Describing a Cone Patch with Height and Angle
A right circular cone has circular cross-sections perpendicular to its axis. Its geometry is especially easy to describe when one parameter measures distance along the axis and another measures angle around it. This choice makes the cone’s radius at each height explicit and gives a direct way to calculate tangent and normal vectors.
We consider one nappe of a cone, extending from its apex in the direction of its axis. Let \(\mathbf{c}\) be the apex, let \(\mathbf{a}\) be a unit vector pointing along the axis, and choose perpendicular unit vectors \(\mathbf{e}_1\) and \(\mathbf{e}_2\) so that \(\mathbf{e}_1\times\mathbf{e}_2=\mathbf{a}\). These three vectors form an orthonormal frame. The semi-vertical angle \(\alpha\) is the angle between a straight generator of the cone and its axis, where \(0<\alpha<\pi/2\).
To see why the formula describes the cone, define the radial unit vector \(\mathbf{u}(\theta)=\cos(\theta)\mathbf{e}_1+\sin(\theta)\mathbf{e}_2\). At height \(h\), the point is \(h\) units along the axis from the apex and \(h\tan(\alpha)\) units away from the axis. The right triangle formed by these distances has opposite side \(h\tan(\alpha)\) and adjacent side \(h\), so its angle with the axis is \(\alpha\).
A restricted interval of angles describes a sector of the cone. A height interval with lower endpoint greater than zero describes a truncated patch that excludes the apex. Taking an angle interval of length \(2\pi\) covers the full circular cross-section, although the two endpoint angles describe the same seam.
Worked Example: A Quarter of a Cone Around the Vertical Axis
Worked Example: A Quarter of a Cone Around the Vertical Axis
Parametrize the portion of \(z=\sqrt{x^2+y^2}\) with \(0\leq z\leq 2\), \(x\geq 0\), and \(y\geq 0\). Find the tangent vectors and their cross product at height \(h=1\) and angle \(\theta=\pi/2\).
The apex is \(\mathbf{c}=(0,0,0)\), the axis vector is \(\mathbf{a}=(0,0,1)\), and we take \(\mathbf{e}_1=(1,0,0)\), \(\mathbf{e}_2=(0,1,0)\). The cone has \(\tan(\alpha)=1\), so \(\alpha=\pi/4\). The conditions on \(x\) and \(y\) select \(0\leq\theta\leq\pi/2\). Thus $$ \mathbf{r}(h,\theta)=(h\cos(\theta),h\sin(\theta),h), \qquad 0\leq h\leq 2,\quad 0\leq\theta\leq\frac{\pi}{2}. $$
Differentiate with respect to each parameter: $$ \mathbf{r}_h=(\cos(\theta),\sin(\theta),1), \qquad \mathbf{r}_\theta=(-h\sin(\theta),h\cos(\theta),0). $$
At \(h=1\) and \(\theta=\pi/2\), the point and tangent vectors are $$ \mathbf{r}\left(1,\frac{\pi}{2}\right)=(0,1,1),\qquad \mathbf{r}_h\left(1,\frac{\pi}{2}\right)=(0,1,1),\qquad \mathbf{r}_\theta\left(1,\frac{\pi}{2}\right)=(-1,0,0). $$
Their cross product, in the stated parameter order, is $$ \mathbf{r}_h\times\mathbf{r}_\theta = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 0&1&1\\ -1&0&0 \end{vmatrix} =(0,-1,1). $$ Its magnitude is \(\sqrt{0^2+(-1)^2+1^2}=\sqrt{2}\), so it is nonzero. The point lies on the cone because \(z=1=\sqrt{0^2+1^2}\). Also, the cross product is perpendicular to both tangent vectors: \((0,-1,1)\cdot(0,1,1)=0\) and \((0,-1,1)\cdot(-1,0,0)=0\).
Tangent Vectors and the Normal Cross Product
For a parametrized surface, the partial derivatives with respect to its parameters give tangent vectors wherever the parametrization is differentiable. For the cone, write \(\mathbf{u}(\theta)=\cos(\theta)\mathbf{e}_1+\sin(\theta)\mathbf{e}_2\). Differentiating this unit radial vector gives \(\mathbf{v}(\theta)=-\sin(\theta)\mathbf{e}_1+\cos(\theta)\mathbf{e}_2\). The cone’s tangent vectors are therefore $$ \mathbf{r}_h=\mathbf{a}+\tan(\alpha)\mathbf{u}(\theta), \qquad \mathbf{r}_\theta=h\tan(\alpha)\mathbf{v}(\theta). $$
The first vector points along a generator as height increases. The second points around a circular cross-section; its length depends on the cross-section’s radius, which explains the factor \(h\tan(\alpha)\).
