P(r,θ) | ≈
| P(3,π⁄4)
+ DP(3,π⁄4)(Δr, Δθ)
| =
|
| +
|
cos(π⁄4)
| −3 sin(π⁄4)
| sin(π⁄4)
| 3 cos(π⁄4)
|
|
| •
|
|
Lin approx of polar grid near (ro,θo):
DP(ro,θo)
= Rotθo∘ Dil(1,ro)
HW 17:
- Parametrize a triangle
- Consider a (u,v)-grid in the (x,y)-plane
R2 whose u-lines (red) are
parallel to the vector (2,1), and whose v-lines (blue)
are parallel to (1,3).
Find a (linear) mapping L(u,v) = (x(u,v), y(u,v))
which takes the standard grid in the (u,v)-plane
to the above grid in the (x,y)-plane.
- Consider the triangular region R (light blue) with vertices (0,0), (2,1), (1,3). Find a simple parameter region R* in the (u,v) plane which is taken to R by the mapping L. This gives a parametrization of R. (Here R* should be a simple triangle easy to describe in terms of inequalities such as u ≥ 0 and v ≤ ℓ(u) for some function ℓ.)
- Find the inverse linear mapping L−1(x,y) = (u(x,y), v(x,y)). Hint: Solve for u,v in terms of x,y
in the system of equations x = x(u,v), y = y(u,v).
Equivalently, find the inverse matrix [L]−1.
- Now let R be any triangular region in the (x,y) plane with vertices a, b, c. Generalize the above
exercise to define an affine linear function A(u,v) and a parameter region R* which parametrize R; recall that affine
means constant + linear. (It may be convenient to define A(u,v) in terms of the vectors b−a, c−a, rather than in coordinates.)
- Parametrizing rotated regions
- Parametrize the region R bounded by the ellipse x2⁄4 + y2⁄9 = 1.
That is, find a function F(u,v) which takes some simple (u,v)-region R*
to R.
Hint: The extreme points of the ellipse are (±2,0) and (0,±3).
Stretch the x and y coordinates of the usual polar parametrization of the unit disk to get the correct height and width.
- Now parametrize the same ellipse, but rotated 45° and with its center shifted to (1,2).
Also find the inverse mapping.
Hint: Take the the previous parametrization mapping, then compose it with rotation and translation mappings.
- With colored pens, plot straight grid lines in the (r,θ) plane corresponding to r = 0, 1, 2, 3, and θ = 0, π⁄4, π⁄2, 3π⁄4, . . . , 7π⁄4, and the corresponding grid curves
in the (x,y) plane, under the polar mapping
P(r,θ) = (x(r,θ), y(r,θ))
= (r cos(θ), r sin(θ)).
Each grid curve is a parametrized curve in the (x,y)-plane, whose parameter is either r or θ. For example, for r = 2, the (r,θ)-gridline (2,θ) is taken to the curve c(θ) = P(2,θ) = (2 cos(θ), 2 sin(θ)).
- Compute the derivative of P at the point (r,θ) = (2,π⁄4), the Jacobian matrix:
| [DP(2,π⁄4)]
| =
|
∂x⁄∂r(2,π⁄4)
| ∂x⁄∂θ(2,π⁄4)
| ∂y⁄∂r(2,π⁄4)
| ∂y⁄∂θ(2,π⁄4)
|
|
|
Note that once we substitute (r,θ) = (2,π⁄4), this becomes a matrix of explicit constants, with no variables.
- Consider the affine approximation to P(r,θ) near (r,θ) = (2,π⁄4):
P(2+Δr, π⁄4+Δθ)
≈
P(2,π⁄4)
+ DP(2,π⁄4)(Δr, Δθ),
or equivalently:
P(r,θ)
≈
P(2,π⁄4)
+ DP(2,π⁄4)(r−2, θ−π⁄4).
Here DP(2,π⁄4) is the linear mapping defined by the Jacobian matrix, and Δr, Δθ are small increments of r, θ.
Problem: Take the colored (r,θ) grid lines above, and map them to grid lines in the (x,y)-plane under the affine linear approximation. (Recall that a parametrized line c(t) = a + tv starts at a and moves in direction v.) Note how these lines approximate the polar grid curves near (r,θ) = (2,π⁄4).
- Inversion mapping
- Let Inv : R2 → R2
be the mapping which takes a non-zero vector (u,v) to a vector (x,y) with the same direction and the reciprocal radius: that is, Inv turns the unit circle "inside out" so that its interior is stretched to cover the whole plane outside the circle, and region outside the circle is squashed inside. Find a vector formula for Inv(v), as well as
a coordinate formula Inv(u,v) = (x,y).
- Sketch the Inv(u,v)-grid produced by the above mapping in the (x,y)-plane. Hint: The integer grid lines in the (u,v) plane are all taken to small circles inside the unit circle. Since each (u,v)-grid line goes to infinity in the (u,v)-plane, its image goes to the origin in the (x,y)-plane. These are called Apollonian circles or a parabolic pencil of circles.
- Find the inverse mapping Inv−1(x,y).
Hint: Duh!
- Extra Credit: Inverting familiar curves in the (u,v) plane gives interesting new curves in the (x,y) plane.
The inversion of the parabola
v = u2 − 1
is called the cardioid: you can parametrize it as c(u) = (x(u), y(u)) = Inv(u, u2−1). Similarly, the inversion of the hyperbola u2 − v2 = 1 is called the lemniscate of Bernoulli.
Explore these curves, and say as much as you can about them,
showing how the inversion definition is equivalent
to the usual ones you can find online.
For example, find (x,y) equations defining these curves,
or polar coordinate equations r = f(θ).
Does the inversion mapping give insight on their properties?
⊞
Solutions
1a. We just need the (u,v) coordinate axis vectors to map as L( i) = (2,1) and L( j) = (1,3), and the rest of the grid follows.
Thus L(u,v) = (2,1)u + (1,3)v = (2u+v, u+3v).
Note that the matrix of L is [].
1b. The parameter region is the standard triangle
with vertices (u,v) = (0,0), (1,0), (0,1).
That is, R* = {(u,v) with 0 ≤ u and 0 ≤ v ≤ 1−u} =
{(u,v) with u ≥ 0, v ≥ 0, u+v ≤ 1}.
1c. [ L ]−1 =
1⁄5
[ ],
so:
L−1 =
(3⁄5 x − 1⁄5 y ,
−1⁄5 x + 2⁄5 y).
1d. A(u,v) = a + ( b− a)u + ( c− a)v will take the standard (u,v) grid
to the grid generated by the edge-vectors b− a
and c− a, with the origin taken to a.
Use the previous parameter region R* = {(u,v) with u ≥ 0, v ≥ 0, u+v ≤ 1}.
2a. F(r,θ) = (1⁄2 r cos(θ),
1⁄3 r sin(θ)) over the region (r,θ) ∈ R* = [0,1]×[0,2π].
2b. The rotation map Rotπ⁄4
is defined by a rotation matrix which works out to
Rot(x,y) = 1⁄√2(x−y, x+y).
Applying this to the output of the previous F(r,θ), then translating by (1,2), gives:
G(r,θ) =
( 1 + r⁄2√2 cos(θ)
− r⁄3√2 sin(θ) ,
2 + r⁄2√2 cos(θ)
+ r⁄3√2 sin(θ) ).
Here we should think of r as measuring an "elliptically scaled" radius measured from the center of the tilted, translated ellipse; and θ as measuring a squashed "elliptical angle".
3. How P maps the (r,θ)-gridlines to the (x,y)-plane:
For example, the (r,θ)-gridline (2,θ) maps to the (x,y)-plane as the grid curve:
c(θ) = P(2,θ) = (2 cos(θ), 2 sin(θ)), a circle of radius 2.
4. Specializing the Jacobian matrix in the above Notes to the point (r,θ) = (2,π⁄4) gives:
| [DP(2,π⁄4)]
| =
|
cos(π⁄4)
| −2 sin(π⁄4)
| sin(π⁄4)
| 2 cos(π⁄4)
|
|
| =
|
|
5. The affine approximation function to P near the base point (r,θ) = (2,π⁄4) is:
A(r, θ)
=
P(2,π⁄4)
+ DP(2,π⁄4)(r-2, θ−π⁄4)
= (√2, √2) +
(√2⁄2(r-2) − √2(θ−π⁄4) ,
√2⁄2(r-2) + √2(θ−π⁄4)).
The corresponding gridlines are c(θ) = A(2,θ), A(3,θ), and parallel lines; and c(r) = A(r, π⁄ 4), A(r, π⁄ 8), and parallel lines:
The faint dotted lines are the polar grid curves.
