Final Flashcards
dy/dt = k(N - y)
y = N - (N - y0)e^(-kt)
N is a specified upper bound
- always CD
dy/dt = ky
y = y0e^(kt)
- exponential growth or decay
dy/dt = -k(y - N)
y = N + (y0 - N)e^(-kt)
N is a specified lower bound
- always CU
dy/dt = (k/N)y(N - y) = ky - ky^2/N
y = (Ny0)/(y0 + (N - y0)e^(-kt)) N is a specified upper bound only inflection point at N/2 < N/2 is CU > N/2 is CD
Time of IP Logistic Equation
T = (1/k)ln((N-y0)/y0)
dy/dt + @y = f(t)
Solve with integrating factor
u = e^(S(P(t)dt)))
y = e^(@t)y0 + e^(@t)S(fe^(-@t)dt
x^2 + y^2 + z^2 = a^2
Sphere
z = +- sqrt(a^2 - x^2 - y^2)
Sphere
x^2/a^2 + y^2/b^2 + z^2/c^2 = 1
Ellipsoid
x^2/a^2 + y^2/b^2 - z^2/c^2 = 1
One sheet Hyperboloid
x^2/a^2 + y^2/b^2 - z^2/c^2 = -1
Two sheet Hyperboloid
x^2/a^2 + y^2/b^2 - z^2/c^2 = 0
Conical Hyperboloid
z = sqrt(x^2 + y^2)
Circular Cone
x^2/a^2 + y^2/b^2 = z/c
Paraboloid
z = 100 - x^2 - y^2
Paraboloid
origin at 100
facing downwards
x/a + y/b + z/c = 1
Plane
Level Curves
Use x^2 + y^2 = r^2
Limits (multivariable)
if = 0 it is inconclusive
lines, parabola
Use x^2 + y^2 = r^2
Parametric with 3 variables
Clairault’s Theorem
if fxy=fyx then both are continuous
Parametric Representation of a Line
in direction v = Aî + Bj
x = x0 + At
y = y0 + Bt
Parametric Representation of a Circle
x = rcost y = rsint
Parametric Representation of an Ellipse
x = acost y = bcost
Parametric Representation of a Hyperbola
x = acosht y = bsinht
Gradient
Normal to curve
Change in position
Tangent to curve
Implicit Function Theorem
dy/dx = -Fx/Fy
Integration by Substitution
Sf(g(x))g’(x)dx
u = g(x)
du = g’(x)
Integration by Parts
Sudv = uv - Svdu
v easier
Cramer’s Rule
Du/Dx = - det(D(F,G)/D(x,v)) / det(D(F,G)/D(u,v))
independent on top
Jacobians
(D(F,G)/D(x,v)) forms a 2 x 2 matrix
take det and use cramer’s rule
Tangent and Normal Lines to Curves
f(x,y) = 0 write as a function of t need point mtan = dy/dx = -Fx/Fy mnor = Fy/Fx y-y0 = m(x-x0)
Tangent Plane and Normal Line
g(x,y,z) = 0
Tangent Plane: gradient and point
Normal Line: directional vector (grad) and point (parametric)
Tangent Line and Normal Plane
Curve of intersection f created by g and h
Tangent Line: directional vector (g x h) and parametric equation
Normal Plane: normal vector (tangent vector) and point
Directional Derivative
Dûf = grad(f) • û = |grad(f)|cos@
Dûf max/min
max: |grad(f)|
when cos@ = 1
min: |grad(f)|
when cos@ = -1
Differentials
(change in)f ~ df = Df/Dx dx + Df/Dy dy
Errors
Use differentials
% = dx/x
Linearization
L = f(x0, y0) + (x - x0)(fx) + (y - y0)(fy)
initial point + tangent line
Extrema Conditions and Proofs
grad(f) = 0 Geometric At extremum, tangent plane horizontal Contradiction If grad(f) =/ 0 move in (same/opposite) direction for extremum
Extrema Sufficiency f(x,y)
D test = (all at extremum) |fxx fxy| = AC - B^2 |fxy fyy| D > 0, A > 0 minimum D > 0, A < 0 maximum D < 0 saddle D = 0 inconclusive
Sufficiency for f(x,y,z)
All at extremum
1 = fxx
2 = D
3 = |fxx fxy fxz|
|fxy fyy fyz|
|fxz fyz fzz|
i) 1, 2, 3 > 0 minimum
ii) 1, 3 < 0 2 > 0 maximum
iii) 3 = 0 inconclusive
iv) all others: neither max nor minLagrange Multipliers and Solving f(x,y)
f extremized, g constraint L = f + ¥g demands grad(L) = 0 Lx = 0 Ly = 0 solve for ¥ and equate, plug into L¥ and solve for variables
Lagrange Sufficiency
H = Lxx gy^2 - 2Lxy gx gy + Lyy gx^2
H > 0 minimum
H < 0 maximum
Lagrange Multipliers and Solving f(x,y,z)
Same as f(x,y)
Lz equate ¥ with Lx or Ly
Simplify and plug into constraint
Extrema on Bounded Domains
Find extrema of f (grad(f) = 0)
Parametrize D and plug in to f
Find extremum of f(t): df/dt = 0
Find values of points
Population Case I: B and D are constant
dy/dt = Ay
y = y0e^At
Exponential growth or decay
Population Case II: Verhulst
B = B0 - B1y D constant
dy/dt = (B0 - D0)y - B1y^2
Logistic: k = B0 - D0 N = B1/k
Population Case III: Limited Environment
A = k(N - y)
dy/dy = Ay = ky(N - y)
Logistic
Population Case IV: Competitive
B constant D = cy
A = B - cy
dy/dt = By - cy^2
Logistic k = B N = B/c
Population Case V: Primitive
B = cy D constant
A = cy - D dy/dt = -D(y - (c/D)y^2) Logistic k = D M = D/c dy/dt = -ky - ky^2/M y = My0/(y0 + (M - y0)e^(kt))
Population Case V: Subcases
i) y0 = M
y(t) = M = y0
stable
ii) y0 < M
lim as t > inf. denominator = inf.
lim as t > inf. y = 0
Extinction
iii) y0 > M When t = T = (1/k)ln(y0/(y0 - M)) lim as t > T denominator = 0 lim as t > T y = inf. Doomsday
F =
-k/(R+x)^2
when x = 0
-k/R^2 = -mg
= m(dv/dt)
Escape Velocity
lim as x > inf
v = 0
