Derivatives of Vector Valued Functions

Section 4.2 Derivatives of Vector Valued Functions

Suppose that \(\vect g\) is a function with domain \(D\subset\mathbb R^N\) and values in \(\mathbb R^k\text{.}\) We denote its component functions by \(g_i\text{,}\) \(i=1,\dots,k\) and define

\begin{equation*} \frac{\partial}{\partial x_j}\vect g(\vect x) = \frac{\partial}{\partial x_j} \begin{bmatrix} g_1(\vect x) \\ \vdots \\ g_k(\vect x) \end{bmatrix} := \begin{bmatrix} \frac{\partial}{\partial x_j}g_1(\vect x) \\ \vdots \\ \frac{\partial}{\partial x_j}g_k(\vect x) \end{bmatrix}\text{.} \end{equation*}

We call such a function a vector valued function.

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If \(N=1\text{,}\) and the variable is \(t\in\mathbb R\) we set

\begin{equation*} \frac{d}{dt}\vect g(t) =\vect g'(t) =\frac{d}{dt} \begin{bmatrix} g_1(t) \\ \vdots \\ g_k(t) \end{bmatrix} := \begin{bmatrix} g_1'(t) \\ \vdots \\ g_k'(t) \end{bmatrix}\text{,} \end{equation*}

that is, we differentiate component-wise. We now want to derive an interpretation of \(\vect g'(t)\text{.}\) To do so it is useful to think of \(\vect g(t)\) to be the position of moving particle at time \(t\text{.}\) Then the difference coefficient

\begin{equation*} \frac{1}{\Delta t}\Bigl(\vect g(t+\Delta t)-\vect g(t)\Bigr) \end{equation*}

points in the direction of the line \(L\) shown in Figure 4.12. The magnitude of that vector is the approximate speed of the particle moving from \(\vect g(t)\) to \(\vect g(t+\Delta t)\text{.}\)

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Figure 4.12. Secants approaching the tangent of a curve.
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Passing to the limit we see that

\begin{equation*} \frac{1}{\Delta t}\Bigl(\vect g(t+\Delta t)-\vect g(t)\Bigr) = \begin{bmatrix} \frac{g_1(t+\Delta t)-g_1(t)}{\Delta t} \\ \vdots \\ \frac{g_k(t+\Delta t)-g_k(t)}{\Delta t} \end{bmatrix} \xrightarrow{\quad\Delta t\to 0\quad} \begin{bmatrix} g_1'(t) \\ \vdots \\ g_k'(t) \end{bmatrix}\text{.} \end{equation*}

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Let us summarise what we found in the following proposition.

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Proposition 4.13. Derivative of vector valued function of a single variable.

Under the above assumption

\begin{equation} \text{the vector }\vect g'(t)\text{ is tangential to the path described by }\vect g(t)\text{.}\tag{4.1} \end{equation}

Moreover,

\begin{equation} \lim_{\Delta t\to 0} \frac{\|\vect g(t+\Delta t)-\vect g(t)\|}{\Delta t} =\|\vect g'(t)\|\text{.}\tag{4.2} \end{equation}

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Remark 4.14. Speed and velocity of a particle.

If we consider \(\vect g(t)\) as the position of a particle at time \(t\text{,}\) then

\begin{equation*} \frac{\|\vect g(t+\Delta t)-\vect g(t)\|}{\Delta t} \end{equation*}

is the approximate speed of that particle travelling from \(\vect g(t)\) to \(\vect g(t+\Delta t)\text{.}\) Now (4.2) tells us that \(\|\vect{g}'(t)\|\) is very close to that speed. The direction in which the particle travels is the tangential direction given by \(\vect g'(t)\text{.}\) In other words, \(\vect g'(t)\) is the velocity of the particle at \(\vect g(t)\text{.}\) The distance travelled in a small time interval is therefore

\begin{equation*} \|\vect g(t+\Delta t)-\vect g(t)\| \approxeq\|\vect g'(t)\|\Delta t\text{.} \end{equation*}

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Note that the above approximate equality holds no matter what \(k\) is.

