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Multivariable Calculus Study Guide:
A LATEX Version
Tyler Silber
University of Connecticut
December 11, 2011
1 Disclaimer
It is not guaranteed that I have every single bit of necessary information for the course. This happened to be some of what I needed to know this specific semester in my course. For example, Stokes’ Theorem is not even mentioned.
2 Vectors Between Two Points
Given : P (x 1 , y 1 ) & Q(x 2 , y 2 )
P Q =
x 2 − x 1 y 2 − y 1
3 Vectors in the Plane
let v =
v 1 v 2
& u =
u 1 u 2
3.1 Simple Operations
cv =
cv 1 cv 2
|v| =
v 12 + v^22
v + u =
v 1 + u 1 v 2 + u 2
3.2 Unit Vectors
i =
& j =
v = v 1 i + v 2 j
3.3 Vectors of a Specified Length
cv |v|
∣ =^ |c|
cv |v| ‖ v
4 Vectors in Three Dimensions
4.1 Notes
Everything in the above section can be expanded to three dimensions. Simply add another component.
k =
4.2 Random Equations
xy-plane {(x, y, z) : z = 0} xz-plane {(x, y, z) : y = 0} yz-plane {(x, y, z) : x = 0}
Sphere: (x − a)^2 + (y − b)^2 + (z − c)^2 = r^2
5 Dot Product
5.1 Definitions
u · v = u 1 v 1 + u 2 v 2 + u 3 v 3 = |u||v| cos θ u ⊥ v ⇔ u · v = 0 u ‖ v ⇔ u · v = ±|u||v|
7.5 Limits
lim t→a r(t) =
lim t→a f (t), lim t→a g(t), lim t→a h(t)
8 Calculus of Vector-Valued Functions
8.1 Derivative and Tangent Vector
r′(t) = f ′(t)i + g′(t)j + h′(t)k
Note: r′(t) is the tangent vector to r(t) at the point (f (t), g(t), h(t)).
8.2 Indefinite Integral
r(t) dt = R(t) + C
Note: C is an arbitrary constant vector and R = F i + Gj + Hk.
8.3 Definite Integral
∫ (^) b
a
r(t) dt =
[∫
b
a
f (t) dt
]
i +
[∫
b
a
g(t) dt
]
j +
[∫
b
a
h(t) dt
]
k
9 Motion in Space
9.1 Definitions
a(t) = v′(t) = r′′(t) Speed = |v(t)|
9.2 Two-Dimensional Motion in a Gravitational Field
Given : v(0) = 〈u 0 , v 0 〉 & r(0) = 〈x 0 , y 0 〉 v(t) = 〈x′(t), y′(t)〉 = 〈u 0 , −gt + v 0 〉
r(t) = 〈x(t), y(t)〉 =
u 0 t + x 0 , −
gt^2 + v 0 t + y 0
9.3 Two-Dimensional Motion
Given : v(0) = 〈|v 0 | cos θ, |v 0 | sin θ〉 & r(0) = 〈 0 , 0 〉
T ime =
2 |v 0 | sin θ g
Range = |v 0 |^2 sin 2θ g
M ax Height = y
T
(|v 0 | sin θ)^2 2 g
10 Planes and Surfaces
10.1 Plane Equations
The plane passing through the point P 0 (x 0 , y 0 , z 0 ) with a normal vector n = 〈a, b, c, 〉 is described by the equations:
a(x − x 0 ) + b(y − y 0 ) + c(z − z 0 ) = 0
ax + by + cz = d, where d = ax 0 + by 0 + cz 0
In order to find the equation of a plane when given three points, simply create any two vectors out of the points and take the cross product to find the vector normal to the plane. Then use one of the above formulae.
10.2 Parallel and Orthogonal Planes
Two planes are parallel if their normal vectors are parallel. Two planes are orthogonal if their normal vectors are orthogonal.
