Calculus

May 20, 2018 | Author: Nahush Bapat | Category: Maxima And Minima, Power Series, Coordinate System, Algebra, Theoretical Physics
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Calculus Formula and Data Overview

Improper Integrals:

∞ 1

 

x p

1

Calculus - Period 1

(9)

dx

This function is convergent for  p >  1 and divergent for  p 1.



Differentiation and Integration

Comparison Theorem:  f (x) g (x) 0 for x a  then: If  f 



Chain Rule: (f (g (x)))  =  f  (g(x))g (x)

• If 

(1)





∞ f (x)dx  is convergent, then ∞ g(x)dx a a

   

   

is convergent. convergent.

• If 

Implicit Differentiation: When applying implicit differentiation for a function y of  x, ever every y term term with a y   should, should, after after normal differentiation (often involving the product rule), be multiplied by  y   because of the chain rule. After that, the equation should be solved for y  .

∞ g (x)dx  is divergent, then ∞ f (x)dx  is a

a

divergent.

Complex Numbers Complex Number Notations:

Lineair Approximations:

i2 =

f (x) − f (a) ≈ f  (a)(x − a)

(2)

− f (a) =  f (c)(b − a)

(10)

z  =  a + bi  =  r (cos θ  + i sin θ ) =  re  r eiθ a  =  r cos θ

Mean Value Theorem: If  f   is continuous on [ a, b] and differentiable on (a, b), then there is a c  in (a, b) such that: f (b)

−1

and

b  =  r sin θ

|z| = r  = √ a2 + b2

θ  = arctan

b a

or

   

θ  = arctan ab  + π

eiθ = cos θ + i sin θ

(3)

(11) (12) (13)

 

(14)

Complex Number Calculation:

Integration: b

 

f (x)dx  =  F (b)

− F (a)

a

(a + bi) + ( c + di) = (a + c) + ( b + d)i (a + bi)(c + di) = (ac bd) + ( ad + bc)i

(4)

Where F   is any antiderivative/primitive function  f , that is, F   =  f . of  f 

z1 z2  =  r 1 r2 (cos(θ1  + θ2 ) + i sin(θ1 + θ2 ))   r1 eiθ1 r2 eiθ2 =  r 1 r2 ei(θ1 +θ2 )

Complex Conjugates:

   

f (g (x))g  (x)dx  =

b

f (g (x))g  (x)dx  =

a

f (u)du

z  =  a + bi

(5)

f (u)du

(6)

g(a)

b

a

f (x)g (x)dx  =  f (x)g (x) −

 

b f (x)g (x)dx  = [f (x)g (x)] − a



z  =  a

− bi

z + w  =  z  + w zw  =  z w z n =  z n zz  = z 2

g (b)

 

(17)

(18)

||

Integration By Parts:

   

(15) (16)

·

Substitution Rule: If  u  =  g (x) then:

   

 



Differential Equations

f  (x)g (x)dx (7)

Separable Differential Equations:

b

  a

f  (x)g (x)dx

Form :

(8)

dy =  y   =  P (x)Q(y) dx

1 Ahilan A,  Noorul Islam Islam College of Engg, [email protected]

Solution :

 

1 dy  = Q(y )

 

P (x)dx

 

number, the series is divergent. Be careful not to an  =  s . confuse the series an  with the series

(19)



(20)

Monotonic Sequence Theorem If a sequence is either increasing ( an+1 > an  for all n  1) or decreasing ( an+1 < an  for all n  1), it is called a monotonic sequence. If there are  c 1 and c2  such that c1  < an < c2  for all  n 1, it is called bounded. Every bounded monotonic sequence is convergent.

Homogeneous second-order linear differential equations:

Test for divergence: If limn→∞ an   does not exist, or if limn→∞ = 0, then the series sn  is divergent.

