Derivative Rules for Vector Valued Functions Examples 1

# Derivative Rules for Vector Valued Functions Examples 1

Recall from the Derivative Rules for Vector-Valued Functions page that if $\vec{u}(t) = (x_1(t), y_1(t), z_1(t))$ and $\vec{v}(t) = (x_2(t), y_2(t), z_2(t))$ are vector-valued functions that are differentiable for $t$ in the interval $I$, and if $k$ be a scalar and $f(t)$ is a real-valued function that is differentiable on $I$, then:

• $( \vec{u}(t) + \vec{v}(t) )' = \vec{u'}(t) + \vec{v'}(t)$ (Sum Rule).
• $( \vec{u}(t) - \vec{v}(t) )' = \vec{u'}(t) - \vec{v'}(t)$ (Difference Rule).
• $(k \vec{u}(t))' = k \vec{u'} (t)$ (Scalar Multiple Rule).
• $(f(t) \vec{u}(t))' = f'(t) \vec{u}(t) + f(t) \vec{u'}(t)$ (Product Rule for Real-Valued and Vector-Valued Functions).
• $(\vec{u} \cdot \vec{v})' = \vec{u'}(t) \cdot \vec{v}(t) + \vec{u}(t) \cdot \vec{v'}(t)$ (Dot Product Rule).
• $(\vec{u} \times \vec{v})' = \vec{u'}(t) \times \vec{v}(t) + \vec{u}(t) \times \vec{v'}(t)$ (Cross Product Rule).
• $\vec{u}(f(t)) = f'(t) \vec{u'}(f(t))$ (Chain Rule).

We will now look at some examples regarding these differentiation rules for vector-valued functions.

## Example 1

Let $\vec{r}(t)$ be a vector-valued function. Prove that $\| \vec{r'}(t) \|$ is constant on an interval $[a, b]$ if and only if $\vec{r'}(t) \cdot \vec{r''}(t) = 0$ for all $t \in [a, b]$.

$\Rightarrow$ Suppose that $\| \vec{r'}(t) \|$ is constant on an interval $[a, b]$. Then for some $C \in \mathbb{R}$ we have that for all $t \in [a, b]$:

(1)
\begin{align} \quad \| \vec{r'}(t) \| = C \end{align}

We note that $\vec{r'}(t) \cdot \vec{r'}(t) = \| \vec{r'}(t) \|^2 = C^2$. Differentiating both sides of this equation and we have that:

(2)
\begin{align} \quad \vec{r''}(t) \cdot \vec{r'}(t) + \vec{r'}(t) \vec{r''}(t) = 0 \\ \quad 2 \vec{r'}(t) \cdot \vec{r''}(t) = 0 \\ \quad \vec{r'}(t) \cdot \vec{r''}(t) = 0 \end{align}

$\Leftarrow$ Now suppose that $\vec{r'}(t) \cdot \vec{r''}(t) = 0$. Then for some $D > 0$ we have that:

(3)
\begin{align} \quad \vec{r'}(t) \cdot \vec{r''}(t) = 0 \\ \quad 2 \vec{r'}(t) \cdot \vec{r''}(t) = 0 \\ \quad \vec{r'}(t) \cdot \vec{r'}(t) = C \quad \| \vec{r'}(t) \|^2 = D \\ \quad \| \vec{r'}(t) \| = \sqrt{D} \end{align}

Therefore $\| \vec{r'}(t) \|$ is constant.

Note that we actually used integration in proving the converse of the problem. We will subsequently look more into integration of vector-valued functions.

## Example 2

Prove that if $\vec{r}(t) \cdot \vec{r'}(t) = 0$ then $\vec{r}(t)$ is a curve lying on a sphere.

To show that $\vec{r}(t)$ lies on a sphere, all we must do is show that $\| \vec{r}(t) \|$ which represents the distance from any point $P$ on $\vec{r}(t)$ to the origin is constant.

If we differentiate $\| \vec{r}(t) \|$ we have that:

(4)
\begin{align} \quad \frac{d}{dt} \| \vec{r}(t) \| = \frac{d}{dt} \left ( \vec{r}(t) \cdot \vec{r}(t) \right ) = 2 \vec{r}(t) \cdot \vec{r'}(t) = 0 \end{align}

Since $\frac{d}{dt} \| \vec{r}(t) \| = 0$, this implies that $\| \vec{r}(t) \| = C$ for some $C > 0$ and $C \in \mathbb{R}$. Therefore, $\vec{r}(t)$ lies on a sphere.