10.3 Surfaces
10.3.1 Ellipsoid
x^2 a^2
y^2 b^2
z^2 c^2
10.3.2 Elliptic Paraboloid
z = x^2 a^2
y^2 b^2 It would be worth it to learn how to derive sections 9.2 and 9.3.
12 Limits and Continuity
12.1 Limits
The function f has the limit L as P (x, y) approaches P 0 (a, b).
lim (x,y)→(a,b)
f (x, y) = lim P →P 0
f (x, y) = L
If f (x, y) approaches two different values as (x, y) approaches (a, b) along two different paths in the domain of f , then the limit does not exist.
12.2 Continuity
The function f if continuous at the point (a, b) provided:
lim (x,y)→(a,b)
f (x, y) = f (a, b)
13 Partial Derivatives
13.1 Definitions
fx(a, b) = lim h→ 0
f (a + h, b) − f (a, b) h
fy (a, b) = lim h→ 0
f (a, b + h) − f (a, b) h
So basically just take the derivative of one (the subscript) given that the other one is a constant.
13.2 Notation for Higher-Order Partial Derivatives
∂x
∂f ∂x
∂^2 f ∂x^2 = (fx)x = fxx
∂ ∂y
∂f ∂y
∂^2 f ∂y^2
= (fy )y = fyy
∂ ∂x
∂f ∂y
∂^2 f ∂x∂y
= (fy )x = fyx
∂ ∂y
∂f ∂x
∂^2 f ∂y∂x
= (fx)y = fxy
Note: fxy = fyx for nice functions.
13.3 Differentiability
Suppose the function f has partial derivatives fx and fy defined on an open region containing (a, b), with fx and fy continuous at (a, b). Then f is differen- tiable at (a, b). This also implies that it is continuous at (a, b).
14 Chain Rule
14.1 Examples
You can use a tree diagram to determine the equation for the chain rule. You can also just think about it. Refer to the following examples. z is a function of x and y, while x and y are functions of t
dz dt
∂z ∂x
dx dt
∂z ∂y
dy dt
w is a function of x, y, and z, while x, y, and z are functions of t
dw dt
∂w ∂x
dx dt
∂w ∂y
dy dt
∂w ∂z
dz dt
z is a function of x and y, while x and y are functions of s and t
∂z ∂s
∂z ∂x
∂x ∂s
∂z ∂y
∂y ∂s
w is a function of z, z is a function of x and y, x and y are functions of t
dw dt
dw dz
∂z ∂x
dx dt
∂z ∂y
dy dt
14.2 Implicit Differentiation
Let F be differentiable on its domain and suppose that F (x, y) = 0 defines y as a differentiable function of x. Provided Fy 6 = 0,
dy dx
Fx Fy
15 Directional Derivatives and Gradient
15.1 Definitions
Let f be differentiable at (a, b) and let u = 〈u 1 , u 2 〉 be a unit vector in the xy-plane. The directional derivative of f at (a, b) in the direction of u is
Duf (a, b) = 〈fx(a, b), fy (a, b)〉 · 〈u 1 , u 2 〉 = ∇f (a, b) · u
Gradient ∇f (x, y) = 〈fx(x, y), fy (x, y)〉 = fx(x, y)i + fy (x, y)j
17.2 Critical Points
A critical point exists if either
- fx(a, b) = fy (a, b) = 0
- one (or both) of fx or fy does not exist at (a, b)
17.3 Second Derivative Test
Let D(x, y) = fxxfyy − f (^) xy^2
- If D(a, b) > 0 and fxx(a, b) < 0, then f has a local maximum at (a, b).
- If D(a, b) > 0 and fxx(a, b) > 0, then f has a local minimum at (a, b).
- If D(a, b) < 0, then f has a saddle point at (a, b).
- If D(a, b) = 0, then the test is inconclusive.
17.4 Absolute Maximum/Minimum Values
Let f be continuous on a closed bounded set R in R^2. To find absolute maximum and minimum values of f on R:
- Determine the values of f at all critical points in R.
