Proving that $\vec{r'}(t)$ is orthogonal to $\vec{r''}(t)$

Question

With a given nonzero vector $\vec{r}(t)$, how do I that $\vec{r'}(t)$ is orthogonal to $\vec{r''}(t)$? The length ($||\vec{r'}(t)||$ is constant.) This is what I have tried so far.

Let $\vec{r}(t)= <f(t),g(t),h(t)>$. Then, $\vec{r'}(t)$ is $<f'(t),g'(t),h'(t)>$. And $\vec{r''}(t)$ is $<f''(t),g''(t),h''(t)>$.

For vectors to be orthogonal to each other, their dot products must be 0, meaning $\vec{r'}(t)\cdot\vec{r''}(t)=0$. This is where I get stuck. How do I prove that $\vec{r'}(t)\cdot\vec{r''}(t)=f'(t)\cdot f''(t)+g'(t)\cdot g''(t)+h'(t)\cdot h''(t) =0$?

Answer 1

\begin{align*} f'(t)\cdot f''(t)+g'(t)\cdot g''(t)+h'(t)\cdot h''(t)&=\frac{1}{2}\frac{d}{dt}(f'^2+g'^2+h'^2)\\ &=\frac{1}{2}\frac{d}{dt}\|r'(t)\|^2 \\ &=0. &&(\because \|r'(t)\|=c) \end{align*}

Answer 2

These problems are usually solved ``backwards'' by taking the derivative of given information. For example since $\|\vec{r'}\|=c$ you have $\vec{r'}\cdot \vec{r'}=c^2$ and then taking derivatives on both sides you obtain $2\vec{r'}\cdot \vec{r''}=0$ which implies that the two are orthogonal.

Answer 3

Since

$\vec r'(t) \cdot \vec r'(t) = \Vert \vec r'(t) \Vert^2 = \text{constant}, \tag 1$

$(\vec r'(t) \cdot \vec r'(t))' = 0; \tag 2$

but

$(\vec r'(t) \cdot \vec r'(t))' = \vec r''(t) \cdot \vec r'(t) + \vec r'(t) \cdot \vec r''(t) = 2\vec r'(t) \cdot r''(t), \tag 3$

whence via (2),

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

that is, $\vec r'(t)$ and $\vec r''(t)$ are orthogonal.