MA2104 NotesVectors

1) Two vectors are equal if they have the same length and direction2) Position Vector u=⟨ a ,b , c ⟩ - where (a ,b , c ) is the terminal point when the initial point is the origin3) Directed Line Segment u=A⃑B=O⃑B−O⃑A- where O⃑B and O⃑A are position vectors of (b1, b2 , b3 ) and (a1 , a2 , a3), respectively4) Length of Vectors- The length of vector u=⟨u1, u2 ,u3 ⟩ is given by||u||=√u12+u22+u325) Unit Vector- A unit vector is a vector with length ¿1

w= u||u||6) Standard Basis of Vectors- And any 3D vector can be written as a linear combination of the standard basis vectors

i= ⟨1,0,0 ⟩ , j= ⟨0,1,0 ⟩ , k=⟨ 0,0,1 ⟩⟨a ,b , c ⟩=ai+b j+c k

Vector Operations1) Addition- If a=⟨ a1 , a2 , a3 ⟩ and b=⟨ b1 , b2 ,b3 ⟩, then

a+b=⟨ a1+b1 , a2+b2 , a3+b3 ⟩2) Scalar Multiplication c u=c ⟨u1 ,u2 , u3 ⟩=⟨ cu1 , cu2, c u3 ⟩

c u=0 iff c=0 or u=0- Length ¿|c|||u||- Direction same as u if c>0 and opposite if c<0 3) Dot Producta ⋅b=⟨a1 , a2 , a3 ⟩ ⋅ ⟨b1 , b2 , b3 ⟩=a1b1+a2b2+a3b3

a ⋅b=0does not imply a=0 or b=0- Dot Product Angle Formulaa ⋅b=||a||||b||cosθ- a ⋅b=b ⋅ a- a ⋅ (b+c )=(a ⋅b )+(a ⋅c )- (d a ) ⋅b=d (a ⋅b )=a ⋅ (d b )- 0 ⋅a=0

- a ⋅a=||a||2=(√a12+a2

2+a32 )2=a1

2+a22+a3

2

- a ⋅b=0 for non-zero a ,b iff they are orthogonal4) Cross Producta×b= ⟨a1 , a2 , a3 ⟩× ⟨b1 , b2 ,b3 ⟩=⟨|a2 b2

a3 b3|,−|a1 b1a3 b3|,|a1 b1

a2 b2|⟩- a×b=−b×a- a× (b+c )= (a×b )+ (a×c )- (a+b )×c= (a×c )+(b×c )- (d a )×b=d (a×b )=a× (d b )- 0×a=a×0=0- a×a=0- a×b=0 for non-zero a ,b iff they are parallel or antiparallel

- Cross Product Angle Formula||a×b||=||a||||b||sin θ or a×b=||a||||b||sin θn

Area/Volume of ShapesArea/Volume Image

Triangle 12||a×b||

Parallelogram ||a||×||b||sin θ=||a×b||

Parallelepiped ||b×c||||a||cosθ=|a ⋅ (b×c )|

Special AnglesDirection θ a ⋅b a×bSame 0 ||a||||b|| 0Opposite π −||a||||b|| 0Orthogonal π 0 ||a×b||=||a||||b||

Coplanar and Collinear1) a ,b , and c are coplanar if |a⋅ (b×c )|=0 2) a ,b , and c are collinear if ||(a−b )× (b−c )||=0 (Any combination of vectors work for both)

Line1) Vector Equation of a line- Any point r on the line can be described as

r=r0+t v=⟨ x0 , y0 , z0 ⟩+t ⟨a ,b , c ⟩

where r0 is a point on the line and v is a vector parallel to the line2) Parametric Equation of a linex=x0+at , y= y0+bt , z=z0+ct3) Symmetric Equation

x−x0a

=y− y0b

=z−z0c

Planes4) Vector Equation of a plane- Any point r on the plane can be described as

n ⋅ (r−r0 )=0

⟨a ,b , c ⟩ ⋅ (r−⟨x0 , y0 , z0 ⟩ )=0where r0 is a point on the plane and n is a vector orthogonal to the plane5) Linear Equation of a planeax+by+cz=d=ax0+b y0+c z0

a (x−x0 )+b ( y− y0 )+c ( z−z0 )=0where ⟨a , b , c ⟩ is a normal to the planeProjections

