$B(t) = \frac{r'(t)\times r''(t)}{||r'(t) \times r''(t)||}$ proof

Question

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

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

$$r'(t) = T(t)\cdot|r'(t)|$$

$$r''(t) = T'(t)\cdot |r'(t)|+T(t)\cdot \frac{d}{dt}(|r'(t)|)$$

Now $$r'(t)\times r''(t) = |r'(t)|^{2}\cdot T(t)\times T'(t)$$

$$T(t)\times T'(t) = \frac{r'(t)\times r''(t)}{|r'(t)|^{2}}$$

Now $T(t)$ is already present. How to get $|T'(t)| \text{ to find } N(t)$?

Answer 1

Hint :

Since $\vec{T}(t)$ is unit tangent vector, It follows that $\vec{T}(t).\vec{T}(t)=1$

differentiate with respect to $t$ you can get $\vec{T^{\prime}}(t).\vec{T}(t)=\vec{0}$

That is $\vec{T}(t)$ and $\vec{T^{\prime}}(t)$ are perpendicular