These two grids do indeed approach each other very close to the base point, on a scale where the grid curves look straight.
6a. The vector parallel to v = (u,v), having length 1⁄|v|, is F(v) = v⁄|v|2, which means Inv(u,v) = ( u⁄u2+v2 , v⁄u2+v2 ).
6b. To see the transformation of the (u,v) grid under the mapping Inv(u,v), apply a parametric curve plotter successively to c(t) = Inv(1,t), Inv(2,t), Inv(3,t) gives red circles labeled u = 1,2,3 with diameter 1, 1⁄2, 1⁄3, all passing through the origin and with centers on the y-axis. Similarly for blue circles c(t) = Inv(t,1), Inv(t,2), Inv(t,3) centered on the x-axis.
Note that each red circle crosses each blue circle at right angles, like the standard grid in the (u,v)-plane. Thus, while Inv(u,v) turns straight edges to circles, it preserves all corner angles. This is called a conformal mapping.
6c. The mapping Inv : R2−{0} → R2−{0} is clearly its own inverse, since Inv(Inv(v)) = v. If we think of the first
R2 as the (u,v)-plane and the second
as the (x,y)-plane, then we may write the inverse mapping as
Inv(x,y) =
( x⁄x2+y2 , y⁄x2+y2 ).
10/17 Lect 21. Review for Midterm Exam: Topics up to Lect 17. Summary Chart,
Old Exam.
10/19 Midterm Exam
Come 5 min early for extra time.
Mid-semester: Last day to drop with no grade reported
(closes 8pm).
10/20 Recitation 8. Extra topic:
⊞
Geometry of complex functions
- Complex field C = R[i]: closure, reciprocals
- Geometry of multiplication, homothety matrix
- Complex analytic function:
Dfα = cpx mult ⇔
Cauchy-Riemann eqns ⇔ conformal mapping
- Conjugate-analytic vect flds F(x,y) = f(x−iy) ⇒ curl, div = 0
10/21 Lect 22.
⊞
Substitution ∬D f(x,y) dx dy
= ∬D* f(F(u,v)) det[DF] du dv.
Det[L] = area.
- Area stretching factor for linear L(u,v) is determinant
Defined by L(1,0) = u = (a,b) , L(0,1) = v = (c,d).
Stretching factor is area of parallelogram D with edges u,v
Area
=
base×height
=
|u| |v| sin θ(u,v)
=
|u⊥| |v| cos θ(u⊥,v)
=
u⊥•v
= −bc+ad = ad−bc
for u⊥ = (−b,a)
=
det[L] = det(u,v)
=
| det | []
| = ad − bc
|
Positive area if v counterclockwise from u,
negative area if clockwise
Properties of det(u,v): bilinear, alternating
det(u,v) = −det(v,u)
⇒ formula
- Substitution Rule for 1-variable integrals: x = g(u)
Local stretching factor:
dx = g'(u) du = dx⁄du du
∫ f(x) dx = ∫ f(g(u)) g'(u) du
- Substitution Rule for G(u,v) = (x,y),
bijective G : D* → D
Assume G is orientation-preserving:
det[DG] ≥ 0 for all (u,v)
Local stretching factor is det of Jacobian matrix of derivatives
∬D f(x,y) dx dy for
=
∬D* f(x(u,v),y(u,v)) det[DG(u,v)] du dv.
Leibnitz notation:
det[DG(u,v)]
=
∂(x,y)⁄∂(u,v)
- Tradeoff in using substitution (x,y) = G(u,v):
simplify region (x,y) ∈ D ⇒ (u,v) ∈ D*
but complicate
integrand f(x,y) ⇒ f(x(u,v),y(u,v))
|∂(x,y)⁄∂(u,v)|
- Ex: Polar P(r,θ) = (x,y) to parametrize roundish regions
Jacobian determinant:
det[DP(r,θ)]
| =
[∂P⁄∂r |
∂P⁄∂θ]
| =
| det |
| =
| det |
cos(θ) | −r sin(θ)
| sin(θ) | r cos(θ)
|
| |
|
= r cos2(θ) + r sin2(θ) = r
Compute volume under surface of revolution z = r2 = x2 + y2
for (x,y) ∈ D = disk of radius 1,
parametrized by (r,θ) ∈ D* = [0,1]×[0,2π].
∬D x2+y2 dx dy
=
∬D* ( (r cos(θ))2 + (r sin(θ))2 ) |det[DP(r,θ)]| dr dθ
=
∫2π0
∫10
r2 r dr dθ
=
∫2π0
1⁄4 dθ
=
1⁄4 2π
=
π⁄2 .
Note: This is exactly half the volume of the enclosing cylinder
under z = 1.
Reading: [MT] Ch 6.2, including odd exercises p 326 (answers p 521).
HW:
- Use an appropriate parametrization to find formulas for
the following quantities:
- The volume under an arbitrary surface of revolution z = f(r) = f(√(x2+y2))
and above a ring-shaped region with a ≤ r ≤ b.
You probably saw the resulting formula in Calculus II: what was it called?
- The integral of an arbitrary f(x,y) over
the elliptical region D defined by
(x−1)2 + y2⁄4 ≤ 1
- The integral of f(x,y) over the triangle with vertices a = (1,1),
b = (3,2),
c = (2,4).
- Gaussian integral
I =
∫
−∞
∞
e−x2 dx.
The integrand e−x2
is the
or bell-curve or normal distribution
which gives the relative probability of a
random variable x with expectation value zero. (For example, x could be the height
of a randomly chosen woman, minus the average height.)
To make this a true probability distribution,
we must scale it so that the total probability
(the integral) is equal to one. That is,
we must divide by the value of the integral I,
but this is difficult to compute since e−x2
has no elementary anti-derivative.
We outline an amazing solution using double integrals
and change of coordinates.
- Compute the double integral
of e−(x2+y2)
over the disk x2 + y2 ≤ R2, for a fixed radius R, and find the limit
as R → ∞.
Use the polar coordinate substitution:
∬D f(x,y) dx dy
=
∬D* f(r cos θ, r sin θ) r dθ dr.
- Show that the integral of
e−(x2+y2)
over the rectangle
[−R,R]×[−R,R]
is equal to
(
∫
−R
R
e−x2)2 dx = I2.
- Compare the integral in (a)
with that in (b), letting R → ∞,
and compute the original integral I.
- Vector geometry of the determinant
- Consider vectors u = (a,b), v = (c,d),
forming an angle from u to v given by α
with −π ≤ α ≤ π
(positive meaning counterclockwise).
Let β be the angle from v to u⊥, the 90° counterclockwise rotation of u.
- Let D be parallelogram with vertices
0, u, v, u+v.
-
Let area(u,v) be the signed area of D,
counted positively if α ≥ 0,
and negatively if α ≤ 0.
Problem:
Considering |u|
as the base length of D, show
that the height of D is |v| sin(α) = |v| cos(β), and hence area(u,v) = u⊥•v = ad−bc.
- The following properties of the area function are easy to see geometrically:
- Bilinear:
area(au1+bu2, v) =
a area(u1,v) + b area(u2,v)
area(u, cv1+dv2) =
c area(u,v1) + d area(u,v2)
- Skew-symmetric: area(u,v) = −area(v,u)
- Unit square: area(i,j) = 1, where i = (1,0) and j = (0,1)
Problem: Assuming the above properties, show that
area(u,v) = det(u,v) = ad−bc
for u = (a,b), v = (c,d).
- p. 348 #15: Use an appropriate change of variables to compute the integral
∬B exp((y−x)⁄(y+x)) dx dy over the triangle B
with vertices (0,0), (1,0), (0,1).
- Consider the change of variables transformation (s,t) = T(x,y) = (y−x, y+x).
Find its matrix [T], and verify that:
- Find the image region B* = T(B): this is a triangle with
vertices T(0,0), T(1,0), T(0,1).
- Given (s,t) = (y−x, y+x), solve the two linear equations to get x,y in terms of s,t.
Use this to write the inverse transformation U = T−1 with (x,y) = U(s,t).
Write the matrix [U] and verify that [U]•[T] = I,
the identity matrix.
Also B = U(B*).
- Write the Change of Variables formula for our
integral, transforming it by U to (s,t) variables over B*.
- Evaluate the integral.
- Routine practice from M&T.
- p. 304 #15(a,b). Do with rectangular coordinates,
then again with polar.
- p. 305 #25. Do with rectangular coordinates,
then again with coordinates (s,t) defined by (x,y)
= s (1,0) + t (2,1).
- p. 305 #27, 37.