6) Scalar Projections (Component)- ||⃑PS|| is the scalar projection of b onto a, or component of b along acom pab=||⃑PS||=||b||cosθ=a⋅ b

||a||- com pλ ab=com pab if λ>0 and com pλ ab=−com pab if λ<0 7) Vector Projections- projab is the vector projection of b onto a

projab=com pab ( a||a||)=a ⋅b

||a|| ( a||a||)=a ⋅b

||a||2a=

a ⋅ba ⋅a a

where a ⋅ba ⋅a is a scalarShortest Distance

1) Point to Point- 2D: (x1 , y1 ) and (x2 , y2 )

d=√(x2−x1 )2+( y2− y1 )2

- 3D: (x1 , y1 , z1 ) and (x2 , y2 , z2)d=√(x2−x1 )2+( y2− y1 )

2+( z2−z2 )22) Point to Line||⃑PQ||sin θ=||⃑PQ×P⃑R||

||⃑PR||3) Point to Plane||⃑PQ||cosθ=

|⃑PQ ⋅n|||n||4) Line to Line- Parallel and non-equal:

Same as point to line- Non-Parallel and Skew:For lines l1=r1+t v1 and l2=r2+t v2

d=(r 1−r2 ) ⋅n

||n||, n=v1×v2

5) Line to Plane (Parallel)- Same as point to plane6) Plane to Plane (Parallel)- Let P and Q be points on each of the planes, and n be the normal to both planes. Then,P1 : ax+by+cd+d1=0 , P2: ax+by+cd+d2=0

θn

P

Q

d=|P⃑Q ⋅n||n|| |=|d1−d2|

||n||

Intersections1) Line and Line- Parallel: Direction vectors parallel or antiparallel

Equal: Share all points Non-equal: Share no points- Non-parallel: Direction vectors are non-parallel Intersect at point: Consistent linear system Skew (non-intersecting): Inconsistent linear system2) Line and Plane- Parallel No intersection: Normal of plane orthogonal to direction vector of line Full intersection: Line contained in plane- Non-parallel Intersect at point: Consistent linear system3) Plane and Plane- Parallel: Normal vectors parallel or antiparallel Equal: Share all lines Non-equal: Share no lines- Non-parallel: Intersect in a straight line Angle θ between planes is angle between normal vectors To find line of intersection, set x , y ,∨z=t and solve linear system (Direction vector is cross of 2 normals)

Partial DerivativesIf f is a function of two variables, its partial derivatives f x and f y are defined by:f x ( x , y )= ∂ f

∂ x=lim

h→0

f ( x+h , y )−f ( x , y )h

f y ( x , y )=∂ f∂ y=lim

h→0

f ( x , y+h )−f (x , y )h

The curve C1 is the graph of the function g ( x )=f (x ,b) which is the intersection of the surface and the plane y=b.The slope of the tangent T 1 at P is: g' (a )=f x (a ,b ) .

The curve C2 is the graph of the function h ( y )= f (a , y ) which is the intersection of the surface and the plane x=a.The slope of the tangent T 2 at P is: h' (b )=f y (a ,b ) .

Higher Partial DerivativesIf z=f (x1, x2 ,…), we use the notation( ( f x1 )x2)⋱=f x1x2…= ∂2 f

…∂x2∂x1=…( ∂

∂x2 ( ∂ f∂ x1 ))

Clairaut’s TheoremSuppose f is defined on a disk D that contains (a ,b ). If the functions f xy and f yx are both continuous on D, then,f xy (a ,b )=f yx(a ,b)As long as the number of variables occurring in the subscript are the same, then the partial derivatives are the same, assuming the continuity of nth order partial derivatives in a disk containing (a ,b)

E.g., for f ( x , y , z ) ,

f=xyzzxyzx= fzyyxyzyx= fzyyxyzyx=…Tangent Plane EquationThe tangent line to the curve y= f (x ) at x=a isy=f (a )+ f ' (a) (x−a )

