Calculus and vectors #rvc

Time-dependent vectors can be differentiated in exactly the same way that we differentiate scalar functions. For a time-dependent vector $\vec{a}(t)$ , the derivative $\dot{\vec{a}}(t)$ is:

Vector derivative definition.#rvc‑ed

$\begin{aligned} \dot{\vec{a}}(t) &= \frac{d}{dt} \vec{a}(t) = \lim_{\Delta t \to 0} \frac{\vec{a}(t + \Delta t) - \vec{a}(t)}{\Delta t}\end{aligned}$

Note that vector derivatives are a purely geometric concept. They don't rely on any basis or coordinates, but are just defined in terms of the physical actions of adding and scaling vectors.

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Increment:	$\Delta t =$ 2 s
Time:	$t =$ 0 s

Vector derivatives shown as functions of $t$ and $\Delta t$ . We can hold $t$ fixed and vary $\Delta t$ to see how the approximate derivative $\Delta\vec{a}/\Delta t$ approaches $\dot{\vec{a}}$ . Alternatively, we can hold $\Delta t$ fixed and vary $t$ to see how the approximation changes depending on how $\vec{a}$ is changing. #rvc‑fd

We will use either the dot notation $\dot{\vec{a}}(t)$ or the full derivative notation $\frac{d\vec{a}(t)}{dt}$ , depending on which is clearer and more convenient. We will often not write the time dependency explicitly, so we might write just $\dot{\vec{a}}$ or $\frac{d\vec{a}}{dt}$ .

Derivatives and vector “positions” #rvc‑sp

When thinking about vector derivatives, it is important to remember that vectors don't have positions. Even if a vector is drawn moving about, this is irrelevant for the derivative. Only changes to length and direction are important.

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Movement: bounce stretch circle twist slider rotate vertical fly

Vector derivatives for moving vectors. Vector movement is irrelevant when computing vector derivatives. #rvc‑fp

Derivatives in components #rvc‑sc

In a fixed basis we differentiate a vector by differentiating each component:

Vector derivative in components#rvc‑ec

$\dot{\vec{a}}(t) = \dot{a}_1(t) \,\hat{\imath} + \dot{a}_2(t) \,\hat{\jmath} + \dot{a}_3(t) \,\hat{k}$

Derivation

Writing a time-dependent vector expression in a fixed basis gives: $\vec{a}(t) = a_1(t)\,\hat{\imath} + a_2(t) \,\hat{\jmath}.$ Using the definition #rvc-ed of the vector derivative gives: $\begin{aligned} \dot{\vec{a}}(t) &= \lim_{\Delta t \to 0} \frac{\vec{a}(t + \Delta t) - \vec{a}(t)}{\Delta t} \\ &= \lim_{\Delta t \to 0} \frac{(a_1(t + \Delta t) \,\hat{\imath} + a_2(t + \Delta t) \,\hat{\jmath}) - (a_1(t) \,\hat{\imath} + a_2(t) \,\hat{\jmath})}{\Delta t} \\ &= \lim_{\Delta t \to 0} \frac{(a_1(t + \Delta t) - a_1(t)) \,\hat{\imath} + (a_2(t + \Delta t) - a_2(t)) \,\hat{\jmath}}{\Delta t} \\ &= \left(\lim_{\Delta t \to 0} \frac{a_1(t + \Delta t) - a_1(t)}{\Delta t} \right) \,\hat{\imath} + \left(\lim_{\Delta t \to 0} \frac{a_2(t + \Delta t) - a_2(t) }{\Delta t}\right) \,\hat{\jmath} \\ &= \dot{a}_1(t) \,\hat{\imath} + \dot{a}_2(t) \,\hat{\jmath} \end{aligned}$ The second-to-last line above is simply the definition of the scalar derivative, giving the scalar derivatives of the component functions $a_1(t)$ and $a_2(t)$ .

Warning: Differentiating each component is only valid if the basis is fixed. #rvc‑wc

Time:	$t =$ 0 s
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Basis:	$\hat\imath,\hat\jmath$ $\hat{u},\hat{v}$

The vector derivative decomposed into components. This demonstrates graphically that each component of a vector in a particular basis is simply a scalar function, and the corresponding derivative component is the regular scalar derivative. #rvc‑fc

Differentiating vector expressions #rvc‑se

We can also differentiate complex vector expressions, using the sum and product rules. For vectors, the product rule applies to both the dot and cross products:

Product rule for dot-product derivatives.#rvc‑ep

$\frac{d}{dt}(\vec{a} \cdot \vec{b}) = \dot{\vec{a}} \cdot \vec{b} + \vec{a} \cdot \dot{\vec{b}}$

Product rule for cross-product derivatives.#rvc‑ex

$\frac{d}{dt}(\vec{a} \times \vec{b}) = \dot{\vec{a}} \times \vec{b} + \vec{a} \times \dot{\vec{b}}$

Example Problem: Differentiating vector product expressions. #rvc‑xe

What is $\displaystyle \frac{d}{dt} \left( \frac{\vec{b}}{\ell} \cdot (\vec{a} + \vec{a} \times \vec{c}) \right)$ ?

