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Parametrized Surfaces · Tutorial 843 of 999

Tangent Planes From Parametric Surface Data

Use parameter derivatives to find tangent directions and equations of tangent planes, including on parametrized spheres and cylinders.

Intermediate 9 min read

What You'll Learn

  • Define regular points of a parametrized surface using its two parameter derivatives
  • Find the tangent plane from two independent tangent directions
  • Compute a normal vector with a cross product and form the plane equation
  • Derive why parameter derivatives span the tangent plane
  • Find and check tangent planes on spherical and cylindrical patches
  • Recognize when a parametrization cannot be used directly to determine a tangent plane

From Surface Parameters to a Tangent Plane

A parametrized surface describes points in space using two parameters. Near a point on a smooth surface, changing one parameter while holding the other fixed traces a curve on the surface; changing the other parameter traces a second curve. The velocities of those curves give two directions lying in the tangent plane.

When the two directions are independent, they determine a plane. A cross product packages the same information in a different form: it gives a vector perpendicular to both directions, which can be used as a normal vector in the plane equation.

Definition: Let \(\mathbf r(u,v)\) be a differentiable parametrization of a surface patch, and let \(P=\mathbf r(u_0,v_0)\). The parameter derivatives at \((u_0,v_0)\) are \(\mathbf r_u(u_0,v_0)\) and \(\mathbf r_v(u_0,v_0)\). The point is regular for this parametrization if these vectors are linearly independent, equivalently if \(\mathbf r_u(u_0,v_0)\times\mathbf r_v(u_0,v_0)\ne\mathbf 0\).

At a regular point, the tangent plane is the plane through \(P\) in the two directions \(\mathbf r_u(u_0,v_0)\) and \(\mathbf r_v(u_0,v_0)\). If \(\mathbf n\) is any nonzero normal vector to that plane, its equation is \(\mathbf n\cdot(\mathbf x-P)=0\), where \(\mathbf x=(x,y,z)\) is a general point in space.

The Tangent-Plane Formulas

Write the point and parameter derivatives as vectors in \(\mathbb R^3\). The two derivative vectors themselves give a parametric description of the plane. Their cross product gives a normal-vector description.

$$ \begin{aligned} P&=\mathbf r(u_0,v_0),\\ \mathbf r_u&=\mathbf r_u(u_0,v_0),\qquad \mathbf r_v=\mathbf r_v(u_0,v_0),\\ \mathbf x&=P+s\mathbf r_u+t\mathbf r_v,\qquad s,t\in\mathbb R,\\ \mathbf n&=\mathbf r_u\times\mathbf r_v,\\ \mathbf n\cdot(\mathbf x-P)&=0. \end{aligned} $$

The order of the cross product matters for the direction of the normal: \(\mathbf r_v\times\mathbf r_u=-(\mathbf r_u\times\mathbf r_v)\). But either nonzero normal gives the same plane equation, because multiplying the entire equation by \(-1\) does not change its solution set.

Method: Evaluate the parametrization at the parameter values to get \(P\). Compute both parameter derivatives and evaluate them at those same values. Check that their cross product is nonzero, then use it as a normal in \(\mathbf n\cdot(\mathbf x-P)=0\). Alternatively, write the plane using the two derivative directions directly.

Why the Parameter Derivatives Give the Tangent Plane

The connection follows from differentiating curves on the surface. Holding \(v=v_0\) fixed and varying \(u\) traces a parameter curve through \(P\); its velocity at \(u_0\) is \(\mathbf r_u(u_0,v_0)\). Holding \(u=u_0\) fixed and varying \(v\) gives a curve with velocity \(\mathbf r_v(u_0,v_0)\). Thus both vectors are tangent directions.

More generally, suppose a differentiable curve on the patch is described by parameter functions \(u=u(t)\) and \(v=v(t)\), with \(u(t_0)=u_0\) and \(v(t_0)=v_0\). The chain rule gives the curve's velocity at \(t_0\) as a linear combination of those two directions. This shows that every such velocity lies in their span.

