Vector Algebra and Calculus Concepts

Laws of Vector Algebra

  • A=a^AA = â A
  • A=a^A=a^nA = â|A| = ân
  • A=Y^Ax+y^Ay+ZAzA = ŶAx + ŷAy + ZAz
  • A=Ax+y^Ay+ZAZA = ✰Ax + ŷAy + ZAZ
  • A2=A<em>x2+A</em>y2+Az2A^2 = A<em>x^2 + A</em>y^2 + A_z^2

Properties of Vector Operations

  • Equality of Two Vectors:
    • A=a^A=Ax+y^Ay+AzA = âA = Ax+ŷAy + Az,
    • B=h^B=Bx+y^By+ı^ВzB =ĥB = Bx +ŷBy +îВz
    • C=A+B=B+AC=A+B=B+A
    • A=BA = B if and only if A=BA = B and \â = ɓ, which requires that Ax=BxAx=Bx, Ay=ByAy = By, and Az=BzAz = B_z.

Position and Distance Vectors

  • Position Vector: From origin to point P
    • R1=OP1=A^x1+y^yı+zz1R₁ = OP₁ = Âx1 + ŷyı + zz₁
  • Distance Vector: Between two points
    • R12=P1P2=R2R1=(x2x1)+y^(y2y1)+(Z2Z1)R12 = P1 P2 = R2 - R1 = (x2 − x1) + ŷ(y2 − y₁) + (Z2 — Z1)
    • the distance dd between P1P₁ and P2P2 equals the magnitude of R12R12:
    • d=R12=[(x2x1)2+(y2y1)2+(Z2Z1)2]1/2d = R12 = [(x2-x1)²+(y2 − y₁)² + (Z2 - Z₁) ^2]^{1/2}

Vector Multiplication: Scalar Product or "Dot Product"

  • AB=ABCOSABA B = AB COS_{AB}
  • A=A=AAA = |A| = \sqrt{A · A}
  • AB=cos1A.BAAB.BAB = cos^{-1} \frac{A. B}{\sqrt{A · A \sqrt{B. B}}}
  • x^y^=y^y^=z^z^=1\hat{x} · \hat{y} = \hat{y} · \hat{y} = \hat{z} · \hat{z} = 1
  • x^y^=y^x^=z^x^=0.\hat{x} · \hat{y} = \hat{y} · \hat{x} = \hat{z} · \hat{x} = 0.
  • AB=BAA B = B A
  • A.(B+C)=AB+ACA. (B+C) = A B+A C
  • If A=(Ax,Ay,Az)A = (Ax, Ay, Az) and B = (Bx, By, Bz), then
    • AB=(Ax+y^A1+zAz)(A^Bx+y^By+ı^Bz)A · B = (Ax +ŷA₁ +zAz) · (ÂBx +ŷBy +îBz).
    • AB=AxBx+AyBy+AzBzA B = AxBx + Ay By + AzBz

Vector Multiplication: Vector Product or "Cross Product"

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

Example 3-1: Vectors and Angles

  • Vector A points from the origin to point P1=(2,3,3)P₁ = (2, 3, 3), and vector B is directed from P1P₁ to point P2=(1,2,2)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)P₁ = (2, 3, 3)
      • A=2+y^3+23A = 2+ŷ3 + 23
      • A=A=22+32+32=22A = |A|= \sqrt{2²+3²+3²} = \sqrt{22}
      • a^=AA=(2+y^3+23)/22\hat{a} = \frac{A}{A} = (2+ŷ3+23)/\sqrt{22}
    • (b)
      • Ay^=Ay^cosβ=AcosβA ŷ= |A||ŷ| cos β = A cos β
      • β=cos1(322)=50.2°.β = cos^{-1} (\frac{3}{\sqrt{22}}) = 50.2°.
    • (c)
      • B=(12)+y^(23)+2(23)=y^52B = (1 − 2) + ŷ(-2-3) + 2(2-3) = − − ŷ5 — 2
    • (d)
      • OAB=cos1(A.BAB)=cos1((2153)2230)=145.1°.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) = B · (C × A) = C · (A × B).
    • A(B×C)=Ax(ByCzBxCy)+Ay(BzCxBxCz)+Az(BxCyByCx)A · (B × C) = Ax(ByCz - BxCy) + Ay(BzCx - BxCz)+Az(BxCy-ByCx). Where,
      A(B×C)=Axamp;Ayamp;Az Bxamp;Byamp;Bz Cxamp;Cyamp;CzA · (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×(BxC)=B(AC)C(AB)A × (B x C) = B(A·C) - C(A · B),

