow t_{0}}\mathbf{r}(t)=\Big[ \lim_{t\rightarrow t_{0}}x(t) \Big] \mathbf{i}+\Big[ \lim_{t\rightarrow t_{0}}y(t)\Big] \mathbf{j}+ \Big[ \lim_{t\rightarrow t_{0}}z(t)\Big] \mathbf{k} \end{equation*}

provided \(\lim\limits_{t\rightarrow t_{0}}x(t),\lim\limits_{t\rightarrow t_{0}}y(t)\), and \(\lim\limits_{t\rightarrow t_{0}}z(t)\) all exist.

NEED TO REVIEW?

Continuity of a real-valued function is discussed in Section 1.3, pp. 93-100.

Continuity of a Vector Function at a Number

A vector function \(\mathbf{r}=\mathbf{r}(t)\) is continuous at a real number \(t_{0}\) if:

• \(\mathbf{r}(t_{0})\) is defined.
• \(\lim\limits_{t\rightarrow t_{0}}\mathbf{r}(t)\) exists.
• \(\lim\limits_{t\rightarrow t_{0}}\mathbf{r}(t)=\mathbf{r}(t_{0})\).

NOW WORK

Problem 49.

spanDEFINITIONspan Derivative of a Vector Function

The derivative of the vector function \(\mathbf{r=r}(t) \), denoted by \(\mathbf{r}^\prime(t) =\dfrac{d\mathbf{r}}{dt}\), is \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\mathbf{r}^\prime (t)=\lim\limits_{h\rightarrow 0} \dfrac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}} \end{equation*}

provided the limit exists. If the limit exists, then we say \(\mathbf{r}\) is differentiable.

IN WORDS

To find the derivative of a vector function, differentiate each of its components.

THEOREM Differentiating a Vector Function

If \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\) and if \(x=x(t)\) and \(y=y(t)\) are both differentiable real functions, then \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\mathbf{r}^\prime (t) =x^\prime (t)\mathbf{ i}+y^\prime (t)\mathbf{j}} \end{equation*}

If \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+z(t)\mathbf{k}\) and if \(x=x(t),\) \(y=y(t)\), and \(z=z(t)\) are each differentiable real functions, then \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\mathbf{r}^\prime (t) =x^\prime (t)\mathbf{ i}+y^\prime (t)\mathbf{j}+z^\prime (t)\mathbf{k}} \end{equation*}

Proof

Use the definition of the derivative of a vector function. \begin{eqnarray*} \mathbf{r}^\prime (t)&=&\lim_{h\rightarrow 0}\frac{\mathbf{r}(t+h)-\mathbf{r} (t)}{h} \notag\\ &=&\lim_{h\rightarrow 0}\frac{[x(t+h)-x(t)]\;\mathbf{i}+[y(t+h)-y(t)] \;\mathbf{j}+[z(t+h)-z(t)]\;\mathbf{k}}{h} \notag \\ &=&\lim_{h\rightarrow 0}\left[ \frac{x(t+h)-x(t)}{h}\right]\!\;\mathbf{i} +\lim_{h\rightarrow 0}\left[ \frac{y(t+h)-y(t)}{h}\right]\!\;\mathbf{j} +\lim_{h\rightarrow 0}\left[ \frac{z(t+h)-z(t)}{h}\right]\!\;\mathbf{k} \notag \\ &=&x^\prime (t)\;\mathbf{i}+ {y}^\prime (t)\;\mathbf{j}+ {z}^\prime (t)\;\mathbf{k} \notag \end{eqnarray*}

Finding the Derivative of a Vector Function

Find the derivative of each vector function.