First-Order Differential Equations Form : y   + P (x)y  =  Q (x)

 ≥

Let Υ(x) be any integral of  P (x). Solution is: y  =  e −Υ(x)

 



eΥ(x) Q(x)dx + C 

 





Form : ay  + by   + cy  = 0

• b2 − 4ac > 0:

Define r  such that

ar2

Integral test: If  f   is a continuous positive decreasing function on [1, ) and an = f (n) for integer n, then the series sn  is convergent if, and only if, the integral ∞ f (x)dx  is convergent. 1



+ br + c  = 0.

Solution : y  =  c 1 er1 x + c2 er2 x

 

(21)

Comparison test: Suppose an and bn  are series with positive terms and an bn  for all  n , then:

b2

• − 4ac = 0:

 ≤

Define r  such that ar2 + br + c  = 0. Solution : y  =  c 1 erx + c2 xerx

• b2 − 4ac   1, then the series

an  is divergent.

Root test:

   | |

•   If lim →∞ |a | = L <   1, then the series n

n

 •

n

f (x) =

an  is absolutely convergent.

n=0

  If limn→∞ an = L >   1, then the series an  is divergent. n

f (x) =

cn (x

n

− a)

where for  n

  k n

a =  a x i + ay j + az k

n=0



n=0

cn ( x

n

=

ncn (x

n=1

n

− a)

dx  =



n=0

cn

(x

− a) −1

(26)

− a) +1  

(27)

n

f (x) =



|a| =

n=0

 

a2x  + a2y  + a2z

 

(33)

Vector addition and subtraction:

n

n+1

a g (x)n+b

·

(32)

Vector length:

a + b = (ax + bx )i + (ay  + by ) j + (az  + bz )k   (34) a

Representation of functions as power series: The first way to represent functions as power series is simple, but doesn’t always work. To find the representation of  f (x), first find a function g (x) such that f (x) = axb 1−g1(x) , where a and b are constants. The power series is then equal to:



 

Where i, j  and  k  are unit vectors.



− a)

= 1.

0

Notation: A vector a  is often written as:

Differentiation and integration: Differentiation and integration of power functions is possible in the interval ( a R, a + R), where the function does not diverge. It goes as follows: cn (x

(31)

n!

k

≥ 1, and

The number  R  is called the radius of convergence, and can often be found using the ratio test.



k (k

=

Vectors

means must point out whether convergence or divergence occurs.



− 1) . . . (k − n +  1)

n=0

•   The series converges only if  x =  a. (R = 0) •   The series converges for all x. (R = ∞) •   The series converges for |x − a|  < R  and diverges for |x − a| > R. For |x − a| =  R  other

  

(30)

(1 + x) =

(25)



 

k xn n

k

where  x  is a variable and the  c n ’s are constant coefficients of the series. When tested for converges, there are only three possibilities:

 ∞

(29)

 | |

n=0

  

n

− a)

Binomial series: If  k   is any real number and x <   1, the power function representation of (1 + x)k is:

Radius of convergence: Power series are written as



f (n) (a) (x n!

This representation is called the Taylor series of  f (x) at a. For the special case that a   = 0, it is called the Maclaurin series.

Power Series







− b = (a − b )i + (a − b ) j + (a − b )k   (35) x

x

y

y

z

z

a b =  a x bx  + ay by  + az bz

 

Dot product:

·

a a = a 2

|| a·b cos θ  = |a||b|

(37)

·

(28)

(36)

 

(38)

3 Ahilan A,  Noorul Islam College of Engg, [email protected]

Cross product: a

Expressing acceleration in unit vectors:

× b =

a

 

i

j

k

ax bx

ay by

az bz

 

a(t) = v(t)  T(t) + κ v(t) 2 N(t)

(39)

× b = (a b − a b )i+ +(a b − a b ) j + (a b − a b )k y z

z x

z y

x z

x y

(53) | r (t) · r (t)  |r(t) × r (t)| N(t) a(t) = T(t) + |r (t)| |r(t)|

|

(54)

y x

Calculus - Period 3 Functions of Multiple Variables

Notation: r(t) =  f (t)i + g (t) j + h(t)k

 

Definitions: The domain D   is the set ( x, y ) for which f (x, y ) exists. The range is the set of values z  for which there are x, y   such that z = f (x, y). The level curves are the curves with equations f (x, y) = k where k  is a constant.

