Chapter 7 Applications to Surfaces

On a curve in two-dimensional space $\mathbb{R}^{2}$, there is:

• a tangent line;

• a normal line.

What are the corresponding features of a surface in three-dimensional space $\mathbb{R}^{3}$?

7.1 Surfaces in Cartesian Coordinates

All surfaces considered in this section are assumed to be described as the set of points in $\mathbb{R}^{3}$ that satisfy some equation of the form $$F(x,y,z)=0.$$

The grad of $F$ is the vector $\nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right)$.

The symbol $\nabla$ is called a nabla or a del.

Let $S$ be a surface described by $F(x,y,z)=0$, and let $P$ be a point on $S$. Then $\nabla F (P)$ is a normal vector to $S$ at $P$.

let $C$ be a curve that passes through $P$, and lies entirely on $S$.

Consider a parametrisation $\mathbf{r}(t) = \big( x(t), y(t), z(t) \big)$ of $C$. The tangent line to $P$ at $C$ is given by $\frac{d \mathbf{r}}{dt} = \big( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \big)$. Since $C$ lies on $S$, it follows that $F \big( x(t), y(t), z(t) \big) =0$, for all $t$. Hence $$\frac{dF}{dt}\big( x(t), y(t), z(t) \big) =0, \qquad \text{for all } t.$$ By the Chain Rule $$\begin{align*} \frac{dF}{dt} &= \frac{dF}{dx} \frac{dx}{dt} + \frac{dF}{dy} \frac{dy}{dt} + \frac{dF}{dz} \frac{dz}{dt} \\[5pt] &= \left( \frac{dF}{dx}, \frac{dF}{dy}, \frac{dF}{dz} \right) \cdot \left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right) \\[5pt] &= \nabla F \cdot \frac{d \mathbf{r}}{dt} \end{align*}$$

Provided $\nabla F \neq \mathbf{0}$ and $\frac{d \mathbf{r}}{dt}\neq \mathbf{0}$, then $\nabla F$ is perpendicular to the tangent to the curve $C$ at $P$.

This discussion applies to all such curves on $S$ through $P$. Therefore $\nabla F$ is perpendicular to all tangent vectors to $S$ at $P$, and is a normal vector to $S$.

The unique plane through $P$ that is perpendicular to $\nabla F$ is the tangent plane to $S$ at $P$.

Find a normal vector and the tangent plane to the surface $z = x + 2y - \frac{1}{2} \left( x^2 + y^2 \right)$ at the point $P$ where $(x,y)= \left( \frac{5}{4}, \frac{5}{2} \right)$.

When $(x,y)= \left( \frac{5}{4}, \frac{5}{2} \right)$, calculate the value of $z$:

$$\begin{align*} \left( \frac{5}{4} \right) + 2 \left( \frac{5}{2} \right) + 4z &= \frac{125}{8} \\ 4z &= \frac{75}{8} \\ z &= \frac{75}{32}. \end{align*}$$

Define $F(x,y,z) = z - x - 2y + \frac{1}{2} \left( x^2 + y^2 \right)$. The surface $S$ is $F(x,y,z)=0$. Calculate

$$\begin{align*} \frac{\partial F}{\partial x} &= -1 + x, \\[3pt] \frac{\partial F}{\partial y} &= -2 + y, \\[3pt] \frac{\partial F}{\partial z} &= 1. \end{align*}$$

Therefore $\nabla F = \left( -1+x, -2+y, 1 \right)$. Evaluating $\nabla F$ at $x= \frac{5}{4}, y = \frac{5}{2}, z= \frac{75}{32}$ shows that the normal to $S$ at $P$ is $\left( \frac{1}{4}, \frac{1}{2}, 1 \right)$. The tangent plane is all the points satisfying

$$\begin{align*} \Bigg( (x,y,z) - \left( \frac{5}{4}, \frac{5}{2}, \frac{75}{32} \right) \Bigg) \cdot \nabla F(P) &= 0 \\[3pt] \left( x - \frac{5}{4}, y - \frac{5}{2}, z-\frac{75}{32} \right) \cdot \left( \frac{1}{4}, \frac{1}{2}, 1 \right) &= 0 \\[3pt] \frac{1}{4} \left( x - \frac{5}{4} \right) + \frac{1}{2} \left( y - \frac{5}{2} \right) + \left( z-\frac{75}{32} \right) &= 0 \\[3pt] x+2y+4z &= \frac{125}{8}. \end{align*}$$

7.2 Surfaces in Parametric Coordinates

Alternatively one can consider a surface $S$ given in parametric form as discussed in Section 2.4.

Let $S$ be a surface defined by $\mathbf{r}(u,v)$ for a set of values of the parameters $u$, $v$, and further that $P = \mathbf{r}(u_0,v_0)$ is a point on $S$. Then the normal vector to $S$ at $P$ is given by $\mathbf{r}_{u} \times \mathbf{r}_v$.

