The derivative $\mathbf{r}'$ of a vector function $\mathbf{r}$ works very similarly to how we are used to thinking of derivatives of one-dimensional functions. For one-dimensional functions, the derivative is defined as
$$ f'(t) = \lim_{h\rightarrow 0} \dfrac{f(t+h) - f(t)}{h} $$
and is represented geometrically as the slope of the tangent line.
This notion is easily extended to vector functions. Last section, we discussed how the limits of functions may be computed component-wise. Because of this, we can also consider derivatives of vector functions component-wise. The derivative $\mathbf{r}'$ of a vector function $\mathbf{r}$ is given by
$$ \dfrac{d\mathbf{r}}{dt} = \mathbf{r}'(t) = \lim_{\Delta t\rightarrow 0} \dfrac{\mathbf{r}(t+h) - \mathbf{r}}{\Delta t} $$
Since we consider $\mathbf{r} = \langle\, f(t),g(t),h(t)\,\rangle$, the definition of the derivative may be written as
$$ \begin{align*}
\mathbf{r}'(t) &= \lim_{\Delta t\rightarrow 0} \dfrac{1}{\Delta t}\Big(\mathbf{r}(t+\Delta t) - \mathbf{r}(t)\Big) \\
\\
&= \lim_{\Delta t\rightarrow 0} \dfrac{1}{\Delta t}\Big(\left\langle\,f(t+\Delta t),g(t+\Delta t),h(t+\Delta t)\,\right\rangle - \left\langle\,f(t),g(t),h(t)\,\right\rangle \Big) \\
\\
&= \lim_{\Delta t\rightarrow 0} \dfrac{1}{\Delta t} \Big\langle\, f(t+\Delta t) - f(t),g(t+\Delta t) - g(t),h(t+\Delta t) - h(t)\,\Big\rangle \\
\\
&= \left\langle\,\lim_{\Delta t\rightarrow 0} \dfrac{ f(t+\Delta t) - f(t)}{\Delta t},\lim_{\Delta t\rightarrow 0}\dfrac{ g(t+\Delta t) - g(t)}{\Delta t},\lim_{\Delta t\rightarrow 0}\dfrac{ h(t+\Delta t) - h(t) }{\Delta t}\,\right\rangle \\
\\
&= \left\langle\, f'(t),g'(t),h'(t)\,\right\rangle
\end{align*} $$
Theorem ¶
Derivative Formula for Vector Functions of a Single Variable
If $\def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \mathbf{r}(t) = \left\langle\, f(t),g(t),h(t)\,\right\rangle = f(t)\ihat + g(t)\jhat + h(t)\khat$, and each of $f$, $g$, and $h$ are differentiable functions, then
$$ \mathbf{r}'(t) = \left\langle\, f'(t),g'(t),h'(t)\,\right\rangle = f'(t)\ihat + g'(t)\jhat + h'(t)\khat $$
Visually, this closely resembles how tangent lines work in the one-dimensional case. If we want the equation of the line tangent to the plane curve for $t = t_0$, the equation is
$$ \mathbf{l}(t) = \mathbf{r}(t_0) + t\mathbf{r}'(t_0) $$
It is more typical to be interested in the
tangent vector
$\mathbf{r}'(t)$ depicted below with the directed line segment from $P = \mathbf{r}(t_0)$ and $Q = \mathbf{r}(t_0 + h)$ (the analogue to the secant line).
For computation purposes, we prefer to have the
unit tangent vector
$$ \mathbf{T}(t) = \dfrac{\mathbf{r}(t)}{\left|\mathbf{r}(t)\right|} $$
to isolate the direction from the scaling from computing the derivative of $\mathbf{r}$.
For the plane curve
$$ \mathbf{r}(t) = e^{2t}\ihat + e^t \jhat $$
The tangent vector in the image above was scaled so that it is actually the unit tangent vector.
Due to the fact that limits work component-wise, the differentiation rules for vector functions are analogous to those for real-valued functions.