- Find the maximum and minimum values of f on the boundary of R.
- The greatest function value found in Steps 1 and 2 is the absolute maxi- mum value of f on R, and the least function value found in Steps 1 and 2 is the absolute minimum values of f on R.
18 Double Integrals
18.1 Double Integrals on Rectangular Regions
Let f be continuous on the rectangular region R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}. The double integral of f over R may be evaluated by either of two iterated integrals:
∫ ∫
R
f (x, y) dA =
∫ (^) d
c
∫ (^) b
a
f (x, y) dx dy =
∫ (^) b
a
∫ (^) d
c
f (x, y) dy dx
18.2 Double Integrals over Nonrectangular Regions
Let R be a region bounded below and above by the graphs of the continuous functions y = g(x) and y = h(x), respectively, and by the lines x = a and x = b. If f is continuous on R, then
∫ ∫
R
f (x, y) dA =
∫ (^) b
a
∫ (^) h(x)
g(x)
f (x, y) dy dx
Let R be a region bounded on the left and right by the graphs of the continuous functions x = g(y) and x = h(y), respectively, and by the lines y = c and y = d. If f is continuous on R, then
∫ ∫
R
f (x, y) dA =
∫ (^) d
c
∫ (^) h(y)
g(y)
f (x, y) dx dy
18.3 Areas of Regions by Double Integrals
area of R =
R
dA
19 Polar Double Integrals
19.1 Double Integrals over Polar Rectangular Regions
Let f be continuous on the region in the xy-plane R = {(r, θ) : 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β}, where β − α ≤ 2 π. Then
∫ ∫
R
f (r, θ) dA =
∫ (^) β
α
∫ (^) b
a
f (r, θ) r dr dθ
19.2 Double Integrals over More General Polar Regions
Let f be continuous on the region in the xy-plane
R = {(r, θ) : 0 ≤ g(θ) ≤ r ≤ h(θ), α ≤ θ ≤ β}
where β − α ≤ 2 π. Then.
∫ ∫
R
f (r, θ) dA =
∫ (^) β
α
∫ (^) h(θ)
g(θ)
f (r, θ) r dr dθ
If f is nonnegative on R, the double integral gives the volume of the solid bounded by the surface z = f (r, θ) and R.
21.4 Integration in Cylindrical Coordinates
D
f (r, θ, z) dV =
∫ (^) β
α
∫ (^) h(θ)
g(θ)
∫ (^) H(r cos θ,r sin θ)
G(r cos θ,r sin θ)
f (r, θ, z) dz r dr dθ
21.5 Rectangular to Spherical
ρ^2 = x^2 + y^2 + z^2
You have to solve for ϕ and θ with trigonometry.
21.6 Spherical to Rectangular
x = ρ sin ϕ cos θ y = ρ sin ϕ sin θ z = ρ cos ϕ
21.7 Integration in Spherical Coordinates
D
f (ρ, ϕ, θ) dV =
∫ (^) β
α
∫ (^) b
a
∫ (^) h(ϕ,θ)
g(ϕ,θ)
f (ρ, ϕ, θ)ρ^2 sin ϕ dρ dϕ dθ
22 Change of Variables
22.1 Jacobian Determinant of a Transformation of Two
Variables
Given a transformation T : x = g(u, v), y = h(u, v), where g and h are differen- tiable on a region of the uv-plane, the Jacobian determinant of T is
J(u, v) =
∂(x, y) ∂(u, v)
∂x ∂u
∂x ∂v
∂y ∂u
∂y ∂v
22.2 Change of Variables for Double Integrals
R
f (x, y) dA =
S
f (g(u, v), h(u, v))|J(u, v)| dA
22.3 Change of Variables for Triple Integrals
I am SO not typing out the expansion of the above into triple integrals. It’s intuitive. Just add stuff where you think it should go.
22.4 YOU have to Choose the Transformation
Just cry.
