A surface $S$ is defined by $x^{2}+z^{2}=25$ and $0\leq y\leq 10.$

Solution (a) Since $S$ is a circular cylinder, we choose a parametrization that uses a variation of cylindrical coordinates. If $x=r\cos \theta ,$ $y=y,$ and $z=r\sin \theta ,$ then $x^{2}+z^{2}=r^{2}=25,$ or equivalently, $r=5$. Then parametric equations for the cylinder are $$x=5\cos \theta \qquad y=y\qquad z=5\sin \theta$$

and a parametrization of $S$ is $$\begin{equation*} \mathbf{r}( \theta ,y) =5\cos \theta\ \mathbf{i}+y\mathbf{j}+5\sin \theta\ \mathbf{k} \end{equation*}$$

where the parameter domain is $0\leq \theta \leq 2\pi$ and $0\leq y\leq 10.$

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Figure 49 The tangent plane $\mathbf{r}( \theta ,y) =5\cos \theta\ \mathbf{i}+y\mathbf{j} +5\sin \theta\ \mathbf{k}$ at the point $( -3,5,4)$.

To determine whether this is a smooth parametrization, we find the partial derivatives of $\mathbf{r=\mathbf{r}}( \theta ,y)$. $$\mathbf{r}_{\theta }( \theta ,y) =-5\sin \theta\ \mathbf{i}+5\cos \theta\ \mathbf{k}\qquad \mathbf{r}_{y}( \theta ,y) =\mathbf{j}$$

Each of these is continuous. Now $$\mathbf{r}_{\theta }\times \mathbf{r}_{y}= \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ -5\sin \theta & 0 & 5\cos \theta \\ 0 & 1 & 0 \end{array}\right| =-5\cos \theta\ \mathbf{i}-5\sin \theta\ \mathbf{k}$$

Since $\left\Vert \mathbf{r}_{\theta }\times \mathbf{r}_{y}\right\Vert =\left\Vert -5\cos \theta\ \mathbf{i}-5\sin \theta\ \mathbf{k}\right\Vert =\sqrt{25\cos ^{2}\theta +25\sin ^{2}\theta }=5,$ then $\mathbf{r}_{\theta }\times \mathbf{r}_{y}\neq \mathbf{0}$. So, $\mathbf{r}=\mathbf{r}( \theta ,y) \mathbf{\ }$ is a smooth parametrization, and $S$ is a smooth surface.

(b) At the point $( -3,5,4) ,$ we have $5\cos \theta =-3,$ $y=5,$ and $5\sin \theta =4.$ Then the normal vector $\mathbf{n}$ to the plane at $( -3,5,4)$ is $\mathbf{r}_{\theta }\times \mathbf{r} _{y}=-5\left( -\dfrac{3}{5}\right) \mathbf{i}-5\left( \dfrac{4}{5}\right) \mathbf{k}=3\mathbf{i}-4\mathbf{k}.$ An equation of the tangent plane is $$3( x+3) -4( z-4) =0\qquad \hbox{or equivalently}\qquad 3x-4z=-25$$

(c) The normal line to the tangent plane contains the point $( -3,5,4)$ and is parallel to $3\mathbf{i}-4\mathbf{k}$. A parametrization of the normal line is $\ \mathbf{r}(t) =(-3+3t) \mathbf{i}+5\mathbf{j}+( 4-4t) \mathbf{k}$. See Figure 49.