vector calculus review

8/13/2019 Vector Calculus Review

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Math 200 Review


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Chapter 1

Important Equations

the Fundamental Theorem of Calculus The integral of a derivativeover a region is equal to the valueof the function at the boundary.

Basic Integrals

dx (length)

dxdy (area)

dxdyzdz (volume) ds (the length of a curve)

dS (surface area)

Hyperbolic Functions

sinh(x) = ex ex

2

cosh(x) =ex +ex

2d

dxsinh(x) = cosh(x)

d

dxcosh(x) = sinh(x)

Common Integral Forms For nan integer.

2n0

sin d= 2n0

cos d= 0 n0

sin2 d=

n0

cos2 d=n

2 a0

ex dx= ea 1

the Ellipse

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x2

a2+

y2

b2 = 1

Area =ab

r(t) = cos t, sin t (oriented positively for t : [0, ])

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Surfaces

Plane Ax+By +Cz = DSphere x2 +y2 +x2 =2

Cylinder x2 +y2 =r2

Saddle z= x2 y2Cone z2 =x2 +y2

Paraboloid z= x2 +y2

Coordinate Surfaces

= k (spheres)

= k (half planes)

= k (cones;= /2 is xy-plane)

Component, Projection, Dot & Cross Products

compba=a b|b| = |a|cos (Component ofa onto b)

projba= a bb b b= (compba)b (Projection ofa onto b)

W =F s (Work W ofF alongs)n= A B (the normaln toA and B)

Area(AB) = |A B| (Area of a Parallelogram) =A (B C) (the flux of A through B C)

Parameterization

r n= P n (equation of a plane)r(t) =P +tPQ (parameterization of a line through pointsP and Q)

r(t) = r cos t, r sin t (parameterization of a positively oriented circle)r(t) = r cos t, r sin t (parameterization of a negatively oriented circle)r(t) = a cos t, b sin t (parameterization of a positively oriented ellipse)r(t) = a cos t, b sin t (parameterization of a negatively oriented ellipse)

T= r(t)|r(t)| (unit tangent vector)

N= a |a| = a av|a av| =

dTdt

dTdt

(unit normal vector)

Arc Length

ds= |r(t)| dt (scalar field)ds= r(t)dt (vector field)

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the Differential

df=fxdx+fydy+fzdz

y y1= m(x x1) (point-slope form)z f(a, b) =fxdx+fydy

f f r= fxx+fyy (approximation using differentials)

Directional Derivative

Du= f u

the Chain Rule

d

dt

f(x(t), y(t)

=

f

x

dx

dt +

f

y

dy

dt

the Gradient

= i x

+ j

y+ k

z (the del operator)

F= Fx, Fy, Fz (the gradient ofF) F= Fx+Fy+Fz (the divergence ofF)

F= Ry QZ, Pz Rx, Qx Py (the curl ofF)

Center of Mass

mass,m=

ds (mass for a given density, )

Myz =

x ds M xz=

y ds M xy=

z ds

x=

Myz

m y=

Mxz

m z=

Mxy

mcenter of mass = x,y ,z

Polar Coordinates

x= r cos

y= r sin

r2 =x2 +y2

tan = y

x

y= ax x2 =

r= a cos

x=

ax y2 = r= a sin

Moments

1. about the x-axis,Ix=R

y2 dA

2. y-axis,Iy=R

x2 dA

3. the origin, Io =R

r2 dA.

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Common Transformations

dA= rdrd (area differential in polar coordinates)

dV =rdzdrd (volume differential in cylindrical coordinates)

dV =2 sin ddd (volume differential in spherical coordinates)

dS= ru rv dudv (general form of the surface differential)dS=

fx,

fy, 1

dxdy (surface differential for a functionz= f(x, y))

J(u, v) =

xu

xv

yu

yv

(the Jacobian Transformation)dS= x,y ,z sin dd (surface differential for a sphere; constant)

= dS= sin dddS= cos , sin , 0rdxd (surface differential of a cylinder; constantr)

= dS=rdxd

Cylindrical Coordinates

x= r cos y= r sin

z= z

x2 +y2 =r2

tan =y

x

Spherical Coordinates

r= sin

x= r cos = sin cos

y= r sin = sin sin

z= cos

1.1 Differential Notation

dr= r(t) dt =

dx

dt,dy

dt,dz

dt

dt

F= P,Q,RF dr= P dx+Q dy+R dz

Line Integrals

C

f ds=

ba

f(r(t))|r(t)|dt (Line Integral of a Scalar Field)

C

F(r) dr= ba

F(r(t)) r(t) dt (Line Integral of a Vector Field)C

F dr=

r(b)

r(a)

(Fundamental Theorem of Calculus for Line Integrals; where F= )

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Integral Calculus

R

Q

x P

y dA=

Bd(R)

F dr (Greens Theorem)

S

F dS=

Bd(S)

F dr (Stokes Theorem)

E

FdV = Bd(E)

F dS (Divergence Theorem)

Optimization

f=c (Lagrange Multiplier; use level curves to classify)fxx fxyfxy fyy

(the Hessian Matrix)

Leading Principal Minor Determinants

minimum IfP is a critical point and all partials exist, then f(P) is a minimum if all LPMDs are positive.

maximum IfP is a critical point, the first LPMD is negative, and succeeding LPMDs alternate in sign,then the f(P) is a maximum.

saddle point IfP is a critical point, the first LPMD is negative and det(H) is negative, then f(P) is asaddle point.

Inconclusive If det(H) = 0, nothing can be concluded.

