Show that the acute angle between the tangent vector to the helix $$\mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k}\quad 0\,{\leq}\,t\,{\leq}\,2\pi$$

and the direction $\mathbf{k}$ is $\dfrac{\pi }{4}$ radian.

Solution A tangent vector at any point on the helix is given by $$\begin{equation*} \mathbf{r}^{\prime} (t)=-\sin t\mathbf{i}+\cos t\mathbf{j}+\mathbf{k} \end{equation*}$$

Figure 10 $\mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k}, \\ 0\leq t\leq 2\pi$

Then $\left\Vert \mathbf{r}^{\prime} (t)\right\Vert\;= \sqrt{(-\sin t)^2 + \cos^2 t +1} = \sqrt{1+1} = \sqrt{2}.$

The cosine of the acute angle $\theta$ between $\mathbf{r}^{\prime} (t)$ and $\mathbf{k}$ is $$\begin{eqnarray*} &&\cos \theta =\frac{\mathbf{r}^{\prime} (t)\,{\cdot}\, \mathbf{k}}{\left\Vert \mathbf{r} ^{\prime} (t)\right\Vert \left\Vert \mathbf{k}\right\Vert }=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\qquad \color{#0066A7}{\hbox{$\mathbf{r}^{\prime} (t)\,{\cdot}\, \mathbf{k} = 1$}}\\ \end{eqnarray*}$$

So, $\theta =\dfrac{\pi }{4}$ radian. See Figure 10.