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Manual for course 26913. Part I: integral calculus
Jos´e M. Gracia-Bond´ıa, Departamento de F´ısica Te´orica
Universidad de Zaragoza, Zaragoza 50009, Spain
April 2, 2016
Abstract If you can, with little difficulty, perform the integral ∫ (^1)
0
tet^ dt,
calculate the partial derivatives of sin(xy + y^2 ), solve linear systems of equations and invert (say) 3 × 3 matrices, then it is likely that you have the prerequisites to attack this course on the multivariate calculus world. More formally, thorough knowledge of linear algebra and one-variable differential and integral calculus will be assumed of the student. In my experience (as a student and as a professor), most people do not follow all the details of a lecture in real time. When you go to a lecture you should expect to witness the big picture of what is going on. You should pay attention to the lecturer’s advice on what is important and what is not. A lecturer spends a long time thinking on how to deliver a presentation of an immense amount of material; they do not expect you to follow every step, but they do expect you to go home and fill in the gaps in your understanding. Not attending lectures really hurts your chances at a deep understanding of the material. Thus I expect you to attend every class, even if there is no formal obligation to do so: it is well nigh impossible to succeeed in this course without attending lessons. Most successful students tend to:
- Attend every class,
- Read textbooks and set of notes and review their own notes daily,
- Work on all the homework as it is assigned,
- Seek help as soon as they encounter trouble.
I encourage you to utilize the review sessions, your classmates and the professor whenever you are having trouble understanding the course material. Get your questions answered as they arise, since waiting until you have many questions (or until the final exam is looming!) will make help in any form less effective. Moreover, it is your responsibility to keep informed of announcements and ad- justments made during scheduled classes. I will try to help you with posting of updates to these notes, policy changes, homeworks, exams...
On the philosophical side: accuracy is important in basic mathematical courses: the student needs to know what he/she is doing. We give their due place thus to conceptual clarity and rigour, even if we do not prove most of the theorems. However, mastery of the techniques of multivariate calculus remains our overriding goal in the first part of the course. For the contents of this manual: a first section, that will not be discussed in class, recalls the formalities of differentiation and Taylor development. The analysis of curves follows, including line integrals almost from the outset, as well as geometric aspects. Next, surface integrals and the Stokes theorem beckon. Triple integrals and the divergence theorem crown the elementary part of the course. An asterisk on an exercise means difficulty. On a section title it means it could be skipped. (Constrained optimization is a “laboratory” for multivariate calculus, involving extremely useful techniques and applications. It constitutes an excellent introduction to its subtleties. In an ideal world we would review it. But it is unlikely that we be able to do so.) A few helpful texts are included in the reference list.
References
[1] M. R. Spiegel, Theory and problems of vector analysis, MacGraw-Hill, New York,
[2] T. M. Apostol, Calculus II, Wiley and Sons, New York, 1969.
[3] A. E. Taylor and W. R. Mann, Advanced calculus, Wiley and Sons, New York, 1983.
[4] J. E. Marsden and A. Weinstein, Calculus III, Springer, New York, 1985.
[5] J. E. Marsden and A. J. Tromba, C´alculo Vectorial, Addison Wesley, M´exico, 1998.
[6] R. C. Wrede and M. R. Spiegel, Advanced calculus, Schaum’s series MacGraw-Hill, New York, 2002.
[7] A. I. Khuri, Advanced calculus with applications in statistics, Wiley and Sons, Hobo- ken, New Jersey, 2003.
[8] C. R. MacCluer, Honors Calculus, Princeton University Press, Princeton, 2006.
[9] M. Moskowitz and F. Paliogiannis, Functions of several real variables, World Scien- tific, Singapore, 2011.
[10] B. Dacorogna and C. Tanteri, Mathematical analysis for engineers, World Scientific, Singapore, 2012.
1 Differentiable functions
Let Rn^3 x = (x 1 ,... , xn) denote a point. Its (Euclidean) norm is
||x|| :=
|x 1 |^2 + · · · + |xn|^2.
The scalar or inner product of x, y ∈ Rn^ is given by
x · y =
∑^ n
i=
xiyi, so that ||x|| =
x · x.