(41)

Differentiation and integration: r (t) =  f  (t)i + g  (t) j + h (t)k

 

(42)

R(t) =  F (t)i + G(t) j + H (t)k + D

 

(43)

Checking for Limits: L1 as (x, y ) If  f (x, y) (a, b) along a path C 1 and f (x, y) L2 as (x, y) (a, b) along a path C 2 , where L1 =  L 2  then lim(x,y)→(a,b) f (x, y ) does not exist. Also f   is continuous at ( a, b) if  lim(x,y)→(a,b) f (x, y) =  f (a, b)



Function dependant unit vectors: r (t) T(t) =  r (t)

B(t) =  T(t)

 

|

T (t) N(t) = T (t)

|

(44)

 

|

(45)

× N(t)

 

(46)

f x (a, b) =  g  (a)

|

t

s(t) =

  | a

r (t) dt

 

|

(48)

 



(55)





The plane described by this equation is the plane tangent to the curve of  f (x, y ) at (x0 , y0 ).

(50)

Differentials:

Trajectory curvature: dT(t) T (t) κ(t) = ==  ds(t) r (t)



g (x) =  f (x, b)

z z0  =  f x (x0 , y0 )(x x0 )+ f y (x0 , y0 )(y y0 ) (56)

(49)

a(t) =  v  (t) =  r  (t)

where

Tangent Planes: For points close to z0 = f (x0 , y0 ) the curve of  f (x, y ) can be approximated by:

Trajectory velocity and acceleration: = |r (t)| |v(t)| = ds(t) dt



In words, to find f x , regard y  as constant and differentiate f (x, y ) with respect to x. f y   is defined similarly. If  f xy  and  f yx  are both continuous on  D , then f xy =  f yx .

(dx)2 + (dy )2 + (dz )2 = r (t) dt   (47)

|



→  

Partial Derivatives: The partial derivative of  f   with respect to x at (a, b) is:

Trajectory length: ds(t) =

|

  (40)

Vector Functions:

|

|



 | |

 |r(t) × r(t)| κ(t) = |r(t)|3

| |

 

dz  =  f x (x, y )dx + f y (x, y )dy  =

(51)

∂z ∂z dx + dy   (57) ∂x ∂y

If  z  =  f (x, y), x  =  g (s, t) and y  =  h (s, t) then:  

(52)

dz ∂z dx ∂z dy = + ds ∂x ds ∂y ds

 

(58)

4 Ahilan A,  Noorul Islam College of Engg, [email protected]

Directional Derivatives: The directional derivative of  f  at (x0 , y0 ) in the direction of a unit vector (meaning, u = 1) u = a, b  is:

If  D 2  is the region such that D2  = (x, y) a b, h1 (y) x h2 (y) , then:

≤ ≤

 | |

 

 

∇f  · u   (59) grad  f  = ∇f  = f  (x, y), f  (x, y )   (60) The maximum value of  D f (x, y ) is |∇f (x, y )| and occurs when the vector u = a, b   has the same direction as ∇f (x, y). x

}

f (x, y )  dA  =

| ≤ y ≤

h2 (y )

b

  

D2

Du f (x0 , y0 ) =  f x (x, y )a + f y (x, y )b  =

{

f (x, y )  dx dy   (65)

h1 (y)

a

Integrating over Polar Coordinates

y

u

Local Maxima and Minima: If  f   has a local maximum or minimum at (a, b), then f x (a, b) = 0 and f y (a, b) = 0. If  f x (a, b) = 0 and f y (a, b) = 0 then (a, b) is a critical point. If  (a, b) is a critical point, then let D  be defined as:

y x

r2 =  x 2 + y2

tan θ  =

x  =  r cos θ

y  =  r sin θ

 

