A parametric surface \mathbf{r}(u,v) traces out a 2D sheet in space. It does this as two parameters, u and v, vary. A normal vector is a vector that points straight out of a surface. It is perpendicular to the surface, at a given point. This surface has two partial-derivative vectors, \mathbf{r}_u and \mathbf{r}_v. Together, they span the tangent plane at each point. Their cross product gives exactly such a normal vector. Chapter 06 rebuilds this same idea using a wedge product, and chapter 07 integrates a 2-form across it.
Picture two threads laid across a curved bedsheet. One thread runs "u-ward," and the other runs "v-ward." Together, they pin down a small parallelogram patch of the sheet, near where they cross. The cross product of those two thread directions points straight out of the sheet there. Its length measures how big that little patch actually is.
Now swap the order of the two threads. The arrow flips, and points out the other side of the sheet instead. Which side counts as "outward" is a choice this course has to track. That choice matters once flux enters the picture — flux is how much stuff flows through a surface, not just across it.
A parametric surface is a map \mathbf{r}:D\to\mathbb{R}^3 from a region D of the (u,v)-plane, \mathbf{r}(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle, sweeping out a 2D sheet as (u,v) ranges over D. Four facts about it carry the rest of this course:
- The partial derivatives \mathbf{r}_u=\partial\mathbf{r}/\partial u
and \mathbf{r}_v=\partial\mathbf{r}/\partial v are tangent vectors,
spanning the tangent plane at each point (the
:::recallbelow). - A normal vector is a vector perpendicular to a surface at a given point. Their cross product \mathbf{r}_u\times\mathbf{r}_v is perpendicular to both \mathbf{r}_u and \mathbf{r}_v — exactly such a normal vector to the surface itself, not just a direction check.
- Its length, |\mathbf{r}_u\times\mathbf{r}_v|, is the local area-scaling factor, exactly the Jacobian idea from section 03 applied to a 2D surface sitting inside 3D space instead of a flat coordinate change. In three lines: a small rectangle du\times dv in the parameter plane maps, to first order, to the parallelogram spanned by \mathbf{r}_u\,du and \mathbf{r}_v\,dv; a parallelogram spanned by two vectors has area equal to the length of their cross product; so that patch's area is |\mathbf{r}_u\times\mathbf{r}_v|\, du\,dv. Summed over D, that's dS=|\mathbf{r}_u\times\mathbf{r}_v| \,du\,dv, the surface-area element.
- The order of the cross product — \mathbf{r}_u\times\mathbf{r}_v versus \mathbf{r}_v\times\mathbf{r}_u — fixes which of the two possible normal directions counts as "outward." Swapping the order flips the sign of dS, the bookkeeping chapter 06 formalizes as orientation.
Formula
Tangent Plane to a Parametric Surface
also known as: the normal vector of a parametric surface
Recalled from Calculus III — Multivariable Calculus · Tangent planes to parametric surfaces — first taught there.
For a parametric surface \mathbf{r}(u,v) with \mathbf{r}_u and \mathbf{r}_v both defined and \mathbf{r}_u\times\mathbf{r}_v \neq \mathbf{0} at (u_0,v_0), the tangent plane at \mathbf{r}(u_0,v_0) has normal vector \mathbf{n} = \mathbf{r}_u(u_0,v_0)\times\mathbf{r}_v(u_0,v_0).
Hold v=v_0 fixed and let u vary: \mathbf{r}(u,v_0) traces a curve lying entirely on the surface and passing through \mathbf{r}(u_0,v_0), with velocity vector \mathbf{r}_u(u_0,v_0) — so \mathbf{r}_u is tangent to the surface there. The same argument with u=u_0 fixed shows \mathbf{r}_v is tangent to the surface too, along the perpendicular grid direction. Two non-parallel vectors tangent to the surface at the same point span its entire tangent plane, and the cross product of two vectors spanning a plane is normal to that plane — so \mathbf{r}_u\times\mathbf{r}_v is a normal vector, and the tangent plane is the plane through \mathbf{r}(u_0,v_0) with that normal.
r_u and r_v span the tangent plane
Drag the (u,v) point around -- watch r_u and r_v change length and direction, and their cross product (the normal) change with them.
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The normal vector, isolated
Set u to <1, 0, 2> and v to <0, 1, 0> (a typical r_u, r_v pair) -- read off u x v as the normal direction the surface widget above draws at its own marked point.
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Worked example
Normal vector to a parametric surface
-
1
Find the normal vector to \mathbf{r}(u,v)=\langle u, v, u^2+v^2\rangle at (u,v)=(1,0).
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2
\mathbf{r}_u = \langle 1, 0, 2u\rangle, \mathbf{r}_v = \langle 0, 1, 2v\rangle. At (1,0): \mathbf{r}_u=\langle 1,0,2\rangle, \mathbf{r}_v=\langle 0,1,0\rangle.
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3
\mathbf{r}_u\times\mathbf{r}_v = \langle (0)(0)-(2)(1),\ (2)(0)-(1)(0),\ (1)(1)-(0)(0)\rangle.
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4
\boxed{\mathbf{r}_u\times\mathbf{r}_v = \langle -2, 0, 1\rangle}
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Where this goes next
Chapters 6 and 7 of this course
\mathbf{r}_u\times\mathbf{r}_v\,du\,dv IS the 2-form dS chapter 06 builds from scratch out of wedge products — this section's normal-vector construction is the concrete computation chapter 06's abstraction is designed to reproduce exactly.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
For \mathbf{r}(u,v) = \langle u,\ v,\ u^2-v^2\rangle, find the z-coordinate at (u,v)=(3,2).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Substitute u=3,v=2: z=u^2-v^2=9-4=5.
Correct answer
Also accepted:
Why
Substitute u=3 and v=2 into z=u^2-v^2 and simplify.
From Calculus III — Multivariable Calculus
Holding v=v_0 fixed and letting u vary in \mathbf{r}(u,v) traces...
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
It's a smooth curve, not a random scatter — a single parameter still varies continuously.
That's the tangent PLANE, a different (and later) object built from BOTH grid directions at once.
Fixing one parameter only traces a curve lying on the surface, not the whole 2D sheet.
Why
Fixing v and letting u vary traces exactly one grid curve lying on the surface, the u-grid curve through that point.
Only one parameter is changing here, so the result has to be a curve, not a 2D patch.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
At a point where \mathbf{r}_u and \mathbf{r}_v happen to be parallel, the surface...
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Parallel tangent vectors give a zero cross product, not a large one — the opposite of what's suggested here.
One singular point says nothing about the shape of the rest of the surface.
Parallel \mathbf{r}_u,\mathbf{r}_v give \mathbf{r}_u\times\mathbf{r}_v=\mathbf{0}, which can't serve as a normal vector.
Why
If \mathbf{r}_u and \mathbf{r}_v are parallel, their cross product is the zero vector, so no normal — and hence no tangent plane — is defined there, the parametric analogue of a non-differentiable point.
Recall what the cross product of two parallel vectors always equals.
From Calculus III — Multivariable Calculus
For the cone \mathbf{r}(u,v)=\langle u\cos v,\ u\sin v,\ u\rangle, the chapter derived |\mathbf{r}_u\times\mathbf{r}_v| = u\sqrt2. Find its value at u=3.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
|\mathbf{r}_u\times\mathbf{r}_v| = u\sqrt2; at u=3 this is 3\sqrt2\approx4.24.
Correct answer
Also accepted:
Why
Substitute u=3 into the formula u\sqrt2 and evaluate numerically.