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Math 344: Calculus III

13.2 Derivatives and Integrals of Vector Functions


13.2.1 Derivatives of Vector Functions

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.
A typical one-dimensional function with its tangent line depicted at one point.
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). A tangent vector for a space curve. Also plotted is the directed line segment and generating vectors for the space curve equivalent to a 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 $\mathbf{r}$.

13.2.2 Derivative Examples

Exercise 13.2.1

For the plane curve

$$ \mathbf{r}(t) = e^{2t}\ihat + e^t \jhat $$

  1. Find $\mathbf{r}'(t)$
  2. Sketch the plane curve using the vector equation, then sketch $\mathbf{r}(0)$ and $\mathbf{r}'(0)$
  3. Determine the unit tangent vector at $t = 0$
Check Your Work Begin by finding $\mathbf{r}'(t)$ component-wise

$$ \mathbf{r}'(t) = \left(e^{2t}\right)'\ihat + \left(e^t\right)'\jhat = 2e^{2t}\ihat + e^t \jhat $$
then plot the function in the $xy$-plane, using the component functions. The function is plotted here for $t\in [-10,0.6]$. The plane curve given by exp(2t) ihat + exp(t) jhat with the vectors r(1) and r'(1) also shown. The lower end of the $t$ range can go in negative infinity since both component functions go to zero for large negative $t$, meaning the plane curve begins at the origin. Overall, the plane curve resembles a square root curve.
The unit tangent vector at $t=1$ is given by

$$ \begin{align*} \mathbf{T}(1) &= \dfrac{\mathbf{r}'(1)}{\left|\mathbf{r}'(1)\right|} \\ \\ &= \dfrac{\left\langle\, 2e^{2(0)},e^0\,\right\rangle}{\left|\left\langle\, 2e^{2(0)},e^0\,\right\rangle\right|} \\ \\ &= \dfrac{\left\langle\, 2,1\,\right\rangle}{\left|\left\langle\, 2,1\,\right\rangle\right|} \\ \\ &= \dfrac{1}{\sqrt{5}}\left\langle\, 2,1\,\right\rangle \end{align*} $$
The tangent vector in the image above was scaled so that it is actually the unit tangent vector.

Exercise 13.2.2

For each of the following functions, find the unit tangent vector at the given value of $t$:
  1. $\mathbf{r}(t) = \left\langle\, \tan^{-1}t,2e^{2t},8te^t\,\right\rangle,\quad t=0$
  2. $\mathbf{r}(t) = \sin^2 t\,\ihat + \cos^2 t\,\jhat + \tan^2 t\,\khat,\quad t=\pi/4$
Check Your Work
  1. $$\mathbf{r}'(t) = \left\langle\, \dfrac{1}{1+t^2},4e^{2t},8e^t + 8te^t\,\right\rangle$$
    $$ \mathbf{T}(0) = \dfrac{1}{9}\left\langle\,1,4,8\,\right\rangle $$
  2. $$\mathbf{r}'(t) = 2\sin t\cos t\,\ihat -2\sin t\cos t\,\jhat + 2\tan t\sec^2 t\,\khat$$
    $$ \mathbf{T}\left(\dfrac{\pi}{4}\right) = \dfrac{1}{3\sqrt{2}}\ihat - \dfrac{1}{3\sqrt{2}}\jhat + \dfrac{4}{3\sqrt{2}}\khat$$

Video Solution

13.2.3 Rules for Differentiating Vector Functions

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

  1. $\dfrac{d}{dt}\Big(\mathbf{u}(t) + \mathbf{v}(t)\Big) = \mathbf{u}'(t) + \mathbf{v}'(t)$

  2. $\dfrac{d}{dt}\Big(c\mathbf{u}(t)\Big) = c\mathbf{u}'(t)$

  3. $\dfrac{d}{dt}\Big(f(t)\mathbf{u}(t)\Big) = f'(t)\mathbf{u}(t) + f(t)\mathbf{u}'(t)$

  4. $\dfrac{d}{dt}\Big(\mathbf{u}(t) \cdot \mathbf{v}(t)\Big) = \mathbf{u}'(t)\cdot\mathbf{v}(t) + \mathbf{u}(t)\cdot\mathbf{v}'(t)$

  5. $\dfrac{d}{dt}\Big(\mathbf{u}(t) \times \mathbf{v}(t)\Big) = \mathbf{u}'(t)\times\mathbf{v}(t) + \mathbf{u}(t)\times\mathbf{v}'(t)$

  6. $\dfrac{d}{dt}\Big(\mathbf{u}(f(t))\Big) = f'(t)\mathbf{u}'(f(t))$

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.