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We can also use the above to find a representation of the tangent line to a curve in space.

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Proposition 4.15. Equation of tangent to curve.

Suppose that \(I\subset\mathbb R\) is an interval, and \(\vect g\colon I\to\mathbb R^N\) is a differentiable function. Then

\begin{equation*} \vect r(s):=\vect g(t)+s\vect g'(t),\qquad s\in\mathbb R \end{equation*}

is an equation for the tangent line to the curve described by \(\vect g(t)\) (\(t\in I\)) \(\vect g(t)\text{.}\)

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For vectors we have three different kinds of products: the multiplication by scalars and the scalar product in \(\mathbb R^N\) for general \(N\text{,}\) and the cross product in \(\mathbb R^3\text{.}\) We conclude this section by proving a ‘product rule’ for these products.

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Theorem 4.16. Generalised product rule.

Suppose that \(\vect f\) and \(\vect g\) are vector valued continuous functions defined on an open set \(D\subset\mathbb R^N\) with values in \(\mathbb R^k\text{,}\) and that \(u\) is a continuous function on \(D\) with values in \(\mathbb R\text{.}\) Then

\(\displaystyle \dfrac{\partial}{\partial x_i} \left(u(\vect x)\vect f(\vect x)\right) =\left(\dfrac{\partial u}{\partial x_i}(\vect x)\right) \vect f(\vect x) +u(\vect x)\dfrac{\partial \vect f}{\partial x_i}(\vect x)\)
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\(\displaystyle \dfrac{\partial}{\partial x_i} \left(\vect f(\vect x)\cdot\vect g(\vect x)\right) =\left(\dfrac{\partial \vect f}{\partial x_i}(\vect x)\right) \cdot \vect g(\vect x) +\vect f(\vect x)\cdot \dfrac{\partial \vect g}{\partial x_i}(\vect x)\)
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If \(k=3\) then

\begin{equation*} \dfrac{\partial}{\partial x_i} \left(\vect f(\vect x)\times\vect g(\vect x)\right) =\left(\dfrac{\partial \vect f}{\partial x_i}(\vect x)\right) \times \vect g(\vect x) +\vect f(\vect x)\times \dfrac{\partial \vect g}{\partial x_i}(\vect x) \end{equation*}

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whenever all relevant derivatives exist.

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Proof.

The proof of all the above formulae are analogous to the classical product rule. As an example we give a proof of the first. By definition of the partial derivative we need to compute the limit of the difference quotient

\begin{equation*} D(t):=\frac{1}{t} \Bigl(u(\vect x+t\vect e_i)\vect f(\vect x+t\vect e_i) -u(\vect x)\vect f(\vect x)\Bigr) \end{equation*}

as \(t\to 0\text{,}\) where \(\vect e_i\) is the vector

\begin{equation*} \vect e_i:=(0,\dots,0,1,0,\dots,0)\text{,} \end{equation*}

where the \(1\) appears in the \(i\)-th coordinate

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By adding and subtracting a term we can rewrite \(D(t)\) by

\begin{equation*} D(t)=\frac{1}{t} \Bigl(u(\vect x+t\vect e_i)-u(\vect x)\Bigr)\vect f(\vect x+t\vect e_i) +u(x)\frac{1}{t} \Bigl(\vect f(\vect x+t\vect e_i)-\vect f(\vect x)\Bigr)\text{.} \end{equation*}

Using the continuity of the functions and the definition of the partial derivatives we get from the above that

\begin{equation*} \lim_{t\to 0}D(t) =\Bigl(\frac{\partial u}{\partial x_i}(\vect x)\Bigr) \vect f(\vect x) +u(\vect x)\frac{\partial \vect f}{\partial x_i}(\vect x) \end{equation*}

as required.

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