The (Euclidean) distance between to points x, y is the norm of their difference.
- A subset U ⊆ Rn^ is open if for each x ∈ U , there is an > 0 such that all points less distant to x than belong to U.
- A subset C ⊆ Rn^ is closed if its complement is open.
- A subset B ⊆ Rn^ is bounded if for some nonnegative number M and for all points x in B, it holds ||x|| ≤ M.
- Closed and bounded sets are compact.
For example, the solid sphere of radius 1 and its surface boundary are compact sets. The solid solid sphere without boundary is bounded, but it is not compact, since it is open. Any open set is a union of open spheres.
1.1 Continuity
Definition 1. A function f : U ⊆ Rn^ → Rm^ is continuous on U if for any x ∈ U and each > 0 exists δ > 0 such that
||f (x) − f (y)|| ≤
for any U 3 y whose distance to x is less or equal than δ.
Definition 2. A function f : C ⊆ Rn^ → R has a maximum on C if there exists a point x 0 such that f (x 0 ) ≥ f (x) for all x in C.
The maximum f (x 0 ), if it exists, is clearly unique; but it may be achieved at more than a point.
Definition 3. A function f : C ⊆ Rn^ → R has a local maximum at x 1 ∈ C if there exists a δ > 0 such that f achieves a maximum in the intersection of C with the sphere of radius δ around x 1. The maximum is strict if x 6 = x 1 ⇒ f (x) < f (x 1 ).
The definitions of minima are analogous, by reversing inequalities.
Theorem 1. A continuous function on a compact set has a maximum and a mini- mum.
1.2 Differentiability. Directional derivatives
To deal with extremal problems in practice one requires more than continuity.
Definition 4. A function f : U ⊆ Rn^ → Rm, with U open, is (one time) differentiable at x ∈ U if there exists a m×n matrix A, > 0 such that the function of h ≡ (h 1 ,... , hn) with ||h|| < given by ||f (x + h) − f (x) − Ah|| ||h||
tends to zero as ||h|| ↓ 0.
Then the matrix A is unique, and we denote it Df (x) or f ′(x), the (total) derivative of f at x. We say that f (x) + f ′(x)h
is the linear approximation to f at x. Differentiability implies continuity. Most important in practice is the chain rule. Let f : U ⊆ Rn^ → Rp^ and g : W ⊆ Rp^ → Rm^ be differentiable. Then g ◦ f is differentiable. Assume f (x) ∈ W. Then
(g ◦ f )′(x) = g′(f (x))f ′(x).
Let (ei)ki=1 denote the standard basis for Rk. The (linear) coordinate functions (e∗ i )ki= from Rk^ into R are given by e∗ i (x) = xi;
they form the dual basis to that of the (ei)ki=1.
Consider differentiable f : U ⊆ Rn^ → Rm^ again. Note that f (x) has m coordinates f 1 (x),... , fm(x), respectively equal to
e∗ 1
f (x)
, e∗ 2
f (x)
,... , e∗ m
f (x)
Often we identify f ≡ (f 1 ,... , fm)t, with each fi a real function of n variables. For v ∈ Rn, let the directional derivative of f in the direction of v:
Dvf (x) := lim t↓ 0
f (x + tv) − f (x) t
In other words, this is the derivative of the map t 7 → f (x + tv) ∈ Rm, a m × 1 matrix. For v = ei, we speak of partial derivatives and write
∂if (x) :=
∂f ∂xi^
(x) := Dei f (x).
Then ∇fi(x) := grad fi(x) :=
∂ 1 fi(x), ∂ 2 fi(x),... , ∂nfi(x)
)t
is the gradient of fi. Naturally, if f is differentiable all its directional derivatives exist and
f ′(x) =
grad f 1 .. . grad fm
∂ 1 f 1 ∂ 2 f 1 · · · ∂nf 1 .. .
∂ 1 fm ∂ 2 fm · · · ∂nfm
A lot of questions in calculus turn around the relation between graphs and level sets. An example is the fact that every graph is a level set. Indeed,
Example 3. Let as usual f : U ⊆ Rn^ → Rm. Define g : U × Rm^ → Rm^ by
g(x, y) = y − f (x).
