# 3.2 Calculus of vector-valued functions

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• Write an expression for the derivative of a vector-valued function.
• Find the tangent vector at a point for a given position vector.
• Find the unit tangent vector at a point for a given position vector and explain its significance.
• Calculate the definite integral of a vector-valued function.

To study the calculus of vector-valued functions, we follow a similar path to the one we took in studying real-valued functions. First, we define the derivative, then we examine applications of the derivative , then we move on to defining integrals. However, we will find some interesting new ideas along the way as a result of the vector nature of these functions and the properties of space curves.

## Derivatives of vector-valued functions

Now that we have seen what a vector-valued function is and how to take its limit, the next step is to learn how to differentiate a vector-valued function. The definition of the derivative of a vector-valued function is nearly identical to the definition of a real-valued function of one variable. However, because the range of a vector-valued function consists of vectors, the same is true for the range of the derivative of a vector-valued function.

## Definition

The derivative of a vector-valued function     $\text{r}\left(t\right)$ is

${r}^{\prime }\left(t\right)=\underset{\text{Δ}t\to 0}{\text{lim}}\frac{\text{r}\left(t+\text{Δ}t\right)-\text{r}\left(t\right)}{\text{Δ}t},$

provided the limit exists. If ${r}^{\prime }\left(t\right)$ exists, then r is differentiable at t. If ${r}^{\prime }\left(t\right)$ exists for all t in an open interval $\left(a,b\right),$ then r is differentiable over the interval $\left(a,b\right).$ For the function to be differentiable over the closed interval $\left[a,b\right],$ the following two limits must exist as well:

${r}^{\prime }\left(a\right)=\underset{\text{Δ}t\to {0}^{+}}{\text{lim}}\frac{\text{r}\left(a+\text{Δ}t\right)-\text{r}\left(a\right)}{\text{Δ}t}\phantom{\rule{0.4em}{0ex}}\text{and}\phantom{\rule{0.4em}{0ex}}{r}^{\prime }\left(b\right)=\underset{\text{Δ}t\to {0}^{-}}{\text{lim}}\frac{\text{r}\left(b+\text{Δ}t\right)-\text{r}\left(b\right)}{\text{Δ}t}.$

Many of the rules for calculating derivatives of real-valued functions can be applied to calculating the derivatives of vector-valued functions as well. Recall that the derivative of a real-valued function can be interpreted as the slope of a tangent line or the instantaneous rate of change of the function. The derivative of a vector-valued function can be understood to be an instantaneous rate of change as well; for example, when the function represents the position of an object at a given point in time, the derivative represents its velocity at that same point in time.

We now demonstrate taking the derivative of a vector-valued function.

## Finding the derivative of a vector-valued function

Use the definition to calculate the derivative of the function

$\text{r}\left(t\right)=\left(3t+4\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left({t}^{2}-4t+3\right)\phantom{\rule{0.1em}{0ex}}\text{j}.$

$\begin{array}{cc}\hfill {r}^{\prime }\left(t\right)& =\underset{\text{Δ}t\to 0}{\text{lim}}\frac{\text{r}\left(t+\text{Δ}t\right)-\text{r}\left(t\right)}{\text{Δ}t}\hfill \\ & =\underset{\text{Δ}t\to 0}{\text{lim}}\frac{\left[\left(3\left(t+\text{Δ}t\right)+4\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left({\left(t+\text{Δ}t\right)}^{2}-4\left(t+\text{Δ}t\right)+3\right)\phantom{\rule{0.1em}{0ex}}\text{j}\right]-\left[\left(3t+4\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left({t}^{2}-4t+3\right)\phantom{\rule{0.1em}{0ex}}\text{j}\right]}{\text{Δ}t}\hfill \\ & =\underset{\text{Δ}t\to 0}{\text{lim}}\frac{\left(3t+3\text{Δ}t+4\right)\phantom{\rule{0.1em}{0ex}}\text{i}-\left(3t+4\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left({t}^{2}+2t\text{Δ}t+{\left(\text{Δ}t\right)}^{2}-4t-4\text{Δ}t+3\right)\phantom{\rule{0.1em}{0ex}}\text{j}-\left({t}^{2}-4t+3\right)\phantom{\rule{0.1em}{0ex}}\text{j}}{\text{Δ}t}\hfill \\ & =\underset{\text{Δ}t\to 0}{\text{lim}}\frac{\left(3\text{Δ}t\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(2t\text{Δ}t+{\left(\text{Δ}t\right)}^{2}-4\text{Δ}t\right)\phantom{\rule{0.1em}{0ex}}\text{j}}{\text{Δ}t}\hfill \\ & =\underset{\text{Δ}t\to 0}{\text{lim}}\left(3\phantom{\rule{0.1em}{0ex}}\text{i}+\left(2t+\text{Δ}t-4\right)\phantom{\rule{0.1em}{0ex}}\text{j}\right)\hfill \\ & =3\phantom{\rule{0.1em}{0ex}}\text{i}+\left(2t-4\right)\phantom{\rule{0.1em}{0ex}}\text{j}.\hfill \end{array}$

Use the definition to calculate the derivative of the function $\text{r}\left(t\right)=\left(2{t}^{2}+3\right)\phantom{\rule{0.1em}{0ex}}\text{i}+\left(5t-6\right)\phantom{\rule{0.1em}{0ex}}\text{j}.$

${r}^{\prime }\left(t\right)=4t\phantom{\rule{0.1em}{0ex}}\text{i}+5\phantom{\rule{0.1em}{0ex}}\text{j}$

Notice that in the calculations in [link] , we could also obtain the answer by first calculating the derivative of each component function, then putting these derivatives back into the vector-valued function. This is always true for calculating the derivative of a vector-valued function, whether it is in two or three dimensions. We state this in the following theorem. The proof of this theorem follows directly from the definitions of the limit of a vector-valued function and the derivative of a vector-valued function.

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