Finding the Principal Unit Normal Vector of a Helix

Show that the principal unit normal vector \(\mathbf{N}(t)\) of the helix \begin{equation*} \mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k} \end{equation*}

is orthogonal to the \(z\)-axis.

Solution We begin by finding \(\mathbf{r^{\prime} }( t) \) and \(\left\Vert \mathbf{r}^{\prime} ( t)\right \Vert.\) \[ \begin{equation*} \mathbf{r}^{\prime} (t)=-\!\sin t\mathbf{i}+\cos t\mathbf{j}+\mathbf{k}\qquad \left\Vert \mathbf{r}^{\prime }(t)\right\Vert =\sqrt{ (-\!\sin t)^{2}+(\cos t)^{2}+1}=\sqrt{2} \end{equation*} \]

Then the unit tangent vector is \(\mathbf{T}(t)=\dfrac{\mathbf{r}^{\prime }(t)}{||\mathbf{r}^{\prime }(t)||}=\dfrac{-\!\sin t\mathbf{i}+\cos t\mathbf{j}+\mathbf{k}}{\sqrt{2}}.\)

Since \[ \begin{equation*} \mathbf{T}^{\prime} (t)=\dfrac{-\!\cos t\mathbf{i}-\sin t\mathbf{j}}{\sqrt{2} }\qquad \hbox{and}\qquad \left\Vert \mathbf{T}^{\prime }(t)\right\Vert =\sqrt{ \left( -\dfrac{\cos t}{\sqrt{2}}\right) ^{2}+\left( -\dfrac{\sin t}{\sqrt{2}} \right) ^{2}}=\dfrac{1}{\sqrt{2}} \end{equation*} \]

we have \begin{equation*} \mathbf{N}(t)=\dfrac{\mathbf{T}^{\prime} (t)}{\left\Vert \mathbf{T}^{\prime} (t)\right\Vert }=-\!\cos t\mathbf{i}-\sin t\mathbf{j} \end{equation*}

Since the direction of the \(z\)-axis is \(\mathbf{k}\), it follows that \(\mathbf{N}(t)\,{\cdot}\, \mathbf{k}=0\) for all \(t.\) So, \(\mathbf{N}(t)\) is always orthogonal to the \(z\)-axis, as shown in Figure 14.