Tangent, Normal and Binormal Vectors — Question 8

Question 8 – Solution

Strategy. Regularity of $\mathbf r$ guarantees $\mathbf T$, but defining $\mathbf N$ additionally requires $\mathbf T'\ne\mathbf 0$.

Step 1: Unit tangent $$\mathbf r'(t)=\left\langle 1,3t^2,0\right\rangle,\qquad \mathbf T(t)=\frac{\left\langle 1,3t^2,0\right\rangle}{\sqrt{1+9t^4}}.$$ Thus $\boxed{\mathbf T(0)=\left\langle 1,0,0\right\rangle}$.

Step 2: Differentiate $\mathbf T$ Differentiation and simplification give $$\mathbf T'(t)=\frac{1}{(1+9t^4)^{3/2}}\left\langle -18t^3,6t,0\right\rangle =\frac{6t}{(1+9t^4)^{3/2}}\left\langle -3t^2,1,0\right\rangle.$$ Therefore $\mathbf T'(0)=\mathbf 0$. The expression $$\mathbf N(0)=\frac{\mathbf T'(0)}{\|\mathbf T'(0)\|}$$ would divide by zero, so $\boxed{\mathbf N(0)\text{ and }\mathbf B(0)\text{ are not defined by the Frenet formulas}}$.

Step 3: One-sided behavior For $t>0$, normalization yields a vector tending to $\left\langle 0,1,0\right\rangle$. For $t<0$, the scalar factor $6t$ reverses direction, giving the limit $\left\langle 0,-1,0\right\rangle$. The unequal limits show that the preferred bending direction flips at the inflection point. Although the curve has a tangent there, it has no single principal-normal direction.