Tangent, Normal and Binormal Vectors — Question 5

Question 5 – Solution

Strategy. Apply the chain rule and keep track of the sign of the parameter derivative during normalization.

Step 1: Increasing reparametrization The chain rule gives $$\mathbf R'(u)=2\mathbf r'(2u+1).$$ Since the factor $2$ is positive, $$\frac{\mathbf R'(u)}{\|\mathbf R'(u)\|} =\frac{2\mathbf r'(2u+1)}{2\|\mathbf r'(2u+1)\|} =\boxed{\mathbf T_{\!r}(2u+1)}.$$ Thus an increasing reparametrization preserves the unit tangent.

Step 2: Principal normal Differentiating $\mathbf T_R(u)=\mathbf T_r(2u+1)$ gives $$\mathbf T_R'(u)=2\mathbf T_r'(2u+1).$$ Again the positive factor cancels upon normalization, so $$\boxed{\mathbf N_R(u)=\mathbf N_r(2u+1)}.$$

Step 3: Reversed parameter If $t=1-2u$, then $dt/du=-2$. Consequently $$\boxed{\mathbf T_R=-\mathbf T_r},$$ at corresponding points. Differentiating this relation with respect to $u$ introduces a second negative factor, so $\mathbf T_R'=2\mathbf T_r'$ and $$\boxed{\mathbf N_R=\mathbf N_r}.$$ Therefore $\mathbf B_R=\mathbf T_R\times\mathbf N_R=-\mathbf B_r$. Reversing orientation changes $\mathbf T$ and $\mathbf B$, but not the principal normal.