The tangent plane to the surface S at the point P(a ,b ,c ) is the plane containing both tangent lines T 1 and T 2 as shown

where P=(a ,b , f (a ,b ) ) and T 1 and T 2 are the tangent lines to the curves C1 and C2, the intersections of S with y=b and x=a

A vector parallel to T 1 is ⟨1,0 , f x (a , b ) ⟩ and T 2 is ⟨0,1 , f y (a ,b ) ⟩. Thus, a normal to the plane is n=⟨0,1 , f y (a ,b ) ⟩ × ⟨1,0 , f x (a ,b ) ⟩=⟨ f x (a ,b ) , f y (a ,b ) ,−1 ⟩* Points in −z direction

Thus, the tangent plane is given byz=f (a ,b )+ f x (a ,b ) ( x−a )+ f y (a ,b ) ( y−b )

General Formula for Tangent planeLet f≔ f (x1 , x2 , x3 ,…). Then, the tangent plane to f at (a1, a2 , a3 ,…) isz=f (a1 , a2 , a3 ,… )+f x1 (a1 ,a2 , a3,…) (x1−a1 )+f x2 (a1, a2 , a3 ,…) (x2−a2 )+…

Chain RuleSuppose that u is a differentiable function of n variables x1 ,…, xn and each x j is a differentiablefunction of m variables t 1 ,…, tm. Then u is a function of t 1 ,…, tmand∂u∂ ti

= ∂u∂ x1

∂ x1∂t i

+ ∂u∂x2

∂ x2∂ t i

+…+ ∂u∂ x2

∂ x2∂ ti

Draw tree diagrame .g . ∂w∂u

= ∂w∂x

∂ x∂u

+ ∂w∂ y

∂ y∂u

+ ∂w∂z

∂ z∂u

+ ∂ w∂t

∂ t∂u

Implicit DifferentiationWe say that z is an implicit function of x and y defined on F ( x , y , z )=0 if for every choice of x and y, the value of z is determined by F ( x , y , z )=0.Suppose the equation F ( x , y , z )=0, where F is differentiable, defines z implicity as a differentiable function of x and y. Then,

∂ z∂x

=

−F x ( x , y , z )F z ( x , y , z )

∧∂ z

∂ y=

−F y ( x , y , z )F z ( x , y , z )

GradientThe gradient of f (x , y ) is the vector-valued function∇ f ( x , y )=⟨ f x , f y ⟩Directional DerivativesThe directional derivative of f ( x , y ) at (x0 , y0) in the direction of unit vector u=⟨ a , b⟩ is

Du f (x0 , y0 )=limh→0

f ( x0+ha , y0+hb )−f (x0, y0 )hwhich is the rate of change of the function at the point (x0 , y0) in the direction given by u.

If f ( x , y ) is a differentiable function, then f has a directional derivative in the direction of any unit vector u=⟨ a ,b ⟩ and Du f ( x , y )=f x ( x , y )a+ f y ( x , y )b=⟨ f x , f y ⟩ ⋅ ⟨ a ,b ⟩=⟨ f x , f y ⟩ ⋅u=∇ f (x , y )⋅u

f x and f y are special cases of directional derivatives where u=i and u= j respectively. (Similar for 3D. Just add one more coordinate value.)

Gradient VectorSuppose f ( x , y ) is a differentiable function of x and y at (x0 , y0). Suppose ∇ f (x0 , y0)≠ 0. Then, ∇ f (x0 , y0)

is the normal to the level curve f ( x , y )=k that contains the point (x0 , y0).Let ⟨ a ,b ⟩ be the normal vector ∇ f (x0 , y0). Then the tangent in rotated clockwise 90 ° is given by ⟨b ,−a ⟩ .

