Tangent, Normal and Binormal Vectors — Question 5

Question 5 – Solution

Strategy Normalize $\vec r'$, differentiate $\mathbf T$, and use the planar geometry as a check.

Tangent and normal With $w=\sqrt{1+4t^2}$, $$\mathbf T=\frac 1w\left\langle 1,2t,0\right\rangle,\qquad \mathbf N=\frac 1w\left\langle-2t,1,0\right\rangle.$$ Consequently $\boxed{\mathbf B=\left\langle 0,0,1\right\rangle}$.

Interpretation The curve is contained in the $xy$-plane, whose oriented unit normal is constant. The normal has positive $y$-component for all $t$, so the upward-opening parabola bends toward its concave side.