Cartesian Coordinate System

  • Differential length vector
    • dl=y^dlx+y^dly+ı^dlz=y^dx+y^dy+zdzdl = ŷ dlx + ŷ dly +î dlz = ŷ dx +ŷ dy + z dz
  • Differential area vectors
    • dsx=y^dlydl2=y^dydzdsx = ŷdly dl₂ = ŷ dy dz
    • dsy=y^dxdzdsy = ŷ dx dz
    • ds2=1dxdyds₂ = 1 dx dy
  • A differential volume equals the product of all three differential lengths:
    • dV=dxdydz.dV = dx dy dz.

Cylindrical Coordinate System

  • The position vector OP shown in Fig. 3-9 has components along and z only.
    • R1=OP=ı^r1+ı^z1R₁ = OP = îr₁ + îz₁
  • The mutually perpendicular base vectors are 1, 6, and 2, with * pointing away from the origin along rr, 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:
    • ı^×6=2î × 6 = 2
    • $ xâ= 1
    • 2×rˊ=o2 × ŕ = o
  • In cylindrical coordinates, a vector is expressed as
    • A = â|A| = îA, +$A$+zAz
    • dlr=drdlr = dr
    • dlo==rdodlo = = r do
    • dlz=dzdlz = 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=OP12OP=frozh,A = OP₁₂- OP = fro― zh,
    • a^=AA=frozh+h2\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°r = 5, 30° ≤ ≤ 60°, and 0Z30 ≤ Z ≤ 3
  • Solution:
    • S=rdφdz=5<em>π/6π/3dφ</em>03dz=5π63=<br/>5π2S=r \int dφ \int dz =5 \int<em>{π/6}^{π/3} dφ \int</em>{0}^{3} dz =5 \frac π6 3 =<br /> \frac {5π}{2}

Spherical Coordinate System

  • RXO^=RXÔ =\, Ô xô = R, $xR=0
  • A vector with components AR,A<em>θAR, A<em>{θ}, and A</em>ɸA</em>{ɸ} is written as
    • A=a^A=A^AR+O^A^<em>θ+O^A^</em>ɸA = â|A| = ÂAR + ÔÂ<em>{θ} + ÔÂ</em>{ɸ}
  • The position vector of point P=(R1,θ<em>1,ɸ</em>1)P = (R1, θ<em>1, ɸ</em>1) is simply
    • R1=OP=RˊR1,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<em>θ=30°60°sinθdθ</em>ɸ=02πdɸ=9[cosθ]30°60°2πS = R² \int<em>{θ=30°}^{60°} sin θ dθ \int</em>{ɸ=0}^{2π} dɸ = 9 \big[ -cos θ \big]_{30°}^{60°} 2π =18π(cos30°cos60°)=20.7cm2=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 ρv\rho_v given by
    • ρv=4cos2θ\rho_v = 4 cos² θ
  • Find the total charge Q contained in the sphere.
  • Solution:
    • Q=ρvdv=(4cos2θ)R2sinθdRdθdφQ= \int \rho_v dv = \int \int \int (4 cos² θ ) R² sinθ dR dθ dφ
    • =323106<em>0πsinθcos2θdθ</em>02πdφ=64910602πdφ=\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
    • =128π9106=44.68µC=\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