Then, clearly,
(x, y) ∈ graph(f ) ⇐⇒ (x, y) ∈ g−^1 (0); this is to say graph(f ) = g−^1 (0) and each graph is a level set.
The “converse” is by no means trivial, and we examine it in following chapters.
1.4 On extrema and the Hessian test
Recall that, for a real function f : U ⊆ Rn^ → R, differentiability at x ∈ U means that it exists a row matrix Df (x), such that the function of h ≡ (h 1 ,... , hn) with ||h|| < given by f (x + h) = f (x) + Df (x)h + δ(h),
where δ(h)/||h|| tends to zero as ||h|| ↓ 0. The directional derivative of f at x in the direction v (with ||v|| = 1) equals v · ∇f (x). Clearly it is greatest when v points in the direction of the gradient and least when it points opposite to the gradient. Obviously, for x 0 to be a local extremum of f differentiable, it needs to be a critical point of f , that is ∇f (x 0 ) = 0. Now, second-order differentiability means
f (x + h) = f (x) + f ′(x)h + 12 Hf (x) + R(h), (1)
where R(h)/||h||^2 tends to zero as ||h|| ↓ 0 —at least if f is differentiable in a neigh- bourhood of x and twice differentiable at x. Here Hf is the quadratic form given by the symmetric Hessian matrix of second derivatives:
Hf (x) = htf ′′(x)h.
This is called the Taylor development of f at x 0 up to second order.
Theorem 2. Let x 0 be a critical point of f. If Hf is positive definite (respectively negative definite), then x 0 is a local minimum of f (respectively maximum). If Hf is indefinite, then x 0 is a saddle point. If Hf is semidefinite, we draw no conclusion.
1.5 The Lagrange formula for the Taylor development
Assuming enough differentiability, more generally than (1) we have, with a notation that ought to be obvious:
f (x + h) = f (x) + f ′(x)h + 12 f ′′(x)(h, h) + · · · +
n!
f (n)(x)(h)(n)^ + R(n)(f ; x; h).
For instance, if f is a function of two real variables,
f (x 1 + h 1 , x 2 + h 2 ) − f (x 1 , x 2 ) = h 1 ∂ 1 f (x 1 , x 2 ) + h 2 ∂ 2 f (x 1 , x 2 ) +
h^21 ∂^21 f (x 1 , x 2 )
- h 1 h 2 ∂ 1 ∂ 2 f (x 1 , x 2 ) +
h^31 ∂^31 f (x 1 , x 2 ) + h^32 ∂ 23 f (x 1 , x 2 )
h^21 h 2 ∂^21 ∂ 2 f (x 1 , x 2 ) + h 1 h^22 ∂ 1 ∂ 22 f (x 1 , x 2 )
For a function of three real variables, taking x = 0 for notational simplicity:
f (h) − f ( 0 ) = h 1 ∂ 1 f ( 0 ) + h 2 ∂ 2 f ( 0 ) + h 3 ∂ 3 f ( 0 ) +
h^21 ∂ 12 f ( 0 ) + h^22 ∂ 22 f ( 0 ) + h^23 ∂^23 f ( 0 )
- 2h 1 h 2 ∂ 1 ∂ 2 f ( 0 ) + 2h 1 h 3 ∂ 1 ∂ 3 f ( 0 ) + 2h 2 h 3 ∂ 2 ∂ 3 f ( 0 )
h^31 ∂^31 f ( 0 ) + h^32 ∂ 23 f ( 0 )
- h^33 ∂^33 f ( 0 ) + 3h^21 h 2 ∂ 12 ∂ 2 f ( 0 ) + 3h 1 h^22 ∂ 1 ∂^22 f ( 0 ) + · · · + 3h 2 h^23 ∂ 2 ∂^23 f ( 0 )
- 6h 1 h 2 h 3 ∂ 1 ∂ 2 ∂ 3 f ( 0 )
The remainder R(n)(f ; x; h) possesses the following exact formula, due to Lagrange:
R(n)(f ; x; h) =
h(n+1) n!
0
(1 − t)nf (n+1)(x + th) dt. (2)
This wonderful formula works when f is n+1 times differentiable with continuous deriva- tives on a sphere of radius ||h|| around x. It deserves proof, at least in a particular case.