⊞
Solutions
1. Use the polar parametrization P(r,θ),
having Jacobian determinant det[DP] = r,
and which maps the parameter region
D* = [a,b]×[0,2π] to
the ring-region
D = {(x,y) | a 2 ≤ x 2+y 2 ≤ b 2}.
The volume is:
Vol = ∬D f(√(x2+y2)) dx dy
=
∫ab ∫02π
f(r) r dθ dr
=
∫ab 2πr f(r) dr.
This is the familiar "Shell Method Formula" for solids of revolution, usually justified by adding
up thin cylindrical shells having
perimeter 2πr, height f(r), and thickness dr.
1b. The region D is an ellipse with center (1,0), height 1, and width 2, so it is parametrized by a shift and scaling of polar coordinates: E(u,v) = (x,y) = (1 + u cos(v), 2u sin(v)), over
the parameter region D* = [0,1]×[0,2π].
This has Jacobian determinant:
det[DE(u,v)]
| =
| det |
| =
| det |
cos(v) | −u sin(v)
| 2sin(v) | 2u cos(v)
|
| |
|
= 2u cos2(v) + 2u sin2(v) = 2u
Thus we can compute the integral:
∬D f(x,y) dx dy
=
∫01 ∫02π
f(1 + u cos(v), 2u sin(v)) 2u dv du.
1c. The given triangle D is half of a parallelogram
with edge-vectors
u = b− a = (2,1) and
v = c− a = (1,3).
We parametrize D by the affine linear mapping
A(u,v) = a + L(u,v) with
L( i) = u, L( j) = v,
so that A(u,v) = (1+2u+v, 1+u+3v),
over the triangular parameter region
D* = {(u,v) | u,v ≥ 0, u+v ≤ 1}.
The Jacobian determinant is:
det[DA(u,v)] =
det[L]
=
| det |
|
=
(2)(3)−(1)(1)
= 5.
|
|
That is, the mapping A(u,v) expands all areas by a
constant factor of 5.
Thus we can compute the integral:
∬D f(x,y) dx dy
=
∫01 ∫01−v
f(1+2u+v, 1+u+3v) 5 du dv.
2. Solution: See M&T p. 332.
3.We have the vectors u = (a,b), v = (c,d), the
projection p of v on u,
and the
projection q of v on u⊥.
By elementary trigonometry, the "height" vector has length:
| q| = | v| sin(α) = | v| cos(β).
Hence:
base×height = |u| |q| =
|u⊥| |v| cos(β) =
u⊥•v
= (−b,a)•(c,d) = ad−bc.
4. Use the linearity properties to expand det( u, v) = det(a(1,0)+b(0,1), c(1,0)+d(0,1)), then use skew-symmetry and unit-square to evaluate the remaining determinants, getting ad(1) + ac(0) + bd(0) + bc(−1). Note that skew-symmetry implies det( u, u) = −det( u, u), so det( u, u) = 0.
5b. B* is the triangle with vertices (s,t) = (0,0), (−1,1), (1,1), given by 0 ≤ t ≤ 1,
−t ≤ s ≤ t.
5c. (x,y) = (−½s + ½t , ½s + ½t).
[U] = [T]−1 =
|
|
This is indeed the inverse of [T], since:
[U] • [T] =
|
| •
|
| =
|
| .
|
5d,e. We have U(B*) = B, one-to-one and onto.
Since U is linear, it is its own derivative
at every point: DU(s,t) = U,
and |∂(x,y)⁄∂(s,t)|
= |det[U]| = ½ .
(Note the det is actually −½, meaning the mapping U
is orientation-reversing, so we have to take the absolute value.)
Also y−x = s, y+x = t.
Thus:
∬B exp((y−x)⁄(y+x)) dx dy
=
∬B* exp(s⁄t)
|∂(x,y)⁄∂(s,t)| ds dt
=
1⁄2 ∫10 ∫t−t es/t ds dt
=
1⁄2
∫10 ∫t−t es/t ds dt
= 1⁄4 (e−e−1).
6. See [MT] pp. 519−520
10/22-25 Fall Break
10/26 Finish up substitution and determinants
10/27 Recitation 9. Weekly HW 6.
Exercises on
substitution and determinants
(answers).
10/28 & 31 Lect 23.
⊞
Vectors in R3 and Rn. Determinant & cross product.
- Vector x = (x1, . . . , xn) ∈ Rn
- Dot product
u•v =
(u1, . . . ,un)•(v1, . . . , vn)
= u1v1 + ··· + unvn
Geometrically: u•v =
|u| |v| cos θuv
- Linear subspace V ⊂ Rn or R3,
k-dimensional with basis {v1, . . . , vk}
Parametrize: V = { v = L(t1, . . . , tk)
| t1, . . . , tk ∈ R },
for linear function L : Rk → Rn,
L(t1, . . . , tk)
= t1v1 + ··· + tkvk
Equations: V = { v ∈ Rn |
q1•v = ··· = qn−k•v = 0 }
for orthogonal vectors q1, . . . , qn−k
⊥ V
- Affine subspace A = V + a ⊂ Rn or R3 for constant vector a
Parametrize by function
P(t1, . . . , tk) =
L(t1, . . . , tk) + a
Equations: A = { v ∈ Rn |
q1•(v−a) = ···
= qn−k•(v−a) = 0 }
- Linear function ℓ: R3 → R
ℓ(x,y,z) = ax+by+cz = (a,b,c)•(x,y,z),
ℓ(v) = q•v
for q = (a,b,c)
Picture ℓ(v) as height of v in direction q,
scaled by |q|
Height ℓ(v) = 0 defines plane orthogonal to q, through origin
Other level surfaces ℓ(v) = c define
affine planes orthogonal to q
Actual height = (length of projection of v
onto direction q) =
q⁄|q| • v
- Determinant det(u,v,w) = signed volume of parallelopiped with edges from the origin parallel to u,v,w.
Sign is positive (+) if u,v,w form a right-handed coordinate system
negative (−) if they form a left-handed one.
- Properties: det(u,v,w)
= −det(v,u,w)
= −det(u,w,v)
det(su+s'u',v,w)
= s det(u,v,w) +
s' det(u',v,w), similarly for variables v, w
det(i,j,k) = 1;
also det(i,i,j) = 0, etc.
- Formula for determinant:
det |
| =
|
| u1v2w3
− u1v3w2
− u2v1w3
+ u2v3w1
+ u3v1w2
− u3v2w1
| =
|
∑σ sgn(σ) uσ(1)vσ(2)wσ(3)
| ,
|
where σ runs over all 3! = 6 permutation functions σ : {1,2,3} → {1,2,3}.
We can denote a permutation by listing σ(1)σ(2)σ(3). For example,
σ(1) = 3, σ(2) = 2, σ(3) = 1 is denoted
σ = 321
- The sign of a permutation sgn(σ) equals +1 if σ is obtained from the identity permutation ε(i) = i by switching an even number of pairs of outputs; and sgn(σ) equals −1 if it is obtained by switching an odd number of pairs of outputs.
For example the permutation σ = 321 is obtained from ε = 123 by switching the single pair
of outputs 1 and 3, so sgn(σ) = −1;
or switch 12, then 23, then 12, again sgn(σ) = −1.
- Cross product on vectors in R3 only (not Rn),
giving vec × vec = vec
Geometrically: u×v = q,
the vector which produces the lin fun ℓ(w)
= det(u,v,w) = q•w
Given u, v, the cross product
u×v is the unique vector
such that for all w.:
det(u,v,w) =
(u×v) • w .
- Formula for u×v
ℓ(x,y,z) = det |
| =
|
| u1v2z
− u1v3y
− u2v1z
+ u2v3x
+ u3v1y
− u3v2x
|
=
(u2v3− u3v2) x
+
(−u1v3+u3v1) y
+
(u1v2− u2v1) z
| =
|
q • (x,y,z)
|
q =
u×v
=
(u1,u2,u3)
×
(v1,v2,v3)
=
(u2v3− u3v2 ,
−u1v3+u3v1 ,
u1v2− u2v1)
- Determinant cofactor expansion
- Properties of u×v come from
properties of det
u×v orthogonal to both u,v: (u×v)•u = (u×v)•v = 0
u, v, u×v form a right-handed basis
anti-commutative: u×v = −v×u,
u×u = 0
distributive: (u+u') × v
= u×v + u'×v
linear: (au)×v =
u×(av) =
a u×v
- Length |u×v| = area of parallelogram spanned by u,v in R3
= |u| |v| sin(θuv)
Proof:
Geometrically, volume(u,v,w)
=
(base area of u,v parallelogram) (height of w perp to u & v).
Algebraically, volume(u,v,w)
=
det(u,v,w)
=
(u×v)•w
=
|u×v| u×v⁄|u×v|•w
=
|u×v|
(height of w along direction q).