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Chapter 2

Vectors

Correspondence between Points and Vectors

A vector with two components corresponds to a point in the xy plane. The components of

c are the coordinates of the point: x= v1 and y =v2. The arrow ends at this point (v1, v2),when it starts from (0, 0). Now we allow vectors to have three components (v1, v2, v3).

Thexy plane is replaced by three-dimensional space. Here are typical vectors (still columnvectors but with three components):

v=

111

and w =

23

4

and v + w=

34

3

The vector v corresponds to an arrow in 3-space. Usually the arrow starts at the origin,where the xyz axes meet and the coordinates are (0, 0, 0). The arrow ends at the point with

the coordinates v1, v2, v3. There is a perfect match between the column vector and thearrow from the origin and the point where the arrow ends.

([Str09,4])

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2.1 Vector Length

|v|is the length or magnitude.

|v| =

v21+ v22+ v

23

=

v v (equal to the square root of the dot product)

2.2 Vector Arithmetic

the sum of two vectors

the difference of two vectors

scalar multiplication Multiplication of a vector by a positive scalaramultiplies the magnitudebut leavesthe direction unchanged. Ifais negative, the direction is reversed.

Scalar multiplication is distributive.

a(A + B) =aA +aB

linear combinations Ifc1, c2 R, then c1a +c2b is a linear combination ofa andb.

Two vectors define a parallelogram.

Three vectors define a box or parallelepiped.

2.3 Define a Vector from Two Points

Given two points P : (x1, y1, z1) and Q: (x2, y2, z2), we can define the vector PQ as

PQ= x2 x1, y2 y1, z2 z1

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2.3.1 Standard Unit Vectors

i,j, and k are the standard unit vectors that for the standard Cartesian coordinate system.

i= 1, 0, 0J= 0, 1, 0

k= 0, 0, 1Standard Position v is in standard position if its tail is at the origin. Then,

v= ai+bj+ck = a,b,c

2.4 Components & Projections

For a1= ab and a2= a,a= a1+ a2= ab+ a

projection a1 = ab is the projection ofa ontob.

a2=a is the part ofa that is perpendicular to b. Ifa ab, then a is the error vector.

component The component ofa onto b is the scalar for which

ab= (compba)b

So,

compba=a b|b| = |a|cos

The projection is given by

projba= (compba)b= a bb b b

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2.5 Vector Operations

2.5.1 Dot Product

also known as the Inner Product. As its result is a scalar, it is also called the Scalar Product. The dotproduct is given by

a b= |a||b|cos = aibiwhere is the angle between the two vectors.

2.5.2 Properties of the Dot Product

a a= |a|2

a b= 0 if and only ifa b= 0

2.5.3 Applications of the Dot Product

Work The Work done by a force vectorF along a direction vector s is

W=F s

Check the Correctness of a Cross Product Because a cross-product is perpendicular to the planedefined by the two vectors being crossed, the dot product of the cross-product and each crossing vector iszero.

2.5.4 Cross Product

also known as the Vector Product. The result of the cross product is a vector perpendicular to both multiplyingvectors. The cross product is given by

a b= |a||b|sin n=

i j ka1 a2 a3b1 b2 b3

where is the angle between the two multiplying vectors and nis a unit vector perpendicular to both vectors,completing a right-handed system.

2.5.5 Properties of the Cross-Product

The cross-product is not commutative, in fact,

A B= (B A)

If two vectors are parallel, their cross product is zero.

A A= 0

2.5.6 Applications of the Cross Product

Area of a Parallelogram The area of a parallelogram defined by two vectors is equal to the magnitude oftheir cross product.

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Equation of a Plane The cross product of two vectors defines a vector that is perpendicular or normalto the the plane defined by the two vectors.

Wheren is the vector normal to the plane, the equation of the plane is given by

r n= P nwherer is any vector in the planex,y,z and P is any specific point in the planex0, y0, z0

2.5.7 Triple Product

also known as the Scalar Triple Product

a (b c)

2.5.8 Applications of the Triple Product

Volume of a Box The absolute value of the triple product is the volume of the box or parallelepiped whichhas edges that are defined by the three vectors.

Flux The signed triple product is the flux through an area defined by two vectors at a rate defined by athird. B C is an area through which A flows.

Important Vectors

Unit Vector a vector whose length is 1

Normal Vector perpendicular to a given curve or surface

Gradient Vector the vector giving the magnitude and direction of maximum increase of a scalar fieldand is orthogonal to a level curve or surface at a point

Conservative Vector Field a vector field which is the gradient of a scalar potential field

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Chapter 3

Parameterization

The vector function r(t) = 2cos t, 4sin t, tand the vector r(19.5).

the Parameter A parameterized vector function is a multi-dimensional vector function often defined by asingle scalar value t, its parameter.

Basic Parameterization For a basic function of one variable, the function can be parameterized by settingthe independent variable equal to the parameter, e.g. t.

Orientation Parametric functions are generally oriented by the Right-Hand-Rule, counter-clockwise. Fora circle this orientation would be

r(t) = cos t, sin t (postively oriented)r(t) =

cos t,

sin t

(negatively oriented)

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The Parameterization of a Line Generally,

r(t) =P +tPQ

The Parameterization of an Ellipse

x= a cos t

y= b sin t

3.1 Derivatives of Parametric Functions

v(t) =r(t) = x(t), y(t), z(t)a(t) =r(t) = x(t), y(t), z(t)

Product Rules

d

dt

f(t)r(t)

=

df(t)

dt r(t) + f(t)

dr(t)

dt

d

dtf(t)

r(t)=df(t)

dt r(t) + f(t)

dr(t)

dtd

dt

f(t) r(t)

=

df(t)

dt r(t) + f(t) dr(t)

dt

3.2 Tangent Vector

v= r is a vector tangent to the curve that points in the direction of particle motion. a= r points into thecave of the curve.

