Tutorials › Calculus III › Singular Points In Standard Surface Parametrizations

Home  /  Tutorials  /  Parametrized Surfaces

Parametrized Surfaces · Tutorial 840 of 999

Singular Points In Standard Surface Parametrizations

Use derivative and cross-product tests to identify singular parameters, then distinguish removable coordinate singularities from geometric singularities.

Intermediate 10 min read

What You'll Learn

  • Apply the rank condition to find singular points of a surface parametrization.
  • Locate the poles where spherical coordinates lose rank.
  • Explain why the sphere remains smooth at its parametrization’s poles.
  • Identify the apex as a geometric singularity of a cone.
  • Distinguish rank loss from a parameter-domain seam or repeated parameter value.

Why a Parametrization Can Fail at a Smooth Point

A parametrization describes a surface by mapping two parameters to points in three-dimensional space. Its derivatives tell us how the image changes as each parameter changes. If those two derivative vectors become dependent, the parametrization loses rank at that parameter value.

That failure does not always mean the surface itself is nonsmooth. At a pole of the usual spherical parametrization, every azimuthal angle gives the same point, so changing the angle produces no first-order movement there. The sphere, however, is smooth at the pole and can be described by a different parametrization. A cone apex presents a different case: the surface itself has no smooth tangent plane there.

Central distinction: A parametrization singularity is a rank failure of a particular map. A geometric singularity is a failure of the image surface itself to be smooth. A point may be singular for one parametrization but regular in another.

Rank Loss: The Basic Test

Let \(\mathbf{X}(u,v)\) be a differentiable parametrization of a surface patch. Its two parameter-direction vectors are \(\mathbf{X}_u\) and \(\mathbf{X}_v\). The parametrization is regular at \((u,v)\) when these vectors are linearly independent; equivalently, the derivative of the map has rank two there.

Definition: A parameter value \((u,v)\) is singular for \(\mathbf{X}\) if \(\mathbf{X}_u(u,v)\) and \(\mathbf{X}_v(u,v)\) are linearly dependent. In three dimensions, this is equivalent to \(\mathbf{X}_u(u,v)\times\mathbf{X}_v(u,v)=\mathbf{0}\).

The cross product gives a convenient test because it is zero exactly when its two input vectors are linearly dependent. Its magnitude also measures the area of the parallelogram spanned by those vectors. A zero magnitude therefore signals that the two parameter directions fail to span a two-dimensional piece of surface at that parameter value.

More explicitly, the identity \(|\mathbf{a}\times\mathbf{b}|^2=|\mathbf{a}|^2|\mathbf{b}|^2-(\mathbf{a}\cdot\mathbf{b})^2\) shows that the squared area is zero precisely when the vectors are linearly dependent. Thus the cross-product test and the rank condition are two forms of the same test.

Worked Example: The Poles of a Sphere

Worked Example: The Poles of a Sphere

Consider a sphere of radius \(3\) parametrized by polar angle \(\phi\) and azimuthal angle \(\theta\):

$$ \mathbf{X}(\phi,\theta) = (3\sin\phi\cos\theta,\;3\sin\phi\sin\theta,\;3\cos\phi), \qquad 0\leq\phi\leq\pi. $$

Differentiate with respect to each parameter:

$$ \mathbf{X}_{\phi} = (3\cos\phi\cos\theta,\;3\cos\phi\sin\theta,\;-3\sin\phi), $$ $$ \mathbf{X}_{\theta} = (-3\sin\phi\sin\theta,\;3\sin\phi\cos\theta,\;0). $$

The cross product is

$$ \mathbf{X}_{\phi}\times\mathbf{X}_{\theta} = (9\sin^2\phi\cos\theta,\;9\sin^2\phi\sin\theta,\;9\sin\phi\cos\phi). $$

Its magnitude is

$$ |\mathbf{X}_{\phi}\times\mathbf{X}_{\theta}| = 9\sqrt{\sin^4\phi(\cos^2\theta+\sin^2\theta)+\sin^2\phi\cos^2\phi} = 9\sin\phi. $$

Here \(\sin\phi\geq 0\) on the stated interval. The magnitude is zero at \(\phi=0\) and \(\phi=\pi\), so the parametrization loses rank at both poles. At \(\phi=0\), the image is \((0,0,3)\) for every \(\theta\); at \(\phi=\pi\), it is \((0,0,-3)\) for every \(\theta\). The azimuthal derivative vanishes at each pole.