Cancelling (height of w), we get area(u,v) = |u×v|.
Reading: [MT] 1.1, 1.2. Cross Product: 1.3, including odd exercises p. 49 (answers p. 494).
HW: Extra Credit HW due Thu 10/31.
- Consider the affine plane A ⊂ R3 containing the points
a = (0,1,1), b = (1,1,2), c = (1,2,0). (As usual, when we define a point by a vector,
we mean the endpoint of the vector starting at the origin.)
- Find two vectors parallel to the plane, for example
from a to b, and from a to c.
- Write a parametric formula P(s,t)
= (x(s,t), y(s,t), z(s,t))
which outputs the points of the plane as we move
s,t ∈ R.
Hint: Start at a and move along the two directions from part (a).
- Find an orthogonal vector q with q•(v−a) = 0 for v ∈ A.
Write a defining linear equation satisfied by all v = (x,y,z) ∈ A,
in the form ax + by + cz = d.
-
Determine the distance from the point
r = (3,4,5) to the plane.
Hint: The distance is the difference in height
with respect to
the unit orthogonal vector q⁄|q|.
- Consider the affine line C ⊂ R3 through the points a = (0,1,1) and b = (1,1,2) from the previous problem.
- Find a parametric form c(t) = (x(t), y(t), z(t)) for C.
- Find two vectors q1, q2 which are orthogonal to the direction of C.
Write two defining equations satisfied by all
v = (x,y,z) ∈ C.
- For each ordered triple of vectors u, v, w:
- Sketch the vectors and the parallelopiped they span.
- Determine whether they form a right-handed triple: that is can you point your right index finger along u, middle finger along v, thumb along w,
so that the thumb is not bent behind or through
the plane of the other two fingers.
- Determine the signed volume of the parallelopiped by evaluating the determinant det(u, v, w).
Verify that a right-handed triple has positive det,
a left-handed one has negative det.
The triples are:
- u = (1,1,0), v = (1,0,1), w = (1,1,2)
- u = (1,0,1), v = (1,1,0), w = (1,1,2)
- u = (1,1,0), v = (1,0,1), w = (2,1,1)
- Again consider the affine plane A
through the points a = (0,1,1), b = (1,1,2),
c = (1,2,0).
- Re-compute the vector q orthogonal to A as the cross product:
q =
(b−a)×(c−a)
- For a general point w = (x,y,z) ∈ A,
expand the vector equation
q•(w−a) = 0,
or equivalently q•w = q•a,
to again get the explicit coordinate equation
ax + by + cz = d defining A.
- Again consider the vectors u = (1,1,0), v = (1,0,1),
w = (1,1,2).
- Compute the cross product q = u×v.
- Compute the volume of the u, v, w parallelopiped by evaluating the triple product (u×v)•w = q•w = det(u,v,w).
- The cross product u×v
can only be defined for three dimensional vectors,
but determinant exists in any dimension Rn,
and the formula
det(u1, . . . , un−1, x)
= ℓ(x) =
q • x
for all x ∈ Rn.
defines a unique vector q, which we may think of as a
"product" of n−1 vectors: q = u1×···×un−1 .
- For n = 2, this operation acts on a single vector u,
giving u×. What familiar operation is this?
- Find a cofactor expansion formula for the cross product of
3 vectors in R4.
⊞
Solutions
1a. Directions along A: v1 = b−a = (1,0,1) ,
v2 =
c−a = (1,1,−1).
1b. P(s,t) = sv1
+ tv2 + a
=
(s+t, 1+t, 1+s−t).
1c. The number of independent orthogonal vectors
is n−k = 3−2 = 1. We want
q = (q1, q2, q3)
with q•v1 = 0
and q•v2 = 0,
giving the linear system:
q1 + q3 = 0 ,
q1 + q2 − q3 = 0.
Solving this by elimination gives q 2 = −2 q 1 and q 3 = −q 1,
or q = q 1(1,−2,−1) for any q 1 ∈ R.
Taking q 1 = 1 gives a specific normal vector q = (1,−2,−1) .
Thus, a defining equation is q•(v−a) = 0, which in coordinates
v = (x,y,z) can be simplified to:
x − 2y − z = − 3.
We can check that the original points a, b, c all satisfy this equation.
1d. Let r − a = (3,3,4) be the vector from the basepoint a = (0,1,1) on V
to the point r = (3,4,5) off of V.
The orthogonal distance from V to r can be obtained by projecting r − a
on the orthogonal vector q;
or simply by dotting r − a
with the unit vector q⁄|q|:
(r−a) • q⁄|q|
=
7⁄√6.
2. The affine line has direction vector v1 = b− a = (1,0,1), parametrized by c(t) = t v1 + a = (t,1,t+1).
There are n−k = 3−1 = 2 independent orthogonal vectors satisfying q• v1 = 0,
i.e. q = (q 1, q 2, −q 1) = q 1(1,0,−1) + q 2(0,1,0), giving q1 = (1,0,−1) and q2 = (0,1,0).
The curve C is the set of points v = (x,y,z) satisfying the two equations qi•( v− a), i.e.:
x − z = −1 , y = 1.
3a. The triple u = (1,1,0), v = (1,0,1), w = (1,1,2):
Notice that we draw the x,y,z axes as a right-handed triple,
and v is in the xz-plane in the foreground, whereas
w recedes into the picture.
This is a left-handed system: model it with your left palm up,
index finger horizontal along u, middle finger almost vertical
along v, thumb pointing away from you along w.
The left-handedness is reflected in the determinant −2,
the signed volume of the corresponding parallelopiped,
as we will see next time.
We compute the determinant,
the signed volume of the parallelepiped spanned by the three vectors,
writing u, v, w as the columns of
a 3×3 matrix, with one term for each permutation of
{1,2,3}:
det(u, v, w)
| =
| det |
| =
|
| u1v2w3
− u1v3w2
− u2v1w3
+ u2v3w1
+ u3v1w2
− u3v2w1
|
=
1·0·2
− 1·1·1
− 1·1·2
+ 1·1·1
+ 0·1·1
− 0·1·1
=
−2.
An easier way to compute this formula is by the cofactor expansion (or minor expansion) along the first column:
if we group the above terms by their first factor
(u 1, u 2, or u 3),
the other factors become 2×2 determinants.
That is, we write v(i) for the vector v with its i th coordinate omitted,
and compute:
det(u,v,w)
=
u1 det(v(1); w(1))
− u2 det(v(2); w(2))
+ u3 det(v(3); w(3))
=
1 det(0,1 | 1,2) − 1 det(1,1 | 1,2) + 0 det(1,0 | 1,1)
=
−1 − 1 + 0
=
−2.
The negative value verifies the left-handedness.
3b. The triple u = (1,0,1), v = (1,1,0), w = (1,1,2),
with the first two vectors switched from part (a),
is right-handed, with the opposite determinant:
det(u,v,w) = 2.
3c. The triple
u = (1,1,0), v = (1,0,1), w = (2,1,1)
has w = u + v, which means the three vectors
all lie in the same plane: they are linearly dependent. Thus, the parallelepiped is flat,
with volume zero, and we may verify det(u,v,w) = 0.
The triple is degenerate, neither right- nor left-handed.
4a. The directions parallel to the plane are
u = b − a = (1,0,1)
and v = c − a = (1,1,−1),
giving the orthogonal direction
q = u × v = (−1,2,1).
4b. The equation q • w = q • a
becomes (−1,2,1) • (x,y,z) = (−1,2,1) • (0,1,1),
or −x + 2y + z = 3. This agrees with the previous HW.
5a. q = (1,1,0) × (1,0,1) = (1,−1,−1)
5b. q • w = (1,−1,−1) • (1,1,2)
= −2. This agrees with the previous HW.
6a. The determinant formula defining
q = u× is:
det(u, w) = q • w.
Writing u = (u1, u2),
w = (x,y), and q = (a,b),
this becomes u1y − u2x = ax + by,
so q = (−u2, u1),
which is the counterclockwise perpendicular vector
to (u1, u2). That is,
u× = u⊥.
6b. For three vectors u, v, w ∈ R4, their cross product
q = u×v×w
is the vector orthogonal to all three and with length equal
to the volume of the three-dimensional box spanned by them.
Its coordinates are the coefficients of x1, . . . , x4 in the determinant:
These can be found using the
cofactor expansion along the last column, in which
the coefficient of each x i
is given by the determinant of the complementary 3×3
submatrix (with alternating ± signs).
11/2 & 4 Lect 24.
⊞
Functions in R3, Gradient Theorem.