Proof. The following identity is valid for functions u of a real variable t with values in Rm:
d/dt
u(t) + (1 − t)u′(t) + 12 (1 − t)^2 u′′(t)
= 12 (1 − t)^2 u′′′(t).
This implies
u(1) = u(0) + u′(0) + 12 u′′(0) + (^12)
0
(1 − t)^2 u′′′(t).
Let now u : [0, 1] → Rm^ be given by t 7 → f (th). We have
u′(t) = f ′(th)h; u′′(t) = f ′(th)h^2 ,
and so on. Thus
f (h) = f ( 0 ) + f ′(th)h + 12 f ′′(th)h^2 +
h^3 2
0
(1 − t)^2 f ′′′(x + th) dt.
It should be clear to the reader how to generalize this argument to any order.
A function is analytic when the Taylor series converges.
Example 4. Prove that differentiable functions are continuous. Indeed, the definition of differentiability implies ∣ ∣ ∣ ∣
f (y) − f (x) ||y − x||
− f ′(x)
for y suitably close to x. Therefore
||f (y) − f (x)|| <
′^ + ||f ′(x)||
||y − x||
is suitably small.
Example 8. Analyze the critical point at the origin for the function f (x, y) = x^2 + y^3. Clearly ∇(x^2 +y^3 )(0, 0) = 0, so we have indeed a critical point. Now ∂yy(x^2 +y^3 ) = 6y vanishes at the origin and ∂xy(x^2 + y^3 ) = 0. Therefore the second derivative test fails. But we can make a concrete analysis. Plot the graph of the function. When z = 0 the level curve is given by y = −x^2 /^3. When x = 0 the level curve is the parabole z = x^2. When y = 0 the level curve is the cubic z = y^3 , with an inflection at the origin. Hence the origin must be a saddle point.
Exercise 4. * Show that the function f : R → R given by
f (x) = 0 for |x| ≥ 1 , f (x) = e−^1 /(1−|x| (^2) ) for |x| ≤ 1 ,
is smooth (that is, indefinitely differentiable) on R, but not analytic.
References
[1] R. Estrada and R. P. Kanwal, Proc. Roy. Soc. London A401 (1985) 281.
[2] K. Drakakis, math.CO/0507439.
2 Curves and line integrals
2.1 Oriented curves
An oriented, parametrized curve C in Rn^ is a smooth map P : [a, b] → Rn^ with the following additional properties and comments.
- a = −∞ or b = ∞ or both may be allowed.
- The velocity P ′(t) 6 = 0 for all t ∈ [a, b].
- P is injective on [a, b[ and ]a, b]. The idea is to avoid self-crossing, although P ((a) = P (b) may be allowed.
- The sense of direction is specified by the vector P ′(a) ∈ Rn. This is particularly important if P (a) = P (b) —a closed curve.
We call ||P ′(t)|| the speed of the curve at “time” t. We may contemplate slightly more complicated cases, like piecewise smooth directed curves, whose components are curves with our definition.
Example 9. Our first example is a curve in the space:
P (t) = cos te 1 + sin te 2 + te 3 ; t ∈ [0, 2 π].
This is a (kink of) a helix, spiralling around a vertical cylinder from (1, 0 , 0) to (1, 0 , 2 π), with speed (^) √
cos^2 t + sin^2 t + 1 =
The first integral we are concerned with in this course is the curve length calculation:
l(C) :=
∫ (^) b
a
||P ′(t)|| dt.
For our example:
l(C) =
∫ (^2) π
0
2 dt = 2
2 π.
A slightly more complicated example: find the length of the curve parametrized by
P (t) = (2 cosh 3t, −2 sinh 3t, 6 t), for t ∈ [0, 5].
We compute:
P ′(t) = (6 sinh 3t, −6 cosh 3t, 6), so that ||P ′(t)|| = 6
2 cosh 3t.
Hence
l(C) = 6
0
cosh 3t = 2
2 sinh 15.
Sometimes it is convenient to parametrize a curve by the length function,
l :=
∫ (^) t
a
||P ′(s)|| ds =: f (t).
Example 13. Find the length of the cardioid
r = a(1 + cos θ).
