Everything

Question 1

dy/dx = f(x)

Answer 1

Direct integration

Question 2

dy/dx = f(x)g(y)

Answer 2

Separation of variables

Question 3

dy/dx = f(y/x)

Answer 3

Substitute v(x)=y(x)/x

Question 4

example: dy/dx = sin(x+y+1)

Answer 4

Substitute to separable form with u(x)=x+y(x)+1

Question 5

dy/dx + p(x)y = q(x)

Answer 5

Integrating factor or, if q(x) not complex, homogeneous, particular integral method as dy/dx + p(x)y = 0 is always separable

Question 6

Linear nice 1st order in general:

Answer 6

Direct integration

Integrating factor

Question 7

Linear 1st order dy/dx = f(x,y)

Answer 7

Substitute RHS to make nicer

Question 8

Non-linear 1st order in general:

Answer 8

Substitution to make linear

Question 9

example: y * dy/dx + sin(x) = y^2

Answer 9

Substitute z=y^2

Question 10

Theorem 3.1 If y1 and y2 are solutions to any homogeneous ODE then a1y1+a2y2 is also a solution.

Answer 10

Take the ODE but with y1 and the ODE but with y2. Multiply them by the appropriate a and then add them together. Differentiation is a linear map so you can combine the derivatives

Question 11

Theorem 3.2 If yp is a solution to the inhomogeneous ODE then y(x) is a solution iff y(x)=yc(x)+yp(x) where yc(x) is a solution to the homogeneous ODE.

Answer 11

Write the ODE with y(x) as a solution (=yc(x)+yp(x)). Use linear map and expand it out and separate it into yc(x) and yp(x). yp(x) but = f(x) so yc(x) bit =0 so its a solution to the homogeneous ODE.

Question 12

Non-linear ODE where a solution is known ( z(x) )

Answer 12

Sub in y(x)=u(x)z(x). ( Remember p(x)z’‘+q(x)z’+r(x)z=0 ). This should give you a differential equation of degree one less than you started. For 2nd order ODES this gives you a (usually) solvable first order ODE

Question 13

Consider the homogeneous linear equation. d2(y)+q d1(y) + ry=0. The equation m^2+qm+r=0 has roots m1,m2. Then we have the 3 cases

Answer 13

Note q=-(m1+m2) and r=m1m2
So the ODE is d2(y) -(m2-m1)d1(y) + m1m2y = 0
e^(m1x) is a solution of this which can be checked.
This gives d2(u) -(m2-m1)d1(u) = 0
If case 1: m2-m1/=0. This gives solution du/dx = ke^(m2-m1) whixh when subbing y back in proves case 1
If case 2: m2-m1=0 so d2(u)=0. Integrating gives case 2 proof
If case 3: m1,m2 complex conjugate pair. Use the solution to case 1 and eulers identity

Question 14

Definition of partial differentiation

Answer 14

Let f:R^n→R be a function of n variables x1,x2,…,xn.
Then the partial derivative ∂f/∂xi is the rate of change of f at a point (p1,…,pn) when we vary only xi and keep other variables constant.
We have
∂f/∂x(p_1,…,p_n ) = lim(h→0)⁡ of
[f(p1,…,p(i-1),pi+h,p(i+1),…,pn ) - f(p_1,…,p_n) ] /h

Question 15

(∂^2f)/∂x∂y

Answer 15

y first then x

Question 16

f sub yx (partial differentiation)

Answer 16

y first then x

Question 17

State the chain rule for dF/dt

Answer 17

Let F(t) = f( u(t), v(t) ). Then
dF/dt= ∂f/∂u * du/dt + ∂f/∂v * dv/dt

Question 18

State the chain rule for ∂F/dx

Answer 18

LEt F(t) = f( u(x,y), v(x,y) ). Then ∂F/∂t = ∂f/∂u * ∂u/∂x + ∂f/∂v * ∂v/∂x

Question 19

Prove the chain rule

∂F/∂t = ∂f/∂u * ∂u/∂x + ∂f/∂v * ∂v/∂x

Answer 19

We know that if F(t) = f( u(t), v(t) ).
Then
dF/dt= ∂f/∂u * du/dt + ∂f/∂v * dv/dt
So if F(t)=f( u(x,y), v(x,y) ) we can say the same. But when we calculate du/dx it’s actually ∂u/∂x because we have to keep y constant.