- Class notes
- Linear Functions on R3
- c : R → R3,
c(t) = tu parametrized line
- L : R2 → R3,
L(t,s) = tu + sv parametrized plane
- ℓ : R3 → R,
ℓ(x) = q•x height of x in dir q, scaled so that ℓ(q) = |q|2
- P : R3 → R2,
Ker(P) = Rq with P(q) = 0, orthogonal plane V = q⊥,
P projects (collapses) R3
along q to V ≅ R2
- P : R3 → R3,
linear transformation, motion of space around origin, possibly with stretching or distortion
Examples: rotation around an axis,
reflection Refq(v) =
v − 2(v•q⁄q•q) q
- Functions f : R3 → R.
- f(x,y,z) can represent temperature, pressure, potential energy, etc.;
or measure some geometric property of v = (x,y,z),
such as radius or height
- Picture f using level surfaces f(x,y,z) = const
like onion or pastry layers slicing through space
- Gradient vector field ∇f
= (∂f⁄∂x,
∂f⁄∂y,
∂f⁄∂z)
direction of fastest increase of f,
length = max rate of increase
- Gradient is ∇f(a)
perpendicular to level surface f(x,y,z) = const
dir of max increase ⊥ directions of no increase
- Ex: linear ℓ(v) = q•v,
ℓ(x,y,z) = ax+by+cz
measures scaled height in direction q
level surfaces are parallel planes
- Ex: f(v) = |v|2, f(x,y,z) = x2+y2+z2 measures
radius
level surfaces are concentric spheres
- Ex: f(x,y,z) = x2+y2−z2
level surface f = 0 is circular cone around z-axis
f = c is hyperboloid of one sheet (c > 0) or two sheets (c < 0)
Reading: [MT] Ch 2.1, 2.6 (esp. p. 139), including odd exercises.
HW:
- Find the 3×3 matrix of each of the following linear transformations of R3
- Orthogonal reflection across the plane V perpendicular to q = (1,2,3)
- Rotation by 90° around the x-axis
- Rotation by 90° around the y-axis
- The composition of the previous two rotations.
Describe the resulting motion geometrically.
- The shear transformation which takes i,j,k to the vectors
i, i+j, i+j+k.
Does this transformation distort volume?
- Extra Credit: Consider the spatial rotation R = Rp,θ around
an axis p with |p| = 1,
by an angle θ ∈ [0,2π] in the right-handed orientation
(thumb along p, fingers in direction of R).
Show that this rotation is given by the formula:
R(v) =
(p•v)p +
cos(θ) (v − (p•v)p)
+ sin(θ) p×v .
Compute the corresponding 3×3 rotation matrix, given
p = (a,b,c).
Hint: The vectors p ,
v−(p•v)p ,
p×v
form an orthogonal right-handed basis
(axis vectors), with the last two of the same length.
In terms of this basis,
v =
(p•v)p +
(v−(p•v)p) .
The axis component (p•v)p remains unchanged,
while the second component gets rotated by θ in the plane of
v−(p•v)p ,
p×v .
- Consider the function f(x,y,z) = xy−z2.
- For each level surface f(x,y,z) = 0, 1, −1,
write the surface as one or more graphs of functions
z = g(x,y), and sketch the surface. Describe a general
level surface f(x,y,z) = c. (If you have trouble visualizing, use a graphing program.)
- Find the gradient ∇f(a) at a = (1,2,3). Write an equation for the tangent plane to the surface at a.
- Consider the surface defined by a graph z = g(x,y).
- Write this as a level surface f(x,y,z) = const.
What do the other level surfaces of f(x,y,z) look like?
- Write the tangent plane of the surface at a point (a,b,c) = (a,b,g(a,b)) in terms of the partial derivatives of g.
Compare this equation to the tangent plane perpendicular to the gradient ∇f(a,b,c).
- Consider the vector field F(x,y,z) = (−x,−y,1).
- Make a rough sketch and verbal description of
the arrows of F.
- Find a potential function f(x,y,z) with ∇f = F, by evaluating a line integral ∮
F(c)•dc over the straight line
c from (0,0,0) to (a,b,c).
- Redo the above line integral over the curve
c = (at, bt2, ct3)
for t ∈ [0,1].
- Write an integral to compute the arclength of the
curve c = (t, t2, t3)
for t ∈ [0,1].
Get a numerical approximation from Wolfram Alpha.
- Consider the vector field F(x,y,z) = (−y,x,1).
- Make a rough sketch and verbal description of
the arrows of F.
- Try to find a potential function f(x,y,z) with
∇f = F, by evaluating a line integral ∮
F(c)•dc over the straight line
c(t) = (at,bt,ct) from (0,0,0) to (a,b,c).
Check to see if the resulting function really has the
desired gradient.
- Redo the above line integral over the curve
c(t) = (at2, bt2, ct2)
for t ∈ [0,1].
This is the same straight line,
but traversed at a different speed.
We say that this is a reparametrization of the original
c(t) = (at,bt,ct).
- Try to express intuitively why the vector field is not conservative. (Compare to the two-variable case of a rotational vector field.)
⊞
Solutions
1a. We know L = Refl q( x) =
x − 2( x•q⁄ q•q) q for q = (1,2,3). Its matrix is:
[L] =
[L(i) | L(j) | L(k)] =
| 1⁄14
|
| .
|
1b. The mapping R1 = Rot90,i
is 90° around the x-axis.
(Here we use a right-hand rule, so that if we
point the thumb along the x-axis, the yz-plane
is rotated in the direction of the fingers.)
The matrix is:
[R1] =
[R1(i) | R1(j) | R1(k)] =
|
| .
|
1c. Get matrix R 2 similarly to 2(b).
1d. The matrix of R3(x) = R2(R1(x)) is gotten
by matrix multiplication: [R3] = [R2]•[R1].
This is a 120° rotation around the axis (1,1,1), taking each axis to the next in a cycle.
The composition of any two rotations is again a rotation.
1e. The shear rotation has a matrix [S] with det[S] = 1, so it does preserve volume, even though it distorts angles.
2a. The level surface f(x,y,z) = xy−z2 = 0 is a circular cone around the axis (1,−1,0).
Its other level surfaces are hyperboloids of one or two sheets:
In fact, a linear change of coordinates
(x,y,z) = (u+v,u−v,w) transforms
f(x,y,z) into f*(u,v,w) = u 2+v 2−w 2,
which we discussed in class.
2b. We have ∇f(x,y,z) =
(∂⁄∂x,
∂⁄∂y,
∂⁄∂z)(xy−z2)
=
(y, x, −2z),
so ∇f(a) = ∇f(1,2,3) = (2,1,−6).
The level surface through a = (1,2,3) is f(x,y,z) = f(1,2,3) = −7,
and the tangent plane to this surface is
orthogonal to the gradient vector at a.
The tangent plane equation is
∇f(a)•(v−a) = 0,
i.e.
(2, 1, −6)•(x−1, y−2, z−3)
=
2(x−1) + (y−2) − 6(z−3)
=
0,
or: 2x + y − 6z = 14.
3a. The function f(x,y,z) = g(x,y) − z has
the level surface f(x,y,z) = 0 equal to the graph z = g(x,y). Any other level surface f = k is the graph
z = g(x,y) + k, an up or down translate of z = g(x,y).
3b. The tangent plane of z = g(x,y) at the point (a,b,g(a,b)) is the graph of the linear approximation:
z = g(a,b) +
∂g⁄∂x(a,b) (x−a) +
∂g⁄∂y(a,b) (y−b).
Alternatively, the tangent plane is orthogonal
to the gradient ∇f(x,y,z) =
( ∂g⁄ ∂x,
∂g⁄ ∂y,
−1) at (x,y,z) = (a,b,g(a,b)):
(∂g⁄∂x(a,b),
∂g⁄∂y(a,b),
−1)•(x−a, y−b, z−g(a,b))
=
∂g⁄∂x(a,b)
(x−a) +
∂g⁄∂y(a,b)
(y−b)
− (z−g(a,b))
= 0,
which reduces to the previous equation.
4a. The vector field F(x,y,z) = (−x,−y,1)
points in toward the z-axis and tilts upward. Nearer the z-axis, the vectors get shorter and tilt up more, pointing straight up along the axis itself.
4b. Using the Gradient Theorem,
f(a,b,c) − f(0,0,0) = ∮ F(c)•dc =
∫10 F(at,bt,ct) • (at,bt,ct)' dt
=
∫10 (−at,−bt,1) • (a,b,c) dt
=
∫10 −a2t − b2t + c dt
=
−½a2t2 − ½b2t2 + ct |1t=0
=
−½a2 − ½b2 + c.
