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)
{ ≤ ≤
E
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 =
E
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
S
f (x(u, v ), y(u, v ))
(83)
Line Integrals: The line integral of f along C is:
b
f (x, y )ds =
C
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 =
C
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.
C
F dr =
·
a
F(r(t)) · r (t) dt =
C
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
C
≤ 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 =
·
S
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 =
S
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 =
S
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:
S
D
v
v
D
+
S
F dS =
S
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 =
C
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 =
C
(curl F) k dA
·
D
F n ds =
C
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:
C
F dr =
·
curl F dS
·
S
(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:
S
F dS =
·
div F dV
(105)
E
8
Ahilan A, Noorul Islam College of Engg,
[email protected]