Reference sheet

Multivariable and Vector Calculus reference sheet

The main definitions, formulas and conditions of the multivariable and vector calculus course on one page, unit by unit, with links to the lessons that explain each topic in full.

This sheet follows the six units of the course in order. Each unit starts with the big idea in a few sentences, then gives the definitions, formulas and theorems you will look up most often, with the conditions they need. When you want the full explanation, a proof or more examples, follow the lesson links under each topic.

Coordinates in space are right-handed and angles are in radians. Spherical coordinates are written (ρ,θ,ϕ)(\rho, \theta, \phi), with the azimuth θ\theta measured in the xyxy-plane from the positive xx-axis and the polar angle ϕ\phi measured from the positive zz-axis. A formula holds wherever both sides are defined unless a condition says otherwise.

Vectors and the Geometry of Space

Multivariable calculus happens in space, so it starts with a precise way to describe points, directions and flat or curved surfaces. This unit sets up coordinates and vectors, the dot and cross products that measure angles, areas and volumes, the matrices that describe linear maps, and the equations of lines, planes, quadric surfaces and round coordinate systems.

Points and vectors

The axes are right-handed. Curling the fingers of the right hand from the positive xx-axis toward the positive yy-axis leaves the thumb along the positive zz-axis. With Δx=x2−x1\Delta x = x_2 - x_1, Δy=y2−y1\Delta y = y_2 - y_1 and Δz=z2−z1\Delta z = z_2 - z_1, the distance between P1(x1,y1,z1)P_1(x_1, y_1, z_1) and P2(x2,y2,z2)P_2(x_2, y_2, z_2) is

d=(Δx)2+(Δy)2+(Δz)2,d = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2},

and the sphere with center (h,k,l)(h, k, l) and radius r>0r > 0 is

(x−h)2+(y−k)2+(z−l)2=r2.\begin{aligned} &(x - h)^2 + (y - k)^2 \\[0.4em] &\quad + (z - l)^2 = r^2. \end{aligned}

An equation x2+y2+z2+Dx+Ey+Fz+G=0x^2 + y^2 + z^2 + Dx + Ey + Fz + G = 0 becomes this form after completing the square in each variable, and it is a sphere, a single point or empty as the right side comes out positive, zero or negative. For example x2+y2+z2−4x+6y−2z=2x^2 + y^2 + z^2 - 4x + 6y - 2z = 2 is the sphere with center (2,−3,1)(2, -3, 1) and radius 44.

The vector from A(x1,y1,z1)A(x_1, y_1, z_1) to B(x2,y2,z2)B(x_2, y_2, z_2) is AB→=⟨x2−x1,y2−y1,z2−z1⟩\overrightarrow{AB} = \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle, and a vector a=⟨a1,a2,a3⟩\mathbf{a} = \langle a_1, a_2, a_3 \rangle equals a1i+a2j+a3ka_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} in the standard basis.

QuantityFormulaConditions
Magnitude∣a∣=a12+a22+a32\lvert \mathbf{a} \rvert = \sqrt{a_1^2 + a_2^2 + a_3^2}zero only for a=0\mathbf{a} = \mathbf{0}
Suma+b=⟨a1+b1,a2+b2,a3+b3⟩\mathbf{a} + \mathbf{b} = \langle a_1 + b_1, a_2 + b_2, a_3 + b_3 \rangletail of b\mathbf{b} at the head of a\mathbf{a}
Scalar multiplec a=⟨ca1,ca2,ca3⟩c\,\mathbf{a} = \langle c a_1, c a_2, c a_3 \rangle with length ∣c∣ ∣a∣\lvert c \rvert \, \lvert \mathbf{a} \rvertreversed direction when c<0c < 0
Unit vectora∣a∣\dfrac{\mathbf{a}}{\lvert \mathbf{a} \rvert}a≠0\mathbf{a} \ne \mathbf{0}
Parallelb=c a\mathbf{b} = c\,\mathbf{a} for a scalar cca\mathbf{a} and b\mathbf{b} nonzero

Forces add as vectors, and an object is in equilibrium when the resultant of all forces on it is 0\mathbf{0}.

Lessons Three-dimensional coordinates and vectors

The dot product

For a=⟨a1,a2,a3⟩\mathbf{a} = \langle a_1, a_2, a_3 \rangle and b=⟨b1,b2,b3⟩\mathbf{b} = \langle b_1, b_2, b_3 \rangle, and with θ\theta in [0,π][0, \pi] the angle between them when both are nonzero,

a⋅b=a1b1+a2b2+a3b3=∣a∣ ∣b∣cos⁡θ.\begin{aligned} \mathbf{a} \cdot \mathbf{b} &= a_1 b_1 + a_2 b_2 + a_3 b_3 \\[0.4em] &= \lvert \mathbf{a} \rvert \, \lvert \mathbf{b} \rvert \cos\theta. \end{aligned}

The product is commutative and distributive, and a⋅a=∣a∣2\mathbf{a} \cdot \mathbf{a} = \lvert \mathbf{a} \rvert^2.

QuantityFormulaConditions
Anglecos⁡θ=a⋅b∣a∣ ∣b∣\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{\lvert \mathbf{a} \rvert \, \lvert \mathbf{b} \rvert}a\mathbf{a}, b\mathbf{b} nonzero, 0≤θ≤π0 \le \theta \le \pi
Orthogonala⋅b=0\mathbf{a} \cdot \mathbf{b} = 00\mathbf{0} is orthogonal to every vector
Direction cosinescos⁡α=a1∣a∣\cos\alpha = \dfrac{a_1}{\lvert \mathbf{a} \rvert}, cos⁡β=a2∣a∣\cos\beta = \dfrac{a_2}{\lvert \mathbf{a} \rvert}, cos⁡γ=a3∣a∣\cos\gamma = \dfrac{a_3}{\lvert \mathbf{a} \rvert}a≠0\mathbf{a} \ne \mathbf{0}, squares add to 11
Scalar projectioncomp⁡ab=a⋅b∣a∣\operatorname{comp}_{\mathbf{a}} \mathbf{b} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{\lvert \mathbf{a} \rvert}a≠0\mathbf{a} \ne \mathbf{0}, negative for an obtuse angle
Vector projectionproj⁡ab=a⋅b∣a∣2 a\operatorname{proj}_{\mathbf{a}} \mathbf{b} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{\lvert \mathbf{a} \rvert^2}\,\mathbf{a}a≠0\mathbf{a} \ne \mathbf{0}
WorkW=F⋅PQ→W = \mathbf{F} \cdot \overrightarrow{PQ}constant force, straight motion from PP to QQ

The remainder b−proj⁡ab\mathbf{b} - \operatorname{proj}_{\mathbf{a}} \mathbf{b} is orthogonal to a\mathbf{a}. Projecting b=⟨1,1,2⟩\mathbf{b} = \langle 1, 1, 2 \rangle onto a=⟨1,0,1⟩\mathbf{a} = \langle 1, 0, 1 \rangle gives ⟨32,0,32⟩\left\langle \frac{3}{2}, 0, \frac{3}{2} \right\rangle, and the remainder has length 62\frac{\sqrt{6}}{2}, the distance from (1,1,2)(1, 1, 2) to the line through the origin along a\mathbf{a}.

The Cauchy-Schwarz inequality ∣a⋅b∣≤∣a∣ ∣b∣\lvert \mathbf{a} \cdot \mathbf{b} \rvert \le \lvert \mathbf{a} \rvert \, \lvert \mathbf{b} \rvert holds for all vectors, with equality exactly when one vector is a scalar multiple of the other. It implies the triangle inequality ∣a+b∣≤∣a∣+∣b∣\lvert \mathbf{a} + \mathbf{b} \rvert \le \lvert \mathbf{a} \rvert + \lvert \mathbf{b} \rvert.

Lessons The dot product

The cross product

For vectors in space, the cross product is the vector with the components listed below as a column,

a×b=[a2b3−a3b2a3b1−a1b3a1b2−a2b1].\mathbf{a} \times \mathbf{b} = \begin{bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{bmatrix}.

It is perpendicular to both factors, b×a=−(a×b)\mathbf{b} \times \mathbf{a} = -(\mathbf{a} \times \mathbf{b}), it is distributive, and it is not associative. In right-handed coordinates i×j=k\mathbf{i} \times \mathbf{j} = \mathbf{k}, j×k=i\mathbf{j} \times \mathbf{k} = \mathbf{i} and k×i=j\mathbf{k} \times \mathbf{i} = \mathbf{j}. For nonzero a\mathbf{a} and b\mathbf{b} with angle θ\theta,

∣a×b∣=∣a∣ ∣b∣sin⁡θ,\lvert \mathbf{a} \times \mathbf{b} \rvert = \lvert \mathbf{a} \rvert \, \lvert \mathbf{b} \rvert \sin\theta,

and when they are not parallel the product points where the right thumb points as the fingers curl from a\mathbf{a} toward b\mathbf{b}. Two nonzero vectors are parallel exactly when their cross product is 0\mathbf{0}.

QuantityFormulaConditions
Parallelogram area∣a×b∣\lvert \mathbf{a} \times \mathbf{b} \rvertadjacent sides a\mathbf{a} and b\mathbf{b}
Triangle area12∣a×b∣\dfrac{1}{2}\lvert \mathbf{a} \times \mathbf{b} \rvertsides a\mathbf{a} and b\mathbf{b} from one vertex
Torque about OOτ=r×F\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}r\mathbf{r} from OO to where F\mathbf{F} acts
Scalar triple producta⋅(b×c)=det⁡\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \det of the matrix with rows a\mathbf{a}, b\mathbf{b}, c\mathbf{c}unchanged by a cyclic shift, sign change on a swap
Parallelepiped volume∣a⋅(b×c)∣\lvert \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \rvertedges a\mathbf{a}, b\mathbf{b}, c\mathbf{c} from one vertex
Tetrahedron volume16∣a⋅(b×c)∣\dfrac{1}{6}\lvert \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \rvertthe same three edges

The triple product is zero exactly when the three vectors are coplanar, positive when (a,b,c)(\mathbf{a}, \mathbf{b}, \mathbf{c}) is a right-handed triple and negative when it is left-handed. For the triangle with vertices P(1,0,0)P(1, 0, 0), Q(0,2,0)Q(0, 2, 0) and R(0,0,3)R(0, 0, 3), the product PQ→×PR→=⟨6,3,2⟩\overrightarrow{PQ} \times \overrightarrow{PR} = \langle 6, 3, 2 \rangle has length 77, so the area is 72\frac{7}{2} and the plane through the three points is 6x+3y+2z=66x + 3y + 2z = 6.

Lessons The cross product

Matrices, linear maps and determinants

An m×nm \times n matrix AA with columns a1,…,an\mathbf{a}_1, \dots, \mathbf{a}_n acts on x\mathbf{x} in Rn\mathbb{R}^n by

Ax=x1a1+⋯+xnan.A\mathbf{x} = x_1 \mathbf{a}_1 + \cdots + x_n \mathbf{a}_n.

Every linear map is multiplication by exactly one matrix, whose jjth column is the image of ej\mathbf{e}_j. The product ABAB is defined when AA has as many columns as BB has rows, it applies BB first and then AA, so (AB)x=A(Bx)(AB)\mathbf{x} = A(B\mathbf{x}), and in general AB≠BAAB \ne BA. A square matrix is invertible exactly when det⁡A≠0\det A \ne 0, and then Ax=bA\mathbf{x} = \mathbf{b} has the single solution x=A−1b\mathbf{x} = A^{-1}\mathbf{b}. For ad−bc≠0ad - bc \ne 0,

[abcd]−1=1ad−bc[d−b−ca].\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.

A 3×33 \times 3 determinant expands along the first row as

det⁡A=a11(a22a33−a23a32)−a12(a21a33−a23a31)+a13(a21a32−a22a31).\begin{aligned} \det A &= a_{11}(a_{22}a_{33} - a_{23}a_{32}) \\[0.4em] &\quad - a_{12}(a_{21}a_{33} - a_{23}a_{31}) \\[0.4em] &\quad + a_{13}(a_{21}a_{32} - a_{22}a_{31}). \end{aligned}

The rules det⁡(AB)=det⁡Adet⁡B\det(AB) = \det A \det B and det⁡AT=det⁡A\det A^{T} = \det A hold for square matrices of one size. Swapping two rows changes the sign, scaling one row by kk scales the determinant by kk, and adding a multiple of one row to another changes nothing. The same holds for columns.

  • Area and volume. A 2×22 \times 2 matrix multiplies every area by ∣det⁡A∣\lvert \det A \rvert, and a 3×33 \times 3 matrix multiplies every volume by ∣det⁡A∣\lvert \det A \rvert.
  • Orientation. An invertible map preserves orientation when det⁡A>0\det A > 0 and reverses it when det⁡A<0\det A < 0. An ordered basis (v1,v2,v3)(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3) is right-handed when v1⋅(v2×v3)>0\mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3) > 0.
  • Change of basis. If the columns of an invertible PP form a basis B\mathcal{B}, the coordinates of x\mathbf{x} are [x]B=P−1x[\mathbf{x}]_{\mathcal{B}} = P^{-1}\mathbf{x}, and the map with standard matrix AA has matrix P−1APP^{-1}AP in those coordinates.

For example AA with rows ⟨2,1⟩\langle 2, 1 \rangle and ⟨1,2⟩\langle 1, 2 \rangle and PP with columns ⟨1,1⟩\langle 1, 1 \rangle and ⟨1,−1⟩\langle 1, -1 \rangle give P−1AP=diag⁡(3,1)P^{-1}AP = \operatorname{diag}(3, 1), so the map stretches by 33 along ⟨1,1⟩\langle 1, 1 \rangle and fixes ⟨1,−1⟩\langle 1, -1 \rangle.

Lessons Matrices, linear maps and determinants

Lines and planes

The line through the point with position vector r0=⟨x0,y0,z0⟩\mathbf{r}_0 = \langle x_0, y_0, z_0 \rangle with direction v=⟨a,b,c⟩≠0\mathbf{v} = \langle a, b, c \rangle \ne \mathbf{0} is

r(t)=r0+tv,t∈R.\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}, \qquad t \in \mathbb{R}.