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3.3 Arc Length

The tangent vector can be rewritten as dx,dy,dzdt

allowing us to rewrite the arc length formula as

ds= |r(t)| dt

We can think of this equation as analogous to the classic distance, ds, equals rate,|r(t)|, times time, dt.

3.4 Curvature

Not on final exam.

3.5 Unit Vectors

Unit Tangent Vector

T= r(t)|r(t)|

Unit Normal Vector

N=dTdt

|dTdt

|

Nis in the same direction as the error vector between the accelerationa and the velocityv, in other words,

N= a|a| =

a av|a av|

Frenet-Serret Formulas Not on final exam, but super cool. Wikipedia: Frenet-Serret Formulas

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http://http//en.wikipedia.org/wiki/Frenet-Serret_formulashttp://http//en.wikipedia.org/wiki/Frenet-Serret_formulas


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Chapter 4

Partial Derivatives & Gradient

4.1 Ordinary Derivatives

The derivativedf/dxdescribes how rapidly the functionf(x) varies when its argument x is changed by a tinyamountdx.

Geometrically,df/dx is the slopeof the graph f versus x.

4.2 Geometric Interpretation of Multi-Dimensional Derivatives

Contour Plot Thecontour plotis the projectionof level curves onto the xy-plane.

A contour line or level curve of a function of twovariables is a curve along which the function has aconstant value.

A contour map is a map illustrated with con-tour lines, for example a topographic map, which thusshows valleys and hills, and the steepness of slopes.Thecontour intervalof a contour map is the differ-

ence in elevation between successive contour lines.

Level Curve A level curve is the intersection of aplane z-constant with the graph ofz= f(x, y).

Consider a functionfwhose graph looks like a hill.The blue curves are then the level sets. The red curvesfollow the direction of the gradient. In other words,

the cautious hiker follows the blue paths, while thebold one the red paths.

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4.3 Partial Derivative

f(x, y), the partial derivative with respect to x is given by

f

x= lim

h0f(x+h, y) f(x)

h

A partial derivative holds all other variables constant, then for f(x,y ,z)

df=

f

x

dx+

f

y

dy+

f

z

dz

Alternate Notationf

x =fx

fx

(a,b)

is the slope of the tangent line to a curve at P where m= riserun

= fx .

Second Partial Derivative For a function of two variables, there are three second partial derivatives.

fxx=

x

f

x

fxy=

y

f

x =

x

f

y =fyx

fyy =

y

f

y

4.4 Directional Derivatives

Given a multi-variable function, an ordinary variable can not be used to describe how fast a function is

changing. For a multi-variable function, the change depends upon which directionwe move.

Estimate the directional derivative fxandfy can be estimated using slopes of secant lines on a contourplot.

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Definition: Directional Derivative The directional derivative, ascalar, is the slope of the line that lieson the tangent plane in the direction of u an arbitrary direction vector.

Duf= f u

4.5 Find the Tangent Plane

Find the equation for the tangent plane to a function f(x, y) at the point P(a,b,f(a, b)). The two first partialderivatives allow us to construct two vectors parallel to the tangent plane.

z f(a, b) =fx dx+fy dyCompare this to the point-slope formula

y y1 = m(x x1)

Tangent Plane to a General Surface For a general surface, defined in terms of x, y, and z, set theequation of the surface equal to zero. Then

f(x,y ,z) = 0

is a level surface for the surface equal to 0. The gradient off atP(x1, y1, z1),f(x1, y1, z1) is the normal tothe level surface at the point P. The equation for a plane defined by a normal,n can be used to define theequation for the plane, where r is any point on the tangent plane r= (x,y ,z).

n r= n P

4.6 the Differential

Define the change in the output on the tangent plane to be the differential df where

df=fx dx+fy dy

Compare this with the change in output on the graph off

f=f(x, y) f(a, b)

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For a differentiable function of one variabley = f(x), we define the differential dxto be anindependent variable; that is, dxcan be given the value of any real number. The differentialofy is then defined as

dy= f(x)dx

Figure 6 shows the relationship between the increment y and the differential dy: y repre-sents the change in height of the curve y = f(x) and dy represents the change in height of the

tangent line when xchanges by an amount dx= x

For a differentiable function of two variables, z = f(x, y), we define the differentials dxand dy to be independent variables; that is, they can be any values. Then the differential dz,also calle the total differential, is defined by

dz= fx(x, y) +fy(x, y) = z

xdx+

z

ydy

([Ste12,919])

4.6.1 Approximation using differentials

The change in a function f is approximately given by

f r= fxx+fyy

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4.7 the Chain Rule

The chain rule is a formula for computing the derivative of the composition of two or more functions.

The change in f is a sum of terms, one for every input term.

For f(x, y), where x= x(t) and y= y(t).

df

dt =

f

x

dx

dt +

f

y

dy

dt

4.8 Gradient

the del operator Also known as the nabla. (the nabla symbol) denotes the vector differential operator,del.

