Tricky Topics (Pure/Mech) Flashcards
(22 cards)
Finding Invariant Points
Points that don’t move under a given transformation.
(a b) (x) - (x)
(c d) (y) - (y)
Finds line of invariant points
Finding Invariant Lines
Finds points on a line that, when transformed, move to another point on the line.
(a b) (x) - (X)
(c d) (mx + c) - (mX + c)
Simultaneous Equations
(sub in top equation to remove big X)
Proof by Induction - Final Statement
- Since true for n = 1
- And true for n = k + 1
- WHEN ASSUMED true for n = k
- Thus true for all n ∈ ℤ*
Acceleration = 0
- No resultant force -> Resolve to = 0
- Constant Velocity (could be 0) (N’s 1st Law)
→ Remember vice versa - Use GCSE Speed = D/T
Acceleration = Constant
SUVAT
Acceleration = Variable
Differentiate given expression
(usually dx/dt)
Vector Projection of a onto b
(Vectors in Further Mechanics)
(a.b / |b|^2) x b
Forgotten Equation: Further Mech
Vectors, Impulse, e
-e(u . I) = v . I
Elasticity Question Techniques
Work Energy Principle
- Asking for energy
- Distance
- Rest
Resolve Forces
- Acceleration
Trig Identities (De Moivre’s)
e.g. cos^5(θ) = …
(z + 1/z)^5 = (2cosθ)^5 = 32cos^5(θ)
(z + 1/z)^5 = BINOMIAL EXPANSION
e.g. cos5θ = …
(cosθ + isinθ)^5 = cos5θ + isin5θ
REAL part of BINOMIAL EXPANSION
Oblique Collisions - N’s Law of Restitution
v(sinβ) = eu(sinα)
v(cosβ) = u(cosα)
tanβ = e(tanα)
Proof by Induction - Mathematical (Σ)
e.g. nΣ (2r-1) = n^2
Basis, Assumption…
(k+1)Σ = (k+1)^2 → AIM
(k+1)Σ = kΣ + (2(k+1) -1)
→ Sub in (k+1) to formula
Proof by Induction - Divisibility Results
e.g. Divisible by 4
f(k+1) - f(k) = 4(x)
f(k+1) -f(k) = af(k) +4(x)
f(k+1) = (a+1)f(k) + 4(x)
Proof by Induction - Matrices
(M)^k+1 = (M)^k x (M)^1
Vector ACUTE ANGLE Formulae
cos/sinθ = | (a .b) / |a||b||
Two Lines = cos
(Direction vectors)
Line & Plane = sin
(Directions vector & Normal)
Two Planes = cos
(Normals)
Sums of Series (Complex No. - Yr2)
+/-1e^aiθ +/- 1 → xe^-1/2iθ
Types of Integration
- Inspection
- Parts
- Substitution
- (Partial Fractions)
∫ ln(x) dx
x(lnx) - x [+ c]
Differentiating & Integrating: a^x
∫dx = 1/lna (a^x) [+ c]
d/dx = a^x(lnx)
Improper Integrals
∫ f(x) dx = Improper if…
- One/both of the limits is infinite
- If f(x) is undefined at some point
between the limits
∫ f(x) dx between ∞ and a
lim t → ∞
∫ f(x) dx between t and a
Either 1/∞ tends towards 0, which is then ignored in the final answer, or the entire graph tends towards ∞.