When aa, bb and cc are all nonzero it has the symmetric form

x−x0a=y−y0b=z−z0c.\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}.

A zero component drops its fraction, so b=0b = 0 with a,c≠0a, c \ne 0 gives y=y0y = y_0 together with x−x0a=z−z0c\frac{x - x_0}{a} = \frac{z - z_0}{c}. The plane through r0\mathbf{r}_0 with normal n=⟨a,b,c⟩≠0\mathbf{n} = \langle a, b, c \rangle \ne \mathbf{0} is

n⋅(r−r0)=0,that isax+by+cz=d\begin{gathered} \mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0, \\[0.4em] \text{that is} \quad ax + by + cz = d \end{gathered}

with d=n⋅r0d = \mathbf{n} \cdot \mathbf{r}_0. Three points PP, QQ, RR not on one line lie in exactly one plane, with normal PQ→×PR→\overrightarrow{PQ} \times \overrightarrow{PR}.

QuestionFormulaConditions
Angle between linescos⁡θ=∣v1⋅v2∣∣v1∣ ∣v2∣\cos\theta = \dfrac{\lvert \mathbf{v}_1 \cdot \mathbf{v}_2 \rvert}{\lvert \mathbf{v}_1 \rvert \, \lvert \mathbf{v}_2 \rvert}0≤θ≤π20 \le \theta \le \dfrac{\pi}{2}, the same with normals for planes
Angle between a line and a planesin⁡φ=∣v⋅n∣∣v∣ ∣n∣\sin\varphi = \dfrac{\lvert \mathbf{v} \cdot \mathbf{n} \rvert}{\lvert \mathbf{v} \rvert \, \lvert \mathbf{n} \rvert}0≤φ≤π20 \le \varphi \le \dfrac{\pi}{2}
Line meets the plane n⋅r=d\mathbf{n} \cdot \mathbf{r} = dt=d−n⋅r0n⋅vt = \dfrac{d - \mathbf{n} \cdot \mathbf{r}_0}{\mathbf{n} \cdot \mathbf{v}}n⋅v≠0\mathbf{n} \cdot \mathbf{v} \ne 0, otherwise parallel
Two planes meet in a linedirection n1×n2\mathbf{n}_1 \times \mathbf{n}_2normals not parallel
Point to planeD=∣ax1+by1+cz1−d∣a2+b2+c2D = \dfrac{\lvert a x_1 + b y_1 + c z_1 - d \rvert}{\sqrt{a^2 + b^2 + c^2}}point (x1,y1,z1)(x_1, y_1, z_1)
Skew linesD=∣(r2−r1)⋅(v1×v2)∣∣v1×v2∣D = \dfrac{\lvert (\mathbf{r}_2 - \mathbf{r}_1) \cdot (\mathbf{v}_1 \times \mathbf{v}_2) \rvert}{\lvert \mathbf{v}_1 \times \mathbf{v}_2 \rvert}v1×v2≠0\mathbf{v}_1 \times \mathbf{v}_2 \ne \mathbf{0}

Two lines are skew when they are not parallel and do not meet, which happens exactly when v1×v2≠0\mathbf{v}_1 \times \mathbf{v}_2 \ne \mathbf{0} and (r2−r1)⋅(v1×v2)≠0(\mathbf{r}_2 - \mathbf{r}_1) \cdot (\mathbf{v}_1 \times \mathbf{v}_2) \ne 0. For example the point (1,2,3)(1, 2, 3) lies at distance ∣2−2+6−1∣3=53\frac{\lvert 2 - 2 + 6 - 1 \rvert}{3} = \frac{5}{3} from the plane 2x−y+2z=12x - y + 2z = 1.

Lessons Equations of lines and planes

Cylinders and quadric surfaces

A trace is the intersection of a surface with a plane, found by fixing one coordinate. An equation that does not contain zz is a cylinder with rulings parallel to the zz-axis, and the same holds for the other variables. With a,b,c>0a, b, c > 0, the standard quadric surfaces with axis along the zz-axis are

SurfaceEquationTraces in z=kz = k
Ellipsoidx2a2+y2b2+z2c2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1ellipses for ∣k∣<c\lvert k \rvert < c, points at k=±ck = \pm c
Elliptic paraboloidzc=x2a2+y2b2\dfrac{z}{c} = \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}ellipses where kc>0\dfrac{k}{c} > 0, here with any c≠0c \ne 0
Hyperbolic paraboloidzc=x2a2−y2b2\dfrac{z}{c} = \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2}hyperbolas for k≠0k \ne 0, crossing lines for k=0k = 0, here with any c≠0c \ne 0
Elliptic conex2a2+y2b2=z2c2\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \dfrac{z^2}{c^2}ellipses for k≠0k \ne 0, the vertex for k=0k = 0
Hyperboloid of one sheetx2a2+y2b2−z2c2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{z^2}{c^2} = 1ellipses for every kk
Hyperboloid of two sheetsz2c2−x2a2−y2b2=1\dfrac{z^2}{c^2} - \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1ellipses for ∣k∣>c\lvert k \rvert > c, empty for ∣k∣<c\lvert k \rvert < c

To identify an equation without cross terms, first look for a missing variable, which means a cylinder. Then complete the square in every variable that appears both squared and to the first power. With all three variables squared and the right side scaled to 11, no minus sign on the left gives an ellipsoid, one gives a hyperboloid of one sheet and two give a hyperboloid of two sheets, and the axis belongs to the variable whose sign differs. With right side 00 and mixed signs it is a cone. Two squared variables and one to the first power give a paraboloid, elliptic for equal signs and hyperbolic for opposite signs. For example x2−2x+y2−z2=3x^2 - 2x + y^2 - z^2 = 3 becomes (x−1)2+y2−z2=4(x - 1)^2 + y^2 - z^2 = 4, a hyperboloid of one sheet whose axis is the line x=1x = 1, y=0y = 0.

Lessons Cylinders and quadric surfaces

Cylindrical and spherical coordinates

Cylindrical coordinates (r,θ,z)(r, \theta, z) use polar coordinates in the xyxy-plane and keep the height,

x=rcos⁡θ,y=rsin⁡θ,z=z,x = r\cos\theta, \quad y = r\sin\theta, \quad z = z,

with r≥0r \ge 0 the distance to the zz-axis. Spherical coordinates (ρ,θ,ϕ)(\rho, \theta, \phi) use the distance ρ\rho to the origin, the same azimuth θ\theta, and the angle ϕ\phi from the positive zz-axis,

x=ρsin⁡ϕcos⁡θ,y=ρsin⁡ϕsin⁡θ,z=ρcos⁡ϕ,\begin{aligned} x &= \rho\sin\phi\cos\theta, \\[0.4em] y &= \rho\sin\phi\sin\theta, \\[0.4em] z &= \rho\cos\phi, \end{aligned}

with ρ≥0\rho \ge 0, 0≤ϕ≤π0 \le \phi \le \pi and usually 0≤θ<2π0 \le \theta < 2\pi. The two systems are linked by r=ρsin⁡ϕr = \rho\sin\phi and z=ρcos⁡ϕz = \rho\cos\phi. Many physics books swap the letters θ\theta and ϕ\phi, so check which angle a source means.

ConversionFormulaConditions
rr and ρ\rhor=x2+y2r = \sqrt{x^2 + y^2}, ρ=x2+y2+z2\rho = \sqrt{x^2 + y^2 + z^2}always
Azimuthcos⁡θ=xr\cos\theta = \dfrac{x}{r} and sin⁡θ=yr\sin\theta = \dfrac{y}{r}r>0r > 0
Polar angleϕ=arccos⁡zρ\phi = \arccos\dfrac{z}{\rho}ρ>0\rho > 0

The formula θ=arctan⁡yx\theta = \arctan\frac{y}{x} gives the right direction only for x>0x > 0, and a negative value then needs 2π2\pi added, so choose θ\theta in the quadrant given by the signs of xx and yy. The point (−1,1,6)(-1, 1, \sqrt{6}) has r=2r = \sqrt{2}, θ=3π4\theta = \frac{3\pi}{4}, ρ=22\rho = 2\sqrt{2} and ϕ=π6\phi = \frac{\pi}{6}, while arctan⁡(−1)=−π4\arctan(-1) = -\frac{\pi}{4} would point the wrong way.

EquationSurface
r=cr = c with c>0c > 0cylinder of radius cc about the zz-axis
θ=β\theta = \betahalf-plane with edge on the zz-axis
ρ=c\rho = c with c>0c > 0sphere of radius cc about the origin
ϕ=α\phi = \alpha with 0<α<π20 < \alpha < \dfrac{\pi}{2} or π2<α<π\dfrac{\pi}{2} < \alpha < \piupper or lower nappe of a cone with vertex at the origin
ϕ=π2\phi = \dfrac{\pi}{2}the xyxy-plane

Off the zz-axis the coordinates in these ranges are unique. On the zz-axis the angle θ\theta is arbitrary, and at the origin ϕ\phi is arbitrary too.

Lessons Cylindrical and spherical coordinates

Vector-Valued Functions

A vector function traces a curve in space as its parameter runs, the way a moving object traces its path. Its calculus works one component at a time. From the derivatives of the position vector come tangent lines, the length of the curve, how sharply it bends and twists, and the velocity and acceleration of the motion.

Vector functions and their derivatives

A vector function r(t)=⟨f(t),g(t),h(t)⟩\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle has the intersection of the domains of its components as its natural domain. Limits, continuity, derivatives and integrals are all taken component by component, so

r′(t)=lim⁡h→0r(t+h)−r(t)h=⟨f′(t),g′(t),h′(t)⟩\begin{aligned} \mathbf{r}'(t) &= \lim_{h \to 0} \frac{\mathbf{r}(t + h) - \mathbf{r}(t)}{h} \\[0.4em] &= \langle f'(t), g'(t), h'(t) \rangle \end{aligned}

exists exactly when all three component derivatives exist. A continuous r\mathbf{r} on an interval traces a space curve, oriented in the direction of increasing tt. The circular helix

r(t)=⟨acos⁡t,asin⁡t,bt⟩\mathbf{r}(t) = \langle a\cos t, a\sin t, bt \rangle

with a>0a > 0 and b≠0b \ne 0 lies on the cylinder x2+y2=a2x^2 + y^2 = a^2 and rises 2πb2\pi b per turn. For b>0b > 0 it turns counterclockwise as seen from above while it rises.

A parametrization on an interval II is regular when it is C1C^1 on II, with one-sided derivatives at endpoints that belong to II, and r′(t)≠0\mathbf{r}'(t) \ne \mathbf{0} for every tt in II. Where r′(t0)≠0\mathbf{r}'(t_0) \ne \mathbf{0}, the tangent line and the unit tangent are

L(s)=r(t0)+s r′(t0),T(t)=r′(t)∣r′(t)∣.\begin{gathered} \mathbf{L}(s) = \mathbf{r}(t_0) + s\,\mathbf{r}'(t_0), \\[0.4em] \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\lvert \mathbf{r}'(t) \rvert}. \end{gathered}

For example r(t)=⟨cos⁡t,sin⁡t,t⟩\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle has r′(0)=⟨0,1,1⟩\mathbf{r}'(0) = \langle 0, 1, 1 \rangle, so its tangent line at (1,0,0)(1, 0, 0) is ⟨1,s,s⟩\langle 1, s, s \rangle. For differentiable u\mathbf{u}, v\mathbf{v} and ff,

(fu)′=f′u+fu′,(u⋅v)′=u′⋅v+u⋅v′,(u×v)′=u′×v+u×v′,\begin{aligned} (f\mathbf{u})' &= f'\mathbf{u} + f\mathbf{u}', \\[0.4em] (\mathbf{u} \cdot \mathbf{v})' &= \mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}', \\[0.4em] (\mathbf{u} \times \mathbf{v})' &= \mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}', \end{aligned}

with the order of the factors kept in the cross product rule. The chain rule reads ddsu(f(s))=f′(s) u′(f(s))\frac{d}{ds}\mathbf{u}(f(s)) = f'(s)\,\mathbf{u}'(f(s)). If r\mathbf{r} is differentiable and ∣r(t)∣\lvert \mathbf{r}(t) \rvert is constant on an interval, then r′(t)\mathbf{r}'(t) is perpendicular to r(t)\mathbf{r}(t). Integrals also go component by component, and for r\mathbf{r} continuous on [a,b][a, b] with antiderivative R\mathbf{R},

∫abr(t) dt=R(b)−R(a).\int_a^b \mathbf{r}(t)\,dt = \mathbf{R}(b) - \mathbf{R}(a).

Lessons Vector functions and space curves, Derivatives and integrals

Arc length and reparametrization

For r\mathbf{r} of class C1C^1 on [a,b][a, b], the speed is ∣r′(t)∣\lvert \mathbf{r}'(t) \rvert and the length is

L=∫ab∣r′(t)∣ dt.L = \int_a^b \lvert \mathbf{r}'(t) \rvert \, dt.

For a piecewise C1C^1 parametrization, add the lengths of the pieces. The length counts multiplicity, so a curve traced twice has twice the length. The arc length function from aa is s(t)=∫at∣r′(u)∣ dus(t) = \int_a^t \lvert \mathbf{r}'(u) \rvert \, du, with dsdt=∣r′(t)∣\frac{ds}{dt} = \lvert \mathbf{r}'(t) \rvert.

For a regular r\mathbf{r} the function ss is strictly increasing with a C1C^1 inverse, and R(s)=r(t(s))\mathbf{R}(s) = \mathbf{r}(t(s)) has ∣R′(s)∣=1\lvert \mathbf{R}'(s) \rvert = 1, the parametrization by arc length. A reparametrization r(φ(u))\mathbf{r}(\varphi(u)) with φ\varphi of class C1C^1 and φ′≠0\varphi' \ne 0 keeps a regular curve regular and leaves its length unchanged. It preserves the orientation when φ′>0\varphi' > 0 and reverses it when φ′<0\varphi' < 0, which replaces T\mathbf{T} by −T-\mathbf{T}.

