Show that particle whose position satisfies $\frac{d \mathbf{r}}{dt}= \mathbf{c} \times \mathbf{r}$ moves in a circular path with constant speed

Question

A particle P moves so that its position vector r at time $t$ satisfies the differential equation

$$\frac{d \mathbf{r}}{dt}= \mathbf{c} \times \mathbf{r},$$

where c is a constant vector. Show that P moves with constant speed on a circular patch.

taking the dot product on the left with c, then using cyclic permutation we obtain;

$$ \mathbf{c} \cdot \frac{d \mathbf{r}}{dt} = \mathbf{c} \cdot \mathbf{c} \times \mathbf{r} = 0$$

similarly with r

$$ \mathbf{r} \cdot \frac{d \mathbf{r}}{dt} = \mathbf{r} \cdot \mathbf{c} \times \mathbf{r} = 0 $$

At this point I did not know where to turn, so I consulted the solution manual.

in the solution manual it says that the above equation $ \mathbf{r} \cdot \frac{d \mathbf{r} }{dt} $ can be integrated to give $ \mathbf{r} \cdot \mathbf{r} = R^2 $.

I cant quite see how.

Answer 1

With summation implied, $$ r \cdot \frac{dr}{dt} = 0 \\ \Rightarrow r_i \frac{dr_i}{dt} = 0 \\ \Rightarrow 2 r_i \frac{dr_i}{dt} = 0 \\ \Rightarrow \frac{d}{dt}(r_i r_i) = 0 \\ \Rightarrow r.r = k^2 $$

The constant must be positive because of the property of the dot product that $r.r \geq 0$, and equality is attained only when $r=0$.

ETA: You don't really need coordinates for this, it is pretty much obvious when you see $r \cdot \frac{dr}{dt}=0$ and recall $r=r(t)$ only.