Alternative way to define the Normal Vector

Question

My calculus book defines the Normal vector in terms of the Tangent vector like so:

$\overrightarrow{r}(t) =$ the original vector function

$$\overrightarrow{T}(t) = \frac{\overrightarrow{r'}(t)}{\|\overrightarrow{r'}(t)\|}$$

$$\overrightarrow{N}(t) = \frac{\overrightarrow{T'}(t)}{\|\overrightarrow{T'}(t)\|}$$

Since $\overrightarrow{T}(t)$ is just $\overrightarrow{r'}(t)$ normalized, can I say that $$\overrightarrow{N}(t) = \frac{\overrightarrow{r''}(t)}{\|\overrightarrow{r''}(t)\|}$$ without changing anything?

Answer 1

The specific reason why this does not work in the simple way you attempted is that $r''$ includes both the normal acceleration (which is in the direction you want) and the tangential acceleration (which is perpendicular to the direction you want).

You can separate out these two components by projecting $r''$ onto the tangential direction (that is, onto the direction of $r'$). This gives the tangential component of $r''$: $$ \frac{r' \cdot r''}{\lVert r'\rVert^2} r'.$$

Subtracting this from the total acceleration, what remains is the normal acceleration: $$ r'' - \frac{r' \cdot r''}{\lVert r'\rVert^2} r'.$$

And if we want a unit vector then we need to divide by the magnitude: $$ \frac{r'' - \frac{r' \cdot r''}{\lVert r'\rVert^2} r'} {\left\lVert r'' - \frac{r' \cdot r''}{\lVert r'\rVert^2} r' \right\rVert}.$$

This may help to explain why you might want to use $T$ as an intermediate step.

Answer 2

No, you cannot. Yes, you will get an unit vector but, in general, it will not be orthogonal to $\overrightarrow{T}(t)$.

Answer 3

No. $\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} = \frac{\frac{\mathbf{r}''(t)\|\mathbf{r}'(t)\| - \mathbf{r}'(t) (\|\mathbf{r}'\|)'(t)}{\|\mathbf{r}'(t)\|^2}}{\|\mathbf{T}'(t)\|}$ (using the quotient rule of differentiation). Since the denominator is just a scalar, analyze the numerator: $\frac{\mathbf{r}''(t)\|\mathbf{r}'(t)\| - \mathbf{r}'(t) (\|\mathbf{r}'\|)'(t)}{\|\mathbf{r}'(t)\|^2} = \frac{\mathbf{r}''(t)}{\|\mathbf{r}'(t)\|} - \mathbf{r}'(t)\frac{(\|\mathbf{r}'\|)'(t)}{\|\mathbf{r}'(t)\|^2}$. The expression reduces to a vector in the direction of $\mathbf{r}''(t)$ minus a vector in the direction of $\mathbf{r}'(t)$. Since these vectors are not in general pointing in the same direction, the resulting vector will not generally be in the direction of $\mathbf{r}''(t)$. Therefore, it won't be $\frac{\mathbf{r}''(t)}{\|\mathbf{r}''(t)\|}$.

Note: by $(\|\mathbf{r}'(t)\|)'(t)$ I mean the function $\frac{d\|\mathbf{r}'(t)\|}{dt}$.