(66)

 

(67)

If  R  is the polar rectangle such that  R  = (r, θ ) 0 a r b, α θ β  where 0 β  α 2π , then:

≤ ≤

 

{ ≤ − ≤

≤ ≤ }

β

f (x, y )  dA  =

R

| ≤

b

    α

f (r cos θ, r sin θ )  r dr dθ

a

• If  D > 0 then:

(68) If D is the polar rectangle such that D  = (r, θ) 0 h1 (θ) r h2 (θ ), α θ β   where 0 β  α 2π, then:



 

D  =  D (a, b) =  f xx (a, b)f yy (a, b)

2

− (f 

xy (a, b))

(61)

≤  ≤

– If  f xx (a, b)  >  0, then f (a, b) is a minimum. – If  f xx (a, b)  <  0, then f (a, b) is a maximum. If  D <  0, then f (a, b) is a saddle point.

≤  ≤ }

f (x, y)  dA  =

   

b

R

m  =

ρ(x, y )  dA

 

(70)

The x-coordinate of the center of mass is: x  =

     D

 x ρ(x, y )  dA

 ρ(x, y )  dA D

 

(71)

The moment of inertia about the x-axis is:

| ≤x≤

I x  =

y 2 ρ(x, y )  dA

 

(72)

d

f (x, y )  dA  =

c

f (x, y )  dy dx   (62)

The moment of inertia about the origin is:

c

I 0  =

b

f (x, y )  dx dy   (63)

{

}

b

f (x, y )  dA  =

a

(x2 + y 2 )ρ(x, y )  dA  =  I x  + I y   (73)

Triple Integrals If  E   is the volume such that E  = (x,y,z ) a x b, g1 (x) y g2 (x), h1 (x, y) z h2 (x, y) , then:

| ≤x≤



g2 (x)

  

 

D

a

Integrals over Regions: If  D 1  is the region such that D1  = (x, y) a b, g1 (x) y g2 (x) , then:

D1

 

D

d

       a

R

 

(69)

D

f (x, y )  dA  =

≤  ≤

f (r cos θ, r sin θ )  r dr dθ

h1 (θ )

Applications: If  m  is the mass, and ρ(x, y ) the density, then:

Integrals over Rectangles: If  R  is the rectangle such that  R  = (x, y) a b, c y d , then:

≤  ≤ }

    α

Multiple Integrals

{

h 2 (θ )

β

R

Absolute Maxima and Minima: To find the absolute maximum and minimum values of a continuous function  f  on a closed bounded set D , first find the values of  f  at the critical points of  f  in D . Then find the extreme values of  f  on the boundary of  D. The largest of these values is the absolute maximum. The lowest is the minimum.

{ | ≤ ≤ − ≤

≤  ≤

 

f (x, y )  dy dx   (64)

g1 (x)

 { ≤  ≤



b

f (x,y,z )dV   =

g2 (x)

h2 (x,y )

     a g1 (x)

|  ≤ }

f (x,y,z )dzdydx

h1 (x,y )

(74)

5 Ahilan A,  Noorul Islam College of Engg, [email protected]

Calculus - Period 4

•   A piecewise-smooth curve - A union of a finite number of smooth curves. •  A closed curve - A curve of which its terminal point coincides with its initial point. •  A simple curve - A curve that doesn’t intersect itself anywhere between its endpoints. •  An open region - A region which doesn’t contain any of its boundary points. •   A connected region - A region D  for which

Three-Dimensional Integrals Cylindrical Coordinates: x  =  r cos θ

y  =  r sin θ

r2 =  x 2 + y2

tan θ  =

z  =  z

y x

 

(75)

z  =  z   (76)



Integrating Over Cylindrical Coordinates:

...