Suppose F ( x , y , z ) is a differentiable function of x,y and z at (x0 , y0 , z0). Suppose ∇F (x0 , y0 , z0)≠0. Then, ∇F (x0 , y0, z0)

is the normal to the level curve F ( x , y , z )=k that contains the point (x0 , y0 , z0).∇F (x0 , y0 , z0) is also the normal vector to the tangent plane to the level surface at (x0 , y0 , z0).Tangent Plane to Level Surface

∇F (x0 , y0 , z0) ⋅ ⟨x−x0 , y− y0 , z−z0 ⟩=0

Maximum/Minimum Rate of ChangeSuppose f is a differentiable function of two or three variables. Let P be a point. Assume ∇ f (P)≠0. ∇ f (P) points in the direction of max rate of change of f at P. Max value of Du f (P ) is |∇ f (P)|

−∇ f (P) points in the direction of min rate of change of f at P. Min value of Du f (P ) is −|∇ f (P)|

Thus, f increases fastest in the direction of the gradient vector, and minimum in the opposite.

Maximum/Minimum Points4) Local MaximumLet f ( x , y ):D→R. Then f has a local maximum at (a ,b ) if

f ( x , y )≤ f (a ,b ) for all points close to (a ,b)

5) Local MinimumLet f ( x , y ):D→R. Then f has a local maximum at (a ,b ) if f ( x , y )≥ f (a ,b ) for all points close to (a ,b)

If f has a local maximum or minimum at (a ,b ) and the first-order derivatives of fexist there, then f x (a ,b )=f y (a ,b )=0

A maximum and minimum may not always exist on the domain, but it will exist on a closed and bounded set D⊆R2 at some points (x1 , y1 ) and (x2 , y2).Saddle PointsLetf ( x , y ):D→R. Then a point (a ,b) is called a saddle point of f if f x (a ,b )=f y (a ,b )=0 and every neighbourhood at (a ,b ) contains points ( x , y )∈D for which f ( x , y )< f (a ,b ) and points ( x , y )∈D for which f ( x , y )> f (a ,b )

Second Derivative Test for Extreme Valuesi) f has a local maximum at (a ,b ) if f xx<0 and f xx f yy− f xy

2 >0 at (a ,b )

ii) f has a local minimum at (a ,b ) if f xx>0 and f xx f yy−f xy2 >0 at (a ,b )

iii) f has a saddle point at (a ,b ) if f xx f yy− f xy2 <0 at (a ,b )

iv) The test is inconclusive at (a ,b ) if f xx f yy−f xy2 =0 at (a ,b ). Another method is required.

Closed and Bounded Sets in R21) Closed Set- A set R⊆R2 is closed if it contains all its boundary points. A boundary point of R is a point (a ,b) such that every disk with centre

(a ,b) contains points in R and also points in R2 ¿

2) Bounded Set- A set R⊆R2 is bounded if it is contained within some disk, said to be finite in extent.

Optimisation ProblemsThe extremum of f ( x , y ) can occur only at

i) Boundary points of the domain of fii) Critical points (interior points where f x=f y=0¿ or points where f x and f y fail to existMethod to find maximum and minimumStep 1: Find the values of f at critical points in DStep 2: Find the extreme values of f on the boundary of DStep 3: Take the largest and smallest values across both stepsLagrange MultiplierSuppose f ( x , y ) and g ( x , y ) are differentiable functions such that ∇ g ( x , y )≠0 on the constraint curve g ( x , y )=k. Suppose that the minimum/maximum value of f (x , y ) subject to the constraint g ( x , y )=k occurs at (x0 , y0 ). Then,

∇ f (x0 , y0 )=λ∇ g (x0 , y0 )

for some constant λ, the Lagrange multiplier.(Similar for 3D. Just add one more coordinate value.)

Double Integrals1) Volume- If f ( x , y )≥0, then the volume V of the solid that lies above the region D and below the surface z=f (x , y ) is

V=∫∫Df ( x , y )dA

2) Area- If f ( x , y )=1, then the area A of the plane region that lies above the region D is given by

A=∫∫D1dA

3) Iterated Double Integral- If f is continuous on the rectangle R=[a ,b ]× [c ,d ], then ∫∫

Rf ( x , y ) dA=∫

a

b

∫c

d

f ( x , y ) dy dx=∫c

d

∫a

b

f ( x , y )dx dy

2 Types of RegionsA plane region is said to be of Type 1 if it lies between the graphs of two continuous functions of x. That is,

D= {( x , y ) :a≤x ≤b ,g1 ( x )≤ y≤ g2 ( x ) }

∫∫Df ( x , y )dA=∫

a

b

∫g1 ( x )

g2 (x)

f ( x , y )dy dx

A plane region is said to be of Type 2 if it lies between the graphs of two continuous functions of y. That is,