1
Differentiate a surface curve.
Write \(\boldsymbol\gamma(t)=\mathbf r(u(t),v(t))\), where \(\boldsymbol\gamma(t_0)=P\).
2
Apply the chain rule.
At \(t_0\), \(\boldsymbol\gamma'(t_0)=\mathbf r_u(u_0,v_0)u'(t_0)+\mathbf r_v(u_0,v_0)v'(t_0)\).
3
Identify the tangent plane.
Every surface-curve velocity lies in the span of the two parameter derivatives, and the coordinate curves supply those two directions. At a regular point they are independent, so their span is a plane.

This argument explains why the construction works, rather than merely providing a calculation rule. Regularity is essential to this parametrization-based method: if the derivatives are dependent, their span is not a two-dimensional plane, and their cross product is zero. The surface itself may still be smooth there, but a different regular parametrization may be needed to find its tangent plane.

Worked Example: Tangent Plane to a Graph Patch

Worked Example: Tangent Plane to a Graph Patch

Find the tangent plane to \(\mathbf r(u,v)=(u,v,u^2+2v^2)\) at the parameter values \((u_0,v_0)=(1,-1)\).

First evaluate the parametrization:

$$ P=\mathbf r(1,-1)=(1,-1,1^2+2(-1)^2)=(1,-1,3). $$

Differentiate with respect to each parameter:

$$ \mathbf r_u(u,v)=(1,0,2u), \qquad \mathbf r_v(u,v)=(0,1,4v). $$

At the specified parameter values, the tangent directions are \(\mathbf r_u=(1,0,2)\) and \(\mathbf r_v=(0,1,-4)\). Their cross product is

$$ \mathbf r_u\times\mathbf r_v = \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ 1&0&2\\ 0&1&-4 \end{vmatrix} =(-2,4,1). $$

Since this vector is nonzero, the point is regular for this parametrization. Use it as the plane normal:

$$ (-2,4,1)\cdot\big((x,y,z)-(1,-1,3)\big)=0. $$

Expanding gives \(-2(x-1)+4(y+1)+(z-3)=0\), or \(-2x+4y+z+3=0\).

Check the result at the point: \(-2(1)+4(-1)+3+3=0\). Also, the normal is perpendicular to each tangent direction: \((-2,4,1)\cdot(1,0,2)=-2+2=0\), and \((-2,4,1)\cdot(0,1,-4)=4-4=0\). Both checks agree with the plane equation.

Worked Example: A Tangent Plane to a Sphere

Worked Example: A Tangent Plane to a Sphere

Use the spherical parametrization \(\mathbf r(\phi,\theta)=(2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi)\) to find the tangent plane at \((\phi_0,\theta_0)=(\pi/3,\pi/2)\). Here \(\phi\) is the angle measured down from the positive \(z\)-axis, and \(\theta\) is the angle in the \(xy\)-plane.

The point is

$$ P=\mathbf r\left(\frac{\pi}{3},\frac{\pi}{2}\right) =(0,\sqrt{3},1), $$

since \(\sin(\pi/3)=\sqrt{3}/2\), \(\cos(\pi/3)=1/2\), \(\cos(\pi/2)=0\), and \(\sin(\pi/2)=1\).

Differentiate in the stated parameter order, first \(\phi\), then \(\theta\):

$$ \begin{aligned} \mathbf r_\phi&=(2\cos\phi\cos\theta,2\cos\phi\sin\theta,-2\sin\phi),\\ \mathbf r_\theta&=(-2\sin\phi\sin\theta,2\sin\phi\cos\theta,0). \end{aligned} $$

At the selected parameters these become

$$ \mathbf r_\phi=(0,1,-\sqrt{3}), \qquad \mathbf r_\theta=(-\sqrt{3},0,0). $$

Taking the cross product in that same order gives

$$ \mathbf r_\phi\times\mathbf r_\theta = \big(0,\;(-\sqrt{3})(-\sqrt{3})-0,\;0-1(-\sqrt{3})\big) =(0,3,\sqrt{3}). $$

This is nonzero, so the parametrization is regular at this point. The normal is \((0,3,\sqrt{3})\), and the plane equation is

$$ (0,3,\sqrt{3})\cdot\big((x,y,z)-(0,\sqrt{3},1)\big)=0, \qquad\text{so}\qquad \sqrt{3}y+z-4=0. $$

Check by substitution: \(\sqrt{3}(\sqrt{3})+1-4=3+1-4=0\). The normal also points in the radial direction: \((0,3,\sqrt{3})=\sqrt{3}(0,\sqrt{3},1)\). Reversing the cross-product order would give the negative normal, but the same tangent plane.

