Let $\mathbf{r}(t)$ be a parametrisation of a curve $\mathcal{C}$ in $\mathbb{R}^3$ such that $\mathbf{r}(0) = (R, −R, R)$, where $R \in \mathbb{R}$. Suppose $\mathbf{r}(t)\neq \mathbf{0}$ and $\mathbf{r}(t)\cdot\mathbf{r'}(t) = 0$ for all points of $\mathcal{C}$. Show that any such $\mathcal{C}$ must lie on the surface of a sphere. Find the position of the sphere’s centre and determine its radius.
I've tried integrating giving, $$ \int \mathbf{r}(t)\cdot\mathbf{r'}(t)\text{dt} = \frac{\mathbf{r}(t)^2}{2} \\ \implies t_0 = \frac{\mathbf{r}(t)^2}{2} $$ where $t_0 \in \mathbb{R}$ but I don't see how this helps. Also differentiating $||\mathbf{r}(t)-s_0||^2$ where $s_0$ is the centre of our sphere $\mathcal{S}$ should equal zero as its a constant radius but I don't see how that helps.
It's pretty much what you said. Let $f(t) = \mathbf{r}(t)^2 \equiv \langle \mathbf{r}(t), \mathbf{r}(t)\rangle$. Then, differentiating the function $f$ yields: \begin{align} f'(t) &= \mathbf{r}(t) \cdot \mathbf{r}'(t) \\ &= 0 \tag{by hypothesis} \end{align} Since $f'$ is identically zero, it follows that $f$ is identically equal to a constant. In other words, for all $t \in \Bbb{R}$, we must have that \begin{align} f(t) &= f(0) \\ &= \mathbf{r}(0) \cdot \mathbf{r}(0) \\ &= (R,-R,R) \cdot (R,-R,R) \\ &= 3R^2 \end{align} Now, recall that $f(t) = \mathbf{r}(t) \cdot \mathbf{r}(t) = x(t)^2 + y(t)^2 + z(t)^2$ is the squared length of the position vector $\mathbf{r}(t)$. This squared length being constant for all time $t$ means that the curve lies on some sphere. In particular, it lies on the sphere centered at the origin, having radius $\sqrt{3} |R|.$
What you did by integration is also equivalent to what I wrote. To figure out what the constant $t_0$ must be, note that if you evaluate both sides at $t=0$ then you find that $t_0 = \dfrac{3R^2}{2}$.