This formula follows from the orthonormal-frame identities \(\mathbf{a}\times\mathbf{v}=-\mathbf{u}\) and \(\mathbf{u}\times\mathbf{v}=\mathbf{a}\). For completeness, \(\mathbf{a}\times\mathbf{e}_1=\mathbf{e}_2\) and \(\mathbf{a}\times\mathbf{e}_2=-\mathbf{e}_1\), so $$ \mathbf{a}\times\mathbf{v} =-\sin(\theta)\mathbf{e}_2-\cos(\theta)\mathbf{e}_1 =-\mathbf{u}. $$ Also, expanding \(\mathbf{u}\times\mathbf{v}\) and using \(\mathbf{e}_1\times\mathbf{e}_2=\mathbf{a}\) gives $$ \mathbf{u}\times\mathbf{v} = \bigl(\cos^2(\theta)+\sin^2(\theta)\bigr)\mathbf{a} =\mathbf{a}. $$
Now set \(t=\tan(\alpha)\). By bilinearity of the cross product, $$ \mathbf{r}_h\times\mathbf{r}_\theta =(\mathbf{a}+t\mathbf{u})\times(ht\mathbf{v}) =ht\bigl(\mathbf{a}\times\mathbf{v}+t(\mathbf{u}\times\mathbf{v})\bigr) =ht(t\mathbf{a}-\mathbf{u}). $$ The vectors \(\mathbf{a}\) and \(\mathbf{u}\) are perpendicular unit vectors. Hence \(\lVert t\mathbf{a}-\mathbf{u}\rVert=\sqrt{t^2+1}=\sec(\alpha)\), since \(0<\alpha<\pi/2\). This proves the stated magnitude.
The nonzero cross product means the two tangent vectors are linearly independent when \(h>0\). They span the tangent plane there. The displayed cross product gives one normal direction for the chosen parameter order; its negative gives the opposite direction. In this convention, the cross product points toward the cone’s interior, and its negative points outward.
Worked Example: A Cone with a Tilted Axis
Worked Example: A Cone with a Tilted Axis
Let the apex be \(\mathbf{c}=(1,-1,2)\), the axis be \(\mathbf{a}=(0,1,0)\), and the semi-vertical angle satisfy \(\tan(\alpha)=1/2\). Use \(\mathbf{e}_1=(1,0,0)\) and \(\mathbf{e}_2=(0,0,-1)\). Parametrize the patch \(1\leq h\leq 2\), \(0\leq\theta\leq\pi/2\), and calculate the tangent vectors and cross product at \(h=2,\theta=0\).
The frame vectors have unit length and are mutually perpendicular. Further, \(\mathbf{e}_1\times\mathbf{e}_2=(0,1,0)=\mathbf{a}\). Substitution into the parametrization gives $$ \mathbf{r}(h,\theta) =(1,-1,2)+h(0,1,0)+\frac{h}{2} \bigl(\cos(\theta)(1,0,0)+\sin(\theta)(0,0,-1)\bigr). $$
At \(h=2,\theta=0\), the radial vector is \(\mathbf{u}(0)=(1,0,0)\), so $$ \mathbf{r}(2,0)=(1,-1,2)+(0,2,0)+(1,0,0)=(2,1,2). $$ The tangent vectors are $$ \mathbf{r}_h(2,0)=\mathbf{a}+\frac12\mathbf{u}(0)=\left(\frac12,1,0\right), \qquad \mathbf{r}_\theta(2,0)=2\cdot\frac12\mathbf{v}(0)=(0,0,-1). $$ Thus $$ \mathbf{r}_h(2,0)\times\mathbf{r}_\theta(2,0) =\left(-1,\frac12,0\right), \qquad \left\lVert\mathbf{r}_h\times\mathbf{r}_\theta\right\rVert =\sqrt{1+\frac14}=\frac{\sqrt{5}}{2}. $$ The general magnitude formula gives the same result: \(h\tan(\alpha)\sec(\alpha)=2\cdot\frac12\cdot\frac{\sqrt5}{2}=\frac{\sqrt5}{2}\). The point’s displacement from the apex is \((1,2,0)\); its axial component has length \(2\), and its perpendicular component has length \(1\), equal to \(h\tan(\alpha)=1\).