Solution

$\begin{aligned} \frac{d}{dt} \left( \frac{\vec{b}}{\ell} \cdot (\vec{a} + \vec{a} \times \vec{c}) \right) &= \left(\frac{d}{dt} \frac{\vec{b}}{\ell}\right) \cdot (\vec{a} + \vec{a} \times \vec{c}) + \frac{\vec{b}}{\ell} \cdot \frac{d}{dt} (\vec{a} + \vec{a} \times \vec{c}) \\ &= \left(\frac{\ell \dot{\vec{b}} - \dot{\ell}\vec{b}}{\ell^2}\right) \cdot (\vec{a} + \vec{a} \times \vec{c}) + \frac{\vec{b}}{\ell} \cdot (\dot{\vec{a}} + \dot{\vec{a}} \times \vec{c} + \vec{a} \times \dot{\vec{c}}) \end{aligned}$

The chain rule also applies to vector functions. This is helpful for parameterizing vectors in terms of arc-length $s$ or other quantities different than time $t$ .

Chain rule for vectors.#rvc‑er

$\frac{d}{dt} \vec{a} (s(t)) = \frac{d\vec{a}}{ds} (s(t)) \frac{ds}{dt}(t) = \frac{d\vec{a}}{ds} \dot{s}$

Example Problem: Chain rule. #rvc‑er

A vector is defined in terms of an angle $\theta$ by $\vec{r}(\theta) = \cos\theta\,\hat\imath + \sin\theta\,\hat\jmath$ . If the angle is given by $\theta(t) = t^3$ , what is $\dot{\vec{r}}$ ?

Solution

We can use the chain rule to compute: $\begin{aligned} \frac{d}{dt} \vec{r} &= \frac{d}{d\theta} \vec{r} \frac{d}{dt} \theta \\ &= \frac{d}{d\theta}\Big( \cos\theta\,\hat\imath + \sin\theta\,\hat\jmath \Big) \frac{d}{dt} (t^3) \\ &= \Big( -\sin\theta\,\hat\imath + \cos\theta\,\hat\jmath \Big) (3 t^2) \\ &= -3 t^2 \sin(t^3)\,\hat\imath + 3 t^2 \cos(t^3)\,\hat\jmath. \end{aligned}$

Alternatively, we can evaluate $\vec{r}$ as a function of $t$ first and then differentiate it with respect to time, using the scalar chain rule for each component: $\begin{aligned} \frac{d}{dt} \vec{r} &= \frac{d}{dt} \Big( \cos(t^3)\,\hat\imath + \sin(t^3)\,\hat\jmath \Big) \\ &= -3 t^2 \sin(t^3)\,\hat\imath + 3 t^2 \cos(t^3)\,\hat\jmath. \end{aligned}$

Did you know?#rvc‑il

Gottfried Leibniz, one of the inventors of calculus, got the product rule wrong [Child, 1920, page 100; Cirillo, 2007]. In modern notation he computed the example $\frac{d}{dx}(x^2 + bx)(cx + d) = (2x + b)c$ and he stated that in general it was obvious that $\frac{d}{dx} (f g) = \frac{df}{dx} \frac{dg}{dx}.$ He later realized his error and corrected it [Cupillari, 2004], but at least we know that product rules are tricky and not obvious, even for someone smart enough to invent calculus.

References

J. M. Child. The Early Mathematical Manuscripts of Leibniz. Open Court Publishing, 1920. (Google ebook, local copy).
M. Cirillo. Humanizing Calculus. The Mathematics Teacher, 101(1):23–27, 2007. (NCTM version, local copy)
A. Cupillari. Another look at the rules of differentiation. Primus: Problems, Resources, and Issues in Mathematics Undergraduate Studies, 14(3):193–200, 2004. DOI: 10.1080/10511970408984087.

