Xc5–Xc7 - Normal lines and tangent planes | OCR A-Level Further Mathematics B (MEI)

Xc5 extends surface equations to $g(x,y,z)=c$ . Xc6 turns first-order partial derivatives into a normal vector. Xc7 combines this vector with the equations of lines and planes to describe the surface near a point.

One equation can describe more than one height

Can a curved surface have one direction parallel to every tangent plane? The surface

(y-x)(y+z)^2=4

does. We will find that direction by differentiating, then see why whole straight lines lie on this curved surface.

You already know

For $z=f(x,y)$ , a partial derivative holds the other input constant. A vector line through position vector $\boldsymbol a$ , in direction $\boldsymbol d\ne\boldsymbol0$ , is $\boldsymbol r=\boldsymbol a+\lambda\boldsymbol d$ . A plane through $\boldsymbol a$ , with normal $\boldsymbol n\ne\boldsymbol0$ , is $\boldsymbol n\cdot(\boldsymbol r-\boldsymbol a)=0$ .

A graph $z=f(x,y)$ gives one height for each allowed input pair. An equation involving all three coordinates need not do that.

Implicitly defined surface

The surface $g(x,y,z)=c$ consists of the triples satisfying that equation, where $c$ is constant.

For example, $x^2+y^2+z^2=9$ is a sphere. Above $(x,y)=(0,0)$ , it has both $z=3$ and $z=-3$ , so the whole sphere is not one graph $z=f(x,y)$ .

Return to $g(x,y,z)=(y-x)(y+z)^2$ , with $g=4$ . Taking $x=0$ , $y=1$ gives

\begin{aligned} (1+z)^2&=4\\ z&=1\quad\text{or}\quad z=-3. \end{aligned}

Thus $A=(0,1,1)$ and $C=(0,1,-3)$ are two different points above the same input pair.

Because the right-hand side is positive, every point on this surface has $y-x>0$ . We can describe its two separate branches by

z=\frac{2}{\sqrt{y-x}}-y \quad\text{or}\quad z=-\frac{2}{\sqrt{y-x}}-y, \qquad y>x.

The positive choice describes a neighbourhood of $A$ ; the negative choice describes a neighbourhood of $C$ . Choosing a branch is a real restriction, not merely an algebraic tidying step.

Key idea

A single relation $g(x,y,z)=c$ can describe a surface even when solving for $z$ requires several branches. Each point still needs all three coordinates and must satisfy the original relation.

The representation of a surface is worth checking before using any normal formula.

First derivatives point normally to the surface

The two coordinate slopes of a graph carry enough information to find a vector perpendicular to its tangent plane. For an implicit surface, include the derivative with respect to $z$ as well.

A normal vector from first derivatives

For a differentiable surface $g(x,y,z)=c$ , at a point $P$ on the surface where $\operatorname{grad}g\ne\boldsymbol0$ , a normal vector is

\boldsymbol n=\left.\operatorname{grad}g\right|_P =\left. \begin{pmatrix} \dfrac{\partial g}{\partial x}\\[3pt] \dfrac{\partial g}{\partial y}\\[3pt] \dfrac{\partial g}{\partial z} \end{pmatrix} \right|_P.

For a graph $z=f(x,y)$ , take $g(x,y,z)=f(x,y)-z$ . A normal vector is

\boldsymbol n= \begin{pmatrix} \dfrac{\partial f}{\partial x}\\[3pt] \dfrac{\partial f}{\partial y}\\[3pt] -1 \end{pmatrix}

evaluated at the point.

In the formula booklet.

Evaluate the derivatives at the specified point before forming a numerical line or plane. Any non-zero scalar multiple of a normal vector gives the same normal direction. A zero gradient does not supply a usable normal vector by this rule.

Worked example 1: differentiate the implicit relation

Question. For $g(x,y,z)=(y-x)(y+z)^2$ , find a normal vector to $g=4$ at $A=(0,1,1)$ .