We naturally parametrize by the angle:
x(θ) = r cos θ = a cos θ + a cos^2 θ; y(θ) = r sin θ = a sin θ + a sin θ cos θ. Then x′(θ) = −a sin θ − 2 a cos θ sin θ; y′(θ) = a cos θ + a cos^2 θ − a sin^2 θ.
We conclude
l =
∫ (^) π
−π
x′(θ)^2 + y′(θ)^2 dθ = a
∫ (^) π
−π
2 + 2 cos θ dθ = 2a
∫ (^) π
−π
cos(θ/2) dθ = 8a.
We have used that cos α is greater than zero for −π/ 2 ≤ α ≤ π/2.
2.2 Vector fields and line integrals
For now, a vector field is simply a function f : Rn^ ⊇ U → Rn. The definition is unsatisfactory in the manifold context, but it will do for now. The concept is uniquitous in physics: think of the flow of a fluid, of an electric current... We have seen already a class of examples: gradients are vector fields. The line integral, of a vector field on a directed curve, is a fundamental tool. The definition is as follows. Let P : [a, b] → C denote as usual a parametrization for its image. Then (^) ∫
C
f =
∫ (^) b
a
f
P (t)
· P ′(t) dt.
The definition appears to be clear: both f
P (t)
and P ′(t) —the latter is a called a vector field along a curve, by the way— are vectors, consequently they can be scalar-multiplied. For the left hand side to be meaningful, however, the definition must be independent of the parametrization used. The reason for this is simple: no matter the parametrization, we can always reduce to the length function, which is unique. Note as well
∫
C
f =
∫ (^) b
a
f
P (t)
· P ′(t) dt =
∫ (^) b
a
f
P (t)
P ′(t) ||P ′(t)||
||P ′(t)|| dt =
∫ (^) b
a
f
P (t)
· T (t) dl(t) =:
C
f · T dl.
Here T is the unit tangent vector to the directed curve, obviously independent of the parametrization. The last equality introduces a notation also often used. Let us gain some experience with this integral. Let n = 3, say. Then if f = (f, g, h) and P (t) =
x(t), y(t), z(t)
C
f :=
∫ (^) b
a
[
f
(x(t), y(t), z(t)
x′(t) + g
(x(t), y(t), z(t)
y′(t)
(x(t), y(t), z(t)
z′(t)
)]
dt =:
C
f dx + g dy + h dz.
The last expression is just a standard notation. The definition does depend on the orientation, and if C˜ is the directed curve obtained from C by changing the orientation, then (^) ∫
C
f = −
C^ ˜
f.
Example 14. Evaluate ∫
C
y^2 dx + x^2 dy where C is the upper half of the ellipse
x(t) = a cos t; y(t) = b sin t, going clockwise.
We compute ∫
C
y^2 dx + x^2 dy =
π
(−ab^2 sin^3 t + a^2 b cos^3 t)dt =
4 ab^2 3
Example 15. Evaluate (^) ∫
C
xy dx + xz^2 dy + xyz dz,
where C is parametrized by (t, t^2 , t^3 ), 0 ≤ t ≤ 1. Clearly, we have to compute: ∫ (^1)
0
(t^3 , t^7 , t^6 ) · (1, 2 t, 3 t^2 ) dt =
0
(t^3 + 5t^8 ) dt =
Example 16. Evaluate
C f^ , where^ C^ is a curl of a helix parametrized by (sin^ t,^ cos^ t, t) between t = 0 and t = 2π and f = (x, y, z). This is quite simple: ∫ (^2) π
0
(sin t cos t − sin t cos t + t) dt = 2π^2.
It is also possible to associate a line integral with scalar functions, by the simple trick of associating to one of them f the vector field f T. Thus ∫
C
f =
∫ (^) b
a
f
P (t)
||P ′(t)|| dt =:
C
f dl.
The last notation is most often used in this case.
Let now ponder the special vector field f = ∇g. The real function g is called a scalar potential for f. If we integrate such a vector field over a directed curve, ∫
C
∇g :=
∫ (^) b
a
∇g
P (t)
· P ′(t) dt =
∫ (^) b
a
d dt
g ◦ P
(t) dt = g(P (b)) − g(P (a)).
