# Calculus

## Differentiation of Functions # Derivatives of Vector-Valued Functions

## Definition of Vector-Valued Functions

A vector-valued function of one variable in Cartesian $$3D$$ space has the form

${\mathbf{r}\left( t \right) = f\left( t \right)\mathbf{i} + g\left( t \right)\mathbf{j} + h\left( t \right)\mathbf{k}}\;\;\text{or}\;\;\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle ,$

where $$f\left( t \right),$$ $$g\left( t \right),$$ $$h\left( t \right)$$ are called the component functions.

Similarly, a vector-valued function in in Cartesian $$2D$$ space is given by

$\mathbf{r}\left( t \right) = f\left( t \right)\mathbf{i} + g\left( t \right)\mathbf{j}\;\;\text{or}\;\;\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right)} \right\rangle.$

## Limits and Continuity of Vector-Valued Functions

Suppose that $$\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle.$$ The limit of $$\mathbf{r}\left( t \right)$$ as $$t$$ approaches $$a$$ is given by

$\lim\limits_{t \to a} \mathbf{r}\left( t \right) = \lim\limits_{t \to a} \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle = \left\langle {\lim\limits_{t \to a} f\left( t \right),\lim\limits_{t \to a} g\left( t \right),\lim \limits_{t \to a} h\left( t \right)} \right\rangle ,$

provided the limit of the component functions exist.

The vector-valued function $$\mathbf{r}\left( t \right)$$ is continuous at $$t = a$$ if

$\lim\limits_{t \to a} \mathbf{r}\left( t \right) = \mathbf{r}\left( a \right).$

## Derivative of a Vector-Valued Function

The derivative $$\mathbf{r}^\prime\left( t \right)$$ of the vector-valued function $$\mathbf{r}\left( t \right)$$ is defined by

$\frac{{d\mathbf{r}}}{{dt}} = \mathbf{r}^\prime\left( t \right) = \lim\limits_{\Delta t \to 0} \frac{{\mathbf{r}\left( {t + \Delta t} \right) - \mathbf{r}\left( t \right)}}{{\Delta t}}$

for any values of $$t$$ for which the limit exists.

The vector $$\mathbf{r}^\prime\left( t \right)$$ is called the tangent vector to the curve defined by $$\mathbf{r}.$$

If $$\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle$$ where $$f, g$$ and $$h$$ are differentiable functions, then

$\mathbf{r}^\prime\left( t \right) = \left\langle {f^\prime\left( t \right),g^\prime\left( t \right),h^\prime\left( t \right)} \right\rangle.$

Thus, we can differentiate vector-valued functions by differentiating their component functions.

## Physical Interpretation

If $$\mathbf{r}\left( t \right)$$ represents the position of a particle, then the derivative is the velocity of the particle:

$\mathbf{v}\left( t \right) = \frac{{d\mathbf{r}}}{{dt}} = \mathbf{r}^\prime\left( t \right).$

The speed of the particle is the magnitude of the velocity vector:

$\left\| {\mathbf{v}\left( t \right)} \right\| = \sqrt {{{\left( {f^\prime\left( t \right)} \right)}^2} + {{\left( {g^\prime\left( t \right)} \right)}^2} + {{\left( {h^\prime\left( t \right)} \right)}^2}.}$

In a similar way, the derivative of the velocity is the acceleration:

$\mathbf{a}\left( t \right) = \frac{{d\mathbf{v}}}{{dt}} = \mathbf{v}^\prime\left( t \right) = \mathbf{r}^{\prime\prime}\left( t \right).$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

A particle moves along the curve

$$\mathbf{r}\left( t \right) = \left( {{t^2} - 1} \right)\mathbf{i} \;+$$ $$\left( {2{t^3} + 3} \right)\mathbf{j} \;+$$ $$\left( {3t - 4} \right)\mathbf{k}.$$

Find the speed of the particle at $$t = 1.$$

### Example 2

Compute the derivative of the vector-valued function $\mathbf{r}\left( t \right) = \left\langle {\sin 2t,{e^{{t^2}}}} \right\rangle .$

### Example 3

Evaluate the derivative of the function $\mathbf{r}\left( t \right) = \left\langle {\ln t,3t + 1,{t^2}} \right\rangle$ at $$t = 1.$$

### Example 4

Differentiate the vector-valued function $\mathbf{r}\left( t \right) = \left\langle {\sin \left( {{t^2}} \right),4{t^2} - t} \right\rangle.$

### Example 1.

A particle moves along the curve

$$\mathbf{r}\left( t \right) = \left( {{t^2} - 1} \right)\mathbf{i} \;+$$ $$\left( {2{t^3} + 3} \right)\mathbf{j} \;+$$ $$\left( {3t - 4} \right)\mathbf{k}.$$

Find the speed of the particle at $$t = 1.$$

Solution.

We differentiate the vector function $$\mathbf{r}\left( t \right)$$ to find the velocity:

${\mathbf{r}^\prime\left( t \right) = \mathbf{v}\left( t \right) = 2t\mathbf{i} + 6{t^2}\mathbf{j} + 3\mathbf{k}.}$

Compute the speed of the particle at $$t = 1:$$

$\left\| {\mathbf{v}\left( t \right)} \right\| = \sqrt {{2^2} + {6^2} + {3^2}} = \sqrt {49} = 7.$

### Example 2.

Compute the derivative of the vector-valued function $\mathbf{r}\left( t \right) = \left\langle {\sin 2t,{e^{{t^2}}}} \right\rangle .$

Solution.

We differentiate the vector function on a component-by-component basis using the chain rule. This yields:

$\left( {\sin 2t} \right)^\prime = \cos 2t \cdot \left( {2t} \right)^\prime = 2\cos 2t;$
$\left( {{e^{{t^2}}}} \right)^\prime = {e^{{t^2}}} \cdot \left( {{t^2}} \right)^\prime = 2t{e^{{t^2}}}.$

So the derivative is given by

$\mathbf{r}^\prime\left( t \right) = \left\langle {2\cos 2t,2t{e^{{t^2}}}} \right\rangle .$

### Example 3.

Evaluate the derivative of the function $\mathbf{r}\left( t \right) = \left\langle {\ln t,3t + 1,{t^2}} \right\rangle$ at $$t = 1.$$

Solution.

We differentiate each component of the vector function. This yields:

$\mathbf{r}^\prime\left( t \right) = \left\langle {\frac{1}{t},3,2t} \right\rangle .$

Substitute the parameter value $$t = 1:$$

$\mathbf{r}^\prime\left( 1 \right) = \left\langle {1,3,2} \right\rangle = \mathbf{i} + 3\mathbf{j} + 2\mathbf{k}.$

### Example 4.

Differentiate the vector-valued function $\mathbf{r}\left( t \right) = \left\langle {\sin \left( {{t^2}} \right),4{t^2} - t} \right\rangle.$

Solution.

We take the derivative of each component of the function:

$\left( {\sin \left( {{t^2}} \right)} \right)^\prime = \cos \left( {{t^2}} \right) \cdot \left( {{t^2}} \right)^\prime = 2t\cos \left( {{t^2}} \right);$
$\left( {4{t^2} - t} \right)^\prime = 8t - 1.$

Hence, the derivative of the vector function is given by

$\mathbf{r}^\prime\left( t \right) = \left\langle {2t\cos \left( {{t^2}} \right),8t - 1} \right\rangle .$

See more problems on Page 2.