Deriving The Normal Formula For Graphs

Why a Graph Has a Simple Normal Formula

A surface given by an equation \(z=f(x,y)\) is determined by a function of two coordinates. At each point where \(f\) is differentiable, small changes in \(x\) and \(y\) produce two natural directions along the surface: one follows a coordinate curve while \(y\) is fixed, and the other follows a coordinate curve while \(x\) is fixed.

These directions are tangent to the surface. A vector perpendicular to both is normal to the surface. Computing that vector gives a compact formula that is useful whenever a surface is presented as a graph. The sign in the formula also records which of the two possible normal directions points upward.

Parametrizing the Graph and Finding Its Tangents

Let \(D\) be an open subset of the \(xy\)-plane, and suppose \(f:D\to\mathbb{R}\) is differentiable. The graph of \(f\) consists of points \((x,y,z)\) with \(z=f(x,y)\). It can be parametrized by using \(x\) and \(y\) themselves as parameters:

Holding \(y\) fixed and varying \(x\) traces a coordinate curve on the graph. Its tangent direction is the partial derivative of \(\mathbf{r}\) with respect to \(x\). Holding \(x\) fixed and varying \(y\) gives the other coordinate curve and its tangent. Differentiating each component gives

Both vectors are tangent to the surface at \(\mathbf{r}(x,y)\). They are also linearly independent at every point: if \(a\mathbf{r}_x+b\mathbf{r}_y=\mathbf{0}\), the first coordinate gives \(a=0\), and the second gives \(b=0\). Thus they span the tangent plane of this graph at the point.

Deriving the Upward Normal

A normal vector must be perpendicular to every tangent direction, so it must in particular be perpendicular to both coordinate tangent vectors. The cross product of two vectors is perpendicular to each of them. Taking the cross product in the order \(\mathbf{r}_x\times\mathbf{r}_y\) gives

The last component is \(1\), which is positive. Therefore this normal points upward, meaning that its vertical component points in the positive \(z\)-direction. The reversed cross product \(\mathbf{r}_y\times\mathbf{r}_x\) has the opposite direction and points downward.

This is a normal vector, not necessarily a unit vector. Its length is \(\sqrt{f_x^2+f_y^2+1}\), so it is never the zero vector. If a unit normal is needed, divide by that positive length:

Dividing by a positive number changes the length but not the direction, so the normalized vector is still upward. The formula applies wherever the graph function is differentiable; at points where the relevant derivatives do not exist, this derivation does not establish a tangent plane or this normal formula.

Why the Formula Is the Only Upward Normal Direction

The cross product gives a normal, but orthogonality also shows why its components must have this form. Write a general vector as \(\mathbf{m}=(a,b,c)\). For it to be perpendicular to both tangent vectors, it must satisfy two equations:

Solving gives \(a=-c f_x\) and \(b=-c f_y\), so every normal vector is \(c(-f_x,-f_y,1)\). If the normal is upward, its vertical component satisfies \(c>0\); hence it is a positive multiple of \((-f_x,-f_y,1)\). This proves that the formula gives the unique upward normal direction, even though infinitely many upward normal vectors of different lengths are possible.

Worked Examples

Common Pitfall: Reversing the Cross Product

The cross product is order-sensitive: reversing its two inputs reverses the resulting vector. For the tilted plane above, a common error is to calculate \(\mathbf{r}_y\times\mathbf{r}_x\) but then report the result as upward:

This vector is perpendicular to both tangent directions, but its third component is \(-1\), so it points downward, not upward. The correction is to use \(\mathbf{r}_x\times\mathbf{r}_y=(-2,3,1)\), or to multiply the reversed result by \(-1\). Perpendicularity alone does not determine orientation; the sign of the vertical component does.

When the Formula Is Useful

The formula turns partial derivatives of a graph function directly into a normal direction. It also gives a quick consistency check: the first two components are the negatives of the slopes in the coordinate directions, while the third component is \(1\). A sign error in either derivative, or a reversed cross-product order, can be detected by checking orthogonality and the sign of the third component.

Keep the distinction between direction and length clear. The vector \((-f_x,-f_y,1)\) is convenient because it is simple and always nonzero; the unit normal has length one and is obtained by normalization. Which form is needed depends on the calculation, but both describe the same upward direction.