Question 20

Partial Differential equations general

Answer 20

Just integrate and consider functions
Separate the variables straight away
Separable solutions

Question 21

Partial Differential equation in one variable

Answer 21

Treat as full derivatives then the constants at the end are actually functions of the other variables

Question 22

Jacobian ∂(x,y)/∂(u,v)

Answer 22

Determinant of the matrix
( ∂x/∂u ∂x/∂v )
( ∂y/∂u ∂y/∂v )

Question 23

Does
dx dy = |∂(x,y)/∂(u,v)| du dv
or
du dv = |∂(x,y)/∂(u,v)| dx dy

Answer 23

dx dy = |∂(x,y)/∂(u,v)| du dv
Remember it by diving by du dv.
The x and y goes on top the u and v goes on the bottom

Question 24

Proposition 5.1.
Let r and s be functions of u and v which in turn are functions of x and y.
Then ∂(r,s)/(∂(x,y) = ∂(r,s)/(∂(u,v) * ∂(u,v)/∂(x,y)
As a corollary ∂(x,y)/∂(u,v) * ∂(u,v)/∂(x,y)=1

Answer 24

Write ∂(r,s)/∂(x,y) as the determinant of the Jacobian matrix. Expand out each element using the chain rule. This can then be written as a multiplication of the (r,s)/(u,v) and the (u,v)/(x,y) Jacobian matrices
Expand out the determinants

For the corollary set r=x, s=y

Question 25

Plane Polar Coordinates

Answer 25

Transformation given by: x=rcosθ, y=rsinθ Inverse given by: r=sqrt(x^2+y^2), θ=tan-1(y/x)

Question 26

Circular symmetric solutions

Answer 26

Solutions independent of θ in plane polar coordinates

Question 27

Cylindrical polar coordinates

Answer 27

Transformation fiven by: x=rcosθ, y=rsinθ, z=z Inverse given by: r=sqrt(x^2+y^2), tanθ=y/x, z=z

Question 28

Parabolic Coordinates

Answer 28

Transformation given by: | x=0.5(u^2-v^2), y=uv

Question 29

Spherical polar coordinates

Answer 29

Transformation given by: x=rsinθcosϕ, y=rsinθsinϕ, z=rcosθ Inverse given by: r=sqrt(x^2+y^2+z^2), tanθ=sqrt(x^2+y^2)/z, tanϕ=y/x

Question 30

Area formula with double integrals

Answer 30

Let A⊂ R^2. Then the area of A is given by | ∫∫(x,y)∈A dx dy

Question 31

Conics: Circle

Answer 31

``` x^2+y^2=a^2 acosθ, asinθ Eccentricity: 0 Directrices: x=±a/e Area: πa^2 ```

Question 32

Conics: Ellipse

Answer 32

x^2/a^2 + y^2/b^2 = 1 acosθ, bsinθ Eccentricity: sqrt(1 - b^2/a^2 ) 0

Question 33

Conics: Parabola

Answer 33

