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One Sheet Vector Review
Vectors are numerical objects characterized by a mag- nitude and a direction. Vectors can be moved around as long as their length (magnitude) and direction/orientation do not change.
Unit Vectors
Unit vectors are vectors of length 1 in the fundamental perpendicular directions of your coordinate system. There are two common conventions for representing the unit vec- tors in the x, y, and z directions:
x ˆ, yˆ, zˆ or ˆi, ˆj, ˆk
We can stretch these unit vectors to any length we like by multiplying by scalar magnitudes, e.g.
Ax xˆ or Byˆj or Cz zˆ
2D Vector Representations
A (^) x x
Ay
A
y
θ A cos( θ)
A sin( θ)
(Quadrants)
Q
Q2 Q
Q
Cartesian: A~ = Ax xˆ + Ay yˆ or A~ = (Ax, Ay ) Polar: A~ = (A, θ) where: Ax = A cos(θ) Ay = A sin(θ)
A =
√ A^2 x + A^2 y θ = tan−^1 (Ay /Ax)
θ is measured positive as the angle between the positive x-axis and the vector in the counterclockwise direction.
Adding or Subtracting Vectors
Given vectors A~ and B~ with components (Ax, Ay ) and (Bx, By ) respectively:
C^ ~ = A~ + B~ = (Ax + Bx)xˆ + (Ay + By)yˆ
similarly:
D^ ~ = A~ − B~ = (Ax − Bx)xˆ + (Ay − By)yˆ
Triangles for Vector Addition
Vectors can graphically be visualized or added by putting the head of one vector at the tail of the other and drawing the resultant as the triangle connecting the first tail to the second head:
A
B
B
(a) C = A + B^ (b) A D = A − B
Useful Trig/Triangles You Should Know
sin(30◦) = cos(60◦) = 1/ 2 cos(30◦) = sin(60◦) =
sin(45◦) = cos(45◦) =
(30◦= π/6, 45◦= π/4) sin(37◦) ≈ 0. 6 cos(37◦) ≈ 0. 8
(3-4-5 triangle angles are 37◦and 53◦) Kinds of Products of (3D) Vectors
Inner or Scalar or Dot Product: A^ ~ · B~ = AxBx + Ay By + AzBz = AB cos(θ)
The (scalar) length of a vector is defined to be:
A = +
A^2 = +
√ A^ ~ · A~ = +
√ A^2 x + A^2 y + A^2 z
Cross or Vector Product: | A~ × B~| = AB sin(θ)
and direction from right hand rule, align fingers of right hand with A~, rotate through the smaller angle in the plane into B~, thumb indicates the direction of the cross product, or use the Cartesian representation: C^ ~ = A~ × B~ = (AyBz − Az By ) ˆx + (Az Bx − AxBz ) yˆ + (AxBy − Ay Bx) ˆz To easily remember this last form, note that there are three cyclic permutations of the coordinates in alphabet- ical order: x y z, y z x, z x y. Observe that the positive term in each parentheses and the unit vector contain x y z in cyclic order : for example AxBy zˆ. The negative terms in each parentheses simply have the first two indices swapped, as in −AyBx ˆz.
One Sheet Calculus Review
Derivatives
dun du
= n un−^1
deu du
= eu
d sin(u) du = cos(u)
d cos(u) du = − sin(u)
d ln(u) du
u Indefinite Integrals
for (n 6 = −1)
∫ un^ du = un+ n + 1
for (n = −1)
∫ u−^1 du =
∫ (^) du u = ln |u| ∫ eu^ du = eu ∫ cos(u) du = sin(u) ∫ sin(u) du = − cos(u)
In all cases a constant of integration must be added if the integral is not used to evaluate a definite integral (one with explicit limits).