Setting the constant of integration as f(0,0,0) = 0,
we get f(x,y,z) = −½x2 − ½y2 + z, which indeed gives
∇f(x,y,z) = (−x,−y,1) = F(x,y,z).
4c. f(a,b,c) =
∫10 F(at, bt2, ct3) • (at, bt2, ct3)' dt
=
∫10 (−at, −bt2, 1)•(a, 2bt, 3ct2) dt
=
∫10 −a2t − 2b2t3 + 3ct2 dt
=
−1⁄2a2t2 − 2⁄4b2t4 + ct3 |1t=0
=
−½a2 − ½b2 + c,
giving the same f(x,y,z) =
−½x2 − ½y2 + z
as before. This is expected because of the Gradient Theorem,
which guarantees that a gradient vector field is path-independent,
depending only on the enpoints of the path of integration.
4d. L = ∮ |dc|
= ∫10 |c'(t)| dt
= ∫10 |(1, 2t, 3t2)| dt
= ∫10 √(1 + 4t2 + 9t4) dt
≈ 1.86 ,
slightly longer than the straight-line path
with length |(1,1,1)| = √3 ≈ 1.73 .
5a. The vector field F(x,y,z) = (−y,x,1)
rotates around the z-axis (with right-handed orientation) and tilts upward. Nearer the z-axis, the vectors get shorter and tilt up more, pointing straight up along the axis itself.
5b. F(c)•dc =
∫10 F(at,bt,ct) • (at,bt,ct)' dt
=
∫10 (−bt,at,1) • (a,b,c) dt
=
∫10 −bat + abt + c dt
=
ct |1t=0
= c.
Thus f(x,y,z) = z gives
∇f(x,y,z) = (0,0,1),
which is not F(x,y,z) = (−y,x,1).
Thus, F cannot be a conservative vector field: there
is no potential function.
5c.
∫10 F(at2, bt2, ct2) • (at2, bt2, ct2)' dt
=
∫10 (−bt2, at2, 1)•(2at,2bt,2ct) dt
=
∫10 2ct dt
=
ct2|1t=0
= c.
This gives the same function f(x,y,z) = z as in part (b).
In fact, even if the vector field is non-conservative,
and the line integral changes upon changing the path
between fixed endpoints,
nevertheless the line integral is
unchanged upon reparametrizing the same curve.
We say that line integrals are always
independent of parametrization, even if they are
not path-independent.
5d. The vector field is not conservative because its
vortex pattern has non-zero circulation,
and thus gives different line integrals
along the two halves of a loop.
In the next lecture, we will define curl F,
and see it is non-zero.
11/3 Recitation 10.
11/7 Lect 25.
⊞
Curl in 3 dim. Conservative vector fields.
- Class notes
- Geometric meaning of line integral ∮ F(c)•dc :
How strongly F pulls along curve c
Negative if F pulls against direction of c
- Physical meaning of line integral:
Work done by force F in moving particle along c
= negative of work done by particle moving along c against force F
If F is conservative, the work done by F
between two points
is the increase of potential between the points, independent of path.
- Conservative Vector Field F:
Defined by four equivalent conditions:
(i) Exists potential f(x,y,z) with F = ∇f;
(ii) Path-independent; (iii) Circulation-Free;
(iv) Irrotational: curl F = 0 everywhere.
Question: How to define curl F(x,y,z) in 3-dim case?
Should be rate of circulation of F around a loop c(t) near (x,y,z),
per area enclosed by c.
- Vector field F(x,y,z) = (p(x,y,z), q(x,y,z), r(x,y,z))
3-dim curl of F is the vector whose coordinates are
2-dim curls of projections of F
parallel to coordinate planes yz (perp to x-axis), xz (⊥y),
xy (⊥z)
curl F = (curlyz(q,r), −curlxz(p,r), curlxy(p,q))
=
(∂r⁄∂y−∂q⁄∂z ,
∂p⁄∂z−∂r⁄∂x ,
∂q⁄∂x−∂p⁄∂y)
= ∇×F
= (∂⁄∂x ,
∂⁄∂y , ∂⁄∂z) ×
(p, q, r)
- The signs in definition of curl come from the Righthand Rule for rotations.
To represent a spatial rotation by an axis vector,
curve right-hand fingers in rotation direction: thumb points along axis
Then x-axis represents rotation from y-axis toward z-axis;
y-axis ⇔ z-axis toward x-axis;
z-axis ⇔ x-axis toward y-axis.
- Rate of circulation of F in plane
pependicular to a unit vector q is (curl F) • q
Maximum rate of circulation is |curl F|
in direction q =
curl F⁄|curl F|
- Direction of curl F(x,y,z) = axis of rotation
of multi-directional paddlewheel
with center fixed at (x,y,z),
under the flow F
Length |curl F(x,y,z)| = strength of rotation
-
If F is conservative, having zero circulation,
then rate of circulation is zero in all directions,
curl F = 0 = (0,0,0)
Reading: [MT] Ch 4.4 pp. 249−253.
HW: Start WHW 7 below.
- For the two vector fields in the previous HW,
compute the curl and visually verify the geometric meaning.
- [MT] Exercises p. 259
#13−20, 21(b), 22(a), 23(b), 29−35.
⊞
Soln
1a. Using the determinant formula,
curl(−x,−y,1) = (0,0,0).
This makes sense from the picture, since for any small rectangle in a coordinate plane, F is either perpendicular to it, or pushes along opposite sides equally, making zero circulation near every point.
1b. We have curl(−y,x,1) = (0,0,2), which matches the rotation of F around the z-axis in the picture.
Also, I can use the right-hand rule to determine
the polarity of the rotation axis: if I curl my fingers
in the direction of rotation, my thumb points upward
in the positive z-direction,
agreeing with curl F = +2k.
2. Odd-numbered exercise answers [MT] p. 515.
11/9 Lect 26.
⊞
Cylind & spherical coords, param surfaces
- Class notes
- Cylindrical coords: Cyl(r,θ,z) =
(r cos(θ), r sin(θ), z)
r = radius from z-axis, θ = turn from xz-plane, z = height
- Spherical coords: Sph(ρ,θ,φ)
= (ρ sin(φ) cos(θ),
ρ sin(φ) sin(θ),
ρ cos(φ))
ρ = radius from origin, θ = turn from xz-plane, φ = tilt from z-axis
z = ρ cos(φ), r = ρ sin(φ),
x = r cos(θ) = ρ sin(φ) cos(θ),
y = r sin(θ) = ρ sin(φ) sin(θ)
- Toroidal coords:
Tor(r,θ,φ) = ((R+r cos(φ))cos(θ),
(R+r cos(φ))sin(θ), r sin(φ))
Decompose v = (x,y,z) = Ru + rw
with |u| = |w| = 1
R = fixed radius of core ring,
Ru = (R cos(θ), R sin(θ), 0)
w points from core point Ru to
v on sectional circle of radius r
φ = tilt of w from xy-plane,
w = r cos(φ)u + r sin(φ)k
- Frenet frame ([MT] 4.2, Ex 16, 20, p. 235;
curve c(t) = (x(t), y(t), z(t))
u(t) = c'(t)⁄|c'(t)|,
unit tangent vector;
n(t) = u'(t)⁄|u'(t)|, unit normal vector (normal acceleration)
b(t) = u(t) × n(t), binormal vector
{u(t), n(t), b(t)} forms orthogonal unit axes at each point of c(t)
- Change of variables for parametrized solid region D.
Parametrization G(u,v,w) = (x(u,v,w), y(u,v,w), z(u,v,w)) for (u,v,w) ∈ D* parameter region
Jacobian derivative, 3 × 3 matrix [DG(u,v,w)].
Integral of f(x,y,z) over region:
∫∫∫D f(x,y,z) dx dy dz
=
∫∫∫D*
f(x(u,v,w), y(u,v,w), z(u,v,w))
det[DG(u,v,w)] du dv dw
- Pappus Centroid Theorem: Vol of solid of revolution = area × distance traveled by center of gravity
Reading: [MT] 1.4 & 7.3.
HW: Finish WHW 6 due 3/23 below.
- Parametrizing the torus with core circle radius 2,
thickness radius 1.
- Compute the Frenet frame u(t), n(t),
b(t) for the core circle
c(t) = (2 cos(t), 2 sin(t), 0),
simplifying with trig identities.
Write out the Frenet coordinate parametrization:
F(t,p,q) =
c(t) + p n(t) + q b(t)
=
(x(t,p,q), y(t,p,q), z(t,p,q)).
- Write out the toroidal coordinate map by
converting the Frenet frame to
polar coordinates (r,φ) in the n(t) &
b(t) plane at each point of the core circle.