Why the Apex Is Singular
At \(h=0\), every angle gives the same point: $$ \mathbf{r}(0,\theta)=\mathbf{c}. $$ The angular tangent vector also becomes zero, because \(\mathbf{r}_\theta(0,\theta)=\mathbf{0}\). Consequently, the cross product is zero there. The two parameter derivatives no longer span a two-dimensional tangent plane: the parametrization has rank one at the apex.
This loss of rank reflects the geometry of the cone, not merely a poor choice of notation. Each angle gives a different generator leaving the apex. For example, on a cone with \(\tan(\alpha)=1\) and vertical axis, the directions from the apex along generators at angles \(0,\pi/2,\pi\) are respectively $$ (1,0,1),\qquad (0,1,1),\qquad (-1,0,1). $$ These three vectors are linearly independent: the determinant of the matrix with these vectors as columns is \(2\). Therefore, no single plane can contain all three directions. The cone has no unique tangent plane at its apex, which is why the apex is a singular point of the surface.
Worked Example: Checking the Apex Derivatives
For the same vertical cone \(z=\sqrt{x^2+y^2}\), use \(\mathbf{r}(h,\theta)=(h\cos(\theta),h\sin(\theta),h)\) to calculate the parameter derivatives at \(h=0,\theta=0\), and compare them with the generator directions at \(\theta=0\) and \(\theta=\pi/2\).
Differentiate before substituting the parameter values: $$ \mathbf{r}_h=(\cos(\theta),\sin(\theta),1), \qquad \mathbf{r}_\theta=(-h\sin(\theta),h\cos(\theta),0). $$ At the apex and angle \(0\), these are $$ \mathbf{r}_h(0,0)=(1,0,1), \qquad \mathbf{r}_\theta(0,0)=(0,0,0). $$ The cross product is therefore \((0,0,0)\), not a nonzero normal.
At angle \(\pi/2\), the generator direction is instead $$ \mathbf{r}_h\left(0,\frac{\pi}{2}\right)=(0,1,1). $$ Both generator directions lie on the cone and leave the same apex, but they are different. Together with the direction \((-1,0,1)\) at angle \(\pi\), they span three dimensions as shown above, so they cannot all belong to one tangent plane. This confirms geometrically why the apex has no unique tangent plane.
A Common Mistake: Losing the Height Factor
A frequent error is to differentiate the angular part as though its size were independent of height. Since the radial displacement is \(h\tan(\alpha)\mathbf{u}(\theta)\), differentiating with respect to angle gives \(h\tan(\alpha)\mathbf{v}(\theta)\), not just \(\tan(\alpha)\mathbf{v}(\theta)\). For the vertical cone above, the incorrect calculation would give $$ \mathbf{r}_\theta^{\text{wrong}}=(-\sin(\theta),\cos(\theta),0), \qquad \mathbf{r}_\theta^{\text{wrong}}(0,0)=(0,1,0). $$ That nonzero answer incorrectly suggests that changing the angle moves the apex. But \(\mathbf{r}(0,\theta)=(0,0,0)\) for every \(\theta\), so the derivative with respect to angle at \(h=0\) must be zero. The correct derivative includes the factor \(h\), making it zero at the apex.
Using the Parametrization Reliably
Height-and-angle coordinates separate the cone’s axial and circular geometry. Away from the apex, their partial derivatives provide two independent tangent directions, and their cross product supplies a normal whose direction depends on parameter order. At the apex, every angle collapses to the same point and the angular derivative vanishes. A useful check is therefore to confirm both that the chosen parameter bounds describe the intended patch and that \(h>0\) wherever a regular tangent plane or nonzero normal is required.
Check Your Understanding
Use the definitions and calculations in this tutorial to answer the following questions.
- What geometric distance from the axis does the cone parametrization give at height \(h\)?
- Why does changing the angle move a point around a circle when \(h>0\), but not when \(h=0\)?
- What does the nonzero cross product of the tangent vectors tell you about the patch away from the apex?
- How does reversing the order of the two tangent vectors affect the normal direction?
- Why does the cone’s apex fail to have a unique tangent plane?