Calculus

Differentiation of Functions

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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.

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