For example the helix ⟨3cos⁡t,3sin⁡t,4t⟩\langle 3\cos t, 3\sin t, 4t \rangle has speed 55, length 10π10\pi over 0≤t≤2π0 \le t \le 2\pi, and arc length parametrization

R(s)=⟨3cos⁡s5,3sin⁡s5,4s5⟩.\mathbf{R}(s) = \left\langle 3\cos\frac{s}{5}, 3\sin\frac{s}{5}, \frac{4s}{5} \right\rangle.

Lessons Arc length and reparametrization

Curvature, torsion and the moving frame

Each object needs its own hypotheses, and each row adds its condition to the rows above it.

ObjectNeeds
T\mathbf{T} and the normal planeC1C^1 and r′≠0\mathbf{r}' \ne \mathbf{0}
κ\kappaalso C2C^2
N\mathbf{N}, B\mathbf{B}, osculating plane and circlealso κ≠0\kappa \ne 0
τ\tau and the Frenet-Serret formulasalso C3C^3

The curvature measures turning per unit length,

κ=∣dTds∣=∣r′×r′′∣∣r′∣3.\kappa = \left\lvert \frac{d\mathbf{T}}{ds} \right\rvert = \frac{\lvert \mathbf{r}' \times \mathbf{r}'' \rvert}{\lvert \mathbf{r}' \rvert^3}.

For the graph y=f(x)y = f(x) of a C2C^2 function this becomes κ=∣f′′∣(1+f′2)3/2\kappa = \frac{\lvert f'' \rvert}{(1 + f'^2)^{3/2}}. Where κ≠0\kappa \ne 0, the principal unit normal and the binormal are

N=T′∣T′∣,B=T×N=r′×r′′∣r′×r′′∣.\begin{gathered} \mathbf{N} = \frac{\mathbf{T}'}{\lvert \mathbf{T}' \rvert}, \\[0.4em] \mathbf{B} = \mathbf{T} \times \mathbf{N} = \frac{\mathbf{r}' \times \mathbf{r}''}{\lvert \mathbf{r}' \times \mathbf{r}'' \rvert}. \end{gathered}

The frame T,N,B\mathbf{T}, \mathbf{N}, \mathbf{B} is right-handed and orthonormal. The osculating plane is spanned by T\mathbf{T} and N\mathbf{N}, and the osculating circle lies in it with radius 1κ\frac{1}{\kappa} and center r+1κN\mathbf{r} + \frac{1}{\kappa}\mathbf{N}. The torsion is the signed number

τ=(r′×r′′)⋅r′′′∣r′×r′′∣2,\tau = \frac{(\mathbf{r}' \times \mathbf{r}'') \cdot \mathbf{r}'''}{\lvert \mathbf{r}' \times \mathbf{r}'' \rvert^2},

and with this sign a helix that turns counterclockwise as it rises in right-handed coordinates has τ>0\tau > 0. The frame moves by the Frenet-Serret formulas

dTds=κN,dNds=−κT+τB,dBds=−τN.\begin{aligned} \frac{d\mathbf{T}}{ds} &= \kappa \mathbf{N}, \\[0.4em] \frac{d\mathbf{N}}{ds} &= -\kappa \mathbf{T} + \tau \mathbf{B}, \\[0.4em] \frac{d\mathbf{B}}{ds} &= -\tau \mathbf{N}. \end{aligned}

A curve with κ≠0\kappa \ne 0 on an interval lies in one plane exactly when τ=0\tau = 0 there. The helix ⟨acos⁡t,asin⁡t,bt⟩\langle a\cos t, a\sin t, bt \rangle has

κ=aa2+b2,τ=ba2+b2,\kappa = \frac{a}{a^2 + b^2}, \qquad \tau = \frac{b}{a^2 + b^2},

so ⟨3cos⁡t,3sin⁡t,4t⟩\langle 3\cos t, 3\sin t, 4t \rangle has κ=325\kappa = \frac{3}{25} and τ=425\tau = \frac{4}{25}.

Lessons Curvature, torsion and the moving frame

Velocity and acceleration

For a position r(t)\mathbf{r}(t) the velocity is v=r′\mathbf{v} = \mathbf{r}', the speed is v=∣v∣v = \lvert \mathbf{v} \rvert and the acceleration is a=r′′\mathbf{a} = \mathbf{r}''. For r\mathbf{r} of class C2C^2 with v≠0\mathbf{v} \ne \mathbf{0},

a=aTT+aNN,aT=dvdt=v⋅a∣v∣,aN=κv2=∣v×a∣∣v∣,\begin{gathered} \mathbf{a} = a_T \mathbf{T} + a_N \mathbf{N}, \\[0.4em] a_T = \frac{dv}{dt} = \frac{\mathbf{v} \cdot \mathbf{a}}{\lvert \mathbf{v} \rvert}, \\[0.4em] a_N = \kappa v^2 = \frac{\lvert \mathbf{v} \times \mathbf{a} \rvert}{\lvert \mathbf{v} \rvert}, \end{gathered}

where the N\mathbf{N} term appears only when κ≠0\kappa \ne 0. Also aT2+aN2=∣a∣2a_T^2 + a_N^2 = \lvert \mathbf{a} \rvert^2, and the acceleration has no binormal part.

A projectile under gravity alone, with no air resistance, launched from height h≥0h \ge 0 at speed v0v_0 and angle α\alpha above the horizontal, has

x(t)=v0tcos⁡α,y(t)=h+v0tsin⁡α−12gt2\begin{aligned} x(t) &= v_0 t\cos\alpha, \\[0.4em] y(t) &= h + v_0 t\sin\alpha - \tfrac{1}{2} g t^2 \end{aligned}

until it lands. For h=0h = 0 and 0<α<π20 < \alpha < \frac{\pi}{2}, the flight time, range and maximum height are

Tf=2v0sin⁡αg,D=v02sin⁡2αg,H=(v0sin⁡α)22g.\begin{gathered} T_f = \frac{2v_0\sin\alpha}{g}, \quad D = \frac{v_0^2 \sin 2\alpha}{g}, \\[0.4em] H = \frac{(v_0\sin\alpha)^2}{2g}. \end{gathered}

With a launch speed of 2020 meters per second, α=π6\alpha = \frac{\pi}{6} and g=9.8g = 9.8 meters per second squared, these are about 2.042.04 seconds, 35.335.3 meters and 5.105.10 meters.

For a particle of constant mass mm the angular momentum L=r×mv\mathbf{L} = \mathbf{r} \times m\mathbf{v} has derivative r×F\mathbf{r} \times \mathbf{F}. A central force, parallel to r\mathbf{r}, keeps L\mathbf{L} constant, and if L≠0\mathbf{L} \ne \mathbf{0} the motion stays in one plane and sweeps area at the rate ∣L∣2m\frac{\lvert \mathbf{L} \rvert}{2m}. For a planet moving only under the inverse-square gravity of a sun of mass MM fixed at the origin, with L≠0\mathbf{L} \ne \mathbf{0}, Kepler’s laws hold. The orbit is a conic with a focus at the sun, an ellipse when bounded, equal areas are swept in equal times, and an elliptical orbit with semimajor axis aa has period TT with

T2=4π2GM a3.T^2 = \frac{4\pi^2}{GM}\, a^3.

Lessons Velocity and acceleration in space

Differentiation of Functions of Several Variables

A function of several variables can change in infinitely many directions at once. This unit defines limits and derivatives that account for every direction, organizes the derivative as a gradient or a Jacobian matrix, and uses it to approximate functions, to solve equations locally and to find largest and smallest values.

Functions, domains and level sets

When only a formula is given, the domain is the largest set where it gives a real number. The graph of f(x,y)f(x, y) is the surface z=f(x,y)z = f(x, y), the level curve at level cc is the set where f(x,y)=cf(x, y) = c, and for three variables the level surface is the set where F(x,y,z)=kF(x, y, z) = k. A contour map draws level curves for equally spaced values. For example f(x,y)=9−x2−y2f(x, y) = \sqrt{9 - x^2 - y^2} has the closed disk x2+y2≤9x^2 + y^2 \le 9 as its domain, the range [0,3][0, 3], and circles centered at the origin as level curves for 0≤c<30 \le c < 3, while the level curve for c=3c = 3 is the origin alone.

A point PP is an interior point of a set SS when some open ball about PP lies in SS, and a boundary point when every open ball about PP meets both SS and its complement. The set is open when all its points are interior, closed when it contains all its boundary points, bounded when it lies in some ball, and compact when it is closed and bounded. Many sets are neither open nor closed.

Lessons Functions of several variables

Limits and continuity

Let (a,b)(a, b) be a limit point of the domain DD of ff. Then f(x,y)→Lf(x, y) \to L as (x,y)→(a,b)(x, y) \to (a, b) when for every ε>0\varepsilon > 0 there is δ>0\delta > 0 with ∣f(x,y)−L∣<ε\lvert f(x, y) - L \rvert < \varepsilon for every (x,y)(x, y) in DD with

0<(x−a)2+(y−b)2<δ.0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta.
  • Paths can disprove a limit. Two paths in DD toward (a,b)(a, b) with different limits show that the limit does not exist. Along y=mxy = mx the function xyx2+y2\frac{xy}{x^2 + y^2} tends to m1+m2\frac{m}{1 + m^2}, which depends on mm.
  • Paths cannot prove a limit. The function x2yx4+y2\frac{x^2 y}{x^4 + y^2} tends to 00 along every line through the origin, but equals 12\frac{1}{2} along y=x2y = x^2, so it has no limit at the origin.
  • Squeeze by distance. If ∣f(x,y)−L∣≤g(x,y)\lvert f(x, y) - L \rvert \le g(x, y) near (a,b)(a, b) and g→0g \to 0, then f→Lf \to L. Since x2≤x2+y2x^2 \le x^2 + y^2, the bound ∣x2yx2+y2∣≤∣y∣≤r\left\lvert \frac{x^2 y}{x^2 + y^2} \right\rvert \le \lvert y \rvert \le r shows that this function tends to 00.

The function ff is continuous at a point (a,b)(a, b) of its domain when the limit there equals f(a,b)f(a, b), with only points of DD tested. Sums, products, quotients with nonzero denominator and compositions of continuous functions are continuous, so polynomials are continuous everywhere and rational functions on their domains.

Lessons Multivariable limits and continuity

Partial derivatives

At an interior point (a,b)(a, b) of the domain, the partial derivatives vary one coordinate and hold the other fixed,

fx(a,b)=lim⁡h→0f(a+h,b)−f(a,b)h,fy(a,b)=lim⁡h→0f(a,b+h)−f(a,b)h.\begin{aligned} f_x(a, b) &= \lim_{h \to 0} \frac{f(a + h, b) - f(a, b)}{h}, \\[0.4em] f_y(a, b) &= \lim_{h \to 0} \frac{f(a, b + h) - f(a, b)}{h}. \end{aligned}

They are the slopes of the traces in the planes y=by = b and x=ax = a. In the subscript notation the order of differentiation reads from left to right, and in the ∂\partial notation from right to left,

fxy=(fx)y=∂2f∂y ∂x.f_{xy} = (f_x)_y = \frac{\partial^2 f}{\partial y \, \partial x}.

Clairaut’s theorem. If fxyf_{xy} and fyxf_{yx} exist and are continuous on an open disk containing (a,b)(a, b), then fxy(a,b)=fyx(a,b)f_{xy}(a, b) = f_{yx}(a, b). More generally, for ff of class CkC^k on an open set, every partial derivative of order at most kk is independent of the order of differentiation. For example f=x3y2+exyf = x^3 y^2 + e^{xy} has fx=3x2y2+yexyf_x = 3x^2 y^2 + y e^{xy} and fxy=fyx=6x2y+(1+xy)exyf_{xy} = f_{yx} = 6x^2 y + (1 + xy)e^{xy}. The existence of both partial derivatives at a point does not even make ff continuous there.

The classical partial differential equations, with constants α≠0\alpha \ne 0 and c≠0c \ne 0, are Laplace’s equation uxx+uyy=0u_{xx} + u_{yy} = 0, the heat equation ut=α2uxxu_t = \alpha^2 u_{xx} and the wave equation utt=c2uxxu_{tt} = c^2 u_{xx}.

Lessons Partial derivatives

Differentiability and tangent planes

Let UU be open in Rn\mathbb{R}^n, ff map UU to Rm\mathbb{R}^m and a\mathbf{a} lie in UU. The function is differentiable at a\mathbf{a} when there is a linear map LL, the total derivative Df(a)Df(\mathbf{a}), with

lim⁡h→0∥f(a+h)−f(a)−L(h)∥∥h∥=0.\lim_{\mathbf{h} \to \mathbf{0}} \frac{\lVert f(\mathbf{a} + \mathbf{h}) - f(\mathbf{a}) - L(\mathbf{h}) \rVert}{\lVert \mathbf{h} \rVert} = 0.

Then every partial derivative exists at a\mathbf{a}, Df(a)Df(\mathbf{a}) is multiplication by the m×nm \times n Jacobian matrix [∂fi∂xj(a)]\left[ \frac{\partial f_i}{\partial x_j}(\mathbf{a}) \right], with row ii for the component fif_i, and ff is continuous at a\mathbf{a}. The converse fails. Partial derivatives can exist where ff is not differentiable, but if all partial derivatives exist near a\mathbf{a} and are continuous at a\mathbf{a}, then ff is differentiable at a\mathbf{a}. In particular every C1C^1 function on an open set is differentiable there.