D= {( x , y ) :c≤ y ≤d ,h1 ( y )≤ x≤h2 ( y ) }

∫∫Df ( x , y )dA=∫

c

d

∫h1( y)

h2( y)

f ( x , y )dxdy

Polar RegionsIf f is continuous on a polar rectangle R given byR={(r , θ ) :0≤a≤ r ≤b ,α ≤θ≤ β

where 0≤ β−α ≤2π , then∫∫

Rf ( x , y )dA=∫

α

β

∫a

b

f (r cosθ , r sin θ ) r dr dθ

2 Types of Non-Polar Rectangular RegionsA polar region is said to be of Type 1 if lies in region D ,

D={(r , θ ) : 0≤a≤ r ≤b ,g1 (r )≤θ≤ g2 (r )}

∫∫Df ( x , y )dA=∫

a

b

∫g1 (r )

g2 (r )

f (r cosθ ,r sin θ ) r dθdr

A polar region is said to be of Type 2 if lies in region D ,

D={(r , θ ): α ≤θ≤ β ,h1 (θ )≤r≤h2 (θ ) }

∫∫Df ( x , y )dA=∫

α

β

∫h1 (θ)

h2 (θ)

f (r cosθ , rsin θ )r dr dθ

To express r in term of θ, given a circle of radius r0 centred at (x0 , y0 ),(x−x0 )2+( y− y0 )2=r 0

2

solve for(r cosθ−x0 )

2+(r sin θ− y0 )2=r02

Polar Parametrisation of Ellipse and CircleFor an ellipse centred at (x0 , y0) given by( x−x0

a )2

+( y− y0b )

2

=1

the parametrisation is given by r (t )=⟨acos t+x0, b sin t+ y0 ⟩ ,0≤ t ≤ π2

Triple Integral1) Type 1

A solid region E is of Type 1 if it lies between the graphs of two continuous functions of x and y. That is,

E={( x , y , z ) : ( x , y )∈D ,u1 (x , y )≤ z≤u2 ( x , y ) }

where D is the projection of E onto the xy-plane.

∭E

f (x , y , z )dV=∬D

∫u1 (x , y )

u2 (x , y )

f (x , y , z )dz dA

2) Type 2A solid region E is of Type 2 if it lies between the graphs of two continuous functions of y and z. That is,

E={( x , y , z ) : ( y , z )∈D,u1 ( y , z )≤x≤u2 ( y , z ) }

where D is the projection of E onto the yz-plane.

∭E

f (x , y , z )dV=∬D

∫u1 ( y , z )

u2 ( y , z )

f ( x , y , z)dx dA

3) Type 3

A solid region E is of Type 3 if it lies between the graphs of two continuous functions of x and z. That is,E={( x , y , z ) : ( x , z )∈D ,u1 ( x , z )≤ y≤u2 (x , z ) }

where D is the projection of E onto the xz-plane.

∭E

f (x , y , z )dV=∬D

∫u1 (x , z )

u2 (x , z )

f ( x , y , z )dy dA

Cylindrical CoordinatesRectangular to Cylindrical Cylindrical to Rectangular

x=r cosθ

y=rsin θ

z=z

r2=x2+ y2

tanθ= yx, x≠0

z=z

If the projection D of the solid E onto the xy-plane can be conveniently described in terms of polar coordinates, then we can use cylindrical coordinates when computing the triple integral.This is a Type 1 solid given by

E={( x , y , z ) : ( x , y )∈D ,u1 (x , y )≤ z≤u2 ( x , y ) }

where D can be described in terms of polar co-ordinates:D={(r , θ ): α ≤θ≤ β ,h1 (θ )≤r≤h2 (θ ) }

Combining, we haveE={(r , θ , z ) :α≤θ≤β ,h1 (θ )≤r ≤h2 (θ ) , u1 (r cosθ , rsin θ )≤ z≤u2(r cosθ , r sin θ)}

Suppose E is a solid described as above. Then,∭

Ef (x , y , z )dV=∫

α

β

∫h1 ( θ)

h2 ( θ)