= i x

+ j

y+ k

z.

f= fx, fy

The gradient has the formal appearance of a vector,, multiplying a scalar T:

T = x

x + y

y + z

z

T

(For once I write the unit vectors to the left, just so no one will think this means i/x, andso on - which would be zero, since i is constant.) The term in parenthesis is called del:

=x x

+ y

y+ z

z

Of course, del is nota vector, in the usual sense. Indeed, it is without specific meaning untilwe provide it with a function to act upon. Furthermore, it does not multiply T; rather itis an instruction to differentiatewhat follows. To be precise, then, we should say that is avector operator that acts upon T, not a vector that multiplies T.

With this qualification, though, mimics the behavior of an ordinary vector in virtuallyevery way; almost anything can be done with other vectors can also be cone with, if wemerely translate multiply to act upon. So by all means take the vector appearance ofseriously: it is a marvelous piece of notational simplification, as you will appreciate if you everconsult Maxwells original work on electromagnetism, written without the benefit of.

([Gri99, 16])

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Definition: Gradient The gradient offis defined as the unique vector field whose dot product with anyvector v at each point x is the directional derivative off alongv.

f(x) v= Dvf(x).

Geometric Interpretation of the Gradient Like any vector, the gradient has both magnitude anddirection.

Imagine you are standing on a hillside. Look all around you, and find the direction of thesteepest ascent. That is the directionof the gradient. Now measure the slopein that direction(rise over run). That is the magnitudeof the gradient.

What would it mean for the gradient to vanish? IfT = 0 a t (x,y ,z), then dT = 0for small displacements about the point (x,y ,z). This is, then, a stationary pointof thefunctionT(x,y,z). It could be a maximum (a summit), a minimum (a valley), a saddle point(a pass), or a shoulder. This is analogous to the situation for functions of one variable,where a vanishing derivative signals a maximum, a minimum, or an inflection. If particular,

if you want to locate the extrema of a function of three variables, set its gradient equal tozero. ([Gri99, 14])

Gradient of the 2-d function f(x, y) = xex2y2 is

plotted as blue arrows over the pseudocolor plot ofthe function.

The gradient of the function f(x, y) = (cos2 x+cos2 y)2 depicted as a projected vector field on thebottom plane.

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4.9 Properties of the Gradient

f(P) is perpendicular to the level curve (or surface or 3-space or n-space iff has n inputs) offthat passes through P. The dot product is 0 when the vectors are perpendicular, and the direction that isparallel to the level curve gives no change in f. That direction then has a 0 derivative.

f(P) points from P in the direction of maximal rate of increase of f. The directional derivativeis largest when the angle between the directional derivative and the gradient is = 0 i.e. u is in the samedirection as the gradient.

The magnitude off(P) is equal to the maximal rate

A real-valued function z = f(x, y) whose partial derivatives f/x and f/y exist andare continuous is called continuously differentiable. Assume thatf(x, y) is such a function andthatf= 0. Let c be a real number in the range offand letv be a unit vector in R2 whichis tangent to the level curve f(x, y) =c (see Figure 2.4.1).

([Cor08, 80])

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4.10 Divergence & Curl

Divergence The divergence off is the flux density at a point.

div(F) = F= Fx+Fy+ Fz

From the definition ofwe construct the divergence: v=

x

x+ y

y+ z

z

(vxx +vyy +vzz)

=vx

x +

vyy

+ vz

z

Observe that the divergence of a vector functions v is itself a scalar v.

Geometrical Interpretation: The name divergence is well chosen, for v is a measureof how much the vector v spreads out (diverges) from the point in question. For example,the vector function in Fig. 1.18a has a large (positive) divergence (if the arrows pointed in,

it would be a large negativedivergence), the function in Fig. 1.18b has zero divergence, andthe function in Fig. 1.18c again has a positive divergence. (Please understand thatv here isa function- theres a different vector associated with every point in space. In the diagrams,

of course, I can only draw the arrows at a few representative locations.) Imagine standingat the edge of a pool. Sprinkle some sawdust or pine needles on the surface. If the materialspreads out, then you dropped it at a point of positive divergence; if it collects together, youdropped it at a point of negative divergence. (The vector function v in this model is thevelocity of the water - this is a two-dimensional example, but it helps give one a feel forwhat the divergence means. A point of positive divergence is a source, or faucet; a point ofnegative divergence is a sink, or drain.) ([Gri99, 17])

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Curl F(P) is the circulation density ofF atP.

curl(F) = F=

i j k

x

y

z

P Q R

=

Ry

Qz, Pz

Rx, Qx

Py

Notice that the curl of a vector function v is, like any cross product, a vector.

Geometrical Interpretation: The name curl is also well chosen, for v is a measure ofhow much the vector v curls around the point in question. Thus the three functions in Fig.1.18 all have zero curl (as you can easily check for yourself), whereas the functions in Fig.1.19 have a substantial curl, pointing in the z-direction, as the natural right-hand rule wouldsuggest. Imagine (again) you are standing at the edge of a pond. Float a small paddlewheel(a cork with toothpicks pointing out radially would do); if it starts to rotate then you placed

it at a point of nonzero curl. A whirlpool would be a region of large curl.

([Gri99, 19])

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Chapter 5

Solids and Surfaces in R3 with Multiple

Integrals

5.1 Fundamental Surfaces

Plane

Ax+By +Cz = D

Coordinate Planes

x= k y= k z= k

Sphere

x2 +y2 +z2 =R2

Cylinderx2 +y2 =R2

Paraboloids

z = x2 +y2 (centered at origin, oriented up)

z = x2y2 (centered at origin, oriented down)z = 4 x2 y2

(centered at0, 0, 4, oriented down)

Conicsz2 =x2 +y2

Saddlesz= x2 y2

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Center of Mass Given , a density function.

mass,m=

ds (mass for a given density, )

Myz =

x ds M xz=

y ds M xy=

z ds

x=

Myzm y=

Mxzm z=

Mxym

center of mass = x,y ,z

5.2 Double Integrals

WheredA = dxdy, we can find

R

f dA

using an iterated integral.

