Vector Algebra and Calculus Concepts

Laws of Vector Algebra

• A = â A
• A = â|A| = ân
• A = ŶAx + ŷAy + ZAz
• A = ✰Ax + ŷAy + ZAZ
• A^2 = Ax^2 + Ay^2 + A_z^2

Properties of Vector Operations

• Equality of Two Vectors:
  • A = âA = Ax+ŷAy + Az,
  • B =ĥB = Bx +ŷBy +îВz
  • C=A+B=B+A
  • A = B if and only if A = B and \â = ɓ, which requires that Ax=Bx, Ay = By, and Az = B_z.

Position and Distance Vectors

• Position Vector: From origin to point P
  • R₁ = OP₁ = Âx1 + ŷyı + zz₁
• Distance Vector: Between two points
  • R12 = P1 P2 = R2 - R1 = (x2 − x1) + ŷ(y2 − y₁) + (Z2 — Z1)
  • the distance d between P₁ and P2 equals the magnitude of R12:
  • d = R12 = [(x2-x1)²+(y2 − y₁)² + (Z2 - Z₁) ^2]^{1/2}

Vector Multiplication: Scalar Product or "Dot Product"

• A B = AB COS_{AB}
• A = |A| = \sqrt{A · A}
• AB = cos^{-1} \frac{A. B}{\sqrt{A · A \sqrt{B. B}}}
• \hat{x} · \hat{y} = \hat{y} · \hat{y} = \hat{z} · \hat{z} = 1
• \hat{x} · \hat{y} = \hat{y} · \hat{x} = \hat{z} · \hat{x} = 0.
• A B = B A
• A. (B+C) = A B+A C
• If A = (Ax, Ay, Az) and B = (Bx, By, Bz), then
  • A · B = (Ax +ŷA₁ +zAz) · (ÂBx +ŷBy +îBz).
  • A B = AxBx + Ay By + AzBz

Vector Multiplication: Vector Product or "Cross Product"

• A × B = în AB sin O_{AB}
• AxB=-B × A
• A× (B+C) = A× B+A× C
• AxB =nn AB sin DAB
• A×A=0
• \hat{x} × \hat{y} = \hat{z},
• \hat{y} × \hat{z} = \hat{x},
• \hat{z} × \hat{x} = \hat{y}.
• Note the cyclic order (xyzxyz…).
• \hat{x} × \hat{ŷ} = \hat{y} × \hat{ŷ} = \hat{z} × \hat{z} = 0
• If A = (Ax, Ay, A_z) and B = (Bx, By, Bz), then

Example 3-1: Vectors and Angles

• Vector A points from the origin to point P₁ = (2, 3, 3), and vector B is directed from P₁ to point P2 = (1, -2, 2).
  • (a) vector A, its magnitude A, and unit vector â,
  • (b) the angle between A and the y-axis,
  • (c) vector B,
  • (d) the angle AB between A and B, and
  • (e) the perpendicular distance from the origin to vector B.
• Solution:
  • (a) Vector A is given by the position vector of P₁ = (2, 3, 3)
    • A = 2+ŷ3 + 23
    • A = |A|= \sqrt{2²+3²+3²} = \sqrt{22}
    • \hat{a} = \frac{A}{A} = (2+ŷ3+23)/\sqrt{22}
  • (b)
    • A ŷ= |A||ŷ| cos β = A cos β
    • β = cos^{-1} (\frac{3}{\sqrt{22}}) = 50.2°.
  • (c)
    • B = (1 − 2) + ŷ(-2-3) + 2(2-3) = − − ŷ5 — 2
  • (d)
    • O_{AB} = cos^{-1} (\frac{A.B}{||A||B|}) = cos^{-1} (\frac{(-2-15-3)}{\sqrt{22} \sqrt{30}}) = 145.1°.

