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[Beginning calculus question.] I saw in a calculus lecture online that for a position vector $\boldsymbol{r}$

$$\left|\frac{d\boldsymbol r}{dt}\right| \neq \frac{d\left| \boldsymbol r \right|}{dt}$$

but I don't understand exactly how to parse this.

It's my understanding that:

  1. $\frac{d\boldsymbol r}{dt}$ refers to the rate of change in the position over time (speed?)
  2. $|\boldsymbol r|$ refers to the magnitude of the position, i.e. the distance (from what to what?)
  3. $\frac{d\left| \boldsymbol r \right|}{dt}$ refers to the rate of change in distance traveled over time, (a different kind of speed?)

Is there a good way to understand what both of these expressions mean?

Hatshepsut
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    Regarding the $\not=$ part: One can show using the Cauchy-Schwartz inequality that $\frac{d|{\bf r}|}{dt} \leq \left|\frac{d{\bf r}}{dt}\right|$ with equality only when the motion is along a straight line (i.e. no change in direction). – Winther Dec 21 '15 at 07:27
  • Find a definition of $\mathbb r$ first – your question `from what to what?' will be answered immediately then. Additionally find the difference between a velocity (a vector describing the movement at the moment) $\mathbb v = \frac{d \mathbb r}{dt}$, which has a value (length) and a direction, and a speed (which is actually the length of the velocity vector) $v = |\mathbb v|$, which is a scalar. – CiaPan Dec 21 '15 at 07:31

2 Answers2

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The first one is looking at the velocity $\mathbf{v}=\frac{d\mathbf{r}}{dt}$, and taking its norm: it's the value of the speed, i.e. the value of (instantaneous) change in position.

The second is looking at the distance from the point of coordinates $\mathbf{0}$ (the origin), $\lvert \mathbf{r}\lvert$, and taking its derivative: it's the instantaneous change in distance from the origin.

The two indeed need not be equal: imagine you are moving very fast, but staying at the same distance from the origin (that is, you're moving very fast on a circle). Then the speed $\left\lvert \frac{d\mathbf{r}}{dt}\right\rvert$ is big (you're moving fast), but $\lvert \mathbf{r}\lvert$ is constant -- so $\frac{d\lvert\mathbf{r}\rvert}{dt} = 0$.

Clement C.
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  • It's important to note that they are not necessarily equal in the 2D or higher dimensionality cases, but they are always equal in the 1D case. – dberm22 Jan 04 '16 at 13:34
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We can understand this with one example. Consider angular motion problem where position of a particle is given by $\mathbf{r}=\hat{\imath}\cos(\omega t)+\hat{\jmath}\sin(\omega t)$. Now $\frac{d\mathbf{r}}{dt}=-\hat{\imath}\omega\sin(\omega t)+\hat{\jmath}\omega \cos(\omega t)$, clearly $\lvert \frac{d\mathbf{r}}{dt}\rvert = \omega$ , whereas $\lvert \mathbf{r} \rvert =1$, hence $\frac{d\lvert \mathbf{r}\rvert}{dt}=0$. It means that a particle is at constant position, within unit radius, only angular position is changing, hence derivative of $\lvert \mathbf{r}\rvert$ is zero.

Deusovi
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Shahid M Shah
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  • For comparison, the intro part of this article: https://gregordu.wordpress.com/2015/08/23/constant-distance-and-velocity-yield-rotation/ – GDumphart Dec 21 '15 at 10:02