Changing lengths and directions #rvc‑sd

Two useful derivatives are the rates of change of a vector's length and direction:

Derivative of vector length.#rvc‑el

$\dot{a} = \dot{\vec{a}} \cdot \hat{a}$

Derivation

We start with the dot product expression #rvv-ed for length and differentiate it:

$\begin{aligned} a &= \sqrt{\vec{a} \cdot \vec{a}} \\ \frac{d}{dt} a &= \frac{d}{dt} \big( (\vec{a} \cdot \vec{a})^{1/2} \big) \\ \dot{a} &= \frac{1}{2} (\vec{a} \cdot \vec{a})^{-1/2} (\dot{\vec{a}} \cdot \vec{a} + \vec{a} \cdot \dot{\vec{a}}) \\ &= \frac{1}{2\sqrt{a^2}} (2 \dot{\vec{a}} \cdot \vec{a}) \\ &= \dot{\vec{a}} \cdot \hat{a}.\end{aligned}$

Derivative of vector direction.#rvc‑eu

$\dot{\hat{a}} = \frac{1}{a} \operatorname{Comp}(\dot{\vec{a}}, \vec{a})$

Derivation

We take the definition #rvv-eu for the unit vector and differentiate it:

$\begin{aligned} \hat{a} &= \frac{\vec{a}}{a} \\ \frac{d}{dt} \hat{a} &= \frac{d}{dt}\left(\frac{\vec{a}}{a}\right) \\ \dot{\hat{a}} &= \frac{\dot{\vec{a}} a - \vec{a} \dot{a}}{a^2} \\ &= \frac{\dot{\vec{a}}}{a} - \frac{\dot{\vec{a}} \cdot \hat{a}}{a^2} \vec{a} \\ &= \frac{1}{a} \big( \dot{\vec{a}} - (\dot{\vec{a}} \cdot \hat{a}) \hat{a} \big)\\ &= \frac{1}{a} \operatorname{Comp}(\dot{\vec{a}}, \vec{a}).\end{aligned}$ Here we observed at the end that we had the expression #rvv-em for the complementary projection of the derivative $\dot{\vec{a}}$ with respect to $\vec{a}$ itself.

An immediate consequence of the derivative of direction formula is that the derivative of a unit vector is always orthogonal to the unit vector:

Derivative of unit vector is orthogonal.#rvc‑eu

$\dot{\hat{a}} \cdot \hat{a} = 0$

Derivation

From #rvc-eu we know that $\dot{\hat{a}}$ is in the direction of $\operatorname{Comp}(\dot{\vec{a}}, \vec{a})$ , and from #rvv-er we know that this is orthogonal to $\vec{a}$ (and also $\hat{a}$ ).

Recall that we can always write a vector as the product of its length and direction, so $\vec{a} = a \hat{a}$ . This gives the following decomposition of the derivative of $\vec{a}$ .

Vector derivative decomposition.#rvc‑em

$\begin{aligned} \dot{\vec{a}} &= \underbrace{\dot{a} \hat{a}}_{\operatorname{Proj}(\dot{\vec{a}}, \vec{a})} + \underbrace{a \dot{\hat{a}}}_{\operatorname{Comp}(\dot{\vec{a}}, \vec{a})}\end{aligned}$

Derivation

Differentiating $\vec{a} = a \hat{a}$ and substituting in #rvv-el and #rvv-eu gives $\begin{aligned} \dot{\vec{a}} &= \dot{a} \hat{a} + a \dot{\hat{a}} \\ &= ( \dot{\vec{a}} \cdot \hat{a} ) \hat{a} + a \frac{1}{a} \operatorname{Comp}(\dot{\vec{a}}, \hat{a}) \\ &= \operatorname{Proj}(\dot{\vec{a}}, \vec{a}) + \operatorname{Comp}(\dot{\vec{a}}, \vec{a}). \end{aligned}$

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Vector derivatives can be decomposed into length changes (projection onto $\vec{a}$ ) and direction changes (complementary projection). Compare to Figure #rvv-fu. #rvc‑fm

Integrating vector functions #rvc‑si

The Riemann-sum definition of the vector integral is:

Vector integral.#rvc‑ei

$\int_0^t \vec{a}(\tau) \, d\tau = \lim_{N \to \infty} \underbrace{\sum_{i=1}^N \vec{a}(\tau_i) \Delta\tau}_{\vec{S}_N} \qquad \tau_i = \frac{i - 1}{N} \qquad \Delta \tau = \frac{1}{N}$

In the above definition $\vec{S}_N$ is the sum with $N$ intervals, written here using the left-hand edge $\tau_i$ in each interval.

