Tangent, Normal and Binormal Vectors — Question 3

Question 3 – Solution

Strategy. Computing $\mathbf T'(1)$ requires differentiating the normalized tangent, not merely normalizing $\mathbf r''(1)$.

Step 1: Unit tangent $$\mathbf r'(t)=\left\langle 1,2t,0\right\rangle,\qquad \mathbf T(t)=\frac{\left\langle 1,2t,0\right\rangle}{\sqrt{1+4t^2}}.$$ Hence $$\boxed{\mathbf T(1)=\frac 1{\sqrt 5}\left\langle 1,2,0\right\rangle}.$$

Step 2: Differentiate $\mathbf T$ Componentwise differentiation gives $$\mathbf T'(t)=\frac 1{(1+4t^2)^{3/2}}\left\langle -4t,2,0\right\rangle.$$ At $t=1$ this vector is a positive multiple of $\left\langle -2,1,0\right\rangle$, whose length is $\sqrt 5$. Therefore $$\boxed{\mathbf N(1)=\frac 1{\sqrt 5}\left\langle -2,1,0\right\rangle}.$$

Step 3: Binormal $$\mathbf B(1)=\mathbf T(1)\times\mathbf N(1) =\frac 15\left\langle 1,2,0\right\rangle\times\left\langle -2,1,0\right\rangle =\boxed{\left\langle 0,0,1\right\rangle}.$$ The curve lies in the $xy$-plane, so its binormal points perpendicular to that plane. The three vectors are unit length and mutually perpendicular.