Under the topic of differentiation of vectors, it says:

an infinitesimal increment dR of a vector R, does not need to be colinear with vector R.

But why? A vector has magnitude and direction. If you make an increase to that vector in a different direction than the original vector, even if it is infinitesimal, the resultant direction changes. So why or how can they say this?

  • $\begingroup$ Why should the vector stay along the same line after an increment $d\bf R$? $\endgroup$ – Shubham Johri Jan 5 '19 at 9:25

The picture of $R$ and $dR$ is like this:

enter image description here

Here, $R$ is the original vector. $dR$ is the change in $R$ to get the new, slightly different vector.

Notice that $R$ and $dR$ are not collinear -- they could be pointing in completely different directions.

| cite | improve this answer | |

Note that there are two ways to alter a vector: by changing its magnitude or by changing its direction. Thus, the differential increment $d\bf R$ could be collinear with $\bf R$ and change its magnitude, not its direction, or it could be non-collinear with $\bf R$ and bring about a change in its direction and optionally, its magnitude.

| cite | improve this answer | |

Perhaps the following will help you understand. It is how I, definitely not a mathematician, first grasped this.

Let's say we have a point somewhere on the unit circle, rotating at angular velocity $\omega$, crossing the positive $x$ axis at time $t = 0$. (You can basically ignore $\omega$, or assume it is equal to $1$ radian per $1$ time unit; $\omega = 1$. I use it because it is easier to first grasp the derivatives when the parametric unit is time; it is just more intuitive this way.)

We can define its location using a pair of functions, one for each coordinate, $$\bbox{\begin{cases} x(t) = \cos \omega t \\ y(t) = \sin \omega t \\ \end{cases}} \tag{1}\label{NA1}$$ or as a vector-valued function, $$\bbox{\mathbf{p}(t) = \left [ \begin{matrix} \cos\omega t \\ \sin\omega t \end{matrix} \right ]} \tag{2}\label{NA2}$$ These are completely equivalent representations.

Do note that notations vary. I personally usually use $\vec{p}$ for vectors, and $\mathbf{p}$ for matrices; and a parameter list (like $(t)$ above) following a scalar, vector, or matrix makes it a scalar-valued, vector-valued, or matrix-valued function. I find it clearest to me. Others use different notations, so always check what kind of notation is used first, before delving into any equations.

If we take the derivatives of $\eqref{NA1}$, we obtain the tangent; i.e. the velocity, or the speed and direction, of where the point is headed at any specific moment in time: $$\bbox{ \begin{cases} \displaystyle \frac{d x(t)}{d t} = \dot{x}(t) = -\omega \sin \omega t \\ \displaystyle \frac{d y(t)}{d t} = \dot{y}(t) = \omega \cos \omega t \\ \end{cases} } \tag{3}\label{NA3}$$ In vector calculus notation, doing the same to $\eqref{NA2}$, yields the exact same result, of course: $$\bbox{ \frac{d \mathbf{p}(t)}{d t} = \left [ \begin{matrix} -\omega \sin \omega t \\ \omega \cos \omega t \end{matrix} \right ] } \tag{4}\label{NA4}$$ These obviously also correspond to the "infinitesimal increments" OP mentioned, via the definition of a scalar derivative. Contrast $\eqref{NA1}$ and $\eqref{NA3}$ to the definition of a derivative: $$\bbox{ \frac{d x(t)}{d t} = \lim_{\tau \to 0} \frac{x(t + \tau) - x(t)}{\tau} }$$

Sidetrack: To obtain the acceleration of the point, we do the second derivative, $$\bbox{ \begin{cases} \frac{d^2 x(t)}{d t^2} = \ddot{x}(t) = -\omega^2 \cos \omega t \\ \frac{d^2 y(t)}{d t^2} = \ddot{y}(t) = -\omega^2 \sin \omega t \\ \end{cases} } \quad \iff \quad \bbox{ \frac{d^2 \mathbf{p}(t)}{d t^2} = \left [ \begin{matrix} -\omega^2 \cos \omega t \\ -\omega^2 \sin \omega t \end{matrix} \right ]}$$ As an example, if $\mathbf{p}(t)$ describes the location of a spacecraft at time $t$, then its derivative describes its velocity vector, and its second derivative the acceleration.

The next step is to realize that this applies no matter how the location of the point is parametrized. The parameter does not have to be time; it can be anything. You can even use more than one parameter (although then we talk about partial derivatives with respect to one specific parameter, or about gradient, instead of derivatives).

That means that there is no need for the derivative, or the infinitesimal change in the vector, to be collinear -- or indeed depend in any way! -- with the vector value itself: they describe completely different things. In the case of a point moving as a function of time, they describe the position (the vector itself) and its velocity (the derivative of the vector), respectively.

If we think of a game where the vector represents the ethical/moral alignment of your character, its derivative represents the effects of their actions and choices on their alignment or ethical/moral stance.

They are obviously related; but represent different, related things.

If we look at $\eqref{NA2}$ and $\eqref{NA4}$, we find that in the circular case $$\bbox{\mathbf{p}(t) \perp \frac{d \mathbf{p}(t)}{d t}}$$ because $$\bbox{\mathbf{p}(t) \cdot \frac{d \mathbf{p}(t)}{d t} = -\omega (\cos \omega t)(\sin \omega t) + \omega (\sin \omega t)(\cos \omega t) = 0}$$ i.e., the derivative is always perpendicular to the position vector.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.