Time:	$t =$ 0 s
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Segments:	$N =$ 1

Integral of a vector function $\vec{a}(t)$ , together with the approximation using a Riemann sum. #rvc‑fi

Just like vector derivatives, vector integrals only use the geometric concepts of scaling and addition, and do not rely on using a basis. If we do write a vector function in terms of a fixed basis, then we can integrate each component:

Vector integral in components.#rvc‑et

$\int_0^t \vec{a}(\tau) \, d\tau = \left( \int_0^t a_1(\tau) \, d\tau \right) \,\hat\imath + \left( \int_0^t a_2(\tau) \, d\tau \right) \,\hat\jmath + \left( \int_0^t a_3(\tau) \, d\tau \right) \,\hat{k}$

Derivation

Consider a time-dependent vector $\vec{a}(t)$ written in components with a fixed basis: $\vec{a}(t) = a_1(t) \,\hat\imath + a_2(t) \,\hat\jmath.$ Using the definition #rvc-ei of the vector integral gives: $\begin{aligned} \int_0^t \vec{a}(\tau) \, d\tau &= \lim_{N \to \infty} \sum_{i=1}^N \vec{a}(\tau_i) \Delta\tau \\ &= \lim_{N \to \infty} \sum_{i=1}^N \left( a_1(\tau_i) \,\hat\imath + a_2(\tau_j) \,\hat\jmath \right) \Delta\tau \\ &= \lim_{N \to \infty} \left( \sum_{i=1}^N a_1(\tau_i) \Delta\tau \,\hat\imath + \sum_{i=1}^N a_2(\tau_j) \Delta\tau \,\hat\jmath \right) \\ &= \left( \lim_{N \to \infty} \sum_{i=1}^N a_1(\tau_i) \Delta\tau \right) \,\hat\imath + \left( \lim_{N \to \infty} \sum_{i=1}^N a_2(\tau_j) \Delta\tau \right) \,\hat\jmath \\ &= \left( \int_0^t a_1(\tau) \, d\tau \right) \,\hat\imath + \left( \int_0^t a_2(\tau) \, d\tau \right) \,\hat\jmath. \end{aligned}$ The second-to-last line used the Riemann-sum definition of regular scalar integrals of $a_1(t)$ and $a_2(t)$ .

Warning: Integrating each component is only valid if the basis is fixed. #rvc‑wi

Example Problem: Integrating a vector function. #rvc‑xi

The vector $\vec{a}(t)$ is given by $\vec{a}(t) = \Big(2 \sin(t + 1) + t^2 \Big) \,\hat\imath + \Big(3 - 3 \cos(2t)\Big) \,\hat\jmath.$ What is $\int_0^t \vec{a}(\tau) \, d\tau$ ?

Solution

$\begin{aligned} \int_0^t \vec{a}(\tau) \,d\tau &= \left(\int_0^t \Big(2 \sin(\tau + 1) + \tau^2 \Big) \,d\tau\right) \,\hat\imath + \left(\int_0^t \Big(3 - 3 \cos(2\tau)\Big) \,d\tau\right) \,\hat\jmath \\ &= \left[-2 \cos(\tau + 1) + \frac{\tau^3}{3} \right]_{\tau=0}^{\tau=t} \,\hat\imath + \left[3 \tau - \frac{3}{2} \sin(2\tau) \right]_{\tau=0}^{\tau=t} \,\hat\jmath \\ &= \left( -2\cos(t + 1) + 2 \cos(1) + \frac{t^3}{3}\right)\,\hat\imath + \left(3t - \frac{3}{2} \sin(2t)\right) \,\hat\jmath. \end{aligned}$

Warning: The dummy variable of integration must be different to the limit variable. #rvc‑wd

In the coordinate integral expression #rvc-ei, it is important that the component expressions $a_1(t)$ , $a_2(t)$ are re-written with a different dummy variable such as $\tau$ when integrating. If we used $\tau$ for integration but kept $a_1(t)$ then we would obtain $\int_0^t a_1(t) \,d\tau = \left[a_1(t) \, \tau\right]_{\tau = 0}^{\tau = t} = a_1(t) \, t,$ which is not what we mean by the integral. Alternatively, if we leave everything as $t$ then we would obtain $\int_0^t a_1(t) \,dt$ which is a meaningless expression, as dummy variables must only appear inside an integral.

TAM 212: Introductory Dynamics

Calculus and vectors #rvc

Derivatives and vector “positions” #rvc‑sp

Derivatives in components #rvc‑sc

Differentiating vector expressions #rvc‑se

Did you know?#rvc‑il

References

Changing lengths and directions #rvc‑sd

Integrating vector functions #rvc‑si