Triple Products

• Scalar Triple Product
  • A · (B × C) = B · (C × A) = C · (A × B).
  • A · (B × C) = Ax(ByCz - BxCy) + Ay(BzCx - BxCz)+Az(BxCy-ByCx). Where,
    A · (B × C) = \begin{vmatrix} Ax & Ay & Az \ Bx & By & Bz \ Cx & Cy & Cz \end{vmatrix}
• Vector Triple Product
  • A × (B x C) = B(A·C) - C(A · B),

Cartesian Coordinate System

• Differential length vector
  • dl = ŷ dlx + ŷ dly +î dlz = ŷ dx +ŷ dy + z dz
• Differential area vectors
  • dsx = ŷdly dl₂ = ŷ dy dz
  • dsy = ŷ dx dz
  • ds₂ = 1 dx dy
• A differential volume equals the product of all three differential lengths:
  • dV = dx dy dz.

Cylindrical Coordinate System

• The position vector OP shown in Fig. 3-9 has components along and z only.
  • R₁ = OP = îr₁ + îz₁
• The mutually perpendicular base vectors are 1, 6, and 2, with * ↑ pointing away from the origin along r, ❖ pointing in a direction tangential to the cylindrical surface, and pointing along the vertical.
• The base unit vectors obey the following right-hand cyclic relations:
  • î × 6 = 2
  • $ xâ= 1
  • 2 × ŕ = o
• In cylindrical coordinates, a vector is expressed as
  • A = â|A| = îA, +$A$+zAz
  • dlr = dr
  • dlo = = r do
  • dlz = dz
• The differential length dl in cylindrical coordinates is given by
  • d1 = f dl, + $ dlo + n dl, = f dr+ội do+idz

Example 3-3: Distance Vector in Cylindrical Coordinates

• Find an expression for the unit vector of vector A shown in Fig. 3-11 in cylindrical coordinates.
• Solution:
  • A = OP₁₂- OP = fro― zh,
  • \hat{a} = \frac{A}{|A|} = \frac{fro-zh}{+h^2}

Example 3-4: Cylindrical Area

• Find the area of a cylindrical surface described by r = 5, 30° ≤ ≤ 60°, and 0 ≤ Z ≤ 3
• Solution:
  • S=r \int dφ \int dz =5 \int{π/6}^{π/3} dφ \int{0}^{3} dz =5 \frac π6 3 =
    \frac {5π}{2}

Spherical Coordinate System

• RXÔ =\, Ô xô = R, $xR=0
• A vector with components AR, A{θ}, and A{ɸ} is written as
  • A = â|A| = ÂAR + ÔÂ{θ} + ÔÂ{ɸ}
• The position vector of point P = (R1, θ1, ɸ1) is simply
  • R1=OP = ŔR₁,

Example 3-5: Surface Area in Spherical Coordinates

• The spherical strip shown in Fig. 3-15 is a section of a sphere of radius 3 cm. Find the area of the strip.
• Solution:
  • S = R² \int{θ=30°}^{60°} sin θ dθ \int{ɸ=0}^{2π} dɸ = 9 \big[ -cos θ \big]_{30°}^{60°} 2π =18π (cos 30°-cos 60°) = 20.7 cm²

Example 3-6: Charge in a Sphere

• A sphere of radius 2 cm contains a volume charge density \rho_v given by
  • \rho_v = 4 cos² θ
• Find the total charge Q contained in the sphere.
• Solution:
  • Q= \int \rho_v dv = \int \int \int (4 cos² θ ) R² sinθ dR dθ dφ
  • =\frac{32}{3} \cdot 10^{-6} \int{0}^{ π} sin θ cos² θ dθ \int{0}^{2π} dφ = \frac{64}{9} \cdot 10^{-6} \int_{0}^{2π} dφ
  • =\frac{128π}{9} \cdot 10^{-6} = 44.68 µC

Coordinate Transformations: Coordinates

• Select the coordinate system according to the geometry.
• Transform between coordinate systems if necessary.