Plan. Verify the point, take three first partial derivatives, and evaluate the gradient.

\begin{aligned} g(0,1,1)&=(1)(2)^2=4 && \text{point lies on the surface}\\ g_x&=-(y+z)^2 && y,z\text{ held constant}\\ g_y&=(y+z)^2+2(y-x)(y+z) && \text{product rule}\\ g_z&=2(y-x)(y+z) && x,y\text{ held constant}\\ \left.\operatorname{grad}g\right|_A &=\begin{pmatrix}-4\\8\\4\end{pmatrix} =4\begin{pmatrix}-1\\2\\1\end{pmatrix}. \end{aligned}

Why this step? Both factors depend on $y$ ; differentiating only one factor would lose part of the normal direction.

Answer. A convenient normal vector is $\boldsymbol n_A=\begin{pmatrix}-1\\2\\1\end{pmatrix}$ .

Check. Moving by $(t,t,-t)$ leaves both $y-x$ and $y+z$ unchanged. Its direction $\begin{pmatrix}1\\1\\-1\end{pmatrix}$ is therefore along the surface, and its dot product with $\boldsymbol n_A$ is $-1+2-1=0$ .

Change only the representation: keep the same surface near the same point, but write it as a graph.

Worked example 2: use the graph form

Question. Use $z=f(x,y)=2(y-x)^{-1/2}-y$ , for $y>x$ , to find a normal at $A$ .

Plan. Find $f_x,f_y$ , then use the supplied vector $(f_x,f_y,-1)$ .

\begin{aligned} f_x&=2\left(-\frac12\right)(y-x)^{-3/2}(-1) =(y-x)^{-3/2}\\ f_y&=2\left(-\frac12\right)(y-x)^{-3/2}-1 =-(y-x)^{-3/2}-1\\ (f_x,f_y)\big|_{(0,1)}&=(1,-2)\\ \boldsymbol n&=\begin{pmatrix}1\\-2\\-1\end{pmatrix} =-\boldsymbol n_A. \end{aligned}

Why this step? The $-1$ is the derivative of $-z$ in $f(x,y)-z=0$ . It is the third component, not another slope to calculate.

Answer. $\begin{pmatrix}1\\-2\\-1\end{pmatrix}$ , which gives the same normal direction as before.

Check. The two answers are scalar multiples, as they must be for these equivalent local descriptions.

Keep the third component when converting coordinate slopes into a graph's normal.

Watch out

Wrong: use $(f_x,f_y,0)$ as the normal to $z=f(x,y)$ .

The implicit equation is $f(x,y)-z=0$ , whose $z$ -partial is $-1$ . Right: use $(f_x,f_y,-1)$ , evaluated at the specified point, or a non-zero scalar multiple.

The two formulas agree even though one uses three partial derivatives and the other uses two. Try changing the point while keeping the implicit relation.

Use the normal as a line direction and a plane normal

At a smooth point, the normal line passes through the surface in the perpendicular direction. The tangent plane passes through the point and is perpendicular to that normal. The same vector plays two different roles.

Normal line and tangent plane

Let $P$ have position vector $\boldsymbol a$ , and let $\boldsymbol n\ne\boldsymbol0$ be a normal vector to the surface at $P$ .

The normal line is

\boldsymbol r=\boldsymbol a+\lambda\boldsymbol n,\qquad \lambda\in\mathbb R.

The tangent plane is

\boldsymbol n\cdot(\boldsymbol r-\boldsymbol a)=0.

These use the supplied gradient formula and the earlier equations of lines and planes.

Not in the formula booklet: learn it.

The tangent plane describes the surface to first order near the point. It need not touch the surface only once, and the whole surface need not lie on one side of it.

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Xc5–Xc7 - Normal lines and tangent planes

One equation can describe more than one height

First derivatives point normally to the surface

Use the normal as a line direction and a plane normal

Continue this lesson.

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