  • rˊ.=cosɸŕ.✰ = cos ɸ
  • rˊy=sinɸŕ y= sin ɸ
  • y^.=sinɸŷ.✰ = − sin ɸ
  • y^.y^=cosɸŷ. ŷ = cos ɸ
  • ı^=y^cosɸ+y^sinɸî = ŷ cos ɸ + ŷ sin ɸ
  • =y^sinɸ+y^cosɸ= −ŷ sin ɸ + ŷ cos ɸ
  • x=rcosɸ0sindx=rcos ɸ − 0 sind
  • y^=sinɸ+cosɸŷ = sin ɸ + cos ɸ

Example 3-7: Cartesian to Cylindrical Transformations

  • Given point P1=(3,4,3)P₁ = (3,-4, 3) and vector A=82y^3+z4A = 82 - ŷ3 + z4, defined in Cartesian coordinates, express P1P₁ and A in cylindrical coordinates and evaluate A at P1P₁.
  • Solution:
    • r =$\sqrt{x²+y²} =5</li><li></li> <li>ɸ =tan^{-1}(\frac{y}{x}) = tan^{-1}(\frac{-4}{3}) =-53.1° = 306.9°</li><li></li> <li>Ar=Ax cos ɸ + Ay sin ɸ = 2 cos ɸ - 3 sin ɸ</li><li></li> <li>A_{ɸ} = -Ax sin ɸ + Ay cos ɸ = -2 sin ɸ - 3 cos ɸ</li></ul></li></ul><h3id="example38cartesiantosphericaltransformation">Example38:CartesiantoSphericalTransformation</h3><ul><li>Expressvector</li></ul></li> </ul> <h3 id="example38cartesiantosphericaltransformation">Example 3-8: Cartesian to Spherical Transformation</h3> <ul> <li>Express vectorA = (x + y) +ŷ(y-x)+zzinsphericalcoordinates.</li><li>Solution:<ul><li>in spherical coordinates.</li> <li>Solution:<ul> <li>AR = Ax sin θ cos ɸ + Ay sin θ sin ɸ + Az cos θ</li><li></li> <li>=(x + y) sin θ cos ɸ + (y - x) sin θ sin ɸ + z cos θ</li><li></li> <li>A_{θ}=(x+y)cos θ cos ɸ + (y-x) cos θ sin ɸ - z sin θ</li><li></li> <li>A{ɸ}=-(x+y)sin ɸ +(y-x)cos ɸ</li><li></li> <li>A = ÂAR + ÔA{θ} + ɸA{ɸ} = ÂR – ÔR sin θ</li></ul></li><li>leadsto:<ul><li></li></ul></li> <li>leads to:<ul> <li>A_{θ} = 0</li><li></li> <li>A_{ɸ} = -R sin θ</li></ul></li></ul><h3id="gradientofascalarfield">GradientofaScalarField</h3><ul><li>Gradientatpoint<ul><li>rateofchangeofthescalarfield</li><li>directionofchange</li></ul></li><li>ThesymbolViscalledthedelorgradientoperatorandisdefinedas<ul><li></li></ul></li> </ul> <h3 id="gradientofascalarfield">Gradient of a Scalar Field</h3> <ul> <li>Gradient at point <ul> <li>rate of change of the scalar field</li> <li>direction of change</li></ul></li> <li>The symbol V is called the del or gradient operator and is defined as<ul> <li>V = \hat{x} \frac{∂}{∂x} + \hat{y} \frac{∂}{∂y} + \hat{z} \frac{∂}{∂z} </li></ul></li></ul><h3id="example39directionalderivative">Example39:DirectionalDerivative</h3><ul><li>Findthedirectionalderivativeof</li></ul></li> </ul> <h3 id="example39directionalderivative">Example 3-9: Directional Derivative</h3> <ul> <li>Find the directional derivative ofT = x² + y²zalongdirectionalong directionx2+ŷ322andevaluateitatand evaluate it at(1, −1, 2).</li><li>Solution:<ul><li>First,wefindthegradientofT:<ul><li>.