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Parametrized Surfaces · Tutorial 842 of 999

Implicit Surfaces As Local Parametrized Patches

Use the implicit function theorem to turn a regular implicit surface into a local graph, then compute and check a normal from its parametrization.

Intermediate 10 min read

What You'll Learn

  • Distinguish a zero set from a genuine local surface.
  • State the conditions that let an implicit equation be solved locally for one coordinate.
  • Construct a graph parametrization from the resulting function.
  • Calculate a normal by crossing the parametrization’s partial derivatives.
  • Relate the normal’s direction to the gradient of the defining function.
  • Check parametrizations and normals using the defining equation and orthogonality.

From an Equation to a Local Patch

An equation in three variables can describe a surface without presenting it as a graph. For instance, a sphere is naturally specified by an equation involving all three coordinates. Near a suitable point, however, that equation may determine one coordinate as a function of the other two. This turns the surface into a parametrized patch that can be studied with the tools of surface parametrization.

The key condition is that the partial derivative with respect to the coordinate being solved for is nonzero at the point. This condition guarantees a local graph, not necessarily a single graph description of the entire surface. It also gives a direct way to compute a normal vector to the patch.

Definition: The zero set of a function \(F:\mathbb{R}^3\to\mathbb{R}\) is the set of points \((x,y,z)\) for which \(F(x,y,z)=0\). A zero set is not automatically a surface: for example, if \(F\equiv 0\), its zero set is all of \(\mathbb{R}^3\). A regular implicit surface near a point is a zero set that is locally a two-dimensional surface, with the standard sufficient condition that \(F\) is \(C^1\) near the point and \(\nabla F\ne \mathbf{0}\) there.

Here \(C^1\) means that the first partial derivatives exist and are continuous. The condition \(\nabla F\ne \mathbf{0}\) is sufficient for the zero set to be a smooth surface locally. The local graph result below uses a more specific version of this condition: the partial derivative corresponding to the coordinate we want to solve for must be nonzero.

The Local Graph Result

Theorem: Suppose \(F\) is \(C^1\) near \((a,b,c)\), \(F(a,b,c)=0\), and \(F_z(a,b,c)\ne 0\). Then there are neighborhoods of \((a,b)\) and \(c\) and a \(C^1\) function \(z=g(x,y)\) such that, in those neighborhoods, \(F(x,y,z)=0\) exactly when \(z=g(x,y)\). The other coordinates can be solved for in the same way: \(F_x(a,b,c)\ne 0\) permits a local representation \(x=g(y,z)\), and \(F_y(a,b,c)\ne 0\) permits \(y=g(x,z)\).

This is the implicit function theorem in the form needed here. It guarantees a unique local choice of the solved-for coordinate near the specified point. It does not promise that the same choice remains valid across the entire zero set. Once, for example, \(z=g(x,y)\) is available, the surface patch has the graph parametrization \(\mathbf{r}(u,v)=(u,v,g(u,v))\), for \((u,v)\) in a sufficiently small open domain around \((a,b)\).

The word “locally” matters. A sphere can have an upper and a lower graph over much of the same region in the \(xy\)-plane. At a point on the upper half, the theorem selects the upper branch in a neighborhood; it does not combine both branches into one graph.

Worked Example: A Local Graph on a Sphere

Worked Example: A Local Graph on a Sphere

Use \(F(x,y,z)=x^2+y^2+z^2-9\) near \(P=(1,2,2)\) to solve for \(z\), parametrize the local patch, and find a normal.

First check that \(P\) lies on the zero set: \(F(1,2,2)=1^2+2^2+2^2-9=1+4+4-9=0\). The relevant partial derivative is \(F_z=2z\), so \(F_z(1,2,2)=4\ne 0\). The local graph result therefore applies.

Solving the equation for \(z\) gives \(z^2=9-x^2-y^2\). Since the point has positive \(z\)-coordinate, the local branch through it is \(z=g(x,y)=\sqrt{9-x^2-y^2}\). A suitable neighborhood is restricted so that \(9-x^2-y^2>0\) and the positive branch is used.

