Trigonometric identities - Wilkes University

Formula Sheet

Trigonometric identities:

sin2 x + cos2 x = 1 tan2 x + 1 = sec2 x sin 2x = 2 sin x cos x

cos 2x = cos2 x - sin2 x sin2 x = (1 - cos 2x)/2 cos2 x = (1 + cos 2x)/2

The line through the point (x0, y0, z0) parallel to the vector v = v1i + v2j + v3k is given by

r(t) = (x0 + tv1)i + (y0 + tv2)j + (z0 + tv3)k

The plane through the point (x0, y0, z0) with normal vector n = Ai + Bj + Ck is given by the equation

A(x - x0) + B(y - y0) + C(z - z0) = 0.

u ? v = |u||v| cos and |u ? v| = |u||v| sin

The (arc) length of a curve C given by r(t) for a t b is

b

L = ds = |v(t)|dt,

C

a

where

v(t)

=

dr dt

.

Distance from a point S to a line passing through point P , parallel to

v:

d

=

|PS?v| |v|

.

Given a point S in space, and a plane with normal n and point P on

the

plane,

the

distance

from

S

to

the

plane

is

d

=

|PS?n| |n|

.

N = (dT/dt)/|dT/dt| (principal unit normal vector, here T is the unit tangent vector).

= (1/|v|)|dT/dt| (curve curvature).

(acceleration) a = aT T+aN N, where aT

=

d(|v|) dt

and aN

=

|a|2 - a2T .

The area enclosed by a polar curve r = f () between two angles 1 and 2 is given by

1

2 r2d = 1

2

(f ())2 d.

2 1

2 1

Directional Derivative: The derivative of the function f in the direction of a unit vector u at the point P0 is Duf (P0) = f (P0) ? u.

The linearization of f (x, y) at a point P0 = (x0, y0) is L(x, y) = f (P0) + fx(P0)(x - x0) + fy(P0)(y - y0).

 The Second Derivative Test for Functions of Two Variables: Let f (x, y) be a twice differentiable function, and assume its second

partial derivatives are continuous. Let (a, b) be a critical point for f ,

and define the Hessian of f at (a, b) to be H(a, b) = fxx(a, b)fyy(a, b) - [fxy(a, b)]2.

? If H(a, b) < 0, then (a, b) is a saddle point. ? If H(a, b) > 0 and fxx(a, b) > 0, then (a, b) is a local min. ? If H(a, b) > 0 and fxx(a, b) < 0, then (a, b) is a local max. ? If H(a, b) = 0, the test is inconclusive.

Lagrange Multipliers (problems with equality constraints): To find min/max of a function f , subject to a constraint g = 0, solve the system of equations f = g

g=0

for x, y, and . The max and min values of f are attained at the points among these solutions.

Polar Coordinates: ? x = r cos , y = r sin , r = ? dA = r drd

x2 + y2,

tan

=

y x

Cylindrical Coordinates: ? x = r cos , y = r sin , z = z, r = ? dV = dz r drd

x2 + y2,

tan

=

y x

Spherical Coordinates:

? x = sin cos , y = sin sin , z = cos

? = x2 + y2 + z2 = r2 + z2, r = sin ? dV = 2 sin ddd

The center of mass (x?, y?, z?) of a three-dimensional object D with

density (x, y, z): x? =

D xdV D dV

,

y?

=

( D dV is the mass of the object.)

D ydV D dV

,

z?

=

D zdV D dV

.

Change of Variables in Double Integrals: Suppose a transformation T given by x = g(u, v) and y = h(u, v) is one-to-one (except perhaps on the boundary of G), maps a region G in the uv-plane to a region R in the xy-plane, and g and h have continuous partial derivatives in G. If f is a continuous function over the region R, then

f (x, y)dx dy = f (g(u, v), h(u, v)) |J(u, v)|du dv,

R

G

where the Jacobian of the transformation is given by

x x

u v

J(u, v) =

.

y y

u v

 Given a vector field F and a smooth curve C, parameterized by r(t), a t b:

? flow/work/circulation along/around C is given by:

C F ? dr =

b a

F(r(t))

?

dr dt

dt.

? flux of F = M i + N j across C in the plane is given by:

C M dy - N dx =

ab(M

dy dt

-

N

dx dt

)dt.

Component Test for Conservative Vector Fields: A vector field F = M, N, P is conservative if and only if M N M P N P = , =, =. y x z x z y Also: there is a potential function f so that f = F and line integrals

are independent of path between any two points A and B:

C F ? dr = f (B) - f (A).

Green's Theorem (R is a region enclosed by a loop C):

C F ? dr = C M dx + N dy =

R

(

N x

-

M y

)

dxdy

C F ? n ds = C M dy - N dx =

R(

M x

+

N y

)

dxdy

Surface integral of f (x, y, z) over surface S, parameterized by r(u, v) over the region R in the uv-plane:

f (x, y, z) d = f (r(u, v))|ru ? rv| dudv.

S

R

For surface area, use f (x, y, z) = 1.

Flux of a vector field F through surface S parameterized by r(u, v) is

F ? n d = F(r(u, v)) ? (ru ? rv) dudv

S

S

.

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