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### Section 6-7 : Calculus with Vector Functions

In this section we need to talk briefly about limits, derivatives and integrals of vector functions. As you will see, these behave in a fairly predictable manner. We will be doing all of the work in $${\mathbb{R}^3}$$ but we can naturally extend the formulas/work in this section to $${\mathbb{R}^n}$$ (i.e. $$n$$-dimensional space).

Let’s start with limits. Here is the limit of a vector function.

\begin{align*}\mathop {\lim }\limits_{t \to a} \vec r\left( t \right) & = \mathop {\lim }\limits_{t \to a} \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle \\ & = \left\langle {\mathop {\lim }\limits_{t \to a} f\left( t \right),\mathop {\lim }\limits_{t \to a} g\left( t \right),\mathop {\lim }\limits_{t \to a} h\left( t \right)} \right\rangle \\ & = \mathop {\lim }\limits_{t \to a} f\left( t \right)\vec i + \mathop {\lim }\limits_{t \to a} g\left( t \right)\vec j + \mathop {\lim }\limits_{t \to a} h\left( t \right)\vec k\end{align*}

So, all that we do is take the limit of each of the component’s functions and leave it as a vector.

Example 1 Compute $$\mathop {\lim }\limits_{t \to 1} \vec r\left( t \right)$$ where $$\vec r\left( t \right) = \left\langle {{t^3},\displaystyle \frac{{\sin \left( {3t - 3} \right)}}{{t - 1}},{{\bf{e}}^{2t}}} \right\rangle$$.
Show Solution

There really isn’t all that much to do here.

\begin{align*}\mathop {\lim }\limits_{t \to 1} \vec r\left( t \right) & = \left\langle {\mathop {\lim }\limits_{t \to 1} {t^3},\mathop {\lim }\limits_{t \to 1} \frac{{\sin \left( {3t - 3} \right)}}{{t - 1}},\mathop {\lim }\limits_{t \to 1} {{\bf{e}}^{2t}}} \right\rangle \\ & = \left\langle {\mathop {\lim }\limits_{t \to 1} {t^3},\mathop {\lim }\limits_{t \to 1} \frac{{3\cos \left( {3t - 3} \right)}}{1},\mathop {\lim }\limits_{t \to 1} {{\bf{e}}^{2t}}} \right\rangle \\ & = \left\langle {1,3,{{\bf{e}}^2}} \right\rangle \end{align*}

Notice that we had to use L’Hospital’s Rule on the $$y$$ component.

Now let’s take care of derivatives and after seeing how limits work it shouldn’t be too surprising that we have the following for derivatives.

$\vec r'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle = f'\left( t \right)\vec i + g'\left( t \right)\vec j + h'\left( t \right)\vec k$
Example 2 Compute $$\vec r'\left( t \right)$$ for $$\vec r\left( t \right) = {t^6}\,\vec i + \sin \left( {2t} \right)\vec j - \ln \left( {t + 1} \right)\vec k$$.
Show Solution

There really isn’t too much to this problem other than taking the derivatives.

$\vec r'\left( t \right) = 6{t^5}\,\vec i + 2\cos \left( {2t} \right)\vec j - \frac{1}{{t + 1}}\vec k$

Most of the basic facts that we know about derivatives still hold however, just to make it clear here are some facts about derivatives of vector functions.

#### Facts

\begin{align*}& \frac{d}{{dt}}\left( {\vec u + \vec v} \right) = \vec u' + \vec v'\\ & {\left( {c\vec u} \right)^\prime } = c\,\vec u'\\ & \frac{d}{{dt}}\left( {f\left( t \right)\vec u\left( t \right)} \right) = f'\left( t \right)\vec u\left( t \right) + f\left( t \right)\vec u'\left( t \right)\\ & \frac{d}{{dt}}\left( {\vec u\centerdot \vec v} \right) = \vec u'\centerdot \vec v + \vec u\centerdot \vec v'\\ & \frac{d}{{dt}}\left( {\vec u \times \vec v} \right) = \vec u' \times \vec v + \vec u \times \vec v'\\ & \frac{d}{{dt}}\left( {\vec u\left( {f\left( t \right)} \right)} \right) = f'\left( t \right)\vec u'\left( {f\left( t \right)} \right)\end{align*}

There is also one quick definition that we should get out of the way so that we can use it when we need to.