To check whether this is a singularity of the sphere itself, use a local parametrization near the north pole:

$$ \mathbf{P}(x,y)=(x,y,\sqrt{9-x^2-y^2}). $$

For \(x^2+y^2<9\), its derivatives at \((0,0)\) are \(\mathbf{P}_x(0,0)=(1,0,0)\) and \(\mathbf{P}_y(0,0)=(0,1,0)\). Their cross product is \((0,0,1)\), which is nonzero. The sphere is therefore smoothly parametrized near its north pole by this chart, even though the spherical-coordinate parametrization loses rank there. The same reasoning applies at the south pole using the negative square-root graph.

Check: Substituting \(\phi=0\) into the original formula gives \((0,0,3)\), and substituting \(\phi=\pi\) gives \((0,0,-3)\), as required. The cross-product magnitude \(9\sin\phi\) vanishes at exactly those two parameter values.

Why the Spherical Pole Is a Coordinate Singularity

At a spherical pole, the parameter \(\theta\) does not distinguish nearby points along the surface: at the pole, all values of \(\theta\) map to the same point. The vanishing derivative in the \(\theta\)-direction records that collapse. It is a defect of this choice of coordinates, not a failure of the sphere to be smooth.

A useful way to test a suspected coordinate singularity is to find another parametrization of the same image near the point. If the new parametrization has two independent derivative vectors there, it shows that the image surface has a regular local description. The graph parametrization in the example supplies exactly such a description near the north pole.

Worked Example: The Apex of a Cone

Worked Example: The Apex of a Cone

Consider the cone \(\,z=\sqrt{x^2+y^2}\,\) parametrized by distance \(r\geq 0\) from the vertical axis and angle \(\theta\):

$$ \mathbf{C}(r,\theta)=(r\cos\theta,\;r\sin\theta,\;r). $$

The parameter derivatives are

$$ \mathbf{C}_r=(\cos\theta,\sin\theta,1), \qquad \mathbf{C}_{\theta}=(-r\sin\theta,r\cos\theta,0). $$

Their cross product and its magnitude are

$$ \mathbf{C}_r\times\mathbf{C}_{\theta} = (-r\cos\theta,-r\sin\theta,r), \qquad |\mathbf{C}_r\times\mathbf{C}_{\theta}| = \sqrt{r^2\cos^2\theta+r^2\sin^2\theta+r^2} = \sqrt{2}\,r. $$

Thus the parametrization is regular for \(r>0\) and loses rank at \(r=0\). At \(r=0\), every angle maps to the apex \((0,0,0)\), and \(\mathbf{C}_{\theta}=\mathbf{0}\).

In this case, the apex is also a geometric singularity. To see why, the directions of straight lines on the cone from the apex are \((\cos\theta,\sin\theta,1)\). If a smooth tangent plane existed at the apex, all these limiting directions would lie in that plane. A plane through the origin has an equation \(ax+by+cz=0\); containing every such direction would require

$$ a\cos\theta+b\sin\theta+c=0 \qquad\text{for every }\theta. $$

Taking \(\theta=0\) and \(\theta=\pi\) gives \(a+c=0\) and \(-a+c=0\), so \(a=c=0\). Taking \(\theta=\frac{\pi}{2}\) then gives \(b=0\). These values do not define a plane. No plane can contain all the limiting directions, so the cone has no smooth tangent plane at its apex.

Check: At \(r=0\), the parametrization gives \((0,0,0)\) for every \(\theta\), and the calculated cross-product magnitude is \(\sqrt{2}\cdot 0=0\). For \(r>0\), it is positive. Both checks agree with the rank calculation.

This argument illustrates the difference from the sphere. A coordinate change cannot supply a regular smooth surface chart at the cone apex because the image itself fails the smoothness test there. Away from the apex, the same cone parametrization has rank two.