With parameters \(u=x\) and \(v=y\), the parametrization is \(\mathbf{r}(u,v)=(u,v,\sqrt{9-u^2-v^2})\). Write \(w=\sqrt{9-u^2-v^2}\). Differentiating gives \(\mathbf{r}_u=(1,0,-u/w)\) and \(\mathbf{r}_v=(0,1,-v/w)\). Their cross product is \(\mathbf{r}_u\times\mathbf{r}_v=(u/w,v/w,1)\).

At the parameter values \((u,v)=(1,2)\), \(w=\sqrt{9-1-4}=2\), so a normal is \((1/2,1,1)\). Multiplying by \(2\) gives the simpler normal \((1,2,2)\).

Check the parametrization: substituting it into \(F\) gives \(u^2+v^2+(9-u^2-v^2)-9=0\). At \((1,2)\), the tangent vectors are \((1,0,-1/2)\) and \((0,1,-1)\). Their dot products with \((1,2,2)\) are \(1-1=0\) and \(2-2=0\), respectively. Thus the proposed vector is perpendicular to both parameter directions, as required.

Why the Cross Product Gives a Normal

For a regular parametrized patch, the two partial derivative vectors point along the patch’s parameter directions. A vector perpendicular to both is therefore normal to the patch. The cross product provides such a vector. In this setting, the relationship between that cross product and the gradient can be derived directly.

1
Differentiate the implicit equation.
Suppose \(F(x,y,z)=0\) is locally \(z=g(x,y)\). Then \(F(x,y,g(x,y))=0\). Differentiating with respect to \(x\) and \(y\), using the chain rule, gives \(F_x+F_zg_x=0\) and \(F_y+F_zg_y=0\). Since \(F_z\ne 0\), \(g_x=-F_x/F_z\) and \(g_y=-F_y/F_z\).
2
Compute the cross product.
The graph parametrization is \(\mathbf{r}(u,v)=(u,v,g(u,v))\), so \(\mathbf{r}_u=(1,0,g_u)\) and \(\mathbf{r}_v=(0,1,g_v)\). Their cross product is \((-g_u,-g_v,1)\).
3
Identify its direction.
Substituting the chain-rule expressions gives \(\mathbf{r}_u\times\mathbf{r}_v=(F_x/F_z,F_y/F_z,1)=\nabla F/F_z\), with the partial derivatives evaluated at \((u,v,g(u,v))\). Hence the cross product is normal to the patch and is parallel to the gradient.

This derivation explains why the gradient is normal to a regular level surface: as one moves along the surface, the value of \(F\) stays zero, so its directional change along either parameter direction is zero. The gradient is perpendicular to those directions. The cross-product calculation also keeps track of orientation: reversing the order of the parameters reverses the cross product.

Worked Example: Solving for the First Coordinate

Worked Example: Solving for the First Coordinate

For \(F(x,y,z)=x+y^2+z^2-5\), construct a local patch near \(P=(1,1,\sqrt{3})\) by solving for \(x\), then calculate a normal at \(P\).

First, \(F(1,1,\sqrt{3})=1+1^2+(\sqrt{3})^2-5=1+1+3-5=0\). Also, \(F_x=1\), which is nonzero. Solving for \(x\) gives \(x=5-y^2-z^2\).

Use \(u=y\) and \(v=z\). Then \(\mathbf{r}(u,v)=(5-u^2-v^2,u,v)\), and the point \(P\) corresponds to \((u,v)=(1,\sqrt{3})\). The parameter derivatives are \(\mathbf{r}_u=(-2u,1,0)\) and \(\mathbf{r}_v=(-2v,0,1)\). Therefore, \(\mathbf{r}_u\times\mathbf{r}_v=(1,2u,2v)\), and at the specified parameters a normal is \((1,2,2\sqrt{3})\).