By fixing the parameter $u=u_0$ while allowing $v$ to vary describes a curve $C_1$ that both lies on the surface $S$, and passes through $P$. Similarly fixing the parameter $v=v_0$ while allowing $u$ to vary describes another curve $C_2$ that lies on the surface $S$ and passes through $P$.

The tangent vectors to $C_1$ and $C_2$ at $P$ are given respectively by $\frac{\partial \mathbf{r}}{\partial u}(P) = \mathbf{r}_{u}(P)$ and $\frac{\partial \mathbf{r}}{\partial v}(P) = \mathbf{r}_{v}(P)$. Hence provided that these tangent vectors exist and are not parallel, the tangent plane to $S$ at $P$ is uniquely determined by the point $P$ and the direction vectors $\mathbf{r}_u$ and $\mathbf{r}_u$. It follows from the geometric interpretation of the cross product that $\mathbf{r}_u \times \mathbf{r}_v$ is a normal vector to $S$ at $P$.

Having a simple formula for the normal of a parametric surface at a point, also provides a simple method by which to calculate the tangent plane of a parametric surface at a point.

Find the normal vector at a general point on the parametric surface $S$ described by $$\mathbf{r}(u,v) = \big( \text{sech} (u) \cos (v), \text{sech}(u) \sin (v), u - \text{tanh} (u) \big), \qquad \text{where } u \in (- \infty, \infty), v \in [0, 2 \pi).$$ Further find the tangent plane to $S$ at the point $P = \mathbf{r} \left(\frac{3}{2},\frac{\pi}{2} \right).$

Calculate that $$\begin{align*} \mathbf{r}_u &= \big( - \text{sech} (u) \text{tanh} (u) \cos (v), - \text{sech} (u) \text{tanh}(u) \sin (v), 1 - \text{sech}^{2} (u) \big) \\ &= - \text{sech} (u) \text{tanh}(u) \Big( \cos (v), \sin (v), - \text{sinh}(u) \Big), \end{align*}$$ and $$\begin{align*} \mathbf{r}_v &= \Big( -\text{sech} (u) \sin (v), \text{sech}(u) \cos (v), 0 \big) \\ &= \text{sech}(u) \Big( - \sin (v), \cos (v), 0 \Big) \end{align*}$$ The scalars can be ignored since we are only interested in the direction of a normal vector. Therefore a normal vector $\mathbf{N}$ at a general point of $S$ is given by $$\begin{align*} \mathbf{N}(u,v) &= \mathbf{r}_u \times \mathbf{r}_v \\[9pt] &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos (v) & \sin (v) & - \text{sinh} (u) \\ -\sin (v) & \cos (v) & 0 \end{vmatrix} \\[9pt] &= \Big( \text{sinh} (u) \cos (v), \text{sinh} (u) \sin (v), 1 \Big). \end{align*}$$

The point $P$ has position vector

$$\mathbf{r}\left( \frac{3}{2},\frac{\pi}{2} \right) = \left( 0, \text{sech} \left( \frac{3}{2} \right), \frac{3}{2} - \text{tanh} \left( \frac{3}{2} \right) \right),$$

and normal vector

$$\mathbf{N} \left( \frac{3}{2},\frac{\pi}{2} \right) = \left( 0 , \text{sinh} \left( \frac{3}{2} \right) , 1\right).$$

Therefore the tangent plane at $P$ has equation

$$\begin{align*} \Bigg( (x,y,z) - \mathbf{r}\left( \frac{3}{2},\frac{\pi}{2} \right) \Bigg) \cdot \mathbf{N} &= 0 \\[3pt] \left( x , y - \text{sech} \left( \frac{3}{2} \right), z-\frac{3}{2} + \text{tanh} \left( \frac{3}{2} \right) \right) \cdot \left( 0, \text{sinh} \left( \frac{3}{2} \right), 1 \right) &= 0 \\[3pt] 0 \cdot x + \text{sinh} \left( \frac{3}{2} \right) \cdot \left( y - \text{sech} \left( \frac{3}{2} \right) \right) + \left( z-\frac{3}{2} + \text{tanh} \left( \frac{3}{2} \right) \right) &= 0 \\[3pt] \left( \text{sinh} \left( \frac{3}{2} \right) \right) y - \text{tanh} \left( \frac{3}{2} \right) + z - \frac{3}{2} + \text{tanh} \left( \frac{3}{2} \right) &= 0 \\[3pt] \text{sinh} \left( \frac{3}{2} \right) y+z &= \frac{3}{2}. \end{align*}$$

The surface appearing in Example 7.2.2 is known as the psuedosphere.