2) Chain Rules Flashcards
Chain rule for maps of two variables
2 maps of 2 variables (2 smooth maps)
(x,y) =F(u,v) &. (X,Y) =G(x,y) s.t.
G(F(u,v)) is defined
Chain rule :
∂X ∂X∂x ∂X∂y
– = —- + —-
∂u ∂x∂u ∂y∂u
Must be in terms of u and v
CHAIN RULE AS MATRIX PRODUCT
D(G•F)(u,v) = [D(G)(F(u,v)][D(F)(u,v)]
Where D(G)(F(u,v) = D(G) (x(u,v),y(u,v))
Jacobian as product also (as determinant of matrix)
Chain rule for a map (X,Y) = F(x,y),
r(t) =( x(t), y(t) )
s.t. F(x(t),y(t)) is defined
Chain rule :
dX ∂Xdx ∂Xdy
– = —- + —-
dt ∂xdt ∂ydt
Must be in terms of t
CHAIN RULE AS MATRIX PRODUCT
[dX/dt dY/dt] = [D(F)(x(t), y(t))][D(r)(t)]
Where D(F)(r(t)) = D(F)(x(t), y(t))
Coordinate curves
Hold one variable constant and observe tangent to coord curves Eg Coordinate curve for (x,y) =F(u,v) where we have u=u_0 Coordinate curve f(v)= F(u_0,v) Tangent to coordinate curve is: [D(F)(u,v) •e_i ] where i changing [D(F)(u_0,v) •e_2 ]
v=v_0
Coordinate curve f(u)= F(u,v_0)
Tangent to coordinate curve is:
[D(F)(u,v_0) •e_1 ]
Polar coordinates and coordinate curves .
Curves:
r=r_0 circles
Tangent: D(F)(r_0,ϴ) •[0,1] = [-r_0sinϴ, r_0cosϴ]
ϴ=ϴ_0 half lines through O
r(r) =(rcosϴ_0, rsinϴ_0)
Tangent: D(F)(r,ϴ_0) •[1,0] = [cosϴ_0, sinϴ_0]
Parallel to these
Elliptic coordinates
Choose a>0 u bigger than or equal to 0 and v between 0 and 2pi
(x,y) =F(u,v) =(acoshucosv, asinhusinv)
Coordinate curves: v=v_0 r(u) = (acosucosv_0, asinhusinv_0) HYPERBOLA as divide by constants D(F)(u,v_0) •[1,0] = [asinhucosv_0, acoshusinv_0]
u=u_0 r(v) = (acosu_0cosv, asinhu_0sinv) ELLIPSEs as divide by constants D(F)(u_0,v) •[0,1] = [acoshu_0cosv, asinhu_0sinv] Tangents to ellipses
Affine coordinate system
Represented by
[x ,y] = [a,b,c,d] [u,v] + [x_0, y_0]
= A[u v] + [x_0 y_0]
We define as linear map
(x,y) = F(u,v) = (au +bv + x_0, cu+dv + y_0]
Solved as [u v] = A^-1 [ x-x_0 y-y_0]
= (1/detA) [ d, -b, -c, a] [x-x_0 y-y_0]
I.e. If invertible
Don’t forget plus vector
Two maps of 3 variables
Let (x,y,z) =F(u,v,w) and (X,Y,Z) = G(x,y,z) be smooth maps
D(F)(u,v,w) is a 3x3 matrix
Has JACOBIAN ALSO
Chain rule gives 9 expressions of partial derivatives
Similarly all in terms of u,v,w
Find coordinate curves for an Affine coordinate system
[x,y] = [1,1,2,-1][u,v]
Det A= -3 is not 0 so invertible
Solutions
[u,v] = (1/-3)[-1,-1,-2,1][x,y]
Coordinate curves for u=u_0:
vector(r(v))= (u_0 +v, 2u_0 -v)
x+y =3u_0 means line y=-x + 3u_0.