1. Hold xconstant and sum vertically.

2. Hold y constant and sum horizontally.

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5.2.1 Order of Integration

It may be advantageous to change the order of integration. In certain cases, a single iterated integral maybecome two iterated integrals, if one of the parameters of the region of integration is not the same for allvalues in the region.

Example Rewrite the integral by switching the order of integration.

I= 2

0

y

2

yf(x, y) dxdy

Solution The region of integration is describe by

R: 0 y 2y x y2

We can not describe a single region of integration holding x as our outside index of integration. We thendefine the integral over two regions of integration.

R: 2 x 4= 2 x 0

x y 2= 0 x 4

x y 2

I=

02

2x

f(x, y) dydx+

40

2x

f(x, y) dydx

5.3 Polar Coordinates

x= r cos

y= r sin

r2 =x2 +y2

tan = y

x

y=

ax x2 = r= a cos x=

ax y2 = r= a sin

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Area of Integration in Polar Coordinates

dA= rdrd

The change of variables formula can be used to evaluate double integrals in polar coordi-nates. Letting

x= x(r, ) =r cos and y = y(r, ) =r sin

we have

J(u, v) =xr

x

yr

y

= cos

r sin

sin r cos =r cos

2

+r sin2

= r = |J(u, v)|= |r|=r

so we have the following formula:

Double Integral in Polar Coordinates

R

f(x, y)dxdy=

R

f(r cos , r sin )rdrd

where the mappingx = r cos , y = r sin maps the region R in the r-plane onto the regionR in the xy-plane in a one-to-one manner. ([Cor08,121])

5.4 Applications of Double Integrals

Average Value

f= 1

Area,R

R

f dA

Weighted Average1

mass,R

R

f

dA

Center of Mass For M= mass(R) =R

dA

Mx=

R

y dA M y =

R

x dA

and the center of mass or expected value is

(x, y) =

MxM

,My

M

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Moment of Inertia The moment of inertia measures the difficulty of imparting a rotation about a givenreference

1. about the x-axis,Ix=R

y2 dA

2. y-axis,Iy=R

x2 dA

3. the origin, Io =R

r2 dA.

Volume Volume for z = f(x, y) can be found over a given area for z , the height of the solid.

5.5 Change of Variables in Multiple Integrals - the Jacobian Determinant

Theorem 3.1. Change of Variables Formula for Multiple IntegralsLetx = x(u, v)and y =y(u, v) define a one-to-one mapping of a region R? in the uv-plane onto a region R

in the xy-plane such that the determinant

J(u, v) =

xu

xv

yu

yv

is never 0 in R . Then

R

f(x, y)dA(x, y) =

R

f

x(u, v), y(u, v)J(u, v)

dA(u, v)

We use the notation dA(x, y) and dA(u, v) to denote the area element in the (x, y) and(u, v) coordinates, respectively.

The determinantJ(u, v) is called the Jacobianofx and y with respect to u and v , and issometimes written as

J(u, v) =(x, y)

(u, v)

Notice that the formula is sayingdA(x, y) = |J(u, v)|dA(u, v), which you can think of as atwo-variable version of the relation dx = g (u) du in the single-variable case.

([Cor08,119])

5.6 Triple Integrals using Rectangular Coordinates

E

f(x,y ,z)dV = limdiamVi0

ni=1

f(xi, yi, zi)Vi= lim

f(xi, yi, zi)xyz

where Vi forms a partition ofEwith nsubregions.

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5.7 Change of Variables in Multiple Integrals - Cylindrical Coordinates

Cylindrical coordinates are a generalization of po-lar coordinates with a z -coordinate added.

Cylindrical Differential

dV =rdrddx

Conversions

x= r cos

y= r sin

z = z

x2 +y2 =r2

tan = y

x

5.8 Change of Variables in Multiple Integrals - Spherical Coordinates

Spherical Differential

dV =2 sin ddd

Conversions

r= sin

x= r cos = sin cos

y= r sin = sin sin

z= cos

Coordinate Surfaces

= k (spheres)

= k (half planes)

= k (cones; = /2 is xy-plane)

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Chapter 6

Fundamental Theorem of Calculus for

Line Integrals

A line integral is in integral defined over a parameterized curve C :r(t). We pull-back to the t-axis, a line,in order to evaluate the integral.

Arc Length Arc length s =C

ds is a line integral.

6.1 Differential Notation

dr= r(t) dt =

dx

dt,dy

dt,dz

dt

dt

F= P,Q,RF dr= P dx+Q dy+R dz

We use the notation dr = r(t)dt = dxi+ dyj to denote the differential of the vector-valued functionr . . . the quantityP(x, y)dx + Q(x, y)dyis known asdifferential form. For areal-valued function F(x, y), the differential ofF is dF= Fx dx +

Fydy. A differential form

P(x, y)dx+Q(x, y)dy is called exact if it equals dFfor some function F(x, y).([Cor08,139])

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6.2 the Line Integral

An integral where the function to be integrated is evaluated along a curve. The function to be integrated maybe a scalar field or a vector field. The value of the line integral is the sum of values of the field at all pointson the curve, weighted by some scalar function on the curve (commonly arc length or, for a vector field, thescalar product of the vector field with a differential vector in the curve).