``` y^2 = 4ax at^2, 2at Eccentricity: 1 Foci: a,0 Directrices: x=-a ```

Question 34

Conics: Hyperbola

Answer 34

``` x^2/a^2 - y^2/b^2 = 1 a sect, b tant Eccentricity sqrt(1 + b^2/a^2) e>1 Foci: ±ae,0 Directrices: x=±a/e Asymptotes: y=±bx/a ```

Question 35

Conics: Eccentricities

Answer 35

Circle: 0 Ellipse: 01

Question 36

Conics: Parametrisations

Answer 36

Circle: acosθ, asinθ Ellipse: acosθ, bsinθ Parabola: at^2, 2at Hyperbola: a sect, b tant

Question 37

Conics: Foci

Answer 37

Circle: Centre Ellipse: ±ae,0 Parabola: a,0 Hyperbola: ±ae,0

Question 38

Conics: Directrices

Answer 38

Circle: x=±a/e Ellipse: x=±a/e Parabola: x=-a Hyperbola: x=±a/e

Question 39

Quadrics: Sphere

Answer 39

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

Question 40

Quadrics: Ellipsoid

Answer 40

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

Question 41

Quadrics: Hyperboloid of one sheet

Answer 41

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

Question 42

Quadrics: Hyperboloid of two sheets

Answer 42

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

Question 43

Quadrics: Paraboloid

Answer 43

z= x^2 + y^2

Question 44

Quadrics: Hyperbolic paraboloid

Answer 44

z = x^2 - y^2

Question 45

Quadrics: Cone

Answer 45

z^2 = x^2 + y^2

Question 46

Find a tangent to a conic

Answer 46

Implicitly differentiate or differentiate the parametrisation vector to find a tangent vector

Question 47

Smooth parametrised surface

Answer 47

A smooth parametrised surface is a map r, given by the parametrisation r: U→R^3:(u,v)↦( x(u,v),y(u,v),z(u,v) ), from an open subset U⊆R^2 to R^3 such that: x,y,z have continuous partial derivatives with respect to u and v of all orders; r is a bijection, with both r and r^(-1) being continuous; At each point the vectors ∂r/∂u and ∂r/∂v are linearly independent

Question 48

Tangent Plane to a surface at a point

Answer 48

Let r be a smooth parametrised surface and let p be a point on the surface. The plane containing p and which is parallel to the vectors ∂r/∂u(p) and ∂r/∂v(p) is called the tangent plane To find a tangent plane you find the plane parallel to both ∂r/∂u(p) and ∂r/∂v(p)

Question 49

Normal Vector

Answer 49

∂r/∂u(p) ∧ ∂r/∂v(p)

Question 50

Scalar line integral

Answer 50

The scalar line integral of a vector field F(r) along a path C given by r=r(t), from r(t0) to r(t1) is ∫along C of F(r) dr = ∫ from (t0) to (t1) of F(t)⋅dr/dt dt

Question 51

Length of a curve

Answer 51

The length of a curve is given by the integral with respect to t along the curve of |r'(t)| (so the modulus of the velocity of r)

Question 52

The gradient vector

Answer 52

Given a scalar function f≔R^n→R, whose partial derivatives all exist, the gradient vector ∇f is defined as ∇f=( ∂f/∂x1, ∂f/∂x2, …, ∂f/(∂xn) ) So partially differentiate in each variable

Question 53

Scalar function

Answer 53

A scalar function is a function where the output is a scalar (most functions are like this)

Question 54

Level Set

Answer 54

A level set of a function f:R^3→R is a set of points {(x,y,z):f(x,y,z)=c}. Essentially a surface is a level set for f(x,y,z) is the surface has equation f(x,y,z)=c where c is a constant

Question 55

Prove that given that a surface S∈R^3 is a level set for f(x,y,z) and point p∈S then ∇f(p) is normal to S at p Essentially for level sets, the gradient vector is normal to the surface

Answer 55

Let r(u,v)=( x(u,v),y(u,v),z(u,v)) be a parametrisation of S Normal is in direction ∂r/∂u∧∂r/∂v, ie. Perpendicular to both ∂r/∂u and ∂r/∂v Show ∇f⋅∂r/∂u=∇f⋅∂r/∂v=0 so it is perpendicular to both. Remember ∂f/∂u=∂f/∂v=0 (This actually implies that the gradient vector is normal in R^2 too but you don’t need to know that)

Question 56

Vector field

Answer 56

A vector field is an assignment of a vector to each point in a space

Question 57

Conservative Vector field

Answer 57

A vector field v is conservative if there exists a function such that v=∇f

Question 58

Finding if a vector field is conservative

Answer 58

Say v = ( g(x,y,z), h(x,y,z), i(x,y,z) ) If v=∇f that means g=∂f/∂x, h=∂f/∂y and i=∂f/∂z. Integrate these and find a possible function. Make sure you consider all 3 and if there is a constant at the end

Question 59

Theorem 8.5 (Independence of path) If a vector field v is conservative (v=∇f) then ∫ from A to B of v.dr= f(B)-f(A) So line integrals don't depend on the path of the curve

Answer 59

Write out ∇f and dr/dt as full vectors and use chain rule to get df/dt

Question 60

Directional Derivative

Answer 60

Let f:R^n->R be a differentiable scalar function and let u be a unit vector. Then the directional derivative of f at a in the direction of u = the lim as t -> 0 of ( f(a+tu)-f(a) ) / t

Question 61

Prove the directional derivative of a function f at the point a in the direction u is ∇f(a)⋅u

Answer 61

Write F(t)=f(a+tu) then use the limit form of directional derivative. Note this equals F'(0) then use the chain rule

Question 62

When is the directional derivative the maximum | In what direction does f increase the fastest

Answer 62

At the point a the direction derivative in the direction of u is ∇f(a)⋅u=|∇f(a)||u|cosθ so fastest increase is when θ=0 When u is in the same direction as ∇f(a)

Question 63

When is the directional derivative the minimum | In what direction does f decrease the fastest

Answer 63

At the point a the direction derivative in the direction of u is ∇f(a)⋅u=|∇f(a)||u|cosθ so fastest decrease is when θ=π When u is in the opposite direction as ∇f(a)

Question 64

Weird gradient vector rules

Answer 64

∇(fg)=f∇(g)+g∇(f) ∇(f^n )=nf^(n-1) ∇(f) ∇(f/g)=(g∇(f)-f∇(g))/g^2 ∇(f(g(x)))=f'(g(x))∇g(x)

Question 65

Taylor's theorem in one variable

Answer 65

f(x)= f(a)+ f'(a)(x-a)+⋯+f^(n)(a)/n!*(x-a)^n+f^(n+1)(ξ)/(n+1)!*(x-a)^(n+1)

Question 66

1st order Taylor expansion in 2 variables

Answer 66

p1 (x,y)=f(a,b)+fx(a,b)(x-a)+fy(a,b)(y-b)

Question 67

2nd order Taylor expansion in 2 variables

Answer 67

p2 (x,y)=f(a,b)+fx(a,b)(x-a)+fy(a,b)(y-b)+1/2( fxx(a,b)(x-a)^2+2fxy(a,b)(x-a)(y-b)+fyy(a,b)(y-b)^2 )

Question 68

Gaussian Integrals

Answer 68

Gaussian integrals are double integrals of the form e^-(x^2+y^2). You can integrate these by switching to polar coordinates

Question 69