Worked Example: A Cylinder and a Parameter Seam

Worked Example: A Cylinder and a Parameter Seam

A radius-\(2\) cylinder can be parametrized by an angle and height:

$$ \mathbf{Q}(\theta,z)=(2\cos\theta,\;2\sin\theta,\;z). $$

The two derivatives are

$$ \mathbf{Q}_{\theta}=(-2\sin\theta,2\cos\theta,0), \qquad \mathbf{Q}_{z}=(0,0,1). $$

The cross product and its magnitude are

$$ \mathbf{Q}_{\theta}\times\mathbf{Q}_{z} = (2\cos\theta,2\sin\theta,0), \qquad |\mathbf{Q}_{\theta}\times\mathbf{Q}_{z}| = \sqrt{4\cos^2\theta+4\sin^2\theta} = 2. $$

The cross product is nonzero for every angle and height, so this parametrization has no rank-loss points. Yet \(\theta=0\) and \(\theta=2\pi\) map to the same point at any fixed height. That repetition is due to the periodic angle coordinate; it is not a singularity, because the derivative vectors remain independent at both parameter values.

Check: The dot product \(\mathbf{Q}_{\theta}\cdot\mathbf{Q}_{z}=0\), and both vectors are nonzero, so they are independent. The computed cross-product magnitude \(2\) confirms the rank test everywhere, including at the two ends of an angular parameter interval.

A Practical Classification

When a derivative test finds a zero cross product, classify what has failed before drawing conclusions about the surface. The parameter map may have collapsed a direction at a point where the image is still smooth; alternatively, the image itself may not admit a smooth local surface description. Repeated parameter values or a seam in the parameter domain do not by themselves imply rank loss.

1
Find the candidate parameter values.
Compute \(\mathbf{X}_u\), \(\mathbf{X}_v\), and \(\mathbf{X}_u\times\mathbf{X}_v\). Locate where the cross product is zero.
2
Identify the image point.
Evaluate \(\mathbf{X}\) at each candidate. Check whether distinct parameter values map to that same point.
3
Test for a different local description.
If another parametrization is regular at the image point, the original rank loss is a coordinate singularity.
4
Check the image’s smoothness.
If no smooth local surface description is possible, as at the cone apex, the point is a geometric singularity.

A Common Mistake

A common error is to conclude that the sphere is geometrically singular at its poles just because the spherical-coordinate cross product vanishes there. That confuses a parametrization with its image. The correct reasoning checks a second chart: near the north pole, \(\mathbf{P}(x,y)=(x,y,\sqrt{9-x^2-y^2})\) has derivative cross product \((0,0,1)\) at \((0,0)\), which is nonzero. The spherical chart loses rank, but the sphere remains smooth.

The opposite error is to treat every rank failure as removable. At the cone apex, the limiting directions from the apex cannot all lie in one plane, so a smooth local surface chart cannot repair the geometry. The correct conclusion depends on the image, not only on the formula for one parametrization.

Why This Distinction Matters

Recognizing coordinate singularities prevents a poor choice of parameters from being mistaken for a defect in a surface. Recognizing geometric singularities prevents a genuine failure of smoothness from being hidden by calculation alone. The distinction is especially useful when choosing local parametrizations: regular charts are the appropriate tools near smooth points, while a geometric singularity requires separate treatment.

Check Your Understanding

Use the rank test and the distinction between parametrization and image to answer these questions.

  1. What condition on \(\mathbf{X}_u\) and \(\mathbf{X}_v\) makes a parametrization regular?
  2. At which parameter values does the stated spherical parametrization of a radius-\(3\) sphere lose rank?
  3. Why does a regular graph parametrization near a sphere pole show that the pole is not a geometric singularity?
  4. For the cone parametrization, what is the magnitude of \(\mathbf{C}_r\times\mathbf{C}_{\theta}\), and where does it vanish?
  5. Why does the equality of cylinder images at \(\theta=0\) and \(\theta=2\pi\) not imply rank loss?

Previous

Next

Back to Tutorials