For f(x,y)f(x, y) differentiable at (a,b)(a, b), the linearization, whose graph is the tangent plane, is

L(x,y)=f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b),\begin{aligned} L(x, y) = f(a, b) &+ f_x(a, b)(x - a) \\[0.4em] &+ f_y(a, b)(y - b), \end{aligned}

with normal vector ⟨fx(a,b),fy(a,b),−1⟩\langle f_x(a, b), f_y(a, b), -1 \rangle. The total differential dz=fx dx+fy dydz = f_x\,dx + f_y\,dy approximates the actual change Δz\Delta z with an error that is small compared to dx2+dy2\sqrt{dx^2 + dy^2}. For example f=x2+3xyf = x^2 + 3xy at (1,2)(1, 2) has f=7f = 7, fx=8f_x = 8 and fy=3f_y = 3, so f(1.1,1.9)≈7+0.8−0.3=7.5f(1.1, 1.9) \approx 7 + 0.8 - 0.3 = 7.5, against the exact value 7.487.48. The function equal to xyx2+y2\frac{xy}{x^2 + y^2} away from the origin and 00 at the origin has fx(0,0)=fy(0,0)=0f_x(0, 0) = f_y(0, 0) = 0 but is not continuous there, so it has no tangent plane at the origin.

Lessons Total derivative, tangent planes and differentials

The chain rule

If gg is differentiable at a\mathbf{a} and ff is differentiable at g(a)g(\mathbf{a}), then f∘gf \circ g is differentiable at a\mathbf{a} and

D(f∘g)(a)=Df(g(a)) Dg(a),D(f \circ g)(\mathbf{a}) = Df\big(g(\mathbf{a})\big)\, Dg(\mathbf{a}),

a product of Jacobian matrices. For z=f(x,y)z = f(x, y) with x=x(t)x = x(t) and y=y(t)y = y(t), and for x=x(s,t)x = x(s, t) and y=y(s,t)y = y(s, t),

dzdt=∂f∂xdxdt+∂f∂ydydt,∂z∂s=∂f∂x∂x∂s+∂f∂y∂y∂s,\begin{aligned} \frac{dz}{dt} &= \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}, \\[0.6em] \frac{\partial z}{\partial s} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial s}, \end{aligned}

with the partial derivatives of ff evaluated at the point (x,y)(x, y) that the inner functions reach. In a dependency diagram, add over every route from the top variable down to the independent variable the product of the derivatives along the route. Differentiability of ff is needed, since for the function xyx2+y2\frac{xy}{x^2 + y^2} above and g(t)=⟨t,t⟩g(t) = \langle t, t \rangle the partial derivatives would predict 00 while f(g(t))=12f(g(t)) = \frac{1}{2} for t≠0t \ne 0. For T(x,y,t)T(x, y, t) along a path, the total derivative

dTdt=∂T∂t+∂T∂xdxdt+∂T∂ydydt\frac{dT}{dt} = \frac{\partial T}{\partial t} + \frac{\partial T}{\partial x} \frac{dx}{dt} + \frac{\partial T}{\partial y} \frac{dy}{dt}

differs in general from the partial derivative ∂T∂t\frac{\partial T}{\partial t}. For example z=x2yz = x^2 y with x=cos⁡tx = \cos t and y=sin⁡ty = \sin t gives dzdt=−2cos⁡tsin⁡2t+cos⁡3t\frac{dz}{dt} = -2\cos t\sin^2 t + \cos^3 t.

Lessons The multivariable chain rule

Directional derivatives and the gradient

For a unit vector u\mathbf{u}, the directional derivative and the gradient are

Duf(a)=lim⁡t→0f(a+tu)−f(a)t,∇f=⟨∂f∂x1,…,∂f∂xn⟩.\begin{gathered} D_{\mathbf{u}} f(\mathbf{a}) = \lim_{t \to 0} \frac{f(\mathbf{a} + t\mathbf{u}) - f(\mathbf{a})}{t}, \\[0.6em] \nabla f = \left\langle \frac{\partial f}{\partial x_1}, \dots, \frac{\partial f}{\partial x_n} \right\rangle. \end{gathered}

If ff is differentiable at a\mathbf{a}, then Duf(a)=∇f(a)⋅uD_{\mathbf{u}} f(\mathbf{a}) = \nabla f(\mathbf{a}) \cdot \mathbf{u} for every unit u\mathbf{u}. Without differentiability this formula can fail even when every directional derivative exists. When ∇f(a)≠0\nabla f(\mathbf{a}) \ne \mathbf{0},

  • the largest rate of increase is ∥∇f(a)∥\lVert \nabla f(\mathbf{a}) \rVert, in the direction of ∇f(a)\nabla f(\mathbf{a}),
  • the smallest is −∥∇f(a)∥-\lVert \nabla f(\mathbf{a}) \rVert, in the opposite direction,
  • the rate is 00 exactly in the directions perpendicular to ∇f(a)\nabla f(\mathbf{a}).

If ff is C1C^1 on an open set and ∇f(a)≠0\nabla f(\mathbf{a}) \ne \mathbf{0}, the level set through a\mathbf{a} is a smooth curve or surface near a\mathbf{a} with normal ∇f(a)\nabla f(\mathbf{a}), and its tangent line or plane is ∇f(a)⋅(x−a)=0\nabla f(\mathbf{a}) \cdot (\mathbf{x} - \mathbf{a}) = 0. At a critical point, where ∇f(a)=0\nabla f(\mathbf{a}) = \mathbf{0}, the gradient gives no direction and no normal. For example f=x2yf = x^2 y at (1,2)(1, 2) has ∇f=⟨4,1⟩\nabla f = \langle 4, 1 \rangle, so in the direction of ⟨3,4⟩\langle 3, 4 \rangle the rate is 12+45=165\frac{12 + 4}{5} = \frac{16}{5}, and the largest rate is 17\sqrt{17}.

Lessons Directional derivatives and the gradient

Implicit and inverse function theorems

Implicit function theorem. Let FF be C1C^1 on an open set WW in Rn+m\mathbb{R}^{n + m} with values in Rm\mathbb{R}^m, and let F(a,b)=0F(a, b) = 0. If the m×mm \times m matrix DyFD_y F of partial derivatives in y1,…,ymy_1, \dots, y_m has det⁡DyF(a,b)≠0\det D_y F(a, b) \ne 0, then near (a,b)(a, b) the solutions of F(x,y)=0F(x, y) = 0 are exactly the points y=g(x)y = g(x) for a unique C1C^1 function gg with g(a)=bg(a) = b, and

Dg(x)=−[DyF]−1DxF,Dg(x) = -\big[ D_y F \big]^{-1} D_x F,

with both matrices evaluated at (x,g(x))(x, g(x)). For one equation F(x,y,z)=0F(x, y, z) = 0 solved for zz this gives

∂z∂x=−FxFz,∂z∂y=−FyFz,Fz≠0.\begin{gathered} \frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}, \\[0.4em] F_z \ne 0. \end{gathered}

If FF is C1C^1 and ∇F(p)≠0\nabla F(p) \ne \mathbf{0}, the level surface F=cF = c through p=(x0,y0,z0)p = (x_0, y_0, z_0) has the tangent plane

Fx(p)(x−x0)+Fy(p)(y−y0)+Fz(p)(z−z0)=0.\begin{aligned} F_x(p)(x - x_0) &+ F_y(p)(y - y_0) \\[0.4em] &+ F_z(p)(z - z_0) = 0. \end{aligned}

For the sphere x2+y2+z2=9x^2 + y^2 + z^2 = 9 at (1,2,2)(1, 2, 2) the partial derivatives of zz are −12-\frac{1}{2} and −1-1, and the tangent plane is x+2y+2z=9x + 2y + 2z = 9.

Inverse function theorem. If ff maps an open set of Rn\mathbb{R}^n to Rn\mathbb{R}^n, is C1C^1, and has det⁡Df(a)≠0\det Df(a) \ne 0, then ff is one-to-one from a neighborhood of aa onto a neighborhood of f(a)f(a), with a C1C^1 inverse and

D(f−1)(f(a))=[Df(a)]−1.D(f^{-1})(f(a)) = \big[ Df(a) \big]^{-1}.

Both theorems are local. The map (u,v)↦(eucos⁡v,eusin⁡v)(u, v) \mapsto (e^u\cos v, e^u\sin v) has Jacobian determinant e2u≠0e^{2u} \ne 0 everywhere, yet it is not one-to-one, since vv and v+2πv + 2\pi give the same point.

Lessons Implicit and inverse function theorems

Taylor approximation

The Hessian of ff at aa is the matrix Hf(a)=[fxixj(a)]Hf(a) = \big[ f_{x_i x_j}(a) \big], symmetric when ff is C2C^2. For ff of class C2C^2 on an open set containing aa and h=x−ah = x - a, the second-order Taylor polynomial is

T2(x)=f(a)+∇f(a)⋅h+12 hTHf(a) h.\begin{aligned} T_2(x) &= f(a) + \nabla f(a) \cdot h \\[0.4em] &\quad + \tfrac{1}{2} \, h^{T} Hf(a) \, h. \end{aligned}

In two variables the quadratic part is 12(fxxh12+2fxyh1h2+fyyh22)\frac{1}{2}\big(f_{xx} h_1^2 + 2 f_{xy} h_1 h_2 + f_{yy} h_2^2\big), with all derivatives at aa. For example excos⁡ye^x\cos y at the origin has T2=1+x+12(x2−y2)T_2 = 1 + x + \frac{1}{2}(x^2 - y^2). In multi-index notation, for ff of class CkC^k,

Tk(a+h)=∑∣α∣≤kDαf(a)α! hα.T_k(a + h) = \sum_{\lvert \alpha \rvert \le k} \frac{D^{\alpha} f(a)}{\alpha!} \, h^{\alpha}.

If ff is Ck+1C^{k+1} on an open set containing the segment from aa to a+ha + h, the remainder Rk(h)=f(a+h)−Tk(a+h)R_k(h) = f(a + h) - T_k(a + h) is the sum of the terms of order k+1k + 1 with every derivative taken at a+θha + \theta h, for some 0<θ<10 < \theta < 1. If also every partial derivative of order k+1k + 1 is at most MM in size on that segment, then

∣Rk(h)∣≤M(k+1)!(∣h1∣+⋯+∣hn∣)k+1.\lvert R_k(h) \rvert \le \frac{M}{(k + 1)!} \big( \lvert h_1 \rvert + \cdots + \lvert h_n \rvert \big)^{k + 1}.

Lessons Multivariable Taylor approximation

Local extrema and saddle points

A critical point is an interior point of the domain where ∇f=0\nabla f = \mathbf{0} or some first partial derivative does not exist. Every local extremum at an interior point is a critical point, but not every critical point is an extremum. A saddle point is a critical point that is neither a local maximum nor a local minimum.

Second derivative test. Let ff be C2C^2 on an open set containing aa with ∇f(a)=0\nabla f(a) = \mathbf{0}. A positive definite Hessian gives a strict local minimum, a negative definite one a strict local maximum, and an indefinite one a saddle point. In two variables, with

D=fxx(a)fyy(a)−fxy(a)2,D = f_{xx}(a) f_{yy}(a) - f_{xy}(a)^2,
ConditionConclusion
D>0D > 0 and fxx(a)>0f_{xx}(a) > 0strict local minimum
D>0D > 0 and fxx(a)<0f_{xx}(a) < 0strict local maximum
D<0D < 0saddle point
D=0D = 0no information

For example f=x3−3x+y2f = x^3 - 3x + y^2 has critical points (1,0)(1, 0), with D=12D = 12 and fxx=6f_{xx} = 6, a local minimum, and (−1,0)(-1, 0), with D=−12D = -12, a saddle. When D=0D = 0, as for both x4+y4x^4 + y^4 and x4−y4x^4 - y^4 at the origin, classify by other means, such as the sign of f(x)−f(a)f(x) - f(a) or restrictions to lines and curves. In three or more variables a degenerate Hessian that is still indefinite gives a saddle, while a semidefinite one gives no conclusion.

Lessons Local extrema and saddle points

Absolute extrema on regions

Extreme value theorem. A function continuous on a nonempty closed and bounded set takes an absolute maximum and an absolute minimum there. The closed region method finds them for a region whose boundary is made of finitely many smooth curves.

  1. Evaluate ff at every critical point inside the region.
  2. On each boundary curve ri(t)\mathbf{r}_i(t) with αi≤t≤βi\alpha_i \le t \le \beta_i, find the extreme values of gi(t)=f(ri(t))g_i(t) = f(\mathbf{r}_i(t)), including the endpoints.
  3. Evaluate ff at every corner.

The largest value is the absolute maximum and the smallest is the absolute minimum. For f=x2+y2−2xf = x^2 + y^2 - 2x on the disk x2+y2≤4x^2 + y^2 \le 4, the only interior critical point gives f(1,0)=−1f(1, 0) = -1, the boundary gives 4−4cos⁡t4 - 4\cos t between 00 and 88, so the maximum is 88 at (−2,0)(-2, 0) and the minimum is −1-1 at (1,0)(1, 0). On a set that is not closed or not bounded an extremum may fail to exist and needs its own argument. For example a continuous ff on a closed nonempty set with f→∞f \to \infty as x2+y2→∞\sqrt{x^2 + y^2} \to \infty has an absolute minimum.

Lessons Absolute extrema on regions

Lagrange multipliers

Let ff and gg be C1C^1 on an open set in Rn\mathbb{R}^n. If ff restricted to g=cg = c has a local extremum at a\mathbf{a} and ∇g(a)≠0\nabla g(\mathbf{a}) \ne \mathbf{0}, then for some number λ\lambda

∇f(a)=λ∇g(a).\nabla f(\mathbf{a}) = \lambda \nabla g(\mathbf{a}).

With m<nm < n constraints g1=c1,…,gm=cmg_1 = c_1, \dots, g_m = c_m whose gradients are linearly independent at a\mathbf{a}, the condition becomes

∇f(a)=λ1∇g1(a)+⋯+λm∇gm(a).\nabla f(\mathbf{a}) = \lambda_1 \nabla g_1(\mathbf{a}) + \cdots + \lambda_m \nabla g_m(\mathbf{a}).