Check the parametrization by substitution: \(F(5-u^2-v^2,u,v)=(5-u^2-v^2)+u^2+v^2-5=0\). At \(P\), the gradient is \(\nabla F=(1,2y,2z)=(1,2,2\sqrt{3})\), agreeing with the computed normal. Directly, the tangent vectors at that parameter pair are \((-2,1,0)\) and \((-2\sqrt{3},0,1)\); their dot products with the normal are \(-2+2=0\) and \(-2\sqrt{3}+2\sqrt{3}=0\).

Parameter Order and the Sign of the Normal

When solving for a different coordinate, the order of the parameters affects the sign of the cross product. For \(x=g(y,z)\), using \((y,z)\) in that order gives a cross product parallel to \(\nabla F\). For \(y=g(x,z)\), using \((x,z)\) in that order gives a cross product antiparallel to \(\nabla F\). In either case it is a valid normal direction; the sign records the chosen parameter order. This is consistent with the fact that swapping two cross-product inputs changes the sign.

Worked Example: Solving for the Second Coordinate

For \(F(x,y,z)=x^2+y+z^2-6\), build a local patch near \(P=(1,1,2)\) by solving for \(y\), and compute a normal.

At the point, \(F(1,1,2)=1^2+1+2^2-6=1+1+4-6=0\). The partial derivative \(F_y=1\ne0\), so solve for \(y\): \(y=6-x^2-z^2\). With parameters \(u=x\) and \(v=z\), the patch is \(\mathbf{r}(u,v)=(u,6-u^2-v^2,v)\).

Its parameter derivatives are \(\mathbf{r}_u=(1,-2u,0)\) and \(\mathbf{r}_v=(0,-2v,1)\). Their cross product is \(\mathbf{r}_u\times\mathbf{r}_v=(-2u,-1,-2v)\). At \((u,v)=(1,2)\), this is the normal \((-2,-1,-4)\).

Substitution checks the patch: \(F(u,6-u^2-v^2,v)=u^2+(6-u^2-v^2)+v^2-6=0\). At \(P\), \(\nabla F=(2,1,4)\), so the computed normal equals \(-\nabla F\). The parameter tangent vectors are \((1,-2,0)\) and \((0,-4,1)\). Their dot products with \((-2,-1,-4)\) are \(-2+2=0\) and \(4-4=0\). The normal is correct; its negative sign relative to the gradient comes from the chosen parameter order.

When the Method Applies—and a Common Mistake

The implicit function theorem gives a local graph only when its hypotheses are satisfied: \(F\) must be \(C^1\) near the point, the point must lie on the zero set, and the partial derivative for the coordinate being solved for must be nonzero at that point. If that particular partial derivative is zero, this theorem does not justify solving for that coordinate there. It does not follow that the zero set is not a surface; another coordinate may have a nonzero partial derivative.

Common mistake: Treating every equation \(F(x,y,z)=0\) as a surface, or dividing by a partial derivative without checking that it is nonzero. For example, \(F(x,y,z)\equiv0\) has zero set \(\mathbb{R}^3\), not a two-dimensional surface. In a local graph calculation, first verify \(F=0\) at the point and then check that the chosen coordinate’s partial derivative is nonzero there. A zero partial derivative only blocks that particular application; it does not by itself settle whether the zero set is a surface.

For a regular implicit surface, at least one component of \(\nabla F\) is nonzero, so at least one coordinate can be solved for locally by this theorem. This is useful when an equation is easier to work with than an explicit parametrization: it converts a neighborhood of the surface into a graph patch, whose parameter derivatives and normal can then be computed directly. The calculations also provide checks: substitution tests whether the patch lies on the surface, and dot products test whether a proposed normal is perpendicular to its parameter directions.

Check Your Understanding

Answer each question using the definitions and local graph result from this tutorial.

  1. Why is the zero set of \(F\equiv0\) not a two-dimensional surface?
  2. What conditions allow \(F(x,y,z)=0\) to be solved locally for \(z\) as a function of \(x\) and \(y\)?
  3. For a graph \(z=g(x,y)\), what parametrization is used, and which two vectors are crossed to obtain a normal?
  4. If \(F_y\ne0\) at a point on the zero set, which coordinate can be represented locally as a function of the other two?
  5. How can substitution and dot products be used to check a local patch and its proposed normal?

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