Graph Surfaces and Their Tangent Geometry

Graphs as Parametrized Surfaces

A function of two variables can be pictured as a surface in three-dimensional space: each input \((x,y)\) determines a height \(z=f(x,y)\). When \(f\) is differentiable, its partial derivatives describe how the height changes in the two coordinate directions. Those changes determine the surface’s tangent vectors and normal direction.

To make this description precise, let \(D\) be a region in the \(uv\)-plane, and let \(f\) be a continuously differentiable function on an open set containing \(D\). The graph over \(D\) consists of points \((u,v,f(u,v))\). Using \(u,v\) as parameters gives a map from the parameter domain into three-dimensional space.

The parameters \(u\) and \(v\) give the horizontal coordinates directly. Holding one parameter fixed and varying the other traces a coordinate curve on the graph. The corresponding tangent vectors describe the instantaneous directions of those curves.

Coordinate Tangent Vectors

Differentiate the parametrization component by component. When \(v\) is held fixed, changing \(u\) changes the first coordinate at rate \(1\), the second at rate \(0\), and the height at rate \(f_u(u,v)\). When \(u\) is held fixed, the corresponding rates are \(0\), \(1\), and \(f_v(u,v)\). Thus

These vectors are linearly independent at every point where they are defined. Indeed, if \(a\mathbf{r}_u+b\mathbf{r}_v=\mathbf{0}\), the first component gives \(a=0\), and the second gives \(b=0\). Consequently their cross product is nonzero, so it supplies a normal direction at every such point.

More generally, if the parameters change instantaneously at rates \(a\) and \(b\), the corresponding tangent direction is the linear combination

This expression shows how the graph’s height changes along a chosen direction in its parameter domain. The tangent vectors and their combinations describe local directions along the surface; the normal vector, by contrast, is perpendicular to those directions.

Deriving the Upward Normal

For vectors in three dimensions, the cross product is perpendicular to both factors. Apply it in the parameter order \(\mathbf{r}_u\times\mathbf{r}_v\). Computing the determinant gives

The third component is \(1\), which is positive. Therefore this cross product points upward and is an upward normal. Reversing the cross-product order changes its sign: \(\mathbf{r}_v\times\mathbf{r}_u=(f_u,f_v,-1)\), which points downward.

Here is a direct verification of perpendicularity. Using the components of the vectors,

Thus the vector is perpendicular to each coordinate tangent vector. Since every tangent direction of the parametrization is a linear combination of those two vectors, it is perpendicular to every such direction as well.

The cross product gives a normal vector, but it need not have length \(1\). Its length is

This length is always positive because its square is at least \(1\). Dividing by it gives the upward unit normal:

The positive third component is essential to the word “upward.” The unnormalized normal is often convenient for calculations, while the unit normal is required whenever a direction vector of length \(1\) is needed.

Worked Example: A Paraboloid

Worked Example: A Saddle-Shaped Graph

Worked Example: A Trigonometric Graph

A Common Sign and Normalization Error

A frequent mistake is to use the cross product in the wrong order and then call the result upward. For the graph \(z=u^2+v^2\), the vectors at \((1,-1)\) are \((1,0,2)\) and \((0,1,-2)\). Taking \(\mathbf{r}_v\times\mathbf{r}_u\) instead gives \((2,-2,-1)\). Although this is a valid normal, its third component is negative, so it points downward. The correction is to keep the stated order \(\mathbf{r}_u\times\mathbf{r}_v\), giving \((-2,2,1)\), whose third component is positive.

A different error is to divide by the wrong length when forming a unit normal. In the trigonometric example, the normal’s length is \(\sqrt{6}/2\). Its second component is \(\sqrt{6}/4\), so the normalized second component is

not \(\sqrt{3}/2\). Checking that all three components have squared lengths summing to \(1\) is a useful way to catch normalization errors.

Why the Formulas Are Useful

For a graph, the horizontal coordinates of the parametrization are simply the parameters themselves. This makes the tangent vectors especially direct: their first two components are \((1,0)\) and \((0,1)\), while their vertical components are the partial derivatives of the height function. The normal formula follows immediately from those vectors and fixes the upward orientation without guesswork.

When applying the formulas, keep the steps distinct: differentiate \(f\), substitute the parameter values, form the cross product in the chosen order, and normalize only if a unit normal is required. At the end, check that the normal has positive third component, that its dot products with both tangent vectors vanish, and—if it is a unit normal—that its length is \(1\).