The conditions are necessary, not sufficient. Collect every candidate, meaning the Lagrange points, the singular points of the constraint where the gradients are dependent, the points where ff or gg is not C1C^1, and any interior critical points, corners and endpoints when there are inequality constraints. On a closed and bounded feasible set, comparing the values of ff at the candidates finds the absolute extrema, and otherwise existence needs a separate argument. For example f=x+2yf = x + 2y on the circle x2+y2=5x^2 + y^2 = 5 gives 1=2λx1 = 2\lambda x and 2=2λy2 = 2\lambda y, so the candidates are (1,2)(1, 2) and (−1,−2)(-1, -2), where the maximum 55 and the minimum −5-5 occur.

Lessons Lagrange multipliers

Multiple Integrals

A multiple integral adds up a quantity spread over a region of the plane or a solid in space, as a limit of sums of small pieces. Fubini’s theorem turns it into repeated single integrals, and a good choice of coordinates, whether polar, cylindrical, spherical or a custom change of variables, makes the bounds simple.

Double integrals and Fubini’s theorem

On a rectangle RR the double integral is the limit of Riemann sums ∑f(xij∗,yij∗) ΔAij\sum f(x_{ij}^*, y_{ij}^*) \, \Delta A_{ij} as the largest subrectangle shrinks. A bounded function whose discontinuities lie on finitely many smooth curves is integrable, and so is every continuous function. When f≥0f \ge 0 the integral is the volume under the graph, and in general a signed volume. The average value over a region of positive area A(D)A(D) is

favg=1A(D)∬Df dA.f_{\text{avg}} = \frac{1}{A(D)} \iint_D f \, dA.

Fubini’s theorem. If ff is continuous on R=[a,b]×[c,d]R = [a, b] \times [c, d], then

∬Rf dA=∫ab∫cdf(x,y) dy dx=∫cd∫abf(x,y) dx dy.\begin{aligned} \iint_R f \, dA &= \int_a^b \int_c^d f(x, y) \, dy \, dx \\[0.4em] &= \int_c^d \int_a^b f(x, y) \, dx \, dy. \end{aligned}

A region of type I is a≤x≤ba \le x \le b, g1(x)≤y≤g2(x)g_1(x) \le y \le g_2(x), and a region of type II is c≤y≤dc \le y \le d, h1(y)≤x≤h2(y)h_1(y) \le x \le h_2(y), with continuous boundary functions in order. For ff continuous on them,

type I∫ab∫g1(x)g2(x)f dy dx,type II∫cd∫h1(y)h2(y)f dx dy.\begin{aligned} \text{type I} \quad &\int_a^b \int_{g_1(x)}^{g_2(x)} f \, dy \, dx, \\[0.6em] \text{type II} \quad &\int_c^d \int_{h_1(y)}^{h_2(y)} f \, dx \, dy. \end{aligned}

To reverse the order, sketch the region from the given bounds and describe it again with the other variable outside, so the new outer bounds are constants. For example

∫01∫x1ey2 dy dx=∫01∫0yey2 dx dy=e−12.\begin{aligned} \int_0^1 \int_x^1 e^{y^2} \, dy \, dx &= \int_0^1 \int_0^y e^{y^2} \, dx \, dy \\[0.4em] &= \frac{e - 1}{2}. \end{aligned}

Integrals add over regions that overlap only along their boundaries.

Lessons Over rectangles, Over general regions

Double integrals in polar coordinates

With x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta and r≥0r \ge 0, the area element is dA=r dr dθdA = r \, dr \, d\theta. For a region α≤θ≤β\alpha \le \theta \le \beta, h1(θ)≤r≤h2(θ)h_1(\theta) \le r \le h_2(\theta), with 0≤β−α≤2π0 \le \beta - \alpha \le 2\pi and continuous 0≤h1≤h20 \le h_1 \le h_2, and ff continuous on it,

∬Df dA=∫αβ∫h1(θ)h2(θ)f r dr dθ,\iint_D f \, dA = \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} f \, r \, dr \, d\theta,

with ff evaluated at (rcos⁡θ,rsin⁡θ)(r\cos\theta, r\sin\theta). The bounds must cover each point once, so the angle interval has length at most 2π2\pi and a curve that retraces itself is used over one tracing only. For example the circle x2+y2=2yx^2 + y^2 = 2y is r=2sin⁡θr = 2\sin\theta for 0≤θ≤π0 \le \theta \le \pi, and

∬x2+y2≤4e−(x2+y2) dA=∫02π∫02e−r2r dr dθ=π(1−e−4).\begin{aligned} &\iint_{x^2 + y^2 \le 4} e^{-(x^2 + y^2)} \, dA \\[0.4em] &\quad = \int_0^{2\pi} \int_0^2 e^{-r^2} r \, dr \, d\theta \\[0.4em] &\quad = \pi\left(1 - e^{-4}\right). \end{aligned}

Lessons Double integrals in polar coordinates

Mass, moments and probability

For a lamina on DD with continuous density σ(x,y)≥0\sigma(x, y) \ge 0,

QuantityFormulaConditions
AreaA=∬D1 dAA = \displaystyle\iint_D 1 \, dAboundary of zero area
Massm=∬Dσ dAm = \displaystyle\iint_D \sigma \, dA
MomentsMx=∬Dy σ dAM_x = \displaystyle\iint_D y\,\sigma \, dA and My=∬Dx σ dAM_y = \displaystyle\iint_D x\,\sigma \, dA
Center of massxˉ=Mym\bar{x} = \dfrac{M_y}{m} and yˉ=Mxm\bar{y} = \dfrac{M_x}{m}m>0m > 0, the centroid when σ\sigma is constant
Moments of inertiaIx=∬Dy2σ dAI_x = \displaystyle\iint_D y^2 \sigma \, dA, Iy=∬Dx2σ dAI_y = \displaystyle\iint_D x^2 \sigma \, dA, I0=Ix+IyI_0 = I_x + I_y
Radius of gyrationR=ImR = \sqrt{\dfrac{I}{m}}m>0m > 0, about the same axis as II

For example the half disk x2+y2≤a2x^2 + y^2 \le a^2, y≥0y \ge 0, has its centroid at (0,4a3π)\left(0, \frac{4a}{3\pi}\right).

A joint density f≥0f \ge 0 of (X,Y)(X, Y) has total integral 11 over the plane, and P((X,Y)∈D)=∬Df dAP\big((X, Y) \in D\big) = \iint_D f \, dA. The marginal density of XX is fX(x)=∫−∞∞f(x,y) dyf_X(x) = \int_{-\infty}^{\infty} f(x, y) \, dy, and XX and YY are independent exactly when f(x,y)=fX(x)fY(y)f(x, y) = f_X(x) f_Y(y) outside a set that does not affect any double integral. Expected values are E[g(X,Y)]=∬gf dAE[g(X, Y)] = \iint g f \, dA when ∬∣g∣f dA\iint \lvert g \rvert f \, dA is finite, and when E[X2]E[X^2] and E[Y2]E[Y^2] are finite

Cov⁡(X,Y)=E[XY]−E[X] E[Y].\operatorname{Cov}(X, Y) = E[XY] - E[X]\,E[Y].

Independent variables have covariance 00, but covariance 00 does not imply independence. For the density x+yx + y on the unit square, E[X]=E[Y]=712E[X] = E[Y] = \frac{7}{12}, E[XY]=13E[XY] = \frac{1}{3} and the covariance is −1144-\frac{1}{144}.

Lessons Applications of double integrals

Triple integrals

Triple integrals are limits of Riemann sums over boxes, and a continuous function on a box can be integrated in any of the six orders. A solid of type 1 lies over a region DD of the xyxy-plane between continuous surfaces u1≤u2u_1 \le u_2, and for ff continuous on it

∭Ef dV=∬D[∫u1(x,y)u2(x,y)f dz]dA.\iiint_E f \, dV = \iint_D \left[ \int_{u_1(x, y)}^{u_2(x, y)} f \, dz \right] dA.

The volume is ∭E1 dV\iiint_E 1 \, dV, the mass is m=∭Eδ dVm = \iiint_E \delta \, dV for a continuous density δ≥0\delta \ge 0, and for m>0m > 0 the center of mass has coordinates 1m∭Ex δ dV\frac{1}{m}\iiint_E x\,\delta \, dV and so on for yy and zz. The moment of inertia about the zz-axis is Iz=∭E(x2+y2) δ dVI_z = \iiint_E (x^2 + y^2)\,\delta \, dV. Round solids are simpler in other coordinates.

CoordinatesPointVolume element
Rectangular(x,y,z)(x, y, z)dV=dx dy dzdV = dx \, dy \, dz
Cylindrical(rcos⁡θ,rsin⁡θ,z)(r\cos\theta, r\sin\theta, z)dV=r dz dr dθdV = r \, dz \, dr \, d\theta
Spherical(ρsin⁡ϕcos⁡θ,ρsin⁡ϕsin⁡θ,ρcos⁡ϕ)(\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi)dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta

The bounds need r≥0r \ge 0, ρ≥0\rho \ge 0, 0≤ϕ≤π0 \le \phi \le \pi and an angle interval for θ\theta of length at most 2π2\pi, and each point of the solid must be covered once. Useful descriptions are

  • the cone ϕ=c\phi = c with 0<c<π20 < c < \frac{\pi}{2}, which opens upward around the positive zz-axis,
  • the plane z=hz = h with h>0h > 0, which is ρ=hsec⁡ϕ\rho = h\sec\phi for 0≤ϕ<π20 \le \phi < \frac{\pi}{2},
  • the ball x2+y2+(z−a)2≤a2x^2 + y^2 + (z - a)^2 \le a^2 with a>0a > 0, which is 0≤ρ≤2acos⁡ϕ0 \le \rho \le 2a\cos\phi for 0≤ϕ≤π20 \le \phi \le \frac{\pi}{2}.

For example the solid between the paraboloid z=x2+y2z = x^2 + y^2 and the plane z=4z = 4 has volume

∫02π∫02(4−r2)r dr dθ=8π,\int_0^{2\pi} \int_0^2 \left(4 - r^2\right) r \, dr \, d\theta = 8\pi,

and the part of the ball ρ≤2\rho \le 2 inside the cone ϕ≤π3\phi \le \frac{\pi}{3} has volume

∫02π∫0π/3∫02ρ2sin⁡ϕ dρ dϕ dθ=8π3.\int_0^{2\pi} \int_0^{\pi/3} \int_0^2 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = \frac{8\pi}{3}.

A solid ball of radius aa and density 11 has Iz=8πa515I_z = \frac{8\pi a^5}{15}. Symmetry helps too. An integrand with f(−x,y,z)=−f(x,y,z)f(-x, y, z) = -f(x, y, z) over a solid unchanged by x↦−xx \mapsto -x integrates to 00.

Lessons Rectangular coordinates, Cylindrical coordinates, Spherical coordinates

Change of variables and the Jacobian

For a C1C^1 map (u,v)↦(x,y)(u, v) \mapsto (x, y), the Jacobian is

∂(x,y)∂(u,v)=xuyv−xvyu,\frac{\partial(x, y)}{\partial(u, v)} = x_u y_v - x_v y_u,

and its absolute value is the local scaling factor for area, while its sign only records orientation. Where it is nonzero, the map has a local C1C^1 inverse whose Jacobian is the reciprocal.

Change of variables theorem. Let SS be a closed bounded region with piecewise smooth boundary, TT a C1C^1 map on an open set containing SS that is one-to-one with nonzero Jacobian on the interior of SS, and R=T(S)R = T(S). For ff continuous on RR,

∬Rf(x,y) dA=∬Sf(T(u,v))∣∂(x,y)∂(u,v)∣du dv.\begin{aligned} &\iint_R f(x, y) \, dA \\[0.4em] &= \iint_S f\big( T(u, v) \big) \left\lvert \frac{\partial(x, y)}{\partial(u, v)} \right\rvert du \, dv. \end{aligned}

Failures of these conditions on the boundary of SS do no harm. In space the same holds with dV=∣∂(x,y,z)∂(u,v,w)∣du dv dwdV = \left\lvert \frac{\partial(x, y, z)}{\partial(u, v, w)} \right\rvert du \, dv \, dw, which gives the factors rr and ρ2sin⁡ϕ\rho^2\sin\phi of cylindrical and spherical coordinates. For example the region with 0≤x−y≤20 \le x - y \le 2 and 0≤x+y≤40 \le x + y \le 4 becomes a rectangle under u=x−yu = x - y and v=x+yv = x + y, with ∂(x,y)∂(u,v)=12\frac{\partial(x, y)}{\partial(u, v)} = \frac{1}{2}, so

∬R(x+y)2 dA=∫02∫04v22 dv du=643.\begin{aligned} \iint_R (x + y)^2 \, dA &= \int_0^2 \int_0^4 \frac{v^2}{2} \, dv \, du \\[0.4em] &= \frac{64}{3}. \end{aligned}

Lessons Change of variables and the Jacobian

Improper integrals and the order of integration

Over an unbounded region, or near points where ff is unbounded, the integral is the limit over a regular exhaustion, closed bounded regions K1⊆K2⊆⋯K_1 \subseteq K_2 \subseteq \cdots inside DD with piecewise smooth boundaries that eventually contain every closed bounded subset of DD. For a continuous f≥0f \ge 0 the limit lies in [0,∞][0, \infty] and is the same for every exhaustion. A continuous ff of any sign is absolutely integrable when ∬D∣f∣ dA\iint_D \lvert f \rvert \, dA is finite.

  • Tonelli. For continuous f≥0f \ge 0, the double integral and both iterated improper integrals are equal, all finite or all infinite.
  • Fubini. For continuous ff of any sign with ∬D∣f∣ dA<∞\iint_D \lvert f \rvert \, dA < \infty, both iterated integrals converge and equal the double integral. Check this with Tonelli applied to ∣f∣\lvert f \rvert.

The course states both with inner integrals that are finite and continuous in the outer variable. Without absolute integrability the two orders can disagree. For x2−y2(x2+y2)2\frac{x^2 - y^2}{(x^2 + y^2)^2} on (0,1]×(0,1](0, 1] \times (0, 1] the order dy dxdy \, dx gives π4\frac{\pi}{4} and the order dx dydx \, dy gives −π4-\frac{\pi}{4}. The Gaussian integral follows from Tonelli and polar coordinates,

(∫−∞∞e−x2 dx)2=∫02π∫0∞e−r2r dr dθ=π,\begin{aligned} \left( \int_{-\infty}^{\infty} e^{-x^2} \, dx \right)^2 &= \int_0^{2\pi} \int_0^\infty e^{-r^2} r \, dr \, d\theta \\[0.4em] &= \pi, \end{aligned}

so the integral of e−x2e^{-x^2} over the real line is π\sqrt{\pi}.