</li> <li>Solution:<ul> <li>First, we find the gradient of T:<ul> <li>VT = \frac{∂}{∂x}(\hat{x}) + \frac{∂}{∂y}(\hat{y})) + \frac{∂}{∂z}(\hat{z}) (x² + y²z)</li><li></li> <li>= 2x + ŷ2yz + y²</li></ul></li><li>WedenoteIasthegivendirection,<ul><li></li></ul></li> <li>We denote I as the given direction,<ul> <li>l = x2 + ŷ3 — 2</li><li></li> <li>\hat{a} = \frac{✰2 + ŷ3 — 2}{\sqrt{17}}</li></ul></li><li>ApplicationofEq.(3.75)gives<ul><li></li></ul></li> <li>Application of Eq. (3.75) gives<ul> <li>dT = VT • â_{l} = (î2x + ŷ2yz + y²). \frac{✰2+ ŷ3 — 2}{\sqrt{17}}</li><li></li> <li>=\frac{4x+6yz - 2y²}{\sqrt{17}}</li><li>At(1,1,2),<ul><li></li> <li>At (1,1,2),<ul> <li>dT = \frac{4-12-2}{\sqrt{17}} = \frac{-10}{\sqrt{17}}</li></ul></li></ul></li></ul></li></ul><h3id="divergenceofavectorfield">DivergenceofaVectorField</h3><ul><li></li></ul></li></ul></li></ul></li> </ul> <h3 id="divergenceofavectorfield">Divergence of a Vector Field</h3> <ul> <li>div E= lim_{AV->0} \frac{\oint E \cdot ds}{AV}</li><li></li> <li>V E = div E = \frac{∂Ex}{∂x} + \frac{∂Ey}{∂y} + \frac{∂Ez}{∂z}</li></ul><h3id="example311calculatingthedivergence">Example311:CalculatingtheDivergence</h3><ul><li>Determinethedivergenceofeachofthefollowingvectorfieldsandthenevaluatethemattheindicatedpoints:<ul><li>(a)</li> </ul> <h3 id="example311calculatingthedivergence">Example 3-11: Calculating the Divergence</h3> <ul> <li>Determine the divergence of each of the following vector fields and then evaluate them at the indicated points:<ul> <li>(a)E = 3x²+ŷ2z+2x²zatat(2, -2, 0);</li><li>(b);</li> <li>(b)E = R(a³ cos θ /R²) – Ô (a³ sin θ/R²)atat(a/2, 0, π).</li></ul></li></ul><h3id="curlofavect">CurlofaVect</h3><ul><li>Circulation<ul><li>.</li></ul></li> </ul> <h3 id="curlofavect">Curl of a Vect</h3> <ul> <li>Circulation<ul> <li>\oint B \cdot dl</li><li></li> <li>V x B = curl B = lim{As ->0} {\frac{1}{As{max}} \oint B \cdot dl}</li><li>Thus,curlBisthecirculationofBperunitarea,withtheareaAsofthecontourCbeingorientedsuchthatthecirculationismaximum.</li></ul></li></ul><h3id="stokesstheorem">StokessTheorem</h3><ul><li>StokesstheoremconvertsthesurfaceintegralofthecurlofavectoroveranopensurfaceSintoalineintegralofthevectoralongthecontourCboundingthesurfaceS.</li><li></li> <li>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.</li></ul></li> </ul> <h3 id="stokesstheorem">Stokes's Theorem</h3> <ul> <li>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.</li> <li>\int (V x B) \cdot ds = \oint B \cdot dl</li></ul><h3id="laplacianoperator">LaplacianOperator</h3><ul><li>LaplacianofaScalarField<ul><li></li> </ul> <h3 id="laplacianoperator">Laplacian Operator</h3> <ul> <li>Laplacian of a Scalar Field<ul> <li>\nabla²V = V · (VV) = \frac{∂²V}{∂x²} + \frac{∂²V}{∂y²} + \frac{∂²V}{∂z²}</li></ul></li><li>LaplacianofaVectorField<ul><li></li></ul></li> <li>Laplacian of a Vector Field<ul> <li>V²E =\frac{∂²}{∂x²} + \frac{∂² }{∂y²} +\frac{∂²}{∂z²}$$