Definite Integral Rule
Given: dF (x) dx = f (x)
or dF = f (x) dx
then
F (x)|ba = F (b) − F (a) =
∫ (^) b
a
dF =
∫ (^) b
a
f (x) dx
Integration by Parts
It’s differentiation of a product, both ways:
d(uv) = v du + u dv
Move one term to the other side, rearrange, and integrate both sides to get: ∫ u dv =
∫ d(uv) −
∫ v du = uv −
∫ v du
The Chain Rule
df dx
df du
du dx u-Substitution (Examples) An application of the chain rule. We wish to do
∫ e−αt^ dt. Let u = −αt. Thus du = −α dt. We convert the integral into a u-form by muliplying and dividing by du dt = −α to turn dt into du: ∫ e−αt^ dt =
( 1 −α
) ∫ e−αt(−α dt) = −
α
e−αt
Similarly:
∫ cos(ωt) dt =
ω sin(ωt)
∫ (3x + 2)^2 dx =
( 1 3
) (3x + 2)^3 3
(3x + 2)^3
∫ dv v − mg/b = ln |v − mg/b|
Taylor Series Works best for “small” x:
f (u) = f (a + x) = f (a) + df dx
∣∣ ∣∣ a
x + d^2 f dx^2
∣∣ ∣∣ ∣ a
x^2 2!
Binomial Expansion A special case of the Taylor Series for f (u) = un^ = (1 + x)n, expanded for u = 1 + x around 1. Requires |x| < 1 to unconditionally converge:
(1 + x)n^ = 1 + nx 1!
n(n − 1)x^2 2!
One Sheet Simultaneous Equation Review
Things to Consider
• You must have as many independent equations
as you have unknowns. If you have three unknowns and only two equations, keep looking, think about constraint equations or missing physics.
• Your “answer” for one unknown may not con-
tain other unknowns. If it does, the system is not solved! This is a common mistake! Find your answer in terms of the given/known quantities only!
- Check the units of your answers! Let me put that more clearly as it applies to ALL of the algebraic work you do on the basis of this guide:
CHECK THE UNITS
OF YOUR ANSWERS!
Simple Substitution or Elimination
Suppose you know (as the result of applying some physical reasoning) that at some given time tf :
xf =
at^2 f + v 0 tf + x 0
and vf = atf + v 0
where x 0 , xf , a and v 0 all are given, but vf and tf are not known. We would like to find vf. To find it, we have to eliminate the unknown tf between the two equations. We could rearrange the first equation (which has only one unknown in it) and solve for tf in terms of the givens, and substitute it into the second, but that involves the quadratic formula. It is simpler to solve the second equa- tion for tf in terms of vf and givens, and substitute this into the first equation:
tf = (vf − v 0 )/a
(xf − x 0 ) =
a(vf − v 0 )^2 /a^2 + v 0 (vf − v 0 )/a
v^2 f − v^20 = 2a(xf − x 0 )
(where you should fill in the missing steps of algebra). Other times it is even simpler:
a = (m 1 − m 2 sin(θ))g/(m 1 + m 2 ) and
α = a/R so α = (m 1 − m 2 sin(θ))g/(m 1 + m 2 )R
Gauss Elimination and Back Substitution This is the meat and potatoes approach for linear prob- lems. It is the way computers often solve the problem (with a few bells and whistles). It involves lining equa- tions up so that their variables are right above one an- other. Then it uses the following reasoning: Multiplying a true equation by a constant produces a true equation. Add two true equations produces a (possibly new) true equation. So we multiply equations by scale factors so that adding or subtracting pairs causes terms with the unknowns to disappear, one at a time, until only one is left and the equation can be solved. One then back sub- stitutes the result into the preceding step (where you had two unknowns, now only one) and solve for the next unknown, repeating until all unknowns are known! I’ll give a single example, corresponding to a falling mass unrolling a rope coiled around a massive disk to make it spin up. The unknowns are a, α and T (don’t worry yet about what these mean). The system is: mg − T = ma
RT = βM R^2 α α = a/R
The knowns are m, M, R, g. Substitute the third equation into the second to eliminate α immediately. The second becomes: T = βM a Now add these two equations: mg − T = ma
+(T = βM a) to cancel T and get: mg = (m + βM )a
Solve for a: a = mg/(m + βM ) Back substitute this into the equation for T above:
T = βM mg/(m + βM ) and α: α = mg/(m + βM )R
Finally, we check units. Hmmm, mass units cancel, g has units of acceleration, T has an extra mass and hence is a force, α is inverse time squared, all correct. We’re good to go!