Theorem ¶
Differentiation Rules for Vector Functions
If $\mathbf{u}$ and $\mathbf{v}$ are differentiable vector functions, $c$ is a scalar, and $f$ is a real-valued function, then
- $\dfrac{d}{dt}\Big(\mathbf{u}(t) + \mathbf{v}(t)\Big) = \mathbf{u}'(t) + \mathbf{v}'(t)$
- $\dfrac{d}{dt}\Big(c\mathbf{u}(t)\Big) = c\mathbf{u}'(t)$
- $\dfrac{d}{dt}\Big(f(t)\mathbf{u}(t)\Big) = f'(t)\mathbf{u}(t) + f(t)\mathbf{u}'(t)$
- $\dfrac{d}{dt}\Big(\mathbf{u}(t) \cdot \mathbf{v}(t)\Big) = \mathbf{u}'(t)\cdot\mathbf{v}(t) + \mathbf{u}(t)\cdot\mathbf{v}'(t)$
- $\dfrac{d}{dt}\Big(\mathbf{u}(t) \times \mathbf{v}(t)\Big) = \mathbf{u}'(t)\times\mathbf{v}(t) + \mathbf{u}(t)\times\mathbf{v}'(t)$
- $\dfrac{d}{dt}\Big(\mathbf{u}(f(t))\Big) = f'(t)\mathbf{u}'(f(t))$
There are some things to notice about these differentiation rules. Properties 1 and 2 are both related to how limits work with sums and scalar multiplication for continuous functions (since differentiable functions are continuous by definition). Properties 3, 4, and 5 are all variations of the product rule . It does not matter if the product is between a real-valued and vector function, the dot product between two vector functions, or the cross product between to vector functions. They all obey the product rule with the appropriate product. Lastly, property 6 is the chain rule computed component-wise.
If a vector $\mathbf{r}$ is
twice differentiable
, that is $\mathbf{r}''$ exists, then
$$ \dfrac{d}{dt}\Big( \mathbf{r}(t)\times\mathbf{r}'(t)\Big) = \mathbf{r}(t)\times\mathbf{r}''(t) $$
A vector's cross product with itself is $\mathbf{0}$.
Find an expression for
$$ \dfrac{d}{dt}\Big[ \mathbf{u}(t)\cdot\Big(\mathbf{v}(t)\times\mathbf{w}(t)\Big)\Big] $$
This is a variation of the triple product rule .
As with derivatives, we can compute the
integrals
of vector functions component-wise due to the fact that limits work component wise. Recall that the definition of the definite integral for a real-valued function is
$$ \int_a^b f(t)\, dt = \lim_{n\rightarrow\infty} \sum_{i=1}^n f(t_i^*)\, \Delta t $$
where each $t_i^*$ is a
sample point
within a subinterval of the
partition
of $[a,b]$. The value $f(t_i^*)$ is the height of the function and $\Delta t$ is the interval width, whose product is a one of the rectangles we sum up.
For vector functions, the computation is very similar
$$ \begin{align*}
\int_a^b \mathbf{r}(t)\,dt &= \lim_{n\rightarrow\infty} \mathbf{r}(t_i^*)\,dt \\
\\
&= \lim_{n\rightarrow\infty} \left[\left(\lim_{n\rightarrow\infty} \sum_{i=1}^n f(t_i^*)\, \Delta t\right)\ihat + \left(\lim_{n\rightarrow\infty} \sum_{i=1}^n g(t_i^*)\, \Delta t\right)\jhat + \left(\lim_{n\rightarrow\infty} \sum_{i=1}^n h(t_i^*)\, \Delta t\right)\khat\right] \\
\\
&= \left(\int_a^b f(t)\,dt\right)\ihat + \left(\int_a^b g(t)\,dt\right)\jhat + \left(\int_a^b h(t)\,dt\right)\khat
\end{align*} $$
We take the last line of this expression as our practical formula for the definite integral of a space curve.
Definition ¶
Definite Integral of a Space Curve
$$ \int_a^b \mathbf{r}(t)\,dt = \left(\int_a^b f(t)\,dt\right)\ihat + \left(\int_a^b g(t)\,dt\right)\jhat + \left(\int_a^b h(t)\,dt\right)\khat $$
Furthermore, if we have a function $\mathbf{R}$ that is that antiderivative of $\mathbf{r}$ $\left(\mathbf{R}'(t) = \mathbf{r}(t)\right)$, we can give a vector function statement of the Fundamental Theorem of Calculus .
Theorem ¶
Fundamental Theorem of Calculus
$$ \int_a^b \mathbf{r}(t)\, dt = \mathbf{R}(t)\,\Big|_a^b = \mathbf{R}(b) - \mathbf{R}(a) $$
Evaluate the integral
$$ \int_0^\frac{\pi}{4} \Big(\sec t\tan t\,\ihat + t\cos 2t\,\jhat + \sin^2 2t\cos 2t\,\khat\Big)\,dt $$
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