Lessons Improper integrals and order of integration

Vector Calculus

Vector calculus integrates fields along curves and across surfaces, then ties those integrals to derivatives inside the region through the theorems of Green, Stokes and Gauss. Each theorem needs its orientation and its smoothness and domain conditions, and the formulas can fail when one of them is dropped.

Vector fields and potentials

A vector field assigns a vector F=⟨P,Q,R⟩\mathbf{F} = \langle P, Q, R \rangle to each point of its domain, and it is continuous or C1C^1 when its components are. A velocity field moves particles along paths with r′(t)=F(r(t))\mathbf{r}'(t) = \mathbf{F}(\mathbf{r}(t)), which are its field lines. For a C1C^1 field exactly one maximal field line passes through each point, so different field lines never cross. The gravitational field of a mass at the origin is

F(x)=−k∣x∣3 x,x≠0,\mathbf{F}(\mathbf{x}) = -\frac{k}{\lvert \mathbf{x} \rvert^3} \, \mathbf{x}, \qquad \mathbf{x} \ne \mathbf{0},

with k=GMm>0k = GMm > 0, an inverse-square law. A field F\mathbf{F} is conservative on an open set DD when F=∇f\mathbf{F} = \nabla f there for a potential function ff. On a connected open set two potentials differ by a constant, and physicists often use the potential energy V=−fV = -f. Every statement about potentials refers to a specific domain, and a singular point, where the field is undefined, can change the answer.

Lessons Vector fields

Line integrals

A curve is regular when it has a regular parametrization r(t)\mathbf{r}(t), a≤t≤ba \le t \le b, that traces it once, and piecewise regular when it is a chain of finitely many regular curves joined end to end. For ff and F\mathbf{F} continuous on CC,

∫Cf ds=∫abf(r(t)) ∣r′(t)∣ dt,∫CF⋅dr=∫abF(r(t))⋅r′(t) dt.\begin{aligned} \int_C f \, ds &= \int_a^b f(\mathbf{r}(t)) \, \lvert \mathbf{r}'(t) \rvert \, dt, \\[0.6em] \int_C \mathbf{F} \cdot d\mathbf{r} &= \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt. \end{aligned}

For the vector integral the parametrization must move in the chosen direction of CC. It equals ∫CF⋅T ds\int_C \mathbf{F} \cdot \mathbf{T} \, ds and, in coordinates, ∫CP dx+Q dy+R dz\int_C P \, dx + Q \, dy + R \, dz. Over a piecewise regular curve, add the pieces.

PropertyScalar ∫Cf ds\int_C f \, dsVector ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r}
Another parametrization tracing CC oncesame valuesame value if the direction is kept
Reversed directionsame valuechanges sign
Meaningmass of a wire with density ff, length when f=1f = 1work done by a force field, circulation when CC is closed

A wire with continuous density δ≥0\delta \ge 0 has mass m=∫Cδ dsm = \int_C \delta \, ds and, when m>0m > 0, center of mass with coordinates 1m∫Cx δ ds\frac{1}{m}\int_C x\,\delta \, ds and so on. Along the helix ⟨cos⁡t,sin⁡t,t⟩\langle \cos t, \sin t, t \rangle for 0≤t≤2π0 \le t \le 2\pi, with speed 2\sqrt{2},

∫Cz ds=∫02πt2 dt=22 π2,∫C⟨−y,x,1⟩⋅dr=4π.\begin{gathered} \int_C z \, ds = \int_0^{2\pi} t\sqrt{2} \, dt = 2\sqrt{2}\,\pi^2, \\[0.4em] \int_C \langle -y, x, 1 \rangle \cdot d\mathbf{r} = 4\pi. \end{gathered}

A simple closed curve in the plane is positively oriented when it runs counterclockwise, with the enclosed region on the left.

Lessons Scalar line integrals, Work and circulation

Conservative fields

Fundamental theorem for line integrals. If ff is C1C^1 on an open set DD and CC is an oriented piecewise regular curve in DD from AA to BB, then

∫C∇f⋅dr=f(B)−f(A).\int_C \nabla f \cdot d\mathbf{r} = f(B) - f(A).

For F\mathbf{F} continuous on a connected open set DD, three statements are equivalent. The field is conservative on DD, its line integrals are path independent in DD, and its integral around every closed piecewise regular curve in DD is 00.

Cross-partial test. If F\mathbf{F} is C1C^1 and conservative on an open set, then Py=QxP_y = Q_x in the plane, and in space

Ry=Qz,Pz=Rx,Qx=Py,R_y = Q_z, \quad P_z = R_x, \quad Q_x = P_y,

that is ∇×F=0\nabla \times \mathbf{F} = \mathbf{0}. The converse holds when the open set is simply connected, which means connected with every closed curve shrinkable to a point inside it. In the plane that means no holes. Space with one point removed is simply connected, while space with the zz-axis removed is not. The test alone is not enough on other sets. The field

F=⟨−yx2+y2,xx2+y2⟩\mathbf{F} = \left\langle \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2} \right\rangle

has Py=QxP_y = Q_x on the plane without the origin, yet its circulation counterclockwise around the unit circle is 2π2\pi, so it is not conservative there.

To find a potential on a simply connected set, integrate PP in xx with a constant g(y,z)g(y, z), match fyf_y with QQ to find gg up to h(z)h(z), match fzf_z with RR, and check that ∇f=F\nabla f = \mathbf{F}. For example F=⟨2xy+z,x2,x⟩\mathbf{F} = \langle 2xy + z, x^2, x \rangle has the potential f=x2y+xzf = x^2 y + xz, so its integral along any path from the origin to (1,2,3)(1, 2, 3) is f(1,2,3)−f(0,0,0)=5f(1, 2, 3) - f(0, 0, 0) = 5.

Lessons Conservative fields and the fundamental theorem

Divergence and curl

With the operator ∇=⟨∂x,∂y,∂z⟩\nabla = \left\langle \partial_x, \partial_y, \partial_z \right\rangle and F=⟨P,Q,R⟩\mathbf{F} = \langle P, Q, R \rangle,

∇⋅F=Px+Qy+Rz,∇×F=⟨Ry−Qz,Pz−Rx,Qx−Py⟩.\begin{gathered} \nabla \cdot \mathbf{F} = P_x + Q_y + R_z, \\[0.6em] \nabla \times \mathbf{F} = \langle R_y - Q_z, P_z - R_x, Q_x - P_y \rangle. \end{gathered}

A plane field ⟨P,Q⟩\langle P, Q \rangle has the scalar curl Qx−PyQ_x - P_y. For a C1C^1 field the divergence is the outward flux per unit volume through small spheres about a point, and the component (∇×F)⋅u(\nabla \times \mathbf{F}) \cdot \mathbf{u} is the circulation per unit area around small circles perpendicular to the unit vector u\mathbf{u}, counterclockwise as seen from its tip. The Laplacian is ∇2f=∇⋅∇f=fxx+fyy+fzz\nabla^2 f = \nabla \cdot \nabla f = f_{xx} + f_{yy} + f_{zz}, and a C2C^2 function with ∇2f=0\nabla^2 f = 0 on an open set is harmonic there.

IdentityConditions
∇×(∇f)=0\nabla \times (\nabla f) = \mathbf{0}ff of class C2C^2 on an open set
∇⋅(∇×F)=0\nabla \cdot (\nabla \times \mathbf{F}) = 0F\mathbf{F} of class C2C^2 on an open set

Both rest on Clairaut’s theorem and say nothing at points where the field is undefined. For example F=⟨xy,yz,zx⟩\mathbf{F} = \langle xy, yz, zx \rangle has divergence x+y+zx + y + z and curl ⟨−y,−z,−x⟩\langle -y, -z, -x \rangle.

Lessons Divergence and curl

Green’s theorem

Let CC be a positively oriented, piecewise smooth, simple closed curve in the plane, DD the closed region made of CC and its inside, and PP and QQ of class C1C^1 on an open set containing DD. Then

∮CP dx+Q dy=∬D(∂Q∂x−∂P∂y)dA.\begin{aligned} &\oint_C P \, dx + Q \, dy \\[0.4em] &\quad = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA. \end{aligned}

With the outward unit normal n\mathbf{n}, the flux form is

∮CF⋅n ds=∮CP dy−Q dx=∬D(Px+Qy)dA.\begin{aligned} \oint_C \mathbf{F} \cdot \mathbf{n} \, ds &= \oint_C P \, dy - Q \, dx \\[0.4em] &= \iint_D \left( P_x + Q_y \right) dA. \end{aligned}

A single point of DD where PP or QQ is undefined breaks the hypothesis. For a region with holes the boundary runs with DD on the left, so the outer curve goes counterclockwise and each inner curve clockwise, and the field may be undefined inside the holes. Taking Qx−Py=1Q_x - P_y = 1 gives the area

area⁡(D)=∮Cx dy=−∮Cy dx=12∮Cx dy−y dx,\begin{aligned} \operatorname{area}(D) &= \oint_C x \, dy = -\oint_C y \, dx \\[0.4em] &= \frac{1}{2} \oint_C x \, dy - y \, dx, \end{aligned}

which gives πab\pi ab for the ellipse x=acos⁡tx = a\cos t, y=bsin⁡ty = b\sin t with a,b>0a, b > 0. Around the unit square counterclockwise, ∮C(x2−y) dx+(x+y2) dy=∬D2 dA=2\oint_C (x^2 - y) \, dx + (x + y^2) \, dy = \iint_D 2 \, dA = 2.

Lessons Green’s theorem

Parametric surfaces and surface area

A parametrization r(u,v)\mathbf{r}(u, v) over a closed bounded region DD with piecewise smooth boundary is regular when r\mathbf{r} is C1C^1 on an open set containing DD, one-to-one on the interior of DD, and

ru×rv≠0\mathbf{r}_u \times \mathbf{r}_v \ne \mathbf{0}

on the interior of DD. Seams and edges that collapse to a point may sit on the boundary of DD. The vectors ru\mathbf{r}_u and rv\mathbf{r}_v are tangent to the grid curves, ru×rv\mathbf{r}_u \times \mathbf{r}_v is normal to the tangent plane, and the surface area is

A(S)=∬D∣ru×rv∣ dA.A(S) = \iint_D \lvert \mathbf{r}_u \times \mathbf{r}_v \rvert \, dA.

For the graph z=g(x,y)z = g(x, y) of a C1C^1 function, rx×ry=⟨−gx,−gy,1⟩\mathbf{r}_x \times \mathbf{r}_y = \langle -g_x, -g_y, 1 \rangle and

A=∬D1+gx2+gy2 dA.A = \iint_D \sqrt{1 + g_x^2 + g_y^2} \, dA.

The sphere of radius aa has the parametrization ⟨asin⁡ϕcos⁡θ,asin⁡ϕsin⁡θ,acos⁡ϕ⟩\langle a\sin\phi\cos\theta, a\sin\phi\sin\theta, a\cos\phi \rangle over 0≤ϕ≤π0 \le \phi \le \pi, 0≤θ≤2π0 \le \theta \le 2\pi, with rϕ×rθ\mathbf{r}_\phi \times \mathbf{r}_\theta pointing outward and of length a2sin⁡ϕa^2\sin\phi, so its area is 4πa24\pi a^2.

Lessons Parametric surfaces and surface area

Surface integrals and flux

For a regular parametrization and gg continuous on SS,

∬Sg dS=∬Dg(r) ∣ru×rv∣ du dv,\iint_S g \, dS = \iint_D g(\mathbf{r}) \, \lvert \mathbf{r}_u \times \mathbf{r}_v \rvert \, du \, dv,

which does not depend on the regular parametrization chosen. With g=1g = 1 this is the area, and with a density g=δg = \delta it is the mass of a thin sheet. An orientation is a continuous choice of unit normal n\mathbf{n}. A connected orientable surface has exactly two, and a Möbius band has none, so flux through it is not defined. The flux of a continuous F\mathbf{F} across an oriented surface is

∬SF⋅dS=∬SF⋅n dS=∬DF(r)⋅(ru×rv) du dv,\begin{aligned} \iint_S \mathbf{F} \cdot d\mathbf{S} &= \iint_S \mathbf{F} \cdot \mathbf{n} \, dS \\[0.4em] &= \iint_D \mathbf{F}(\mathbf{r}) \cdot (\mathbf{r}_u \times \mathbf{r}_v) \, du \, dv, \end{aligned}

where the last form needs ru×rv\mathbf{r}_u \times \mathbf{r}_v to point the same way as n\mathbf{n}. The graph z=f(x,y)z = f(x, y) has the upward unit normal

n=⟨−fx,−fy,1⟩1+fx2+fy2.\mathbf{n} = \frac{\langle -f_x, -f_y, 1 \rangle}{\sqrt{1 + f_x^2 + f_y^2}}.

Reversing the orientation changes the sign of the flux, while scalar surface integrals do not change. A closed surface, the boundary of a bounded solid, is oriented by the outward normal unless stated otherwise. For example the outward flux of ⟨x,y,z⟩\langle x, y, z \rangle through the sphere of radius aa is 4πa34\pi a^3, and the upward flux of ⟨0,0,z⟩\langle 0, 0, z \rangle through z=1−x2−y2z = 1 - x^2 - y^2 over the unit disk is π2\frac{\pi}{2}.

Lessons Scalar surface integrals and flux

Stokes’ theorem

Let SS be a bounded oriented piecewise smooth surface whose boundary ∂S\partial S is made of finitely many simple closed piecewise smooth curves with the induced orientation, and let F\mathbf{F} be C1C^1 on an open set containing SS. Then

∮∂SF⋅dr=∬S(∇×F)⋅n dS.\oint_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{n} \, dS.