That is, T(θ, r, φ)
is at angle θ ∈ [0,2π] of the core circle,
radius r ≥ 0 from the core circle,
and tilt φ ∈ [0,2π] from the horizontal plane.
- Use the above function to draw a parametrized torus surface
T(θ, 1, φ) on Mathematica (install for free)
or Wolfram Alpha.
- Determine the inverse function of the
spherical coordinate map Sph.
That is, write ρ, θ, φ as functions
of x,y,z.
- Sphere D of radius R has volume 4⁄3 π R3.
- Verify the spherical coordinate stretching factor
det[DSph(ρ,θ,φ)] = −ρ2 sin(φ).
Use the cofactor expansion of a determinant:
Simplify using cos2 + sin2 = 1.
- Compute the volume of sphere D by a spherical substitution
D = Sph(D*), where D* is a rectangular parameter domain
in (ρ,θ,φ) space:
Vol(D) =
∫∫∫D*
ρ2 sin(φ) dφ dθ dρ.
- Try to determine this volume in rectangular coordinates,
Vol(D) = ∫∫∫D 1 dz dy dx,
writing the sphere as the set of (x,y,z) with:
−R ≤ x ≤ R
−√(R2−x2)
≤ y ≤ √(R2−x2)
−√(R2−x2−y2)
≤ z ≤ √(R2−x2−z2).
- For the region under a graph curve z = g(x) in the xz-plane and above the interval x ∈ [0,R],
parametrize its solid of revolution around the z-axis,
and compute its volume using cylindrical coordinates (r,θ,z).
Which Calc I method does this correspond to?
- Extra Credit: Prove Pappus' Centroid Theorem:
The volume of a solid of revolution of a region
is equal to the area of the region, times the distance traveled by its center of gravity.
The center of gravity cent(D) of a plane region D
is the average position of a point of D.
To compute cent(D) = (x, y), take the average of each coordinate function over D:
x
=
1⁄area(D)
∫∫∫D x dx dy
and similarly for y. Now assume D is rotated around the x-axis. How far is cent(D) rotated?
Also, the theorem can be generalized to moving D along the Frenet frame of any curve, not just a circle . . . .
- Extra Credit:
For each solid below, determine:
- Volume using rectangular coordinates (taking an appropriate order of integration)
- Volume using a cylindrical (or elliptic cylindrical) parametrization
- Surface area using any convenient parametrization.
The solids are constructed from
specified slices perpendicular to a certain axis.
- Trilobite: The solid is built above the unit circle
x2 + y2 ≤ 1 in the xy-plane.
The vertical slices are perpendicular to the x-axis; the slice
at a given x is a square with base line segment
cutting across the circle: z = 0,
−√(1−x2) ≤ y ≤ √(1−x2).
Hint: This solid is an intersection of a vertical circular
cylinder and a horizontal parabolic cylinder:
its surface can be constructed out of paper.
- Pillow Solid: Silces are perpendicular to z-axis;
the slice at height z is the ellipse
(x⁄(z2−1))2 + y2 ≤ 1.
⊞
Soln
1a. For the circle c(t) = (2 cos(t), 2 sin(t), 0),
we have the tangent vector c'(t) = 2(−sin(t), cos(t), 0)
and the unit tangent vector is:
u(t) = (−sin(t), cos(t), 0).
Then the normal (centripetal) acceleration is
u'(t) = −(cos(t), sin(t), 0),
which has unit length, so it is equal to the unit
normal acceleration:
n(t) = −(cos(t), sin(t), 0).
Finally, the binormal is:
b(t) = u(t) × n(t)
= (0,0,1).
These define the Frenet frame u(t), n(t), b(t), an orthonormal basis of space defined for each point c(t), which we can use just like i, j, k.
The Frenet coordinates of c(t) are thus:
F(t,p,q) =
(2 cos(t) − p cos(t), 2 sin(t) − p sin(t), q)
1b. We trace a circle of radius r around each point
c(t), in the plane with basis n(t), b(t):
P(t,φ)
=
c(t) + r cos(φ) n(t) + r sin(φ) b(t)
=
(2 cos(t) − r cos(φ) cos(t), 2 sin(t) − r cos(φ) sin(t), r sin(φ))
Taking r = 1 gives the torus with thickness radius 1.
1c. Wolfram gives:
Here is a Mathematica file you can use to learn relevant commands
2. Given (x,y,z), the radius from the z-axis is r = √(x2+y2), and the radius from the origin is the hypotenuse of the right triangle with vertices (0,0,0), (x,y,0), (x,y,z):
ρ = √(r2+z2) =
√(x2+y2+z2).
The turn of (x,y,z) from the xz-plane is the same as the
turn of (x,y,0) from the x-axis:
θ = arccos(x⁄√(x2+y2))
=
arcsin(y⁄√(x2+y2))
= arctan(y⁄x).
The tilt of (x,y,z) from the z-axis is φ, satisfying
z = ρ cos(φ) and r = ρ sin(φ), so that
φ
=
arccos(z⁄√(x2+y2+z2))
=
arcsin(√(x2+y2)⁄√(x2+y2+z2))
=
arctan(√(x2+y2)⁄z).
Therefore:
(ρ,θ,φ) =
Sph−1(x,y,z) =
(√(x2+y2+z2,
arccos(x⁄√(x2+y2)),
arccos(z⁄√(x2+y2+z2))
),
assuming 0 ≤ arccos(t) ≤ π and y ≥ 0, 0 ≤ θ ≤ π. For general y, we should replace the above θ by sgn(y) θ, where sgn(y) = +1 for y ≥ 0 and sgn(y) = −1 for y < 0.
3a. Taking 2×2 subdeterminants of the Jacobian DS(ρ,θ,φ) of S = Sph gives the cross product:
∂S⁄∂θ × ∂S⁄∂φ
=
−ρ2 sin(φ) (cos(θ)sin(φ),
sin(θ)sin(φ), cos(φ)),
and taking the dot product with ∂S⁄ ∂ρ gives the
Jacobian determinant:
det[DS] =
∂S⁄∂ρ
•
(∂S⁄∂θ × ∂S⁄∂φ)
=
−ρ2 sin(φ) ( cos2(θ)sin2(φ) +
sin2(θ)sin2(φ) + cos2(φ) )
=
−ρ2 sin(φ).
3b. Since the integrand factors into functions of the individual variables, and the rectangular parameter
domain is D* a Cartesian product
of intervals (ρ,θ,φ) ∈ [0,R]×[0,2π]×[0,π],
the triple integral factors into a product
of single-variable integrals:
Vol(D) =
∫Rr=0
∫2πθ=0
∫2πφ=0
ρ2 sin(φ) dφ dθ dρ
=
(∫Rr=0 ρ2 dρ)
(∫2πθ=0 dθ)
(∫πφ=0 sin(φ) dφ) .
=
(1⁄3R3)
(2π) (2).
3c. Taking advantage of symmetry, we can reduce to the positive side of each integral:
Vol(D) =
8 ∫x=0R∫y=0√(R2−x2)∫z=0√(R2−x2−y2) dz dy dx
=
8 ∫x=0R∫y=0√(R2−x2)
√(R2−x2−y2) dy dx
=
4 ∫x=0R
[ y√(R2−x2−y2) + (R2−x2) arctan(y⁄√(R2−x2−y2)) ]y=0√(R2−x2) dx
=
4 ∫x=0R (R2−x2) π⁄2 dx
=
2π(R3 − R3⁄3).
Here we used the antiderivative
∫ √(a2−y2) dy
=
1⁄2[ y√(a2−y2) + a2 arctan(y⁄√(a2−y2)) ],
and that
arctan(√(R2−x2)⁄0+)
=
limu→∞ arctan(u)
=
π⁄2 .
4. The solid of revolution D is parametrized
by the (r,θ,z) parameter region
D* with 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π,
0 ≤ z ≤ g(r).
The stretching factor of the cylindrical parametrization
Cyl(r,θ,z) = (r cos(θ), r sin(θ), z) is the
same as for polar coordinates, namely r,
so the volume is:
Vol(D) =
∫r=0R∫θ=02π∫z=0g(r) r dz dθ dr
=
(∫θ=02π dθ)
(∫r=0R r g(r) dr)
=
∫r=0R 2πr g(r) dr.
This is the "shell method" volume formula for a solid of revolution.
11/10 Recitation 11
11/11 Lect 27.
⊞
Area of parametric surface.
