Final

Question 0

dy/dt = k(N - y)

Answer 0

y = N - (N - y0)e^(-kt)
N is a specified upper bound
- always CD

Question 1

dy/dt = ky

Answer 1

y = y0e^(kt)

- exponential growth or decay

Question 2

dy/dt = -k(y - N)

Answer 2

y = N + (y0 - N)e^(-kt)
N is a specified lower bound
- always CU

Question 3

dy/dt = (k/N)y(N - y) = ky - ky^2/N

Answer 3

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

Question 4

Time of IP Logistic Equation

Answer 4

T = (1/k)ln((N-y0)/y0)

Question 5

dy/dt + @y = f(t)

Answer 5

Solve with integrating factor
u = e^(S(P(t)dt)))
y = e^(@t)y0 + e^(@t)S(fe^(-@t)dt

Question 6

x^2 + y^2 + z^2 = a^2

Answer 6

Sphere

Question 7

z = +- sqrt(a^2 - x^2 - y^2)

Answer 7

Sphere

Question 8

x^2/a^2 + y^2/b^2 + z^2/c^2 = 1

Answer 8

Ellipsoid

Question 9

x^2/a^2 + y^2/b^2 - z^2/c^2 = 1

Answer 9

One sheet Hyperboloid

Question 10

x^2/a^2 + y^2/b^2 - z^2/c^2 = -1

Answer 10

Two sheet Hyperboloid

Question 11

x^2/a^2 + y^2/b^2 - z^2/c^2 = 0

Answer 11

Conical Hyperboloid

Question 12

z = sqrt(x^2 + y^2)

Answer 12

Circular Cone

Question 13

x^2/a^2 + y^2/b^2 = z/c

Answer 13

Paraboloid

Question 14

z = 100 - x^2 - y^2

Answer 14

Paraboloid
origin at 100
facing downwards

Question 15

x/a + y/b + z/c = 1

Answer 15

Plane

Question 16

Level Curves

Answer 16

Use x^2 + y^2 = r^2

Question 17

Limits (multivariable)

Answer 17

if = 0 it is inconclusive
lines, parabola
Use x^2 + y^2 = r^2
Parametric with 3 variables

Question 18

Clairault’s Theorem

Answer 18

if fxy=fyx then both are continuous

Question 19

Parametric Representation of a Line

Answer 19

in direction v = Aî + Bj
x = x0 + At
y = y0 + Bt

Question 20

Parametric Representation of a Circle

Answer 20

x = rcost
y = rsint

Question 21

Parametric Representation of an Ellipse

Answer 21

x = acost 
y = bcost

Question 22

Parametric Representation of a Hyperbola

Answer 22

x = acosht
y = bsinht

Question 23

Gradient

Answer 23

Normal to curve

Question 24

Change in position

Answer 24

Tangent to curve

Question 25

Implicit Function Theorem

Answer 25

dy/dx = -Fx/Fy

Question 26

Integration by Substitution

Answer 26

Sf(g(x))g'(x)dx u = g(x) du = g'(x)

Question 27

Integration by Parts

Answer 27

Sudv = uv - Svdu | v easier

Question 28

Cramer's Rule

Answer 28

Du/Dx = - det(D(F,G)/D(x,v)) / det(D(F,G)/D(u,v)) | independent on top

Question 29

Jacobians

Answer 29

(D(F,G)/D(x,v)) forms a 2 x 2 matrix | take det and use cramer's rule

Question 30

Tangent and Normal Lines to Curves

Answer 30

``` 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) ```

Question 31

Tangent Plane and Normal Line

Answer 31

g(x,y,z) = 0 Tangent Plane: gradient and point Normal Line: directional vector (grad) and point (parametric)

Question 32

Tangent Line and Normal Plane

Answer 32

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

Question 33

Directional Derivative

Answer 33

Dûf = grad(f) • û = |grad(f)|cos@

Question 34

Dûf max/min

Answer 34

max: |grad(f)| when cos@ = 1 min: |grad(f)| when cos@ = -1

Question 35

Differentials

Answer 35

(change in)f ~ df = Df/Dx dx + Df/Dy dy

Question 36

Errors

Answer 36

Use differentials | % = dx/x

Question 37

Linearization

Answer 37

L = f(x0, y0) + (x - x0)(fx) + (y - y0)(fy) | initial point + tangent line

Question 38

Extrema Conditions and Proofs

Answer 38

``` grad(f) = 0 Geometric At extremum, tangent plane horizontal Contradiction If grad(f) =/ 0 move in (same/opposite) direction for extremum ```

Question 39

Extrema Sufficiency f(x,y)

Answer 39

``` 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 ```

Question 40

Sufficiency for f(x,y,z)

Answer 40

``` 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 min ```

Question 41

Lagrange Multipliers and Solving f(x,y)

Answer 41

``` 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 ```

Question 42

Lagrange Sufficiency

Answer 42

H = Lxx gy^2 - 2Lxy gx gy + Lyy gx^2 H > 0 minimum H < 0 maximum

Question 43

Lagrange Multipliers and Solving f(x,y,z)

Answer 43

Same as f(x,y) Lz equate ¥ with Lx or Ly Simplify and plug into constraint

Question 44

Extrema on Bounded Domains

Answer 44

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

Question 45

Population Case I: B and D are constant

Answer 45

dy/dt = Ay y = y0e^At Exponential growth or decay

Question 46

Population Case II: Verhulst | B = B0 - B1y D constant

Answer 46

dy/dt = (B0 - D0)y - B1y^2 | Logistic: k = B0 - D0 N = B1/k

Question 47

Population Case III: Limited Environment | A = k(N - y)

Answer 47

dy/dy = Ay = ky(N - y) | Logistic

Question 48

Population Case IV: Competitive | B constant D = cy

Answer 48

A = B - cy dy/dt = By - cy^2 Logistic k = B N = B/c

Question 49

Population Case V: Primitive | B = cy D constant

Answer 49

``` 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)) ```

Question 50

Population Case V: Subcases

Answer 50

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 ```

Question 51

F =

Answer 51

-k/(R+x)^2 when x = 0 -k/R^2 = -mg = m(dv/dt)

Question 52

Escape Velocity

Answer 52

lim as x > inf | v = 0