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Given a vector parametrization that describes the position of a particle $\vec{r}(t)$, is it valid to say that $\vec{r} (t)$ is perpendicular to $\vec{r'} (t)$ through the following: $$ \frac{\mathrm{d}(\vec{r}(t) \cdot \vec{r}(t))}{\mathrm{d}t} = 0 $$

so through the product rule $$ \frac{\mathrm{d}\vec{r}(t)}{\mathrm{d}t} \cdot \vec{r}(t) + \vec{r}(t) \cdot \frac{\mathrm{d}\vec{r}(t)}{\mathrm{d}t} = 0 $$ which means $$ 2\frac{\mathrm{d}\vec{r}(t)}{\mathrm{d}t} \cdot \vec{r}(t) = 0 $$ so $\vec{r}(t) \cdot \vec{r'}(t) = 0$, meaning that they are perpendicular to each other? Whenever I draw a curve, if I draw a position vector to a minima in that curve and then the velocity vector, they are not perpendicular to each other.

I feel like I'm making a very simple mistake.

Asco 2
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2 Answers2

2

The first step is already wrong. $\vec r(t)\cdot\vec r(t)$ is the squared length of $\vec r(t)$, which can vary arbitrarily with time.

And so from this incorrect premise, absurd things may be proven, like you just did.

Parcly Taxel
  • 103,344
1

Your mistake is indeed very simple; it is only true that

$\dfrac{d(\vec r(t) \cdot \vec r(t))}{dt} = 0 \tag 1$

if

$\vec r(t) \cdot \vec r(t) = c, \; \text{a constant}; \tag 2$

but this is only true if the curve $\vec r(t)$ is a circle (when $c > 0$) or the single point $O$, the origin of coordinates. For any other curve, (1) will not bind.

Obviously, in the case

$\vec r(t) = O, \tag 3$

we have

$\vec r'(t) = 0, \tag 4$

so

$\vec r'(t) \cdot \vec r(t) = 0 \tag 5$

holds trivially. When $\vec r(t)$ is a circular path, however, this equation may in fact bind with

$\vec r(t) \ne 0 \ne \vec r'(t). \tag 6$

Finally, note that (2) forces

$c \ge 0, \tag 7$

since it is the square of the magnitude of the vector function $\vec r(t)$.

Robert Lewis
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