Worked Example: A Tangent Plane to a Cylinder

Worked Example: A Tangent Plane to a Cylinder

Consider the radius-\(3\) cylinder parametrized by \(\mathbf r(\theta,z)=(3\cos\theta,3\sin\theta,z)\). Find the tangent plane at \((\theta_0,z_0)=(\pi/3,2)\).

The point on the cylinder is

$$ P=\mathbf r\left(\frac{\pi}{3},2\right) =\left(\frac{3}{2},\frac{3\sqrt{3}}{2},2\right). $$

The parameter derivatives are \(\mathbf r_\theta=(-3\sin\theta,3\cos\theta,0)\) and \(\mathbf r_z=(0,0,1)\). At the specified parameters,

$$ \mathbf r_\theta=\left(-\frac{3\sqrt{3}}{2},\frac{3}{2},0\right), \qquad \mathbf r_z=(0,0,1). $$

Their cross product is

$$ \mathbf r_\theta\times\mathbf r_z = \left(\frac{3}{2},\frac{3\sqrt{3}}{2},0\right), $$

which is nonzero. Dividing the normal by \(3/2\) gives the simpler normal \((1,\sqrt{3},0)\). Therefore the plane equation is

$$ (x-\tfrac{3}{2})+\sqrt{3}\left(y-\tfrac{3\sqrt{3}}{2}\right)=0, \qquad\text{or}\qquad x+\sqrt{3}y=6. $$

Substitute the point: \(3/2+\sqrt{3}(3\sqrt{3}/2)=3/2+9/2=6\), as required. The normal is perpendicular to the parameter directions: \((1,\sqrt{3},0)\cdot(-3\sqrt{3}/2,3/2,0)=-3\sqrt{3}/2+3\sqrt{3}/2=0\), and \((1,\sqrt{3},0)\cdot(0,0,1)=0\).

Regularity and a Common Cross-Product Mistake

For a cylinder of radius \(a>0\), the usual parametrization \(\mathbf r(\theta,z)=(a\cos\theta,a\sin\theta,z)\) has derivatives \(\mathbf r_\theta=(-a\sin\theta,a\cos\theta,0)\) and \(\mathbf r_z=(0,0,1)\). Their cross product is \((a\cos\theta,a\sin\theta,0)\), which is nonzero because \(a>0\). By contrast, standard spherical coordinates have \(\mathbf r_\theta=\mathbf 0\) at the poles, where \(\sin\phi=0\). The zero cross product there signals that this particular parametrization is singular; it does not by itself show that the sphere lacks a tangent plane.

A frequent error is to compute the cross product in one order and then use the result for the other order. For example, with \(\mathbf a=(1,0,0)\) and \(\mathbf b=(0,1,0)\), the correct calculation is \(\mathbf a\times\mathbf b=(0,0,1)\), while \(\mathbf b\times\mathbf a=(0,0,-1)\). These are opposite vectors. They produce the same plane normal direction, but only if the chosen vector is used consistently in the plane equation.

Another common error is to omit the point \(P\) when writing the plane. A normal vector alone specifies a direction, not the plane's location. The equation must include the displacement from the point of tangency: \(\mathbf n\cdot(\mathbf x-P)=0\).

Takeaway: At a regular point, the parameter derivatives are independent tangent directions. Their cross product is a nonzero normal, and the tangent plane is the plane through the parametrized point with that normal. Always check both the cross-product order and the point used in the plane equation.

Check Your Understanding

Use the parameter derivatives and tangent-plane formulas to answer these questions.

  1. What condition on \(\mathbf r_u(u_0,v_0)\) and \(\mathbf r_v(u_0,v_0)\) makes a point regular for a parametrization?
  2. Why does differentiating a curve on a parametrized surface produce a linear combination of the two parameter derivatives?
  3. If a cross product used as a plane normal is replaced by its negative, does the tangent plane change? Explain.
  4. For \(\mathbf r(\theta,z)=(a\cos\theta,a\sin\theta,z)\), why is the parametrization regular when \(a>0\)?
  5. What does a zero cross product tell you, and what does it not necessarily tell you, about the surface?

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