Example 13.2.1

If a vector $\mathbf{r}$ is twice differentiable , that is $\mathbf{r}''$ exists, then

$$ \dfrac{d}{dt}\Big( \mathbf{r}(t)\times\mathbf{r}'(t)\Big) = \mathbf{r}(t)\times\mathbf{r}''(t) $$

Solution

$$ \begin{align*} \dfrac{d}{dt}\Big( \mathbf{r}(t)\times\mathbf{r}'(t)\Big) &= \mathbf{r}'(t)\times\mathbf{r}'(t) + \mathbf{r}(t)\times\mathbf{r}''(t) \\ \\ &= \mathbf{0} + \mathbf{r}(t)\times\mathbf{r}''(t) \\ \\ &= \mathbf{r}(t)\times\mathbf{r}''(t) \end{align*} $$

A vector's cross product with itself is $\mathbf{0}$.

Example 13.2.2

Find an expression for

$$ \dfrac{d}{dt}\Big[ \mathbf{u}(t)\cdot\Big(\mathbf{v}(t)\times\mathbf{w}(t)\Big)\Big] $$

Solution

$$ \begin{align*} \dfrac{d}{dt}\Big[ \mathbf{u}(t)\cdot\Big(\mathbf{v}(t)\times\mathbf{w}(t)\Big)\Big] &= \mathbf{u}'(t)\cdot\Big(\mathbf{v}(t)\times\mathbf{w}(t)\Big) + \mathbf{u}(t)\cdot\Big(\mathbf{v}(t)\times\mathbf{w}(t)\Big)' \\ \\ &= \mathbf{u}'(t)\cdot\Big(\mathbf{v}(t)\times\mathbf{w}(t)\Big) + \mathbf{u}(t)\cdot\Big(\mathbf{v}'(t)\times\mathbf{w}(t) + \mathbf{v}(t)\times\mathbf{w}'(t)\Big) \\ \\ &= \mathbf{u}'(t)\cdot\Big(\mathbf{v}(t)\times\mathbf{w}(t)\Big) + \mathbf{u}(t)\cdot\Big(\mathbf{v}'(t)\times\mathbf{w}(t)\Big) + \mathbf{u}(t)\cdot\Big(\mathbf{v}(t)\times\mathbf{w}'(t)\Big) \end{align*} $$


This is a variation of the triple product rule .

13.2.4 Integrating Vector Functions

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

$$ \int_a^b \mathbf{r}(t)\, dt = \mathbf{R}(t)\,\Big|_a^b = \mathbf{R}(b) - \mathbf{R}(a) $$

Exercise 13.2.3

Evaluate the integral

$$ \int_0^\frac{\pi}{4} \Big(\sec t\tan t\,\ihat + t\cos 2t\,\jhat + \sin^2 2t\cos 2t\,\khat\Big)\,dt $$

Check Your Work Since the integrals may be computed component-wise, it is best to consider the integrals separately.

The first is a simple antiderivative.

$$ \int_0^\frac{\pi}{4} \sec t\tan t\,dt = \sec t\,\Big|_0^\frac{\pi}{4} = \sqrt{2}-1 $$
Integration by parts is required for this integral.

$$ \begin{align*} \int_0^\frac{\pi}{4} t\cos 2t\,dt &= \frac{t}{2}\sin 2t\,\Big|_0^\frac{\pi}{4} - \int_0^\frac{\pi}{4} \frac{1}{2}\sin 2t\,dt \\ \\ &= \frac{\pi}{8}\cdot 1 - \dfrac{0}{2}\cdot 0 - \left(-\frac{1}{4}\cos 2t\,\Big|_0^\frac{\pi}{4}\right) \\ \\ &= \frac{\pi}{8} - \frac{1}{4} \end{align*} $$
Lastly, we have a $u$-substitution integral. Setting $u = \sin 2t$ will allow us to evaluate it. Under this substitution, $du = 2\cos 2t$ and the limits of integration become $\sin 2\left(\frac{\pi}{4}\right) = 1$ and $\sin 2(0) = 0$.

$$ \begin{align*} \int_0^\frac{\pi}{4} \sin^2 2t\cos 2t\,dt &= \int_0^1 \frac{u^2}{2}\,du \\ \\ &= \left.\frac{u^3}{6}\,\right|_0^1 \\ \\ &= \frac{1}{6} \end{align*} $$
Therefore,

$$ \int_0^\frac{\pi}{4} \Big(\sec t\tan t\,\ihat + t\cos 2t\,\jhat + \sin^2 2t\cos 2t\,\khat\Big)\,dt = \Big(\sqrt{2}-1\Big)\ihat + \left(\frac{\pi}{8}-\frac{1}{4}\right)\jhat + \frac{1}{6}\khat $$

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