6.2.1 Line Integral of a Scalar FieldThe line integral over a scalar field f can be thought of as the area under the curve Calong a surfacez = f(x, y),described by the field.

C

f ds=

ba

f(r(t))|r(t)|dt

The function f is called the integrand, the curve C is the domain of integration, and the symbol ds maybe intuitively interpreted as an elementary arc length. Line integrals of scalar fields over a curve Cdo notdepend on the chosen parametrization r ofC.

Awesome Wikipedia Line Integral of a Scalar Field Animation Line Integral of Scalar Field

Geometric Interpretatoin Geometrically, when the scalar fieldfis defined over a plane (n= 2), its graphis a surface z =f(x, y) in space, and the line integral gives the (signed) cross-sectional area bounded by thecurve Cand the graph off.

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6.2.2 Line Integral of a Vector Field

A line integral of a scalar field is a line integral of a vector field where the vectors are always tangential tothe line.

C

F(r) dr= ba

F(r(t)) r(t)dt

wherer: [a, b]

Cis a parametrization of the curve C such that r(a) and r(b) give the endpoints ofC.

Awesome Wikipedia Line Integral of a Vector Field Animation Line Integral of Vector Field

Line integrals of vector fields are independent of the parametrizationr in absolute value, but they do dependon its orientation. Specifically, a reversal in the orientation of the parametrization changes the sign of the lineintegral.

6.2.3 Applications of the Line Integral

Work

Mass on a Wire

6.3 Potential Functions

F is a gradient fieldif and only if there exists a potential function for which

F=

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IfC :r(t), a t b, and F= , thenC

F dr=

r(b)

r(a)

(FTCLI)

If the line integral is challenging, check to see if the field is conservative ( F= 0) and if so, apply theFTCLI.

Properties of the Gradient Function If one of these is true, then all of them are true.

1. F is conservative

2.C

F dr= 0 ifCis a closed curve

3. F is path independent.

4. F= 0 i.e. the curl ofF = 0

6.4 Finding a Potential Function

Let (0, 0, 0) = 0

(a,b,c) =

x0

F(a, 0, 0) dx +

b0

F(a,y, 0) dy +

c0

F(a,b,z)dz

Then (x,y ,z) = (a,b,c) (for any arbitrary x, y, z))

6.5 Greens Theorem

Greens Theorem connects a double integral over a region Rto aline integral alongits boundary C.

The integral ofdf/dxequalsf(b)f(a). This connects a one-dimensional integral to a zero-dimensional integral. The boundary only contains two points a and b! The answerf(b) f(a)is some kind of point integral. It is this absolutely crucial idea - to integrate a derivativefrom information at the boundary- that Greens Theorem extends into two dimensions.

([Str91,563])

LetF = P, Q be continuously differentiable with positively oriented boundary B d(R). Then

R

Q

x P

y dA=

Bd(R)

F dr

The boundary is a closed path. To evaluate the line integral, the union of several oriented curves mayneed to be evaluated in order to complete a closed path.

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D

Q

x P

y dA=

C1C2C3C4

F dr

6.6 Surface Integrals

The double integral in Greens Theorem is over a flat surface R. Now the region movesout of the plane. It becomes a curved surfaceS, part of a sphere or cylinder or cone. Whenthe surface has only one z for each (x, y), it is the graph of a function z(x, y). In other casesScan twist and close up - a sphere has an upper z and a lower z .

([Str91,573])

A surface integral is an expression of the formS

v da

wherev is again some vector function, and dais an infinitesimal patch of area, with directionperpendicular to the surface (Fig. 1.22). There are, of course, two directions perpendicular toany surface, so the signof a surface integral is intrinsically ambiguous. If the surface is closed(forming a balloon), in which case I shall again put a circle on the integral sign

v da

then tradition dictates that outward is positive, but for open surfaces its arbitrary. If vdescribes the flow of a fluid (mass per unit area per unit time), then

v da represents the

total mass per unit time passing through the surface - hence the alternative name, flux.

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([Gri99, 26])

Compare:

Surface Differential

dS= ru rv dudvThe sign of the surface differential depends on how S is oriented.

6.7 Surface Integral - Special Cases

S is a function ofx and yIn general, where z= f(x, y),

dS= fx, fy, 1dxdy

Using a Jacobian Transformation

In general,dxdy= dS=

(x, y)(u, v)

dudv

Special ndS= ndS

n=x,y,z

dS=2 sin dd (for a sphere)

n= cos , sin , 0 dS=rdzd (for a cylinder)

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([Str91, 578])

6.8 Stokes Theorem

A generalization of Greens Theorem. Where Sis an oriented surface in R3

S

F dS=

Bd(S)

F dr

In physical applications, for a simple closed curve Cthe line integralC

fdris often calledthe circulationof f around C.

In physics, the curl is interpreted as a measure ofcirculation density.([Cor08,174])

The fundamental theorem for curls, which goes by the special name ofStokes theoremstates that

S

( v) da=P

v dL

As always, the integral of a derivative(here, the curl) over a region(here, a patch ofsurface)is equal to the value of the function at the boundary(here, the perimeter of the patch). Asin the case of the divergence theorem, the boundary term is itself an integral - specifically aclosed line integral.

Geometrical Interpretation: Recall that the curl measures the twist of the vectors v; aregion of high curl is a whirlpool - if you put a tiny paddle wheel there, it will rotate. Now,the integral of the curl over some surface (or, more precisely, the flux of the curl throughthat surface) represents the total amount of swirl, and we can determine that swirl just aswell by going around the edgeand finding how much the flow is following the boundary (Fig.1.31). You may find this a rather forced interpretation of Stokes theorem, but its a helpfulmnemonic, if nothing else.