23 Vector Fields
23.1 Vector Fields in Two Dimensions
F(x, y) = 〈f (x, y), g(x, y)〉
23.2 Radial Vector Fields in R^2
Let r = (x, y). A vector field of the form F = f (x, y)r, where f is a scalar-valued function, is a radial vector field.
F(x, y) =
r |r|p^
〈x, y〉 |r|p
p is a real number. At every point (sans origin), the vectors of this field are
directed outward format he origin with a magnitude of |F| =
|r|p−^1
. You can
also apply all of this to R^3 by just adding a z component.
23.3 Gradient Fields and Potential Functions
Let z = ϕ(x, y) and w = ϕ(x, y, z) be differentiable functions on regions of R^2 and R^3 , respectively. The vector field F = ∇ϕ is a gradient field, and the function ϕ is a potential function for F.
24 Line Integrals
24.1 Evaluating Scalar Line Integrals in R^2
Let f be continuous on a region containing a smooth curve C : r(t) = 〈x(t), y(t)〉, for a ≤ t ≤ b. Then
∫
C
f ds =
∫ (^) b
a
f (x(t), y(t))|r′(t)| dt =
∫ (^) b
a
f (x(t), y(t))
x′(t)^2 + y′(t)^2 dt
24.2 Evaluating Scalar Line Integrals in R^3
Simply add a z component to the above where it obviously belongs.
D. Then, F is a conservative vector field on D (there is a potential function ϕ such that F = ∇ϕ) if and only if
∂f ∂y
∂g ∂x
∂h ∂x
∂g ∂z
∂h ∂y
For vector fields in R^2 , we have the single condition ∂f ∂y
∂g ∂x
25.2 Finding Potential Functions
Suppose F = 〈f, g, h〉 is a conservative vector field. To find ϕ such that F = ∇ϕ, take the following steps:
- Integrate ϕx = f with respect to x to obtain ϕ, which includes an arbitrary function c(y, z.
- Compute ϕy and equate it to g to obtain an expression for cy (y, z).
- Integrate cy (y, z) with respect to y to obtain c(y, z), including an arbitrary function d(z).
- Compute ϕz and equate it to h to get d(z).
Beginning the procedure with ϕy = g or ϕz = h may be easier in some cases. This method can also be used to check if a vector field is conservative by seeing if there is a potential function.
25.3 Fundamental Theorem for Line Integrals
C
F · T ds =
C
F · dr = ϕ(B) − ϕ(A)
25.4 Line Integrals on Closed Curves
Let R in R^2 (or D in R^3 ) be an open region. Then F is a conservative vector field on R if and only if
C F^ ·^ dr^ = 0 on all simple closed smooth oriented curves C in R.
26 Green’s Theorem
26.1 Circulation Form
C
F · dr =
C
f dx + g dy =
R
∂g ∂x
∂f ∂y
dA
26.2 Area of a Plane Region by Line Integrals
C
x dy = −
C
y dx =
C
(x dy − y dx)
26.3 Flux Form
C
F · n ds =
C
f dy − g dx =
R
∂f ∂x
∂g ∂y
dA
27 Divergence and Curl
27.1 Divergence of a Vector Field
div(F) = ∇ · F = ∂f ∂x
∂g ∂y
∂h ∂z
27.2 Divergence of Radial Vector Fields
div(F) = 3 − p |r|p
F =
r |r|p^
〈x, y, z〉 (x^2 + y^2 + z^2 )p/^2
27.3 Curl
curl(F) = ∇ × F
Just derive the curl by doing the cross product.
27.4 Divergence of the Curl
∇ · (∇ × F) = 0
28 Surface Integrals
28.1 Parameterization
28.1.1 z is Explicitly Defined
Use x = x, y = y, and since z is explicitly defined, you already have what z equals.
28.1.2 Cylinder
Simply use cylindrical coordinates to parameterize the surface in terms of θ and z.
