2
$\begingroup$

From "An Introduction to Mechanics" by Kleppner & Kolenkow, SIE-2007, Chapter 1 (Vectors and Kinematics), Section 1.8 - "More about the derivative of a vector".


In this section, towards the end, they have given the derivation for the expression of a vector derivative for two cases -

Case 1: When the magnitude of the vector remains constant, but the direction varies

Case 2: When both the magnitude as well as direction vary

Let the vector be A(t) at time 't' and A(t+$\Delta$t) after time $\Delta$t. Define a vector -

$\Delta$A = A(t+$\Delta$t) - A(t)

Now see the image provided below for case 2:-

vector derivative general case

The vector $\Delta$A has been divided into two components, one parallel to A(t) and the other perpendicular to it, such that-

$\Delta$A = $\Delta$A$_\perp$ + $\Delta$A$_\parallel$

Let |A(t)| = A and |A(t+$\Delta$t)| = A+$\Delta$A. At this point, they write that for sufficiently small $\Delta\theta$, we have -

|$\Delta$A$_\parallel$| = $\Delta$A, which is fine. But the other expression is -

|$\Delta$A$_\perp$| = A$\Delta\theta$. But this expression can be more exactly written as |$\Delta$A$_\perp$| = (A+$\Delta$A)$\Delta\theta$ (I have provided the derivation at the very end). It seems they have assumed $\Delta$A to be small as well. But I suppose that would not be true in general. Take the following case for example -

counter case

The $\Delta$A would be quite large, even though $\Delta\theta$ is small. Then why have they neglected the term '$\Delta$A$\Delta\theta$' from that expression? If it were not neglected, the final formula for the vector derivative of the perpendicular component would come out to be -

|$\frac{\it{d}\bf{A}_\perp}{\it{dt}}$| = A$\frac{\it{d}\theta}{\it{dt}}$ + $\frac{\it{dA}}{\it{dt}}\frac{\it{d}\theta}{\it{dt}}$. Is this expression wrong in that case?


Note: (i) I have denoted vector quantities in bold letters and magnitudes in italic letters, to match the notation used in the book.

(ii)Derivation -

Use cosine rule for $\Delta\theta$ in two triangles, one formed by the sides A(t), $\Delta$A & A(t+$\Delta$t), and the other formed by A(t)+$\Delta$A$_\parallel$, $\Delta$A$_\perp$ & A(t+$\Delta$t), to get the following two expressions -

$\cos{\Delta\theta}$ = $\frac{(\it{A}+\Delta\it{A})^2 + \it{A}^2 - ((\Delta\it{A}_\parallel)^2 + (\Delta\it{A}_\perp)^2)}{2\it{A}(\it{A}+\Delta\it{A})}$; and

$\cos{\Delta\theta}$ = $\frac{(\it{A}+\Delta\it{A})^2 + (\it{A}+\Delta\it{A}_\parallel)^2 - (\Delta\it{A}_\perp)^2}{2(\it{A}+\Delta\it{A})(\it{A}+\Delta\it{A}_\parallel)}$.

Now use the above two equations to eliminate $\Delta$A$_\parallel$ to get

$\Delta$A$_\perp$ = (A+$\Delta$A)$\sin{\Delta\theta}$, which reduces to the desired expression for small $\Delta\theta$.

$\endgroup$

1 Answer 1

0
$\begingroup$

I'd say the premise is the other way around: We assume $\Delta A$ to be small, as we intend to make it infinitesimal later on, and so $\Delta \theta$ is also small.

Also, when you look at $\Delta A\Delta \theta$, this is even smaller, or "quadratically small" so to say. When doing math with small deltas or infinitesimals, you usually only have to take the first order into account, quadratically and higher orders are so small you can ignore them.

$\endgroup$
1
  • $\begingroup$ Can't we take the values of $\Delta$A and $\Delta\theta$ to be small, independent of each other? $\endgroup$
    – seavoyage
    Commented May 13, 2016 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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