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You might have noticed an apparent ambiguity in Stokes theorem: concerning the boundaryline integral, which wayare we supposed to go around (clockwise or counterclockwise)? If wego the wrong way well pick up an overall sign error. The answer is that it doesnt matterwhich way you go as long as as you are consistent, for there is a compensating sign ambiguityin the surface integral: Which way does da point? For a closedsurface (as in the divergencetheorem) da points in the direction of the outwardnormal; but for an open surface, whichway is out? Consistency in Stokes theorem (as in all such matters) is given by the right-hand rule: If your fingers point in the direction of the line integral, then your thumb fixes thedirection ofda (Fig. 1.32).

Now, there are plenty of surfaces (infinitely many) that share any given boundary line.Twist a paper clip into a loop and dip it in soapy water. The soap film constitutes a surface,with the wire loop as its boundary. If you blow on it, the soap dil will expand, making a largersurface, with the same boundary. Ordinarily, a flux integral depends critically on what surfaceyou integrate over, but evidently this is notthe case with curls. For Stokes theorem says that

( v da is equal to the line integral of v around the boundary, and the latter makes noreference to the specific surface you choose.

Corollary 1:

( v dadepends only on the boundary line, not on the particular surfaceused.

Corollary 2:

( v) da = 0 for any closed surface, since the boundary line, like themouth of a balloon shrinks down to a point, and hence the right side vanishes.

([Gri99, 34])

6.9 the Divergence Theorem

The flux out of two adjacent cubes cancel on the common edge:

So if we sum up the fluxes out of a partition of a solid, only the flux through its boundary will contributeto the sum.

E

FdV =

Bd(E)

F dS

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Theorem 4.8. (Divergence Theorem) Let be a closed surface in R3 which boundsa solid S, and let f(x,y,z) =f1(x,y ,z)i +f2(x,y ,z)j +f3(x,y ,z)k be a vector field definedon some subset ofR3 that contains . Then

f d=

S

div f dV

where

divf=f1

x +

f2y

+ f3

z

is called the divergence of f.

In physical applications, the surface integral

f d is often referred to as the flux of fthrough the surface . For example, if frepresents the velocity field of a fluid, then the fluxis the net quantity of fluid to flow through the surface per unit time. A positive flux meansthere is a net flow outof the surface (i.e. in the direction of the outward unit normal vectorn), while a negative flux indicates a net flow inward (in the direction of - n).

The term divergence comes from interpreting div fas a measure of how much a vector fielddiverges from a point. ([Cor08,162])

The fundamental theorem for divergences states that:V

( v)d=S

v da

In honor, I suppose of its great importance, this theorem has at least three special names:Gauss theorem, Greens theorem, or, simply the divergence theorem. Like the otherfundamental theorems, it says that the integral of a derivative(in this case the divergence)over a region (in this case a volume) is equal to the value of the function at the boundary(in this case the surfacethat bounds the volume). Notice that the boundary term is itself anintegral (specifically, a surface integral). This is reasonable: the boundary of a line is justtwo end points, but the boundary of a volumeis a (closed) surface.

Geometrical Interpretation: If v represents the flow of an incompressible fluid, then theflux of v (the right side of the equation) is the total amount of fluid passing out through thesurface, per unit time. Now, the divergence measures the spreading out of the vectors from

a point - a place of high divergence is like a faucet, pouring out liquid. If we have lots offaucets in a region filled with incompressible fluid, an equal amount of liquid will be forcedout through the boundaries of the region. In fact there are two ways we could determine howmuch is being produced: (a) we could count up all the faucets, recording how much each putsout, or (b) we could go around the boundary, measuring the flow at each point, and add it allup. You get the same answer either way:

(faucets within the volume) =

flow out through the surface

([Gri99, 31])

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Chapter 7

Optimization

7.1 Second Partials Test

Critical Points P = (a, b) is a critical point off(x, y) if and only ifP the domain off and fx(P) =0 =fy(P) or at least of the partials does not exist while the others are zero.

Hessian Matrix fxx fxyfxy fyy

Classify Critical Points Use the leading principal minor determinants of the Hessian matrix toclassify the critical points. The first LPMD is fxx. The second LPMD is the determinant of the Hessian,det(H).

minimum IfP is a critical point and all partials exist, then f(P) is a minimum if all LPMDs are positive.

maximum IfP is a critical point, the first LPMD is negative, and succeeding LPMDs alternate in sign,

then the f(P) is a maximum.

saddle point IfP is a critical point, the first LPMD is negative and det(H) is negative, then f(P) is asaddle point.

Inconclusive If det(H) = 0, nothing can be concluded.

7.2 Lagrange Multipliers for Constrained Functions

Lagranges method is equivalent to find the level curves for f(x, y) that share a tangent line with a constantconstraint function.

At these critical pointsf andc will be parallel since both are perpendicular to the same tangent line.Then

f=cwhere is a lagrange multiplier.

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Chapter 8

Review Exercises

8.1 Vectors

Diagonals of a Parallelogram Prove that the diagonals of a parallelogram bisect each other.

Given a = i +j 2kand b = 1, 1, 2, show that the compba= 46 ,ab = 231, 1, 2,ab = 5,1,23 , and

cos = 23 .

Work & Projections What is the work done by a force F onto a displacement s?

Area of a Parallelogram Show that the area of a parallelogram spanned bya = 1, 1, 2and b = 1, 1, 1is

30.