   

 f (x,y,z )dV   =

E  u2 (r cos θ,r sin θ ) u1 (r cos θ,r sin θ)

 

h2 (θ) h1 (θ )

β α

   

...



rf (r cos θ, r sin θ, z )dzdrdθ

(77)

any two points in D  can be connected by a path that lies in D.  A simply-connected region - A region  D  such that every simple closed curve in  D  encloses only points that are in D. It contains no holes and consists of only one piece.   Positive orientation - The positive orientation of a simple closed curve C   refers to a single counterclockwise traversal of  C .

Spherical Coordinates: Vector Field: A vector field on Rn is a function F  that assigns to each point (x, y) in an n-dimensional set an ndimensional vector F(x, y ). The gradient f  is defined by:

x  =  ρ cos θ sin φ y  =  ρ sin θ sin φ z  =  ρ cos φ 2

2

2

ρ =  x + y + z

(78) (79)

2

 ∇

Integrating Over Spherical Coordinates: If E  is the spherical wedge given by E  = (ρ,θ,φ) a ρ b, α θ β, c φ d , then:



≤ ≤

   

{

≤ ≤ }

 f (x,y,z )dV   =



b a

β α

d c

     

| ≤

ρ2 sin φ . . .

. . . f (  ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)dρdθdφ

∂ (x, y ) = ∂ (u, v )

∂x ∂u ∂y ∂u

∂x ∂v ∂y ∂v

 

=

∂x ∂y ∂u ∂v

∂y  − ∂x ∂v ∂u

x

(80)

f (x, y )dA  =

R

 



f (x(u, v ), y(u, v ))

 

 

(83)

Line Integrals: The line integral of  f   along C  is:

 

    

b

 

f (x, y )ds  =



dx dt

f (x(t), y (t))

a

2

+

dy dt

2

dt

(84) The line integral of  f  along  C  with respect to  x  is:

  (81)

 

If the Jacobian is nonzero and the transformation is one-to-one, then:

 

j + ...

y

and is called the gradient vector field. A vector field F  is called a conservative vector field if it is the gradient of some scalar function.

Change of Variables: The Jacobian of the transformation  T  given by x  = g (u, v) and y  =  h (u, v) is:

 

∇f (x , y , . . .) =  f   i + f 

b

f (x, y )dx  =



 

 

f (x(t), y (t))

a

dx dt   (85) dt

∂ (x, y) The line integral of a vector field F along C  is: dudv ∂ (u, v) b

 

(82) This method is similar to the one for triple integrals, for which the Jacobian has a bigger matrix and the change-of-variable equation has some more terms.



F dr =

·

  a

F(r(t)) · r (t)  dt  =

 



F T ds   (86)

·



Where T  = | |  is the unit tangent vector. r



r

Conservative Vector Fields: If  C  is the curve given by  r(t) (a

Basic Vector Field Theorems

  ∇  ·

f  dr =  f (r(b))

Definitions



≤ t ≤ b), then:

− f (r(a))

(87)

6 Ahilan A,  Noorul Islam College of Engg, [email protected]

  

The integral C  F dr is independent of path in  D if and only if  C  F dr = 0 for every closed path  C  in  D . If  F(x, y) = P (x, y)i +  Q (x, y ) j  is a conservative vector field, then:

For a surface graph of  g (x, y), the normal vector is given by:

· ·

∂P  ∂Q = ∂y ∂x

 

n =

∂g ∂x

− i−

 

Surfaces

F dS =

·



Parametric Surfaces: A surface described by r(u, v ) is called a paramet∂  ∂  ric surface. r = ∂u and r = ∂v . For smooth surfaces (r r =  0  for every u and v ) the tangent plane is the plane that contains the tangent vectors r and r , and the vector r r   is the normal vector to the tangent plane. r

u

r

 

v

 ×

v

A  =

u

  |

r

u

D

× r |dA

F n  dS 

 