∫u1 (r cosθ , r sinθ )

u2(r cosθ , r sinθ )

f (r cosθ , r sin θ , z )r dz dr dθ

Spherical CoordinatesRectangular to Spherical Spherical to Rectangular

x=ρ sin ϕ cosθ

y=ρsin ϕsinθ

z=ρ cos ϕ

ρ2=x2+ y2+z2

(ρ sin ϕ )2=x2+ y2

Spherical coordinates simplify equations and integrals for regions involving spheres or cones, and is used for regions where E is a spherical wedge or a more general region in terms of spherical coordinatesA spherical wedge is a solid given by

E={( ρ ,θ ,ϕ ):a≤ ρ≤b ,α ≤θ≤ β , c ≤ϕ≤d }

∭E

f (x , y , z )dV=∫c

d

∫α

β

∫a

b

f (ρsin ϕcosθ , ρ sinϕ sin θ , ρcos ϕ )ρ2 sinϕ dρ dθdϕ

A general spherical region is a solid given byE={( ρ ,θ ,ϕ ) :c ≤ϕ≤d ,α ≤θ≤ β ,g1(θ ,ϕ)≤ ρ≤g2(θ ,ϕ)}

∭E

f (x , y , z ) dV=∫c

d

∫α

β

∫g1(θ ,ϕ )

g2(θ ,ϕ )

f (ρsin ϕcosθ , ρ sinϕ sinθ ,ρ cos ϕ ) ρ2 sinϕ dρ dθdϕ

Max ϕ=π

Change of Variables and Jacobian2DThe Jacobian of the transformation T given by x=x (u , v ) and y= y (u , v ) is∂ (x , y )∂ (u , v )

=|∂ x∂u

∂ x∂v

∂ y∂u

∂ y∂v

|= ∂ x∂u

∂ y∂v

−∂ x∂ v

∂ y∂u

Note that T can be expressed as u=u(x , y ) and v=v (x , y). Then,∂ (u , v )∂ (x , y )

= 1∂ ( x , y )∂ (u , v )Let dA be the image of the rectangle dudv under the plane transformation T given by x=x (u , v ) and y= y (u , v ). Then

dA=|∂ ( x , y )∂ (u , v ) |dudv

Then,∫∫

Rf ( x , y )dA=∫∫

Sf (x (u , v ) , y (u , v ) )|∂ (x , y )

∂ (u , v )|dudv

3DLet dV be the image of the rectangular box dudv dw under the space transformation T given by x=x (u ,v ,w ) , y= y (u , v ,w ) and z=z (u , v ,w ). Then

∭R

f (x , y , z )dV=∭S

f (x (u ,v ,w ) , y (u , v ,w ) , z (u , v ,w ) )|∂ (x , y , z )∂ (u , v ,w )|dudvdw

where

∂ (x , y , z )∂ (u , v , w )

=|∂x∂u

∂ x∂v

∂ x∂w

∂ y∂u

∂ y∂v

∂ y∂w

∂ z∂u

∂ z∂v

∂ z∂w

|=∂ x∂u ( ∂ y

∂v∂ z∂w

− ∂ y∂w

∂z∂v )− ∂ x

∂v ( ∂ y∂u

∂z∂w

− ∂ y∂w

∂z∂u )+ ∂ x

∂w ( ∂ y∂u

∂ z∂v

−∂ y∂ v

∂ z∂u )

and∂ (u , v , w )∂ (x , y , z )

= 1∂ ( x , y , z )∂ (u , v ,w )

Line Integrals1) Scalar function2D:Let f be defined on a smooth curve C given by r (t )=⟨ x (t ) , y (t ) ⟩, a≤ t ≤b. Then, the line integral of f along curve C is

∫Cf ( x , y ) ds=∫

a

b

f (x ( t ) , y ( t ) )||r ' ( t )||dt=∫a

b

f (x ( t ) , y ( t ) )√x t2+ y t

2dt

* Note: When f ( x , y )=1, then this reduces to the arc length of C∫Cf ( x , y )ds=∫

