Derivatives of Vector-Valued Functions

Derivatives of Vector-Valued Functions

We are now going to extend out concept of a derivative to vector-valued functions. Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function defined for $t$ in the interval $I$ that traces out the curve $C$. Let $P$ and $Q$ be points on $C$, and let $\vec{r}(t)$ be the position vector of $P$, and let $\vec{r}(t + h)$ be the position vector of $Q$. Then $\vec{PQ} = \vec{r}(t + h) - \vec{r}(t)$ forms a secant vector to the curve $C$. Now for $h \neq 0$ the vector $\frac{\vec{r}(t + h) - \vec{r}}{h}$ has the same direction as $\vec{r}(t + h) - \vec{r}(t)$ and differs only in length/magnitude. As $h \to 0$, this vector $\frac{\vec{r}(t + h) - \vec{r}(t)}{h}$ approaches a vector that is tangent to the curve and that lies on the tangent line of the point $P$, and we define $\vec{r'}(t) = \vec{v}(t)$ to be the tangent vector to $C$ defined by $\vec{r}(t)$ at $P$.

Definition: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function defined for all $t$ in the interval $I$. Then the Derivative of $\vec{r}(t)$ is denoted $\vec{r'}(t)$ or $\frac{d}{dt} \vec{r}(t)$ is $\vec{r'}(t) = \lim_{h \to 0} \frac{\vec{r}(t + h) - \vec{r}(t)}{h}$ provided this limit exists. If $P$ is a point on the curve with position vector $\vec{r}(a)$ where $a \in I$, then $\vec{r'}(a)$ is the Tangent Vector on $C$ of the point $P$. The Tangent Line to the point $P$ is the line that passes through $P$ are is parallel to $\vec{r'}(a)$.

Computing the derivatives of vector-valued functions is fortunately easy, as the following theorem will tell us.

Theorem 1: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function defined for $t$ in the interval $I$ and where $x(t)$, $y(t)$, and $z(t)$ are differentiable on $I$. Then $\vec{r'}(t) = (x'(t), y'(t), z'(t))$
  • Proof: Let $\vec{r}(t) = (x(t), y(t), z(t))$ be a vector-valued function. Then we have that:
\begin{align} \quad \vec{r'}(t) = \lim_{h \to 0} \frac{1}{h} [ \vec{r}(t + h) - \vec{r}(t)] \\ \quad \vec{r'}(t) = \lim_{h \to 0} \frac{1}{h} [ (x(t + h), y(t + h), z(t + h)) - (x(t), y(t), z(t))] \\ \quad \vec{r'}(t) = \lim_{h \to 0} \frac{1}{h} [x(t + h) - x(t), y(t + h) - y(t), z(t + h) - z(t)] \\ \quad \vec{r'}(t) = \left ( \lim_{h \to 0} \frac{x(t+h) - x(t)}{h}, \lim_{h \to 0} \frac{y(t+h) - y(t)}{h}, \lim_{h \to 0} \frac{z(t+h) - z(t)}{h} \right ) \\ \quad \vec{r'}(t) = (x'(t), y'(t), z'(t)) \quad \blacksquare \end{align}

We note that for $\vec{r}(t)$ to be differentiable at $t$ we must have that $x = x(t)$, $y = y(t)$, and $z = z(t)$ are all differentiable functions at $t$.

Often times, vector-valued functions are described in a physical setting. If we think of $\vec{r}(t)$ as a vector-valued function that tells us the position vector of some point $P$, then $\vec{r'}(t)$ is a vector-valued function that tells us the velocity vector of some point $P$. Hence, we often will write $\vec{r'}(t) = \vec{v}(t)$. All of this is analogous to that of real-valued functions specifying position, and velocity. The following image represents some velocity vectors from an arbitrary vector-valued function $\vec{r}(t)$.


Furthermore, if we imagine $\vec{r}(t)$ as tracing the curve $C$ with point $P$, then the speed of $P$ with position vector $\vec{r}(t)$ and velocity vector $\vec{r'}(t) = \vec{v}(t)$ will have its speed equal to the magnitude of the velocity vector, that is:

\begin{align} \mathrm{speed} = \| \vec{r'}(t) \| = \| \vec{v}(t) \| = \sqrt{ [x'(t)]^2 + [y'(t)]^2 + [z'(t)]^2} \end{align}

As an even further extension, if we take the derivative of $\vec{v}(t)$, we get a vector-valued function $\vec{v'}(t) = \vec{a}(t)$ which gives us an acceleration vector for the point $P$.

We will look more into these concepts later on, but first, let's go back to some fundamentals and look at an example of computing a derivative of a vector-valued function. Consider the vector-valued function $\vec{r}(t) = (t^2, e^t, 2t + 1)$. $x(t) = t^2$, $y(t) = e^t$, and $z(t) = 2t + 1$ are differentiable for all $t \in \mathbb{R}$ and so $\vec{r}(t)$ will be differentiable for all $t \in \mathbb{R}$. We would compute $\vec{v}(t)$ as follows:

\begin{align} \vec{v}(t) = \left ( \frac{d}{dt} t^2, \frac{d}{dt} e^t, \frac{d}{dt} 2t + 1 \right ) = (2t, e^t, 2) \end{align}

Example 1

Compute $\vec{r'}(t)$ given $\vec{r}(t) = \left ( e^t \sin t, 2^t + t^2, \sec t \right )$.

\begin{align} \quad \vec{r'}(t) = \left ( \frac{d}{dt} e^t \sin t, \frac{d}{dt} 2^t + t^2, \frac{d}{dt} \sec t \right ) = \left (e^t \sin t + e^t \cos t, 2^t \ln(2) + 2t, \sec t \tan t \right ) \end{align}
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