One Sheet Line Integral Review
Motivation
In physics we have a number of occasions to integrate quantities along a specific directed path. Sometimes the quantity of interest is a scalar, such as mass density to find the total mass of e.g. a piece of string. More of- ten it is used to integrate forces or fields (both vector quantities) along a vector path. In particular, this occurs when evaluating work, potential energy, and poten- tial, where the latter is the potential energy per unit mass or charge for the gravitational field or electrostatic field. Line integrals of this sort appear in Maxwell’s Equa- tions, which describe the fundamental electromagnetic field.
Definitions
Let C be a curved path of finite length in space. Imag- ine chopping the curve into a large (eventually infinite) number N of pieces of length ∆s. Then:
∫
C
f (x, y, z)ds = lim n→∞
∑^ n i=
f (xi, yi, zi)∆si
is the integral of f (x, y, z) along the curve C. In order to use our existing skills in one dimensional integration, we usually have to express ds in terms of a parameter that plays the role of i in this sum and then integrate over that parameter.
Example: Find the total mass M of a piece of string with uniform mass density λ shaped like a circular arc of radius R from θ = 0 to θ = π/2.
Solution: An infinitesimal chunk of the string has length ds =
√ dx^2 + dy^2 and hence mass dm = λ ds. With con- siderable effort we could express and perform this integral in terms of cartesian coordinates. However, since R is constant, it is much easier to express ds in terms of the single parameter θ: ds = Rdθ
and
M =
∫ dm =
∫ λds
= λ
∫ (^) π/ 2
0
R dθ
= πRλ 2
This makes sense, since the length of the circular arc is πR
Vector Application
Suppose F~ = Fx ˆx + Fy yˆ + Fz zˆ where the Fi(x, y, z) are functions of the general space coordinates x, y, z. In order to evaluate
∫ C F~^ ·d~ℓ^ along the curve^ C^ we once again have to break the curve up into infinitesimal vector chunks of scalar length dℓ, each directed tangent to the curve. Let us write:
d~ℓ = dℓ ˆℓ = dℓx xˆ + dℓy yˆ + dℓz zˆ
Then ∫
C
F^ ~ · d~ℓ =
∫
C
(Fxdℓx + Fydℓy + Fz dℓz )
In order to make this integral doable using ordinary one- dimensional integration techniques, we usually (again) pa- rameterize the pieces in terms of an independent variable e.g. s. That is, let each point (x, y, z) on the curve C be a function of a one-dimensional monotonic parameter s: (x(s), y(s), z(s)). Then we can write the integral in terms of s: ∫
C
F^ ~ · d~ℓ =
∫ (^) s 1
s 0
Fx(x(s), y(s), z(s) dℓx ds
ds
Fy (x(s), y(s), z(s) dℓy ds ds
Fz (x(s), y(s), z(s) dℓz ds
ds
We only evaluate this for simple cases in an introduc- tory class. For example, suppose B~ = μ 2 πr^0 I (+ sin(θ)xˆ − cos(θ)yˆ)) (clockwise and tangent to a circle of radius r) and the vector curve C is a circle of radius R directed counterclockwise. Then we can parameterize C with θ such that d~ℓ = Rdθ(− sin(θ)xˆ + cos(θ)yˆ)) and
∫
C
B^ ~ · d~ℓ = − μ^0 I 2 πR
∫ (^2) π
0
R(sin^2 (θ) + cos^2 (θ))dθ = −μ 0 I
An even simpler example is the change in gravitational po- tential energy of a particle of mass m moving from height y 0 to height y 1 along any path:
∆U = −
∫
C
F^ ~ · d~ℓ = −
∫
C
−mg yˆ · d~ℓ =
∫ (^) y 1
y 0
mgdy
= mg(y 1 − y 0 )