Coordinate Transformations: Unit Vectors

• ŕ.✰ = cos ɸ
• ŕ y= sin ɸ
• ŷ.✰ = − sin ɸ
• ŷ. ŷ = cos ɸ
• î = ŷ cos ɸ + ŷ sin ɸ
• = −ŷ sin ɸ + ŷ cos ɸ
• x=rcos ɸ − 0 sind
• ŷ = sin ɸ + cos ɸ

Example 3-7: Cartesian to Cylindrical Transformations

• Given point P₁ = (3,-4, 3) and vector A = 82 - ŷ3 + z4, defined in Cartesian coordinates, express P₁ and A in cylindrical coordinates and evaluate A at P₁.
• Solution:
  • r =$\sqrt{x²+y²} =5
  • ɸ =tan^{-1}(\frac{y}{x}) = tan^{-1}(\frac{-4}{3}) =-53.1° = 306.9°
  • Ar=Ax cos ɸ + Ay sin ɸ = 2 cos ɸ - 3 sin ɸ
  • A_{ɸ} = -Ax sin ɸ + Ay cos ɸ = -2 sin ɸ - 3 cos ɸ

Example 3-8: Cartesian to Spherical Transformation

• Express vector A = (x + y) +ŷ(y-x)+zz in spherical coordinates.
• Solution:
  • AR = Ax sin θ cos ɸ + Ay sin θ sin ɸ + Az cos θ
  • =(x + y) sin θ cos ɸ + (y - x) sin θ sin ɸ + z cos θ
  • A_{θ}=(x+y)cos θ cos ɸ + (y-x) cos θ sin ɸ - z sin θ
  • A{ɸ}=-(x+y)sin ɸ +(y-x)cos ɸ
  • A = ÂAR + ÔA{θ} + ɸA{ɸ} = ÂR – ÔR sin θ
• leads to:
  • A_{θ} = 0
  • A_{ɸ} = -R sin θ

Gradient of a Scalar Field

• Gradient at point
  • rate of change of the scalar field
  • direction of change
• The symbol V is called the del or gradient operator and is defined as
  • V = \hat{x} \frac{∂}{∂x} + \hat{y} \frac{∂}{∂y} + \hat{z} \frac{∂}{∂z}

Example 3-9: Directional Derivative

• Find the directional derivative of T = x² + y²z along direction x2+ŷ322 and evaluate it at (1, −1, 2).
• Solution:
  • First, we find the gradient of T:
    • VT = \frac{∂}{∂x}(\hat{x}) + \frac{∂}{∂y}(\hat{y})) + \frac{∂}{∂z}(\hat{z}) (x² + y²z)
    • = 2x + ŷ2yz + y²
  • We denote I as the given direction,
    • l = x2 + ŷ3 — 2
    • \hat{a} = \frac{✰2 + ŷ3 — 2}{\sqrt{17}}
  • Application of Eq. (3.75) gives
    • dT = VT • â_{l} = (î2x + ŷ2yz + y²). \frac{✰2+ ŷ3 — 2}{\sqrt{17}}
    • =\frac{4x+6yz - 2y²}{\sqrt{17}}
    • At (1,1,2),
      • dT = \frac{4-12-2}{\sqrt{17}} = \frac{-10}{\sqrt{17}}

Divergence of a Vector Field

• div E= lim_{AV->0} \frac{\oint E \cdot ds}{AV}
• V E = div E = \frac{∂Ex}{∂x} + \frac{∂Ey}{∂y} + \frac{∂Ez}{∂z}

Example 3-11: Calculating the Divergence

• Determine the divergence of each of the following vector fields and then evaluate them at the indicated points:
  • (a) E = 3x²+ŷ2z+2x²z at (2, -2, 0);
  • (b) E = R(a³ cos θ /R²) – Ô (a³ sin θ/R²) at (a/2, 0, π).

Curl of a Vect

• Circulation
  • \oint B \cdot dl
  • V x B = curl B = lim{As ->0} {\frac{1}{As{max}} \oint B \cdot dl}
  • Thus, curl B is the circulation of B per unit area, with the area As of the contour C being oriented such that the circulation is maximum.

Stokes's Theorem

• Stokes's theorem converts the surface integral of the curl of a vector over an open surface S into a line integral of the vector along the contour C bounding the surface S.
• \int (V x B) \cdot ds = \oint B \cdot dl

Laplacian Operator

• Laplacian of a Scalar Field
  • \nabla²V = V · (VV) = \frac{∂²V}{∂x²} + \frac{∂²V}{∂y²} + \frac{∂²V}{∂z²}
• Laplacian of a Vector Field
  • V²E =\frac{∂²}{∂x²} + \frac{∂² }{∂y²} +\frac{∂²}{∂z²}$$