(a) \(\mathbf{r}(t)=2\sin t\mathbf{i}+3\cos t\mathbf{j}\)

(b) \(\mathbf{r}(t)=e^{t}\mathbf{i}+(1+t)\mathbf{j}+\sin t\mathbf{k}\)

Solution

(a) We find the derivative of each component. Since \(\dfrac{d}{dt}\left( 2\sin t\right) =2\cos t\) and \(\dfrac{d}{dt}\left( 3\cos t\right) =-3\sin t,\) \begin{equation*} \dfrac{d\mathbf{r}}{dt}=\mathbf{r}^\prime (t)=2\cos t\mathbf{i}-3\sin t \mathbf{j} \end{equation*}

(b) \(\dfrac{d\mathbf{r}}{dt}=\mathbf{r}^\prime (t)=\dfrac{d}{dt} e^{t}\mathbf{i}+\dfrac{d}{dt}(1+t)\mathbf{j}+\dfrac{d}{dt}\sin t \mathbf{k}=e^{t}\mathbf{i}+\mathbf{j}+\cos t\mathbf{k}\)

Finding \(\mathbf{r}^{\prime}\) and \(\mathbf{r}^{\prime\prime}\) for a Vector Function

Find \(\mathbf{r}^{\prime} (t)\) and \(\mathbf{r}^{\prime \prime} (t)\) for the vector function \(\mathbf{r}(t)=e^{t}\mathbf{i}+\ln t\mathbf{j}-\cos t\mathbf{k}\).

Solution \begin{eqnarray*} \mathbf{r}^{\prime} (t) &=&\dfrac{d}{dt}e^{t}\mathbf{i}+\dfrac{d}{ dt}\ln t\mathbf{j}-\dfrac{d}{dt}\cos t\mathbf{k}=e^{t}\mathbf{i}+\frac{ 1}{t}\mathbf{j}+\sin t\mathbf{k} \notag \\[6pt] \mathbf{r}^{\prime \prime} (t) &=&\dfrac{d}{dt}e^{t}\mathbf{i}+ \dfrac{d}{dt}\frac{1}{t}\mathbf{j}+\dfrac{d}{dt}\sin t\mathbf{k}=e^{t} \mathbf{i}-\frac{1}{t^{2}}\mathbf{j}+\cos t\mathbf{k} \notag \end{eqnarray*}

NOW WORK

Problem 59.

THEOREM Derivative Formulas

If the vector functions \(\mathbf{u}=\mathbf{u}(t)\) and \(\mathbf{v}=\mathbf{v}(t)\) are differentiable and the real function \(y=f(t)\) is differentiable, then

• Sum formula: \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\dfrac{d}{dt}\left[ \mathbf{u}(t)+\mathbf{v}(t)\right] =\dfrac{\ d}{dt}\mathbf{u}(t)+\dfrac{d}{dt}\mathbf{v}(t)=\mathbf{u}^{\prime}(t)+\mathbf{ v}^{\prime} (t)} \end{equation*}
• Scalar multiplication formula: \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\dfrac{d}{dt}\left[ f(t)\mathbf{u}(t)\right] =\left[ \dfrac{d }{dt}f(t) \right] \mathbf{u}(t) +f(t) \left[ \dfrac{d}{dt}\mathbf{u}(t) \right] =f^{\prime} (t)\mathbf{u}(t)+f(t)\mathbf{u}^{\prime} (t)} \end{equation*}
• Dot product formula: \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\dfrac{d}{dt}\left[ \mathbf{u}(t) \,{\cdot}\, \mathbf{v} (t) \right] =\left[ \mathbf{u}(t)\,{\cdot}\, \mathbf{v}(t)\right] ^{\prime} =\mathbf{u}^{\prime}(t)\,{\cdot}\, \mathbf{v}(t)+\mathbf{u}(t)\,{\cdot}\, \mathbf{v} ^{\prime} (t)} \end{equation*}
• Cross product formula: \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\dfrac{d}{dt}\left[ \mathbf{u}(t) \times \mathbf{v} (t) \right] =\left[ \mathbf{u}(t)\times \mathbf{v}(t)\right] ^{\prime} =\mathbf{u}^{\prime}(t)\times \mathbf{v}(t)+\mathbf{u}(t)\times \mathbf{ v}^{\prime} (t)} \end{equation*}

  where \(\mathbf{u}\) and \(\mathbf{v}\) are vector functions in space.