C1ds=∫

a

b

||r ' (t )||dt

Suppose C is a union of a finite number of smooth curves C1 ,C2 ,…,Cn , where the initial point of C i+1 is the terminal point of C i. Then,∫Cf ( x , y )ds=∫

C1

f ( x , y )ds+…+∫Cn

f ( x , y )ds

3D:∫Cf ( x , y , z )ds=∫

a

b

f (x (t ) , y ( t ) , z (t ) )||r ' (t )||dt=∫a

b

f (x ( t ) , y (t ) , z (t ) )√x t2+ y t

2+z t2dt

2) Vector Field2D:Let F be a continuous vector field defined on a smooth curve C given by the vector function r ( t )=⟨ x ( t ) , y (t ) ⟩, a≤ t ≤b. Then, the line integral of F ( x , y ) along curve C is∫CF ⋅ dr=∫

CF ⋅T ds=∫

a

b

F (x ( t ) , y ( t ) )⋅r ' ( t )dt

where T (t )= r ' ( t )

||r ' ( t )||, dsdt

=||r ' ( t )||

3D:∫CF ⋅ dr=∫

CF ⋅T ds=∫

a

b

F (x (t ) , y ( t ) , z (t ) ) ⋅r ' ( t )dt

*Note: Parametrisation is important. Parametrisation must be in direction of curve.∫CF ⋅ dr=−∫

−CF ⋅dr

Conservative Vector FieldsA vector field F is conservative if we can write F=∇ f for some scalar function f on D. This function is called the potential function of F. Additionally, curl F=02D:Suppose F ( x , y )=P (x , y )i+Q ( x , y ) j is a vector field in an open and simply-connected region D and both P and Q have continuous first-order partial derivatives on D. Then,

Q x=P yat each point ( x . y ) in D if and only if F is conservative on D.3D:R y=Qz , Rx=PZ ,Q x=PySteps to find f :

For F ( x , y , z )=⟨ a ,b , c ⟩=∇ f ( x , y , z )=⟨ f x , f y , f z ⟩,Recover f using partial integrations and compare to solveFundamental Theorem for Line IntegralSuppose F is a conservative vector field with potential function f , and C is a smooth curve with initial point A and terminal point B. Then,∫CF ⋅ dr=∫

C∇ f ⋅d r=f (B )−f (A)

If there are two paths with the same initial and terminal points but their line integrals differ, then the vector field is not conservative. Conservative → Independent of pathGreen’s Theorem Let C be a positively oriented, piecewise-smooth, simple closed curve in the plane and let D be the region bounded by C. Let F ( x , y )=P (x , y ) i+Q ( x , y ) j. Thus, if the positively oriented C is given by the vector function r (t ), a≤ t ≤b, then the region D enclosed by C is always on the left as we walk around C.Then,

∫CF ⋅ dr=∫

CPdx+Qdy=∫∫

DQx−P ydA

Area (D )=∫Cx dy=−∫

Cy dx=1

2∫Cx dy− y dx

Surface Integrals of Scalar FieldSuppose a surface S is smooth and has a parametrisation

r (u ,v )= ⟨ x (u , v ) , y (u , v ) , z (u , v ) ⟩ , (u , v )∈D

Then ru (a ,b )×r v (a ,b ) is normal to the tangent plane of S at the point (x (u , v ) , y (u , v ) , z (u , v ) )

∫∫Sf ( x , y , z )dS=∫∫

Df (x (u , v ) , y (u ,v ) , z (u , v ) )||ru×r v||dA

Suppose S is a union of a finite number of smooth surfaces S1 , S2 ,…,Sn , that intersect only along their boundaries then the surface integral of f over S is the sum of surface integrals over each Si

