Vector Algebra and Calculus Concepts

Laws of Vector Algebra

• $A = â A$
• $A = â|A| = ân$
• $A = ŶAx + ŷAy + ZAz$
• $A = ✰Ax + ŷAy + ZAZ$
• $A^2 = A<em>x^2 + A</em>y^2 + A_z^2$

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,
    $A · (B × C) = \begin{vmatrix} Ax &amp; Ay &amp; Az \ Bx &amp; By &amp; Bz \ Cx &amp; Cy &amp; Cz \end{vmatrix}$
• 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 $AR, A<em>{θ}$, and $A</em>{ɸ}$ is written as
  • $A = â|A| = ÂAR + ÔÂ<em>{θ} + ÔÂ</em>{ɸ}$
• The position vector of point $P = (R1, θ<em>1, ɸ</em>1)$ 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:
  • $S = R² \int<em>{θ=30°}^{60°} sin θ dθ \int</em>{ɸ=0}^{2π} dɸ = 9 \big[ -cos θ \big]_{30°}^{60°} 2π$ $=18π (cos 30°-cos 60°) = 20.7 cm²$

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:
  • $Q= \int \rho_v dv = \int \int \int (4 cos² θ ) R² sinθ dR dθ dφ$
  • $=\frac{32}{3} \cdot 10^{-6} \int<em>{0}^{ π} sin θ cos² θ dθ \int</em>{0}^{2π} dφ = \frac{64}{9} \cdot 10^{-6} \int_{0}^{2π} dφ$
  • $=\frac{128π}{9} \cdot 10^{-6} = 44.68 µC$

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}$$