This independence of the path has a host of consequences. To begin with, the integral of a gradient on a closed curve vanishes. The converse is true: if U is connected and if
C 1 f^ =^
C 2 f^ for any two (piecewise smooth, directed) curves^ C^1 , C^2 , then^ f^ is a gradient. This is helpful to compute the potential; the trouble is that verifying it is well nigh impossible. What is needed is a criterion readily allowing to distinguish gradients from other vector fields.
Again, ∂z g = xyez^ + φ′(z).
This means that φ is a constant c, and we have shown
g(x, y, z) = xyez^ + c,
which is the general solution.
Example 19. Find, if they exist, scalar potentials for the vector fields:
- f (x, y, z) = (xy, yz, zx).
- f (x, y, z) =
2 x cos(x^2 + yz), z cos(x^2 + yz), y cos(x^2 + yz)
Clearly (xy, yz, zx) does not have a potential, since for instance
∂y(xy) = y 6 = 0 = ∂x(yz).
For the other item, we have ∫ 2 x cos(x^2 + yz) dx = sin(x^2 + yz) + ψ(y, z)
Differentiating this with respect to y, z, it is seen at once that one must take ψ equal to any constant.
Exercise 5. Let F be the gravitational field of force of a particle at x = (x, y, z) due to a unit mass at the origin:
F (x, y, z) =
(x, y, z) ||x||^3
Prove that the work done by the particle in moving from x 1 to x 2 depends only from r 1 ≡ ||x 1 || and r 2 ≡ ||x 2 ||.
Exercise 6. Compute the work done by the field of force
F (x, y, z) = (2xy^3 z^4 , 3 x^2 y^2 z^4 , 4 x^2 y^3 z^3 ).
in moving a particle from (0, 0 , 0) to (1, 1 , 1) along the curve given by the intersection of the surfaces: 2 z^3 = x^3 + y^3 ; 4 z = x^2 + 2x^4 + y^2.
Remark 1. On R^3 the condition that the vector field f derive from a potential g can be rewritten as ∇ × f =: curl f =: rot f = 0. (4)
We can extract much more mileage from the ∇ operator yet. A natural idea, valid in any dimension is to explore the meaning of ∇ · f , that is
∇ · f = (∂ 1 ,... , ∂n) · (f 1 ,... , fn) =
i
∂fi ∂xi^
=: div f.
This is called the divergence of f. However, a more intrisic view is to recall that the total derivative Df of f is the matrix
Df =
∂f 1 ∂x 1 · · ·^
∂f 1 ∂xn .. .
∂fn ∂x 1 · · ·^
∂fn ∂xn
Then by definition the divergence div f of f is the trace of Df. The trace has well-known invariance properties. The divergence of a gradient vector field is a scalar function, called the Laplacian ∆g of the original scalar function g.
∆g := div grad g = ∇ · (∇g) = tr(D grad g).
2.2.1 Magnetic monopoles
Let us return to (4). It is well known, and alredy dealt with in the exercises, that f (x) = x/||x||^3 has a scalar potential, given by − 1 /||x|| on R^3 \ { 0 }. It is less well- known that it also has a vector potential, although defined only outside an axis —or a half-axis. The assertion is that, with
A(x, y, z) =
(yz, −xz, 0) (x^2 + y^2 )r
we have ∇ × A =
x r^3
Indeed, if we write ∇ × A = (ϕ 1 , ϕ 2 , ϕ 3 ), we find
ϕ 1 =
x (x^2 + y^2 )
∂z
(z r
x (x^2 + y^2 )
r − z^2 /r r^2
x r^3
ϕ 2 =
y r^3
similarly;
ϕ 3 = −z
∂x
x (x^2 + y^2 )r
y (x^2 + y^2 )r
= −z
2(x^2 + y^2 )r − 2(x^2 + y^2 )r − (x^2 + y^2 )^2 /r (x^2 + y^2 )^2 r^2
z r^3
In fact, this formulation is pertinent if (^) rx 3 is thought of as a magnetic field: here we are dealing with the magnetic monopole problem.
Example 20. Verify that A′(x, y, z) = (−(ry,x,+z)0)r is also a good potential vector for (^) rx 3 on the appropriate domain. What could the relation between A and A′^ be?