• Chain rule: \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED] {\dfrac{d}{dt}\mathbf{v}\left( f(t) \right) = \mathbf{v}^{\prime}(f(t) )f^{\prime} (t)} \end{equation*}

Proof (Scalar Multiplication Formula)

Let \(\mathbf{u}(t)=x(t)\kern1pt\mathbf{i}+y(t)\kern1pt\mathbf{j}+z(t)\;\mathbf{k}\). Then \begin{equation*} f(t)\mathbf{u}(t)=[ f(t)x(t)]\kern1pt\mathbf{i}+[ f(t)y(t)]\kern1pt\mathbf{j}+[ f(t)z(t)]\mathbf{k} \end{equation*}

We use the Product Rule and differentiate each component. \begin{eqnarray*} \left[ f(t)\mathbf{u}(t)\right] ^{\prime} &=&[f(t)x(t)]^{\prime} \mathbf{i} +[f(t)y(t)]^{\prime} \mathbf{j}+[f(t)z(t)]^{\prime} \mathbf{k} \notag \\[6pt] &=&[f(t)x^{\prime} (t)+f^{\prime} (t)x(t)]\kern1pt\mathbf{i}+[f(t)y^{\prime} (t)+f^{\prime} (t)y(t)]\kern1pt\mathbf{j}\\ &&+\,[f(t)z^{\prime} (t)+f^{\prime} (t)z(t)] \mathbf{k} \notag \\[6pt] &=&f^{\prime} (t)[x(t)\mathbf{i}+y(t)\mathbf{j}+z(t)\mathbf{k} ]+f(t)[x^{\prime} (t)\mathbf{i}+y^{\prime} (t)\mathbf{j}+z^{\prime} (t)\mathbf{k }] \notag \\[6pt] &=&f^{\prime} (t)\mathbf{u}(t)+f(t)\mathbf{u}^{\prime} (t) \notag \end{eqnarray*}

Finding the Derivative of a Cross Product

Find the derivative of \(\mathbf{u}(t)\times \mathbf{v}(t)\) if \[ \mathbf{u}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k}\qquad \hbox{and}\qquad \mathbf{v}(t)=t\mathbf{i}+\ln t\mathbf{j}+\mathbf{k} \]

Solution The derivatives of \(\mathbf{u}\) and \(\mathbf{v}\) are \[ \mathbf{u}^{\prime} (t)=-\!\sin t\mathbf{i}+\cos \dot{t}\mathbf{j}+\mathbf{k}\qquad \hbox{and}\qquad \mathbf{v}^{\prime}(t) =\mathbf{i}+\dfrac{1}{t} \mathbf{j} \]

We use the cross product formula. \begin{eqnarray*} \left[ \mathbf{u}(t)\times \mathbf{v}(t)\right] ^{\prime} &=&\mathbf{u}^{\prime} (t)\times \mathbf{v}(t)+\mathbf{u}(t)\times \mathbf{v}^{\prime} (t) \notag \\[6pt] &=&\left\vert \begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} &\mathbf{j} &\mathbf{k} \\ -\sin t &\cos t &1 \\ t &\ln t &1 \end{array} \right\vert +\left\vert \begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} &\mathbf{j} &\mathbf{k} \\ \cos t &\sin t &t \\ 1 &\dfrac{1}{t} &0 \end{array} \right\vert \notag \\[6pt] &=&\big[ ( \cos t-\ln t) \mathbf{i}+(\sin t+t)\mathbf{j} -[(\sin t)(\ln t)+t\cos t]\mathbf{k}\big] \\[6pt] &&+\left[ -\mathbf{i}+t\mathbf{ j}+\left( \frac{1}{t}\cos t-\sin t\right) \!\;\mathbf{k}\right] \notag \\[6pt] &=&(\cos t-\ln t-1)\mathbf{i}+(\sin t+2t)\mathbf{j}\\[6pt] &&+\left[ \frac{1}{t}\cos t-\sin t-(\sin t)(\ln t)-t\cos t\right] \!\;\mathbf{k} \notag \end{eqnarray*}

NOW WORK

Problems 73 and 79.

CAUTION

Be careful when using the cross product formula. Since a cross product is not commutative, order is important.