Equation of a Plane Show that an equation for the plane that containsP = (2, 5, 7), Q = (1, 1, 2), andR= (1, 1, 3) is 12x 5y+ 2z= 13.

8.2 ParameterizationParameterization of a Line Segment Parameterize the line segment from1, 2, 3 to4, 8, 10. Onepossible parameterization is r(t) = 1 + 3t, 2 + 6t, 3 + 7t.

Curve on a Surface Show that the curve r(t) = 2t,t, 3t2 lies on a saddle.

8.3 Partial Derivatives & Gradient

Find the equation of a tangent plane Find the tangent plane in standard form to the surface z =g(x, y) = 2x+y+ ln(xy) at the point (1, 1, 3).

Find an equation in standard form for the tangent plane to the graph ofz =

4 x2 2y2 at the point(1, 1, 1).

Find the tangent plane to the surface 5x3 3xy= 3 xyz at P = 1, 4, 5.

Estimate using differentials Use differentials to estimatef(2.9, 0.2) iff(x, y) =

x+e4y.

Use differentials to estimate f(1.9, 4.2) iff(x, y) =x2

y.

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the Chain Rule Find ft

at t= 3 iff(x, y) =xy,x= sin t, y (/3) = 2 and y(/3) = 5.

Use the chain rule to findfx(2, 1) iff(x, y) =g(u,v,w) = ln(u2v) + e4uv whereu = 2x,v = y2, andw = xy.

the Directional Derivative Letg(x, y) = 2x + y +ln(xy). FindDug(1, 1) where uis in the same directionas1, 1 and where u is in the same direction as1, 1.

Gradient Letg(x, y) = 2x+y+ ln(xy). Find the direction vector for which g is decreasing most rapidlyat the point (1, 1) and a direction in which g has a zero rate of change.

Electric Potential Suppose that over a certain region of space the electrical potential is given byV(x,y,z) =5x2 3xy+xyz .

1. Find the rate of change of V at the pointP = 1, 4, 5 in the direction of v = i+ j+k.2. Find the direction in which V changes most rapidly atP.

3. Find the maximum rate of change at P.

4. Find a direction in which V has zero rate of change at P.

8.4 Solids and Surfaces in R3 with Multiple Integrals

Double Integrals Find I=R

x+yx2 dA over the given R.

Volume using a Double Integral Find the volume of the solid bounded by z = y,y= x, and z= 0.

Order of Integration Rewrite I =20

y2yf(x, y) dxdy by switching the order of integration. Show the

new integral to be

I=

20

x2

f(x, y) dydx+

40

x2

f(x, y) dydx

Solve an Integral by Switching the Order Show thatI=R

sin2(xy) dxdy = 1.

Double Integrals in Polar Coordinates

1. ForR: x2 +y2 1; show that I= 1 x2 y2 dA= 22. *challenge* Show that I=

e

x2 dx=

3. Show that the volume of the solid bounded by z = 3x2 + 3y2 andz= 4 x2 y2 is 2.

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Applications of Double Integrals, Center of Mass Given= xy and the givenR, show that the center

of mass is

248225 ,

248225

.

Applications of Double Integrals, Moment of Inertia

1. Show that Io for the disk r a, = 1 is a42 .

2. Show that Io for the same disk translated to r = 2a cos is 3a4

2 .

8.5 Line Integrals

Mass of a wire Show that the mass of a wireCwith linear density (x,y ,z) = 3z2 grams/cm ifC :r(t) =

sin t, 3t, cos t

is

3104

grams

Show that the center of mass is 4

3,2 + 2

4 ,

8

3

Work Along a Curve Show that the work done byF = sin y, cos x on a particle that moves from (0, 0)to (, 2) on y= x2 is

2Show that the work will change if we choose a different path from (0, 0) to (, 2).

Work on a Particle Show that the work done byF= yz,x,y that moves along the path C, where C isthe line segment from (0, 1, 0 to (0, 1, 1) and then the circular arc from (0, 1, 1) to (1, 1, 0) in the y= 1-planeis

1 +

4

Potential Functions F= yx, 2 x22is conservative. Find a potential function for F.

Find the word done byF on a particle moving along a curve C :r(t) = t2 + 2, t2 4t, 0 t 2.

FTCLI Evaluate the line integral using the FTCLI, if possible. If it is not possible to use FTCLI, use

another method.

1. F= tan y, x sec2 y, Cis the line from (1, 0) to (2, 4 ). Show that the potential function is = x tan yand that the line integral evaluates to 2.

2. F =xy,yx, C : r(t) =t2, t2,1 t 1. Show that the potential function is = x tan y andthat the line integral evaluates to 2. Show that F has no potential function and that the line integralevaluate to47 .

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Greens Theorem Verify Greens Theorem showing thatI= 2e1 for

I=

R

yexdA

The Area of an Ellipse Use Greens Theorem to prove that the area of an ellipse is equal toab.

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Bibliography

[Cor08] Michael Corral. Vector Calculus. 2008.

[Gri99] David J. Griffiths. Introduction to Electrodynamics. Prentice Hall, 3rd edition, 1999.

[Sch05] H.M. Schey. Div, grad, curl and all that: an informal text on vector calculus. 4th ed. W W Norton& Company Incorporated, 2005.

[Ste12] James Stewart. Calculus. Cengage Learning, 7th edition, 2012.

[Str91] G. Strang. Calculus. Number v. 1. Wellesley-Cambridge Press, 1991.

[Str09] G. Strang. Introduction to Linear Algebra 4e. Wellesley-Cambridge Press, 2009.

vector calculus review

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