# On the derivative of a vector function

In "An Introduction to Mechanics" by Kleppner and Kolenkow, in the section on the time derivative of a vector:

Given $A(t)$ is a vector valued function, then, $$\Delta A = A(t + \Delta t) - A(t)$$ for some increment of time $\Delta t$.

Since $\Delta A$ is a vector, we can write it as the sum of its components, one of which is parallel to $A$, and one which is perpendicular to $A$:

$$\Delta A = \Delta A_\perp + \Delta A_\parallel\tag{1}$$

where $\theta$ is the angle between $A(t)$ and $A(t + \Delta t)$.

Then they write \begin{align} \lvert \Delta A_\perp\rvert &= \lvert A(t)\rvert\Delta\theta\\ \lvert \Delta A_\parallel\rvert &= \lvert\Delta A\rvert \end{align} for sufficiently small $\theta$ and

What bothers me is that if $\lvert\Delta A_\parallel\rvert = \lvert\Delta A\rvert$ then by (1), $\lvert\Delta A_\perp\rvert = \lvert A(t)\rvert\Delta\theta$ = 0.

How can we assume the magnitude of the component of $\Delta A$ in the direction perpendicular to $A$ be 0, yet still allow for the possibility that the change in the vector $A$ is not a pure change in magnitude, but could also involve a rotation.

• Equation (1) is the addition of vectors so think of a vector triangle with the component vectors adding together to produce the resultant vector. Jun 23, 2018 at 10:00
• Right but $\Delta A$ is the hypotenuse of a right triangle isn't it? So how can it's length equal the length of one of it's legs, unless the other is 0? Jun 23, 2018 at 10:05
• Because you're setting $\Delta \theta \rightarrow 0$ Jun 23, 2018 at 10:16
• If you draw the triangle you also see that the relationship between the magnitude of delta A parallel and the magnitude of A also is true for sufficiently small delta theta ie the triangle is not quite right angled! Jun 23, 2018 at 10:19
• In my judgment, this development only works if $\mathbf{A}$ is the velocity vector. Jun 24, 2018 at 2:49

In general, the components of $\Delta A$ perpendicular and parallel to $A$ are $\Delta A_\perp = |\Delta A|\sin\Delta\theta$ and $\Delta A_\parallel = |\Delta A|\cos\Delta\theta$. Ignoring higher order terms, $\cos\Delta\theta \approx 1-\Delta\theta^2/2$ and $\sin\Delta\theta \approx \Delta\theta(1-\Delta\theta^2/6)$. The quadratic terms become very much smaller than one for small $\Delta\theta$, making $\cos\Delta\theta \approx 1$ and $\sin\Delta\theta \approx \Delta\theta$ for very small values of $\Delta\theta$.