Find the curve, given that $r'(t) = Cr(t)$

Question

I need to find the curve, given that $r'(t) = Cr(t)$ (where $C$ is a constant), for all real $t$ and $r(0)=i+2j+3k$.

To start, I know that the equation I will need is the $$K=\frac{||r'(t) \times r''(t)||}{||r'(t)||^3}.$$

However, I'm not sure how to get what I currently have $(r'(t))$ into $r(t)$ so that I can plug the values into the equation. Any help would be greatly appreciated!

Answer 1

$r'(t)=Cr(t)$ implies $(x'(t),y'(t),z'(t))=C(x(t),y(t),z(t))$ So $x(t)=K_{1}e^{Ct}$, $y(t)=K_{2}e^{Ct}$ $z(t)=K_{3}e^{Ct}$ with $x(0)=1$, $y(0)=2$, $z(0)=3$, where $K_{1}$, $K_{2}$ $K_{3}$ re constants to determine.$x(0)=1$ then $K_{1}=1$, similarly, $K_{2}=2$, $K_{3}=3$. So I think $r(t)=e^{Ct}i+2e^{Ct}j+3e^{Ct}k$

Answer 2

If you only need the curvature, then solving isn't necessary.

Note that $$ \textbf{r}''(t) = C\textbf{r}'(t)$$

Thus $$ K = \frac{||\textbf{r}' \times C\textbf{r}'||}{||\textbf{r}'||^3} = 0 $$