### Differentiating Vectors

#### Definition and Notation

The derivative of a vector is found by differentiating each component of the vector with respect to the variable of interest.

Suppose a vector $\boldsymbol{r}$ with components $x$, $y$ and $z$ is defined in terms of $t$, that is, $\boldsymbol{r}=\bigl(x(t),y(t),z(t)\bigl)$. Then the derivative with respect to $t$ of that vector is:

$\dfrac{\mathrm{d}\boldsymbol{r} }{\mathrm{d} t}=\dfrac{\mathrm{d}}{\mathrm{d} t}\Bigl[\bigl(x(t),y(t),z(t)\bigl)\Bigl]=\left(\dfrac{\mathrm{d} x}{\mathrm{d} t},\dfrac{\mathrm{d} y}{\mathrm{d} t},\dfrac{\mathrm{d} z}{\mathrm{d} t}\right).$

The prime notation used to denote the derivative of a scalar function is also used to denote the derivative of a vector:

$\dfrac{\mathrm{d}\boldsymbol{r} }{\mathrm{d} t}=\boldsymbol{r}' .$

If the variable $t$ is used to denote time, then the dot notation can be used to denote the derivative:

$\dfrac{\mathrm{d}\boldsymbol{r} }{\mathrm{d} t}=\dot{\boldsymbol{r} } .$

#### Applications

Vectors are often used to represent physical quantities. Information about how these quantities change over time can be found through differentiation.

Example: Suppose the position of a particle is given by the position vector $\boldsymbol{x}=\bigl(x(t),y(t),z(t)\bigl)$, where $x(t)$, $y(t)$, $z(t)$ give the distance along the $x$, $y$, $z$ axes from a chosen point of origin at time $t$.

Recall that the velocity of an object is defined to be the rate of change of position with respect to time. The vector that describes the velocity of the particle, denoted $\boldsymbol{v}$, is therefore:

$\boldsymbol{v}=\dot{\boldsymbol{x} }=\dfrac{\mathrm{d}\boldsymbol{x} }{\mathrm{d} t}.$

Similarly, the acceleration of an object is defined to be the rate of change of velocity with respect to time. The vector that describes the acceleration of the particle, denoted $\boldsymbol{a}$, is therefore:

$\boldsymbol{a}=\ddot{\boldsymbol{x} }=\dfrac{\mathrm{d}\boldsymbol{v} }{\mathrm{d}t}=\dfrac{\mathrm{d} }{\mathrm{d}t}\left[\dfrac{\mathrm{d}\boldsymbol{x} }{\mathrm{d}t}\right]=\dfrac{\mathrm{d}^2 \boldsymbol{x} }{\mathrm{d}t^2}.$

#### Worked Example

###### Example 1

Suppose the position of an object at time $t$ is described by $\boldsymbol{x}=\bigl(3t^2,t^3,\cos{t}\bigl)$. Find the velocity vector $\boldsymbol{v}$ and acceleration vector $\boldsymbol{a}$ at time $t=2$.

###### Solution

Recall that the velocity of an object with position $\boldsymbol{x}$ is given by $\dfrac{\mathrm{d}\boldsymbol{x} }{\mathrm{d} t}$.

Here $\boldsymbol{x}=\left(3t^2,t^3,\cos{t}\right)$ so the velocity of the object is given by differentiating each component of the position vector.

\begin{align} \boldsymbol{v}=\dot{\boldsymbol{x} } &= \dfrac{\mathrm{d

{\mathrm{d} t}\left[\left(3t^2,t^3,\cos{t}\right)\right] \\ &= \left(\dfrac{\mathrm{d}}{\mathrm{d} t}\left[3t^2\right],\dfrac{\mathrm{d}}{\mathrm{d} t}\left[t^3\right],\dfrac{\mathrm{d}}{\mathrm{d} t}\left[\cos{t}\right]\right) \\ &= \left(6t,3t^2,-\sin{t}\right). \end{align}

At time $t=2$, the velocity vector of the object is

\begin{align} \boldsymbol{v} &= \left(6\cdot2,3\cdot2^2,-\sin{2}\right) \\ &=\left(12,12,-\sin{2}\right). \end{align}

Now recall that the acceleration of a moving body with velocity $\boldsymbol{v}$ is given by differentiating each component of the velocity vector.

$\boldsymbol{a}=\ddot{\boldsymbol{x} }=\dfrac{\mathrm{d}\boldsymbol{v} }{\mathrm{d}t}=\dfrac{\mathrm{d}^2 \boldsymbol{x} }{\mathrm{d}t^2}.$

Here $\boldsymbol{v}= \left(6t,3t^2,-\sin{t}\right)$ so the acceleration of the object is given by

\begin{align} \boldsymbol{a}=\ddot{\boldsymbol{x} } &= \dfrac{\mathrm{d}}{\mathrm{d} t}\left[\left(6t,3t^2,-\sin{t}\right)\right] \\ &=\left(\dfrac{\mathrm{d}}{\mathrm{d} t}\left[6t\right],\dfrac{\mathrm{d}}{\mathrm{d} t}\left[3t^2\right],\dfrac{\mathrm{d}}{\mathrm{d} t}\left[-\sin{t}\right]\right) \\ &=\left(6,6t,-\cos{t}\right). \end{align}

At time $t=2$, the acceleration vector of the object is

\begin{align} \boldsymbol{a} &= \left(6,6 \times 1,-\cos{1}\right) \\ &= \left(6,6,-\cos{1}\right). \end{align}

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#### Video Examples

###### Example 1

Prof. Robin Johnson states the result for the derivative of a vector and derives the general result for the derivative of the product of a scalar function and a vector.

###### Example 2

Prof. Robin Johnson derives the general result for the derivative of the dot product of two vectors.

#### Workbooks

These workbooks produced by HELM are good revision aids, containing key points for revision and many worked examples. 