The induced orientation is the one in which you walk along ∂S\partial S with your head along n\mathbf{n} and the surface on your left, so n×T\mathbf{n} \times \mathbf{T} points into SS. Reversing only the normal or only the direction breaks the formula. With several boundary curves, add the circulations around all of them. Two oriented surfaces with the same boundary and the same induced orientation have the same curl flux whenever F\mathbf{F} is C1C^1 on an open set containing both, so a flat surface often replaces a curved one. For F=⟨−y,x,0⟩\mathbf{F} = \langle -y, x, 0 \rangle, whose curl is ⟨0,0,2⟩\langle 0, 0, 2 \rangle, the upper unit hemisphere with upward normal and the unit disk both give 2π2\pi, the circulation counterclockwise around the unit circle seen from above.

Lessons Stokes’ theorem

The divergence theorem

Let EE be a bounded solid whose boundary ∂E\partial E is a finite union of closed piecewise smooth surfaces with the outward unit normal n\mathbf{n}, and let F\mathbf{F} be C1C^1 on an open set containing EE and its boundary. Then

∬∂EF⋅n dS=∭E∇⋅F dV.\iint_{\partial E} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_E \nabla \cdot \mathbf{F} \, dV.

Every piece of the boundary counts, including the inner surface around a cavity, whose outward normal points into the cavity. For F=⟨x3,y3,z3⟩\mathbf{F} = \langle x^3, y^3, z^3 \rangle and the ball of radius aa, the divergence is 3ρ23\rho^2 and both sides equal 12πa55\frac{12\pi a^5}{5}.

If F\mathbf{F} is C1C^1 except at one interior point P0P_0 and ∇⋅F=0\nabla \cdot \mathbf{F} = 0 away from it, remove a small ball about P0P_0. The flux through ∂E\partial E then equals the flux outward from P0P_0 through any small sphere about it. The field r∣r∣3\frac{\mathbf{r}}{\lvert \mathbf{r} \rvert^3} has divergence 00 away from the origin and flux 4π4\pi through every sphere about the origin. In the same way Gauss’s law says that a point charge qq at the origin, with field E=q4πε0r∣r∣3\mathbf{E} = \frac{q}{4\pi\varepsilon_0} \frac{\mathbf{r}}{\lvert \mathbf{r} \rvert^3}, has outward flux qε0\frac{q}{\varepsilon_0} through a closed surface around it and 00 through one that leaves it outside.

All the main theorems integrate a derivative over a region and get the original object on the boundary.

TheoremRegion integralBoundary term
Fundamental theorem∫abf′(x) dx\displaystyle\int_a^b f'(x)\, dxf(b)−f(a)f(b) - f(a)
Line integrals∫C∇f⋅dr\displaystyle\int_C \nabla f \cdot d\mathbf{r}f(B)−f(A)f(B) - f(A)
Green∬D(Qx−Py) dA\displaystyle\iint_D (Q_x - P_y) \, dA∮∂DP dx+Q dy\displaystyle\oint_{\partial D} P\,dx + Q\,dy
Stokes∬S(∇×F)⋅dS\displaystyle\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}∮∂SF⋅dr\displaystyle\oint_{\partial S} \mathbf{F} \cdot d\mathbf{r}
Divergence∭E∇⋅F dV\displaystyle\iiint_E \nabla \cdot \mathbf{F} \, dV∬∂EF⋅dS\displaystyle\iint_{\partial E} \mathbf{F} \cdot d\mathbf{S}

Lessons The divergence theorem

Further Differential Equations

This unit continues the differential equations of single-variable calculus. Exact equations reuse potentials from vector calculus, complex exponentials reduce linear equations with constant coefficients to algebra, and power series, the Laplace transform and matrices handle variable coefficients, switched inputs and systems of equations.

Exact equations

The equation M dx+N dy=0M\,dx + N\,dy = 0, with MM and NN continuous on an open set DD, is exact on DD when it has a potential FF there, a C1C^1 function with Fx=MF_x = M and Fy=NF_y = N. Its solutions then lie on the level curves F(x,y)=CF(x, y) = C.

Test for exactness. Let MM and NN be C1C^1 on DD. Exactness implies My=NxM_y = N_x on DD, and when DD is open and simply connected, My=NxM_y = N_x on DD implies exactness. On an open rectangle containing (x0,y0)(x_0, y_0) a potential is

F(x,y)=∫x0xM(s,y0) ds+∫y0yN(x,s) ds.\begin{aligned} F(x, y) &= \int_{x_0}^{x} M(s, y_0)\,ds \\[0.4em] &\quad + \int_{y_0}^{y} N(x, s)\,ds. \end{aligned}

For example (2xy+3) dx+(x2+4y) dy=0(2xy + 3)\,dx + (x^2 + 4y)\,dy = 0 has My=Nx=2xM_y = N_x = 2x and the solutions x2y+3x+2y2=Cx^2 y + 3x + 2y^2 = C.

Integrating factorWorks whenConditions
μ(x)=e∫h(x) dx\mu(x) = e^{\int h(x)\,dx}My−Nx=h(x) NM_y - N_x = h(x)\,Nhh continuous
μ(y)=e∫k(y) dy\mu(y) = e^{\int k(y)\,dy}Nx−My=k(y) MN_x - M_y = k(y)\,Mkk continuous

Both need DD open and simply connected and MM, NN of class C1C^1. A factor that is never zero keeps the same solutions, while one that is zero or undefined somewhere can add or lose solutions there. For example (3xy+y2) dx+(x2+xy) dy=0(3xy + y^2)\,dx + (x^2 + xy)\,dy = 0 has h(x)=1xh(x) = \frac{1}{x}, and μ=x\mu = x makes it exact with potential x3y+12x2y2x^3 y + \frac{1}{2}x^2 y^2 on a region where x≠0x \ne 0.

Lessons Exact differential equations

Complex numbers and complex exponentials

A complex number z=a+biz = a + bi with i2=−1i^2 = -1 has real part aa, imaginary part bb, conjugate z‾=a−bi\overline{z} = a - bi and modulus ∣z∣=a2+b2=zz‾\lvert z \rvert = \sqrt{a^2 + b^2} = \sqrt{z\overline{z}}. To divide, multiply the top and bottom by the conjugate of the denominator, so 3+4i1−2i=−1+2i\frac{3 + 4i}{1 - 2i} = -1 + 2i. Every z≠0z \ne 0 has polar form r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta) with r=∣z∣r = \lvert z \rvert and an argument θ\theta, unique up to multiples of 2π2\pi. Moduli multiply and arguments add, and for a positive integer nn

zn=rn(cos⁡nθ+isin⁡nθ).z^n = r^n (\cos n\theta + i \sin n\theta).

A number R(cos⁡φ+isin⁡φ)R(\cos\varphi + i\sin\varphi) with R>0R > 0 has exactly nn distinct nnth roots,

R1/n(cos⁡φ+2πkn+isin⁡φ+2πkn),k=0,1,…,n−1.\begin{gathered} R^{1/n}\Bigl(\cos \tfrac{\varphi + 2\pi k}{n} + i \sin \tfrac{\varphi + 2\pi k}{n}\Bigr), \\[0.4em] k = 0, 1, \ldots, n - 1. \end{gathered}

For example the cube roots of −8-8 are 1+i31 + i\sqrt{3}, −2-2 and 1−i31 - i\sqrt{3}. The exponential series gives Euler’s formula for real θ\theta and the complex exponential for real aa and bb,

eiθ=cos⁡θ+isin⁡θ,ea+bi=ea(cos⁡b+isin⁡b).\begin{gathered} e^{i\theta} = \cos\theta + i\sin\theta, \\[0.4em] e^{a + bi} = e^a(\cos b + i\sin b). \end{gathered}

So eze^z is never 00, ez+w=ezewe^{z + w} = e^z e^w, and ddtert=rert\frac{d}{dt}e^{rt} = re^{rt} for a complex constant rr. If pp and qq are real, the real and imaginary parts of a complex solution of y′′+py′+qy=0y'' + py' + qy = 0 are real solutions.

Lessons Complex numbers and complex exponentials

Second-order linear equations

On an interval where P(t)≠0P(t) \ne 0, the equation Py′′+Qy′+Ry=GP y'' + Q y' + R y = G has the normalized form

y′′+p(t) y′+q(t) y=g(t).y'' + p(t)\,y' + q(t)\,y = g(t).

If pp, qq and gg are continuous on an open interval II containing t0t_0, the initial value problem with y(t0)=y0y(t_0) = y_0 and y′(t0)=y1y'(t_0) = y_1 has exactly one solution, and it exists on all of II. Solutions of the homogeneous equation can be added and scaled. Two solutions y1y_1 and y2y_2 form a fundamental set, and every solution is c1y1+c2y2c_1 y_1 + c_2 y_2, exactly when their Wronskian

W=y1y2′−y1′y2W = y_1 y_2' - y_1' y_2

is nonzero. By Abel’s formula W(t)=W(t0) e−∫t0tp(s) dsW(t) = W(t_0)\,e^{-\int_{t_0}^{t} p(s)\,ds}, so for two solutions WW is zero everywhere on II or nowhere. For ay′′+by′+cy=0a y'' + b y' + c y = 0 with real constants and a≠0a \ne 0, the roots of the characteristic equation ar2+br+c=0a r^2 + b r + c = 0 decide the general solution.

RootsGeneral solution
real r1≠r2r_1 \ne r_2, when b2−4ac>0b^2 - 4ac > 0c1er1t+c2er2tc_1 e^{r_1 t} + c_2 e^{r_2 t}
double root r=−b2ar = -\dfrac{b}{2a}, when b2−4ac=0b^2 - 4ac = 0(c1+c2t) ert(c_1 + c_2 t)\,e^{rt}
α±iβ\alpha \pm i\beta with β>0\beta > 0, when b2−4ac<0b^2 - 4ac < 0eαt(c1cos⁡βt+c2sin⁡βt)e^{\alpha t}(c_1 \cos \beta t + c_2 \sin \beta t)

For example y′′+2y′+5y=0y'' + 2y' + 5y = 0 with y(0)=1y(0) = 1 and y′(0)=1y'(0) = 1 has roots −1±2i-1 \pm 2i and the solution y=e−t(cos⁡2t+sin⁡2t)y = e^{-t}(\cos 2t + \sin 2t). A boundary value problem, with values at two different points, may have one solution, none or infinitely many.

Lessons Second-order linear equations

Nonhomogeneous equations

With pp, qq and gg continuous on II and a fundamental set y1y_1, y2y_2, every solution of y′′+py′+qy=gy'' + py' + qy = g is

y=c1y1+c2y2+ypy = c_1 y_1 + c_2 y_2 + y_p

for one particular solution ypy_p and exactly one pair of constants.

Undetermined coefficients. For ay′′+by′+cy=ga y'' + b y' + c y = g with constant coefficients and a term Pm(t) eαtcos⁡βtP_m(t)\,e^{\alpha t}\cos\beta t or Pm(t) eαtsin⁡βtP_m(t)\,e^{\alpha t}\sin\beta t, where PmP_m has degree mm, try

yp=tseαt(A(t)cos⁡βt+B(t)sin⁡βt)\begin{aligned} y_p = t^s e^{\alpha t}\big(&A(t)\cos \beta t \\[0.4em] &+ B(t)\sin \beta t\big) \end{aligned}

with polynomials AA and BB of degree mm, both needed even when gg has only a cosine or only a sine. The power ss is the multiplicity of α+iβ\alpha + i\beta as a root of the characteristic equation, 00 when it is not a root. For a sum of such terms, add the particular solutions. For example y′′−y=ety'' - y = e^t has 11 as a simple root, so yp=At ety_p = At\,e^t, which gives A=12A = \frac{1}{2}.

Variation of parameters. For the normalized equation with any continuous gg and W=y1y2′−y1′y2≠0W = y_1 y_2' - y_1' y_2 \ne 0 on II,

yp=−y1∫y2 gW dt+y2∫y1 gW dt.y_p = -y_1 \int \frac{y_2\,g}{W}\,dt + y_2 \int \frac{y_1\,g}{W}\,dt.

The leading coefficient must be 11, so for Py′′+Qy′+Ry=GP y'' + Q y' + R y = G the forcing is g=GPg = \frac{G}{P}. For example y′′+y=sec⁡ty'' + y = \sec t on (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) has yp=cos⁡t ln⁡(cos⁡t)+tsin⁡ty_p = \cos t\,\ln(\cos t) + t\sin t.

Lessons Nonhomogeneous linear equations

Oscillations, damping and resonance

A mass m>0m > 0 on a spring with constant k>0k > 0, damping c≥0c \ge 0 and applied force F(t)F(t), with xx measured from equilibrium, satisfies

mx′′+cx′+kx=F(t),ω0=km.\begin{gathered} m x'' + c x' + k x = F(t), \\[0.4em] \omega_0 = \sqrt{\frac{k}{m}}. \end{gathered}

Without damping or force the motion is x=Rcos⁡(ω0t−δ)x = R\cos(\omega_0 t - \delta) with period 2πω0\frac{2\pi}{\omega_0}. With c>0c > 0 and no force every solution tends to 00, in one of three ways.

CaseSolutionBehavior
Underdamped, c2<4mkc^2 < 4mke−ct/(2m)(c1cos⁡μt+c2sin⁡μt)e^{-ct/(2m)}(c_1 \cos \mu t + c_2 \sin \mu t) with μ=4mk−c22m\mu = \dfrac{\sqrt{4mk - c^2}}{2m}oscillates with quasi period 2πμ\dfrac{2\pi}{\mu}
Critically damped, c2=4mkc^2 = 4mk(c1+c2t) e−ct/(2m)(c_1 + c_2 t)\,e^{-ct/(2m)}crosses equilibrium at most once
Overdamped, c2>4mkc^2 > 4mkc1er1t+c2er2tc_1 e^{r_1 t} + c_2 e^{r_2 t} with r2<r1<0r_2 < r_1 < 0crosses equilibrium at most once

For a force F0cos⁡ωtF_0\cos\omega t without damping, ω≠ω0\omega \ne \omega_0 gives the bounded response F0m(ω02−ω2)cos⁡ωt\frac{F_0}{m(\omega_0^2 - \omega^2)}\cos\omega t, while ω=ω0\omega = \omega_0 gives pure resonance,

xp=F02mω0 tsin⁡ω0t,x_p = \frac{F_0}{2m\omega_0}\,t \sin \omega_0 t,

which grows without bound. With c>0c > 0 and F0>0F_0 > 0 every solution stays bounded, and the steady periodic response A(ω)cos⁡(ωt−γ)A(\omega)\cos(\omega t - \gamma) has amplitude

A(ω)=F0(k−mω2)2+c2ω2.A(\omega) = \frac{F_0}{\sqrt{(k - m\omega^2)^2 + c^2\omega^2}}.