``` Derivatives: tan(kx) sec(kx) csc(kx) cot(kx) ```

Answer 69

tan(kx) -> ksec^2(kx) sec(kx) -> ktan(kx)sec(kx) csc(kx) -> -kcsc(kx)cot(kx) cot(kc) -> -kcsc^2(kx)

Question 70

``` Integrals: tan(kx) cot(kx) csc(kc) sec(kx) ```

Answer 70

tan(kx) -> 1/k ln | sec(kx) | cot(kx) -> 1/k ln | sin(kx) | csc(kc) -> -1/k ln | csc(kx)+cot(kx) | sec(kx) -> 1/k ln | sec(kx)+tan(kx) |

Question 71

Derivatives: arcsin(x) arccos(x) arctan(x)

Answer 71

arcsin(x) -> 1/root(1-x^2) arccos(x) -> - 1/root(1-x^2) arctan(x) -> 1/a+x^2

Question 72

Derivatives: sinhx coshx tanhx

Answer 72

sinhx -> coshx coshx -> sinhx tanhx -> sech^2x

Question 73

Derivatives: arsinhx arcoshx artanhx

Answer 73

arsinhx -> 1/root(1+x^2) arcoshx -> 1/root(x^2-1) artanhx -> 1/1-x^2

Question 74

Local minima

Answer 74

(x0,y0) is called alocal minima if, in a disc around (x0,y0): f(x,y)>=f(x0,y0)

Question 75

Theorem: If (x0,y0) is a local extrema: ∇f(x0,y0)=0

Answer 75

``` Take v(t)=(x0,y0)+tu where u is any unit vector. Then take g(t)=f(v(t)). Consider g'(t) in limit form and show g'(t)=0 g'(t) is the directional derivative in direction u. This=0 for all u so ∇f(x0,y0)=0 ```

Question 76

Critical Point

Answer 76

A critical point is a point where ∇f(x0,y0)=0. Note all local extrema are critical points but not all critical points are local extrema

Question 77

Theorem 10.2 If fxx*fyy-fxy^2>0 and fxx<0 then (x0,y0) is a local maximum. If fxx*fyy-fxy^2>0 and fxx>0 then (x0,y0) is a local minimum

Answer 77

Rewrite taylors series in a disgusting fashion (they usually give you this, hopefully) and show that this expansion is positive/negative giving you the results.

Question 78

Classifying critical points

Answer 78

If fxxfyy - (fxy)^2 > 0 we then have fxx<0 means maximum fxx>0 means mimmum If fxxfyy - (fxy)^2=0 we have no information If fxxfyy - (fxy)^2<0 we have a saddle point

Question 79

Finding critical points

Answer 79

fx=fy=0

Question 80

Lagrange Multipliers system of equations for one constraint.

Answer 80

∇f(x,y,z)=λ∇F(x,y,z) | F(x,y,z)=0

Question 81

Method to solve Lagrange multipliers for one constraint

Answer 81

∇f(x,y,z)=λ∇F(x,y,z) F(x,y,z)=0 So let G(x,y,z,λ)=f(x,y,z)-λF(x,y,z) The solutions are the critical points of G

Question 82

Lagrange multipliers for more than one constant

Answer 82

Say we have constraints F1,...,Fn We let G(x,y,z,λ1,...,λn)=f(x,y,z)-λ1F1(x,y,z)-...-λnFn(x,y,z) And then solve ∂G/∂x=∂G/∂y=∂G/∂z=∂G/∂λ1=⋯=∂G/∂λn=0