2.3 Theory of curves in R^2
We use a unit speed (arc length) parametrization now. Let then P (t) =
x(t), y(t)
for t ∈ [a, b] denote the directed curve C, with unit tangent
T (t) ≡
x′(t), y′(t)
There are just two unit vectors in R^2 perpendicular to them. The reference unit nor- mal N (t) to C at the point P (t) is obtained by rotating T by π/2 in the anticlockwise direction (that is, to the left with respect to the direction of motion along C):
N (t) =
− y′(t), x′(t)
Differentiating the identity T (t) · T (t) = 1 we obtain 0 = T (t) · T ′(t), implying
T ′(t) = κ(t)N (t),
for some scalar κ(t), which is called the curvature of C at P (t). We have a formula for it:
κ(t) = T ′(t) · N (t) = y′′(t)x′(t) − x′′(t)y′(t). (5)
The absolute curvature of C at P (t) is given by
|κ(t)| = ||T ′(t)|| = ||P ′′(t)||.
Example 21. Prove that, with an arbitrary parametrization (maybe by the time in a physical situation), one has
κ(t) =
y′′(t)x′(t) − x′′(t)y′(t) ||T (t)||^3
y′′(t)x′(t) − x′′(t)y′(t) (( x′(t)
y′(t)
) 2 ) 3 / 2.^ (6)
Hint. Let P : [a, b] → R denote now the general parametrization. Then P (t(l)) is a unit speed parametrization, and recall that
dt dl
||P ′(t)||
x′(t)
y′(t)
so that by the chain rule,
x′(l) =
dx dt
dt dl
= x′(t)/
x′(t)
y′(t)
; y′(l) = y′(t)/
x′(t)
y′(t)
Then keep derivating —maybe with an eye on Exercise 2 in Chapter 1— and substitute the results in (5) with t replaced now by l. (You may as well consult the solution to the following problem, in which the whole Frenet–Serret apparatus is developed for general parametrizations. Just take into account that in three dimensions the curvature is always positive by definition, while here according to our conventions it can be negative or zero.)
An important subcase is when C is the graph of a function f : [a, b] → R directed from left to right, so that P (t) = (t, f (t)). The curvature is then clearly given by
κ(t) =
f ′′(t) [ 1 +
f ′(t)
) 2 ]^3 /^2.^ (7)
It should be clear, by constructing the (best) quadratic approximation to the curve at a given point —see the next section— that |κ(t)| is the inverse of the radius of the circle of curvature at P (t). The sign of κ determines whether the circle of curvature and the normal are on the same side (when positive) or on opposite sides.
2.4 The Frenet–Serret equations ∗
Matters become trickier for curves in Rn, for n > 2. Then there are an infinite number of sides to a curve C. The policy is to define a normal associated intrinsically to the curve. We decide on
N (t) =
T ′(t) ||T ′(t)||
This is undefined if the curvature κ(t), now defined as ||T ′(t)||, is zero. That is, now we have the same equation as in the previous section,
T ′(t) = κ(t)N (t), (8)
with a different meaning. For instance, straight lines on the plane do have a normal; but on R^3 do not. The plane closest to the curve near P (t) is called the osculating plane. As the point P (t) moves, the osculating plane may change, the more so the more “twisted” is the curve. Let us consider curves on R^3. Then there are precisely two unit vectors perpendicular to the unit vectors T , N. We pick one of them, called the binormal B, by choosing the standard orientation in three-dimensional space for
T , N , B
. That is to say, B(t) := T (t) × N (t).
The reader is supposed familiar with the main properies of the vector product. The variable orthonormal basis
T , N , B
, adapted to studying the curve, is called a moving frame along C. It is easy to verify that
N (t) × B(t) = T (t).
Indeed,
N (t)×B(t) := N (t)×
T (t))×N (t)
= T (t)
N (t))·N (t)
−N (t)
N (t)·T (t)
= T (t).
We may now use the orthonormal basis (T , N , B) to expand any vector v ∈ R^3 in terms of it: v =
v · T (t)
T (t) +
v · N (t)
N (t) +
v · B(t)
B(t). (9)
