Calculus and vectors
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.
\[\begin{aligned} \dot{\vec{a}}(t) &= \frac{d}{dt} \vec{a}(t) = \lim_{\Delta t \to 0} \frac{\vec{a}(t + \Delta t) - \vec{a}(t)}{\Delta t}\end{aligned}\]
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 = $ s |
| Time: | $t = $ 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.
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”
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.
Derivatives in components
In a fixed basis we differentiate a vector by differentiating each component:
Vector derivative in components
\[\dot{\vec{a}}(t) = \dot{a}_1(t) \,\hat{\imath} + \dot{a}_2(t) \,\hat{\jmath} + \dot{a}_3(t) \,\hat{k}\]
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.
When we differentiate a vector by differentiating each component and leaving the basis vectors unchanged, we are assuming that the basis vectors themselves are not changing with time. If they are, then we need to take this into account as well.
| Time: | $t = $ 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.
Differentiating vector expressions
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.
\[\frac{d}{dt}(\vec{a} \cdot \vec{b}) = \dot{\vec{a}} \cdot \vec{b} + \vec{a} \cdot \dot{\vec{b}}\]
Product rule for cross-product derivatives.
\[\frac{d}{dt}(\vec{a} \times \vec{b}) = \dot{\vec{a}} \times \vec{b} + \vec{a} \times \dot{\vec{b}}\]
Example Problem: Differentiating vector product expressions.
What is $\displaystyle \frac{d}{dt} \left( \frac{\vec{b}}{\ell} \cdot (\vec{a} + \vec{a} \times \vec{c}) \right)$?
\[\begin{aligned} \frac{d}{dt} \left( \frac{\vec{b}}{\ell} \cdot (\vec{a} + \vec{a} \times \vec{c}) \right) &= \left(\frac{d}{dt} \frac{\vec{b}}{\ell}\right) \cdot (\vec{a} + \vec{a} \times \vec{c}) + \frac{\vec{b}}{\ell} \cdot \frac{d}{dt} (\vec{a} + \vec{a} \times \vec{c}) \\ &= \left(\frac{\ell \dot{\vec{b}} - \dot{\ell}\vec{b}}{\ell^2}\right) \cdot (\vec{a} + \vec{a} \times \vec{c}) + \frac{\vec{b}}{\ell} \cdot (\dot{\vec{a}} + \dot{\vec{a}} \times \vec{c} + \vec{a} \times \dot{\vec{c}}) \end{aligned}\]
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.
\[\frac{d}{dt} \vec{a} (s(t)) = \frac{d\vec{a}}{ds} (s(t)) \frac{ds}{dt}(t) = \frac{d\vec{a}}{ds} \dot{s}\]
Example Problem: Chain rule.
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 $\dot{\vec{r}}$?
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: \[\begin{aligned} \frac{d}{dt} \vec{r} &= \frac{d}{dt} \Big( \cos(t^3)\,\hat\imath + \sin(t^3)\,\hat\jmath \Big) \\ &= -3 t^2 \sin(t^3)\,\hat\imath + 3 t^2 \cos(t^3)\,\hat\jmath. \end{aligned}\]
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
Two useful derivatives are the rates of change of a vector's length and direction:
Derivative of vector length.
\[\dot{a} = \dot{\vec{a}} \cdot \hat{a}\]
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.
\[\dot{\hat{a}} = \frac{1}{a} \operatorname{Comp}(\dot{\vec{a}}, \vec{a})\]
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.
\[\dot{\hat{a}} \cdot \hat{a} = 0\]
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.
\[\begin{aligned} \dot{\vec{a}} &= \underbrace{\dot{a} \hat{a}}_{\operatorname{Proj}(\dot{\vec{a}}, \vec{a})} + \underbrace{a \dot{\hat{a}}}_{\operatorname{Comp}(\dot{\vec{a}}, \vec{a})}\end{aligned}\]
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.
Integrating vector functions
The Riemann-sum definition of the vector integral is:
Vector integral.
\[ \int_0^t \vec{a}(\tau) \, d\tau = \lim_{N \to \infty} \underbrace{\sum_{i=1}^N \vec{a}(\tau_i) \Delta\tau}_{\vec{S}_N} \qquad \tau_i = \frac{i - 1}{N} \qquad \Delta \tau = \frac{1}{N} \]
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 = $ s |
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| Segments: | $N = $ |
Integral of a vector function $\vec{a}(t)$, together with the approximation using a Riemann sum.
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.
\[ \int_0^t \vec{a}(\tau) \, d\tau = \left( \int_0^t a_1(\tau) \, d\tau \right) \,\hat\imath + \left( \int_0^t a_2(\tau) \, d\tau \right) \,\hat\jmath + \left( \int_0^t a_3(\tau) \, d\tau \right) \,\hat{k} \]
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.
Integrating a vector function by integrating each component separately is only valid if the basis vectors are not changing with time. If the basis vectors are changing then we must either transform to a fixed basis or otherwise take this change into account.
Example Problem: Integrating a vector function.
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 $\int_0^t \vec{a}(\tau) \, d\tau$?
\[\begin{aligned} \int_0^t \vec{a}(\tau) \,d\tau &= \left(\int_0^t \Big(2 \sin(\tau + 1) + \tau^2 \Big) \,d\tau\right) \,\hat\imath + \left(\int_0^t \Big(3 - 3 \cos(2\tau)\Big) \,d\tau\right) \,\hat\jmath \\ &= \left[-2 \cos(\tau + 1) + \frac{\tau^3}{3} \right]_{\tau=0}^{\tau=t} \,\hat\imath + \left[3 \tau - \frac{3}{2} \sin(2\tau) \right]_{\tau=0}^{\tau=t} \,\hat\jmath \\ &= \left( -2\cos(t + 1) + 2 \cos(1) + \frac{t^3}{3}\right)\,\hat\imath + \left(3t - \frac{3}{2} \sin(2t)\right) \,\hat\jmath. \end{aligned}\]
Warning: The dummy variable of integration must be different to the limit variable.
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.