·

(95)

∂g ∂x

1+

2

+

∂g ∂y

 

f (x,y,z )  dS  =



 

f (r(u, v)) r

u

·

u

×r

v

)  dA   (96)

·

  − D

 ∂ g P  ∂x

 −

∂g Q +R ∂y



dA   (97)

Advanced Vector Field Theorems Curl: If  F =  P i + Q j + Rk, then the curl of  F, denoted by curl F  or also F, is defined by:

 ∇ ×

  (90)



∂R ∂y

 −

∂Q ∂z

  i+

∂P  ∂z



 

∂R  j + ∂x

∂Q ∂x

 −

∂P  ∂y



k

(98)

If  f  is a function of three variables, then: curl( f ) =  0



v

 

(99)

This implies that if F is conservative, then curl F = 0. The converse is only true if  F is defined on all of  n n R . So if F is defined on all of  R and if curl F = 0, then F  is a conservative vector field.

For a surface graph of  g (x, y ), the surface integral is given by:

 

F (r

D

(89)

| × r |dA   (91)

D

F dS =



2

dA

·

 

For a surface graph of  g (x, y), the flux is given by:

Surface Integrals: For a parametric surface, the surface integral is given by:

f (x,y,z )  dS  =

Divergence: If  F = P i +  Q j +  R k, then the divergence of  F, denoted by div F or also F, is defined by:



D

 

v

 

v

        D

 

+



F dS =



For a surface graph of  g (x, y), the surface area is given by:

 

(94)

2

∂g ∂y

v

Surface Areas: For a parametric surface, the surface area is given by:

A  =

2

This integral is also called the flux of  F across S . For a parametric surface, the flux is given by:

×  

u

∂g ∂x

Flux: If F is a vector field on a surface  S  with unit normal vector n, then the surface integral of  F  over  S  is:

Also, if  D  is an open simply-connected region, and if  ∂P  = ∂Q  , then F is conservative in D. ∂y ∂x

u

+k

      1+

(88)

∂g  j ∂y

     

f (x,y,g (x, y ))

1+

∂g ∂x

2

+

∂g ∂y

2

 ∇ ·

dA

(92)

div F  =

Normal Vectors: For a parametric surface, the normal vector is given by: r r   (93) n = r r v

u

v

 

(100)

If  F is a vector field on Rn , then div curl F = 0. If div F = 0, then F is said to be incompressible. Note that curl F returns a vector field and div F returns a scalar field.

× | × | u

∂P  ∂Q ∂R + + ∂x ∂y ∂z

7

Ahilan A,  Noorul Islam College of Engg, [email protected]

Green’s Theorem: Let C   be a positively oriented piecewise-smooth simple closed curve in the plane and D  be the region bounded by C . Now:

 

P dx + Q dy  =



   D

∂Q ∂x

 −

∂P  ∂y



dA   (101)

This can also be useful for calculating areas. To calculate an area, take functions  P  and Q  such that ∂Q ∂P   = 1 and then apply Green’s theorem. ∂x ∂y In vector form, Green’s theorem can also be written as:

 −

  ·     ·   F dr =



(curl F) k  dA

·

D

F n  ds  =



div F(x, y )  dA

 

(102)  

(103)

D

Stoke’s Theorem: Let  S  be an oriented piecewise-smooth surface that is bounded by a simple, closed, piecewise-smooth boundary curve  C  with positive orientation. Let F be a vector field that contains S . Then:

 



F dr =

·

 

curl F dS

·



 

(104)

The Divergence Theorem: Let E   be a simple solid region and let S   be the boundary surface of  E , given with positive (outward) orientation. Let F   be a vector field on an open region that contains E . Then:

 



F dS =

·

 

div F  dV  

 

(105)



8

Ahilan A,  Noorul Islam College of Engg, [email protected]

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