A smooth curve is any curve for which $$\vec r'\left( t \right)$$ is continuous and $$\vec r'\left( t \right) \ne 0$$ for any $$t$$ except possibly at the endpoints. A helix is a smooth curve, for example.

Finally, we need to discuss integrals of vector functions. Using both limits and derivatives as a guide it shouldn’t be too surprising that we also have the following for integration for indefinite integrals

\begin{align*}\int{{\vec r\left( t \right)\,dt}} & = \left\langle {\int{{f\left( t \right)\,dt}},\int{{g\left( t \right)\,dt}},\int{{h\left( t \right)\,dt}}} \right\rangle + \vec c\\ \int{{\vec r\left( t \right)\,dt}} & = \int{{f\left( t \right)\,dt}}\,\,\vec i + \int{{g\left( t \right)\,dt}}\,\,\vec j + \int{{h\left( t \right)\,dt}}\,\,\vec k + \vec c\end{align*}

and the following for definite integrals.

\begin{align*}\int_{{\,a}}^{{\,b}}{{\vec r\left( t \right)\,dt}} & = \left\langle {\int_{{\,a}}^{{\,b}}{{f\left( t \right)\,dt}},\int_{{\,a}}^{{\,b}}{{g\left( t \right)\,dt}},\int_{{\,a}}^{{\,b}}{{h\left( t \right)\,dt}}} \right\rangle \\ \int_{{\,a}}^{{\,b}}{{\vec r\left( t \right)\,dt}} & = \int_{{\,a}}^{{\,b}}{{f\left( t \right)\,dt}}\,\,\vec i + \int_{{\,a}}^{{\,b}}{{g\left( t \right)\,dt}}\,\,\vec j + \int_{{\,a}}^{{\,b}}{{h\left( t \right)\,dt}}\,\,\vec k\end{align*}

With the indefinite integrals we put in a constant of integration to make sure that it was clear that the constant in this case needs to be a vector instead of a regular constant.

Also, for the definite integrals we will sometimes write it as follows,

\begin{align*}\int_{{\,a}}^{{\,b}}{{\vec r\left( t \right)\,dt}} & = \left. {\left( {\left\langle {\int{{f\left( t \right)\,dt}},\int{{g\left( t \right)\,dt}},\int{{h\left( t \right)\,dt}}} \right\rangle } \right)} \right|_a^b\\ \int_{{\,a}}^{{\,b}}{{\vec r\left( t \right)\,dt}} & = \left. {\left( {\int{{f\left( t \right)\,dt}}\,\,\vec i + \int{{g\left( t \right)\,dt}}\,\,\vec j + \int{{h\left( t \right)\,dt}}\,\,\vec k} \right)} \right|_a^b\end{align*}

In other words, we will do the indefinite integral and then do the evaluation of the vector as a whole instead of on a component by component basis.

Example 3 Compute $$\int{{\vec r\left( t \right)\,dt}}$$ for $$\vec r\left( t \right) = \left\langle {\sin \left( t \right),6,4t} \right\rangle$$.
Show Solution

All we need to do is integrate each of the components and be done with it.

$\int{{\vec r\left( t \right)\,dt}} = \left\langle { - \cos \left( t \right),6t,2{t^2}} \right\rangle + \vec c$
Example 4 Compute $$\int_{{\,0}}^{{\,1}}{{\vec r\left( t \right)\,dt}}$$ for $$\vec r\left( t \right) = \left\langle {\sin \left( t \right),6,4t} \right\rangle$$.
Show Solution

In this case all that we need to do is reuse the result from the previous example and then do the evaluation.

\begin{align*}\int_{{\,0}}^{{\,1}}{{\vec r\left( t \right)\,dt}} & = \left( {\left\langle { - \cos \left( t \right),6t,2{t^2}} \right\rangle } \right)_0^1\\ & = \left\langle { - \cos \left( 1 \right),6,2} \right\rangle - \left\langle { - 1,0,0} \right\rangle \\ & = \left\langle {1 - \cos \left( 1 \right),6,2} \right\rangle \end{align*}