When c2<2mkc^2 < 2mk the amplitude peaks at the practical resonance frequency

ωr=km−c22m2<ω0,\omega_r = \sqrt{\frac{k}{m} - \frac{c^2}{2m^2}} < \omega_0,

and when c2≥2mkc^2 \ge 2mk it decreases for all ω>0\omega > 0. For example m=1m = 1, c=2c = 2 and k=10k = 10 give ωr=22\omega_r = 2\sqrt{2} and the peak amplitude F06\frac{F_0}{6}. A series circuit obeys Lq′′+Rq′+1Cq=E(t)L q'' + R q' + \frac{1}{C} q = E(t), the same equation with LL, RR and 1C\frac{1}{C} in place of mm, cc and kk.

Lessons Oscillations, damping and resonance

Series solutions

Write y′′+p(x) y′+q(x) y=0y'' + p(x)\,y' + q(x)\,y = 0 with p=QPp = \frac{Q}{P} and q=RPq = \frac{R}{P}. A point x0x_0 is ordinary when pp and qq are analytic there, and then every solution is a power series ∑an(x−x0)n\sum a_n (x - x_0)^n that converges at least where the series of pp and qq both do. For polynomial coefficients with no common factor, that radius is at least the distance from x0x_0 to the nearest complex zero of PP. Substitute the series, shift indices so every sum shows the same power, and set each coefficient to 00. For example y′′=xyy'' = xy gives a2=0a_2 = 0 and

an+2=an−1(n+2)(n+1),n≥1.a_{n+2} = \frac{a_{n-1}}{(n + 2)(n + 1)}, \qquad n \ge 1.

A singular point x0x_0 is regular when (x−x0) p(x - x_0)\,p and (x−x0)2q(x - x_0)^2 q are analytic there. At a regular singular point 00, with x p(x)=∑pnxnx\,p(x) = \sum p_n x^n and x2q(x)=∑qnxnx^2 q(x) = \sum q_n x^n, the Frobenius series xr∑anxnx^r \sum a_n x^n with a0≠0a_0 \ne 0 needs a root of the indicial equation

r(r−1)+p0r+q0=0.r(r - 1) + p_0 r + q_0 = 0.

For real roots r1≥r2r_1 \ge r_2 there is always a solution for r1r_1. A second one is another Frobenius series when r1−r2r_1 - r_2 is not an integer, it contains y1ln⁡xy_1 \ln x when r1=r2r_1 = r_2, and it may contain y1ln⁡xy_1 \ln x when r1−r2r_1 - r_2 is a positive integer. The Bessel equation of order ν≥0\nu \ge 0,

x2y′′+x y′+(x2−ν2) y=0,x^2 y'' + x\,y' + (x^2 - \nu^2)\,y = 0,

has indicial roots ±ν\pm\nu and the general solution c1Jν(x)+c2Yν(x)c_1 J_\nu(x) + c_2 Y_\nu(x) for x>0x > 0, where

Jν(x)=∑m=0∞(−1)mm! Γ(m+ν+1)(x2)2m+νJ_\nu(x) = \sum_{m=0}^{\infty} \frac{(-1)^m}{m!\,\Gamma(m + \nu + 1)} \Big(\frac{x}{2}\Big)^{2m + \nu}

and YνY_\nu is unbounded as x→0+x \to 0^+.

Lessons Series solutions of differential equations

The Laplace transform

For ff defined for t≥0t \ge 0 and integrable on every [0,T][0, T],

F(s)=L{f}(s)=∫0∞e−stf(t) dtF(s) = \mathcal{L}\{f\}(s) = \int_0^\infty e^{-st} f(t)\, dt

for the real ss where the improper integral converges. A function is of exponential order aa when ∣f(t)∣≤Meat\lvert f(t) \rvert \le M e^{at} for all t≥T0t \ge T_0. If ff is piecewise continuous on every [0,T][0, T] and of exponential order aa, the integral converges absolutely for every s>as > a and F(s)→0F(s) \to 0 as s→∞s \to \infty.

f(t)f(t)F(s)F(s)Valid for
111s\dfrac{1}{s}s>0s > 0
tnt^n, with n=0,1,2,…n = 0, 1, 2, \ldotsn!sn+1\dfrac{n!}{s^{n+1}}s>0s > 0
eate^{at}1s−a\dfrac{1}{s - a}s>as > a
sin⁡bt\sin btbs2+b2\dfrac{b}{s^2 + b^2}s>0s > 0
cos⁡bt\cos btss2+b2\dfrac{s}{s^2 + b^2}s>0s > 0
eatf(t)e^{at} f(t)F(s−a)F(s - a)s−as - a in the range for FF
uc(t)f(t−c)u_c(t) f(t - c) with c≥0c \ge 0e−csF(s)e^{-cs} F(s)the range for FF
δ(t−c)\delta(t - c) with c>0c > 0e−cse^{-cs}all ss
(f∗g)(t)(f * g)(t)F(s) G(s)F(s)\,G(s)both of exponential order aa, s>as > a

Here ucu_c is 00 for t<ct < c and 11 for t≥ct \ge c, and the convolution is (f∗g)(t)=∫0tf(τ) g(t−τ) dτ(f * g)(t) = \int_0^t f(\tau)\, g(t - \tau)\, d\tau. The impulse δ(t−c)\delta(t - c) is not a function but the limit of pulses of area 11 on [c,c+ε)[c, c + \varepsilon), with ∫0∞δ(t−c) g(t) dt=g(c)\int_0^\infty \delta(t - c)\, g(t)\, dt = g(c) for continuous gg.

Derivatives. If ff is continuous on [0,∞)[0, \infty) and of exponential order aa and f′f' is piecewise continuous on every [0,T][0, T], then for s>as > a

L{f′}=sF(s)−f(0).\mathcal{L}\{f'\} = s F(s) - f(0).

If f′f' also satisfies these conditions, then

L{f′′}=s2F(s)−sf(0)−f′(0).\mathcal{L}\{f''\} = s^2 F(s) - s f(0) - f'(0).

Two piecewise continuous functions of exponential order with the same Laplace transform agree wherever both are continuous, so the inverse Laplace transform is read off the table after partial fractions. For ay′′+by′+cy=ga y'' + b y' + c y = g with y(0)=y0y(0) = y_0 and y′(0)=y1y'(0) = y_1,

(as2+bs+c) Y(s)=G(s)+a(sy0+y1)+by0.\begin{gathered} (a s^2 + b s + c)\, Y(s) = G(s) \\[0.4em] {} + a(s y_0 + y_1) + b y_0. \end{gathered}

For example y′′+3y′+2y=2y'' + 3y' + 2y = 2 with zero initial values gives

Y(s)=2s(s+1)(s+2)=1s−2s+1+1s+2,\begin{aligned} Y(s) &= \frac{2}{s(s + 1)(s + 2)} \\[0.4em] &= \frac{1}{s} - \frac{2}{s + 1} + \frac{1}{s + 2}, \end{aligned}

so y=1−2e−t+e−2ty = 1 - 2e^{-t} + e^{-2t}. An impulse k δ(t−t0)k\,\delta(t - t_0) with t0>0t_0 > 0 leaves yy continuous and makes y′y' jump by ka\frac{k}{a} at t0t_0.

Lessons Laplace transforms and initial-value problems

Linear systems

The system x′=A(t) x+g(t)\mathbf{x}' = A(t)\,\mathbf{x} + \mathbf{g}(t) with entries continuous on an open interval II has exactly one solution through each initial value x(t0)=x0\mathbf{x}(t_0) = \mathbf{x}_0, defined on all of II. For a constant AA, an eigenvalue λ\lambda with eigenvector v≠0\mathbf{v} \ne \mathbf{0}, a root of det⁡(A−λI)=0\det(A - \lambda I) = 0, gives the solution eλtve^{\lambda t}\mathbf{v}. For a 2×22 \times 2 matrix with trace TT and determinant DD the characteristic equation is λ2−Tλ+D=0\lambda^2 - T\lambda + D = 0.

  • Diagonalizable. With nn independent eigenvectors, which is the case for nn distinct eigenvalues, the general solution is c1eλ1tv1+⋯+cneλntvnc_1 e^{\lambda_1 t}\mathbf{v}_1 + \cdots + c_n e^{\lambda_n t}\mathbf{v}_n.
  • Complex pair. For real AA and λ=α+iβ\lambda = \alpha + i\beta with β≠0\beta \ne 0, the real and imaginary parts of eλtve^{\lambda t}\mathbf{v} are two independent real solutions.
  • Defective double eigenvalue. For a 2×22 \times 2 matrix A≠λIA \ne \lambda I with one independent eigenvector v\mathbf{v}, choose w\mathbf{w} with (A−λI)w=v(A - \lambda I)\mathbf{w} = \mathbf{v}, and then eλtve^{\lambda t}\mathbf{v} and eλt(tv+w)e^{\lambda t}(t\mathbf{v} + \mathbf{w}) form a fundamental set.

The matrix exponential eAt=∑k≥0(At)kk!e^{At} = \sum_{k \ge 0} \frac{(At)^k}{k!} converges for every AA and gives the solution x(t)=eAtx0\mathbf{x}(t) = e^{At}\mathbf{x}_0 of x′=Ax\mathbf{x}' = A\mathbf{x} with x(0)=x0\mathbf{x}(0) = \mathbf{x}_0. The rule eA+B=eAeBe^{A + B} = e^A e^B holds when AB=BAAB = BA and can fail otherwise. For a real 2×22 \times 2 system with det⁡A≠0\det A \ne 0, the eigenvalues classify the origin.

EigenvaluesOrigin
real with opposite signssaddle
real, distinct, same signnode
real and equal with A=λIA = \lambda Iproper node
real and equal with A≠λIA \ne \lambda Idegenerate node
α±iβ\alpha \pm i\beta with α≠0\alpha \ne 0, β≠0\beta \ne 0spiral
±iβ\pm i\beta with β≠0\beta \ne 0center

The origin is asymptotically stable when every eigenvalue has negative real part and unstable when one has positive real part. When every real part is at most 00 and some is 00, it is stable exactly when each eigenvalue with zero real part has as many independent eigenvectors as its multiplicity. For example the matrix with rows ⟨1,2⟩\langle 1, 2 \rangle and ⟨2,1⟩\langle 2, 1 \rangle has eigenvalues 33 and −1-1, with eigenvectors ⟨1,1⟩\langle 1, 1 \rangle and ⟨1,−1⟩\langle 1, -1 \rangle, so the origin is a saddle and

x=c1e3t[11]+c2e−t[1−1].\mathbf{x} = c_1 e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}.

Lessons Linear systems and eigenvalues

Nonlinear systems

For an autonomous system x′=f(x,y)x' = f(x, y), y′=g(x,y)y' = g(x, y) with ff and gg of class C1C^1 on an open set, the equilibria are the points where f=g=0f = g = 0, the crossings of the nullclines. The linearization at an equilibrium uses the Jacobian

J=[fxfygxgy](x∗,y∗).J = \begin{bmatrix} f_x & f_y \\ g_x & g_y \end{bmatrix}_{(x^*, y^*)}.

An equilibrium is hyperbolic when no eigenvalue of JJ has zero real part.

Hartman-Grobman theorem. If F=⟨f,g⟩\mathbf{F} = \langle f, g \rangle is C1C^1 on an open set containing a hyperbolic equilibrium, a homeomorphism of a neighborhood of the equilibrium onto a neighborhood of 0\mathbf{0} carries the trajectories of the nonlinear system onto those of the linearization and keeps the direction of time. So all eigenvalues with negative real part give an asymptotically stable sink, all with positive real part a source, and real eigenvalues of opposite signs an unstable saddle. If F\mathbf{F} is C2C^2, real distinct eigenvalues of one sign give a node and α±iβ\alpha \pm i\beta with α≠0\alpha \ne 0 and β≠0\beta \ne 0 give a spiral. The conclusion is local, in a neighborhood whose size the theorem does not give.

For example x′=yx' = y, y′=−sin⁡x−12yy' = -\sin x - \frac{1}{2}y has a stable spiral at the origin, where the eigenvalues have real part −14-\frac{1}{4}, and a saddle at (π,0)(\pi, 0), where det⁡J=−1\det J = -1. A nonhyperbolic equilibrium needs other tools, such as a conserved quantity or a radial equation. The system x′=−y−x(x2+y2)x' = -y - x(x^2 + y^2), y′=x−y(x2+y2)y' = x - y(x^2 + y^2) has a linear center at the origin, yet r′=−r3r' = -r^3, so every trajectory spirals into it.

The Lotka-Volterra model with a,b,c,d>0a, b, c, d > 0,

x′=ax−bxy,y′=−cy+dxy,\begin{aligned} x' &= ax - bxy, \\[0.4em] y' &= -cy + dxy, \end{aligned}

has a saddle at (0,0)(0, 0) and a linear center at (cd,ab)\left(\frac{c}{d}, \frac{a}{b}\right) with eigenvalues ±iac\pm i\sqrt{ac}. In the open first quadrant the quantity

H=dx−cln⁡x+by−aln⁡yH = dx - c\ln x + by - a\ln y

is constant along trajectories, so every trajectory there other than the equilibrium is a closed curve and the populations are periodic.

Lessons Nonlinear systems and phase-plane models