Coordinate curves for v=v_0:
vector(r(u))= (u+v_0, 2u -v_0)
2x-y =3v_0 means line y=2x -3v_0
Drawn at various constant values
For SPHERICAL POLARS
Coordinate surfaces and derivative matrix
Holding for coordinate curves: r=r_0 bigger than 0: Vector r( θ, φ) = (r_0sinθcosφ, r_0sinθsinφ, r_0cosθ) Is a sphere centre O radius r_0 as x^2 + y^2 + z^2 = r_0 ^2
r=r_0 bigger than 0 and θ=θ_0: Vector s(φ) = (r_0sinθ_0cosφ, r_0sinθ_0sinφ, r_0cosθ_0) Is a circle radius r_0sinθ_0 in the plane z = r_0cosθ_0. A parallel of latitude. X^2 +y^2 +z^2 = r_0^2
r=r_0 bigger than 0 and θ=θ_0 Vector s(φ) = (r_0sinθ_0cosφ, r_0sinθ_0sinφ, r_0cosθ_0) Is a circle radius r_0sinθ_0 in the plane z = r_0cosθ_0. A parallel of latitude. X^2 +y^2 +z^2 = r_0^2 also.
Derivative of vector has 0 in z component so parallel to xy plane and is an ellipse (rearranges using constants)
r=r_0 bigger than 0 and φ=φ_0 Vector r( θ) = (r_0sinθcosφ_0, r_0sinθsinφ_0, r_0cosθ) A meridian of longitude latitude. Is a semicircle of radius r_0
Oblate spheroid also coordinates
x=acoshucosθcosφ
y= acoshucosθsinφ
z= asinhusinθ
For u bigger than or equal to 0. And θ in [-pi/2, pi/2].
u=u_0 gives a coord surface which is a suspected sphere/ ellipsis compressed along z.
z = sqrt( a’^2 -1)*sinθ where a’= acoshu_0 is bigger than 0. Hence not a sphere.
θ=θ_0 gives another
(bcoshucosφ , bcoshusinφ , sqrt(1-b^2)sinhu) where b=a*cosθ_0
As is a curvilinear coordinate system has derivative matrix experience
Cylindrical polars
Partial derivatives
and JACOBIAN is r
Inverse of a map
Suppose smooth map (x,y) = F(u,v) has a smooth inverse
That is..
There exists a smooth map (u,v) = G(x,y) such that ( G•F)(u,v) =(u,v) for all (u,v) and (F•G)(x,y) = (x,y) for all (x,y)
IF A MAP LIEK THIS EXISTS IYS UNIQUE
Jacobian and inverse theorem for three variables/ two variables
Is the JACOBIAN of an inverse
Suppose that (x,y,x) =F(u,v,w) is a smooth map with a smooth inverse
Suppose that (x,y,x) =F(u,v,w) is a smooth map with a smooth inverse
(u,v,w) = F^-1(x,y,z). Then ∂(x,y,z)/ ∂(u,v,w) is not equal to 0. Further
[∂(u,v,w)/ ∂(x,y,z)] ( F(u,v,w)) = ( ∂(x,y,z)/ ∂(u,v,w) )^-1
Inverse functions and maps smooth map and inverse theorem
2 variables
Suppose that (x,y) = F(u,v) is a smooth map with a smooth I nverse then the JACOBIAN is non zero
∂(x,y) / ∂(u,v) not equal to 0
And D(F^-1) (F(u,v)) = [D(F)(u,v)] ^-1
Further [∂(u,v) / ∂(x,y) ] (F(u,v)) = [∂(x,y) / ∂(u,v)]^-1
Taking determinants
Lhs means substitution of F(u,v) into partial
JACOBIAN of the inverse is the inverse of the JACOBIAN of original
Finding the inverse of the derivative matrix requires the JACOBIAN also
If a smooth map has a smooth inverse
Then it’s JACOBIAN is non zero and it’s derivative matrix is invertible.
Converse is not true and there are many smooth maps for which JACOBIAN is non 0 and don’t have universes.
SOMETIMES RESTRICTED e.g. Square roots
If the JACOBIAN is non zero for any point in the domain then we can say that there is a subset of the domain i.e. Restriction of F that does have a smooth inverse