∫Sf ( x , y , z )ds=∫

S 1

f ( x , y , z )ds+…+∫Sn

f ( x , y , z )ds

Surface AreaA (S )=∫∫

S1dS=∫∫

D||ru×rv||dA

Special Case:If S is given by r ( x , y )= ⟨x , y , g ( x , y ) ⟩ , ( x , y )∈D, then the surface integral becomesThen, r x=⟨ 1,0 , gx ⟩ and r y=⟨ 0,1, gy ⟩

r x×r y=⟨−gx ,−g y ,1 ⟩

∫∫Sf ( x , y , z )dS=∫∫

Df (x , y , g ( x , y ) )||r x×r y||dA=∫∫

Df (x , y , g ( x , y ) )√gx

2+gy2+1dA

Surface Integrals of Vector FieldSuppose a surface S is smooth and has a parametrisation

r (u ,v )= ⟨ x (u , v ) , y (u , v ) , z (u , v ) ⟩ , (u , v )∈D

Define the unit normal vector n of the surface n=

ru×rv

||ru×rv||

For a closed surface that is the boundary of a solid region E, the positive orientation is the one for the normal points outwards, and negative orientation if it points inwards.If F ( x , y , z )=⟨ P (x , y , z ) ,Q ( x , y , z ), R ( x , y , z ) ⟩ is a continuous vector field defined on an oriented surface S has a with unit normal vector n, the surface integral of F over S is

∫∫SF ⋅ d S=∫∫

SF ⋅ndS=∫∫

SF ⋅

ru×r v

||ru×r v||dS=∫∫

DF ⋅ ( ru×rv )dA

which is the flux of F across S.*Note: Ensure that the orientation is correct. Use −n if necessary.Special Case:If S is given by r ( x , y )= ⟨x , y , g ( x , y ) ⟩ , ( x , y )∈D, then the surface integral becomes

- Upward Orientation∫∫

SF ⋅ d S=∫∫

D(−Pgx−Q gy+R )dA

- Downward Orientation∫∫

SF ⋅ d S=∫∫

D(Pgx+Qg y−R )dA

Divergence and Curl1) Divergence2D:Let F ( x , y )=⟨P ( x , y ) ,Q ( x , y ) ⟩ be a vector field in a space where P and Q have first order partial derivatives in a region D. Then, the divergence of Fis a scalar function defined by¿ F=P x+Q y

or, the differential operator ∇= ∂∂ x

i+ ∂∂ x

j+ ∂∂ x

k

∇⋅F=⟨ ∂∂ x

, ∂∂ y ⟩⋅ ⟨P ,Q ⟩=Px+Q y=¿F

3D:¿ F=P x+Q y+R zGauss’ TheoremLet E be a solid region where the boundary surface S of E is piecewise smooth with positive orientation. Then,

∫∫SF ⋅ d S=∫∫∫

E¿F dV

2) CurlLet F ( x , y )=⟨P ( x , y ) ,Q ( x , y ) ⟩ be a vector field in a space where P and Q have first order partial derivatives in a region D. Then, the curl of Fis a scalar function defined bycurl F=(R y−Qz )i+(P z−Rx ) j+(Q x−Py )k

or, the differential operator ∇= ∂∂ x

i+ ∂∂ x

j+ ∂∂ x

k

∇× F=⟨ ∂∂ x

, ∂∂ y

, ∂∂ z ⟩× ⟨ P ,Q ,R ⟩=curlF

Stokes’ TheoremLet C be the boundary curve (simple closed curve) of a surface S with unit normal n. Suppose that C is positively oriented with respect to n. Let F be a vector field whose components have partial derivatives on an open region in R3 that contains S. Then,

∫CF ⋅ dr=∫∫

Scurl F ⋅d S

*Note: This is regardless of the surface, as long as they have the same boundarySpecial Case: If S is flat and lies in thexy-plane with upward orientation n=k, then this reduces to Green’s Theorem.Fundamental Theorem of CalculusLet ∫ f ( t )dt=F (t). Then,ddx [∫

a

g (x )

f ( t )dt ]¿ ddx

g ( x ) ⋅ ddg ( x ) ∫a

g ( x )

f ( t )dt¿ddx

g ( x ) ⋅ ddg ( x ) [F (g ( x )−F (a ) ) ]¿ d

dxg ( x ) ⋅ f (g ( x ) )

¿ g' (x ) ⋅ f (g ( x ) )

ddx [∫

g (x )

h ( x )

f ( t )dt ]¿ ddx [∫

a

h ( x )

f (t )dt ]− ddx [∫

a

g ( x )

f ( t )dt ]¿h' ( x ) ⋅ f (h (x ) )−g ' ( x )⋅ f (g ( x ))