- Class notes
- Parameter region S* ⊂ R2
for surface S ⊂ R3
Parametrization P : S* → S, P(u,v) = (x(u,v), y(u,v), z(u,v))
- Find stretching factor of P(u,v) near a point (u,v) = (uo,vo)
Same as stretching factor of linear approx
DP(uo,vo) = [∂P⁄∂u | ∂P⁄∂v ] Jacobian matrix 3×2
with column vector ∂P⁄∂u
= (∂x⁄∂u,
∂y⁄∂u,
∂z⁄∂u),
and similar for ∂P⁄∂v
Linear mapping DP(uo,vo) : R2 → R3 stretches unit square
to parallelogram in R3 spanned by
∂P⁄∂u ,
∂P⁄∂v
Area of parallelogram spanned by a, b is
length |a×b| = |a| |b| sin(θab)
Stretching factor is: |∂P⁄∂u × ∂P⁄∂v|
- Area = limit of sum of areas of small parallelograms in S
Area(S) = ∬S* |∂P⁄∂u×∂P⁄∂v| du dv
Reading: [MT] Ch 7.4.
HW: Start WHW 7 due 3/30 below.
- Consider the sadddle surface given by the graph z = x2 − y2 above the unit square
(x,y) ∈ [0,1] × [0,1]; that is
0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.
- Find a parametrization P : S* → S.
Hint: What parameter variables control the position of a point on S?
- Find the Jacobian matrix dP(u,v),
and the stretching factor
|∂P⁄∂u×∂P⁄∂v|.
- Set up a double integral to evaluate the area.
Use Wolfram Alpha to find numerical approximation.
- Find the lengths of the edge curves of S,
above the x- and y-axes.
(Parametrize the curves and compute the length
L = ∮ |dc|.)
How does the area compare to the product of the edge-lengths,
and to the area of the parameter region S*?
- Write and simplify the area integrals for
- A general graph z = g(x,y) for (x,y) in a plane region D ∈ R2
- A general surface of revolution, rotating a curve y = f(x),
for x ∈ [a,b], around the x-axis.
(Assume f(x) ≥ 0.)
- Check your answer in (b) by applying it to
the section of a right circular cone
with height a and base radius b: this surface
has equation z = a⁄b √(x2+y2). We can unroll the surface to get a circular sector
with radius √(a2+b2) and
outer circumference 2πb, and we can find its area
without calculus.
- Compute the following integrals
- The sphere of radius r.
Hint: The vector ∂P⁄∂u×∂P⁄∂v
for a spherical parametrization P(θ, φ)
was computed in the previous HW solutions, as a
step in computing det[DSph(ρ, θ, φ)].
In fact, the volume and area stretching factors are the
same, since dρ is not stretched by the Sph mapping.
- Compute the surface area and the enclosed volume
of the torus with core radius R, sectional radius r
(see previous lecture).
Hint: Again, the volume and area stretching factors
will be the same, since dr is not stretched by the toroidal coordinate mapping (see previous lecture).
- Review of cross product. Let a = (1,2,−1),
b = (1,0,1).
- Compute a×b
- Verify the defining property of
cross product: det(a,b,v) = (a×b) • v.
- Verify that a×b is orthogonal to
a, b.
- Verify that |a×b| = |a| |b| sin(θab),
and that this is the area of the parallelogram (actually a rectangle) spanned by a, b.
Hint: What is the familiar way to determine
cos(θab)?
⊞
Solutions
1. The parametrization P(u,v) = (u, v, u 2−v 2)
gives A = ∫ 10∫ 10
√(1+4u 2+4v 2) du dv
≈ 1.86 . The edge above the y-axis is
c(t) = (0, t, −t 2), with length
∫ 10√(1+4t 2) dt
≈ 1.48. The other edge curve has the same length,
and their product is 1.49 2 ≈ 2.19 , bigger
than the actual area, which is in turn bigger than 1, the parameter region area.
2a. The graph z = f(x,y) is parametrized by
T(x,y) = (x, y, f(x,y)), giving stretching factor:
|∂T⁄∂x
×
∂T⁄∂y|
=
|(1, 0, ∂f⁄∂x)
×
(0, 1, ∂f⁄∂y)|
=
|(−∂f⁄∂x ,−∂f⁄∂y , 1)|
=
√(1 + (∂f⁄∂x)2 + (∂f⁄∂y)2).
The area is:
A =
∬ √(1 + (∂f⁄∂x)2 + (∂f⁄∂y)2) dx dy.
This will usually be over some finite region (x,y) ∈ R* in the xy-plane.
2b. The surface of revolution z = f(√(x2+y2) is parametrized by
the cylindrical coordinate mapping
T(r,θ) = (r cos(θ), r sin(θ), f(r)), giving stretching factor:
|∂T⁄∂r
×
∂T⁄∂θ|
=
|(cos(θ), sin(θ), f '(r))
×
(−r sin(θ), r cos(θ), 0)|
=
|r (cos(θ) f '(r) , sin(θ) f '(r) , 1)|
=
r √(1 + f '(r)2) .
The area is:
A =
∬ r √(1 + f '(r)2) dr dθ
Again, this is over a parameter region (r, θ) ∈ R*.
2c. The right circlular cone with height a and
base radius b has a vertex-to-rim radius of ρ = √(a2+b2), and a rim
circumference C = 2πb.
We can unroll the lateral
surface into a sector of a circle
of radius ρ, with circular arc C.
The area A of a sector is proportional to its
circular arc, so C⁄2πρ
= A⁄πρ2 .
That is: A = πb√(a2+b2).
The cone is parametrized by T(r,θ) =
(r cos(θ), r sin(θ), a⁄b r)
for 0 ≤ r ≤ b and 0 ≤ θ ≤ 2π.
The formula of part (b) becomes:
A =
∬ r √(1 + g'(r)2) dr dθ
=
∫02π ∫0b r √(1 + (a⁄b)2) dr dθ
=
[ 2π√(1 + (a⁄b)2)
r2⁄2 ]r=0b
=
πb√(a2+b2).
3a. The sphere of radius ρ = R is parametrized by
P(θ,φ) = Sph(R,θ,φ), with
∂P⁄∂θ × ∂P⁄∂φ
=
−ρ2 sin(φ)
(cos(θ)sin(φ),
sin(θ)sin(φ), cos(φ)),
and area stretching factor
|∂P⁄∂θ × ∂P⁄∂φ|
=
R2 sin(φ).
The area is thus:
Area =
∫θ=02π∫φ=0π R2 sin(φ) dθ dφ
= 4πR2.
Note that the area formula is the derivative of the volume formula: A(R) = d⁄ dRV(R).
Geometrically, this means
that an increment of Δr produces a
layer of uniform thickness Δr
over the toric surface, so that the rate
of change of volume is the area.
3b. Consider the toroidal coordinate mapping:
T(r,θ,φ) =
R (cos(θ), sin(θ), 0)
+ r cos(φ) (cos(θ), sin(θ), 0)
+ (0, 0, r sin(φ)).
The first term parametrizes the core circle of radius R in the xy-plane by the angle θ, similar to the usual polar coordinate mapping.
On a vertical circle of radius r,
a radial vector tilted by angle φ from horizontal has horizontal component r cos(φ) and vertical component r sin(φ). This horizontal component should be in direction (cos(θ), sin(θ), 0), giving the second term, and the vertical component should be in direction (0, 0, 1), giving the third term.
The Jacobian derivative matrix is:
[DT] =
|
cos(φ) cos(θ)
| cos(φ) sin(θ)
| sin(φ)
|
| −sin(θ)(r cos(φ)+R)
| cos(θ)(r cos(φ)+R)
| 0
|
| −r sin(φ) cos(θ)
| −r sin(φ) sin(θ)
| r cos(φ)
|
|
| .
|
The Jacobian determinant simplifies to:
det[D T] = r(R + r cos(φ)).
W|A can compute this if you input
the above 3×3 matrix in the format {{a,b,c},{d,e,f},{g,h,i}}. The determinant remains
the same if you transpose rows and columns.
For fixed r,
the torus surface parametrization T(θ,φ)
= T(r,θ,φ) gives the tangent vectors ∂T⁄∂θ ,
∂T⁄∂φ
of T(θ,φ)
are the same as the corresponding columns
of T(r,θ,φ) above.
The normal vector is:
n =
∂T⁄∂θ
×
∂T⁄∂φ =
r (R + r cos(φ)) (cos(φ) cos(θ), cos(φ) sin(θ), sin(φ)).
This is a scalar times the unit vector
(cos(φ) cos(θ), cos(φ) sin(θ), sin(φ) ),
which happens to be the same as
∂T⁄ ∂r :
that is, the radial derivative vector of T
is the unit orthogonal to the other two derivative vectors,
which is evident from the geometry.
Therefore the length of the cross product
is the same as the volume stretching factor above:
| n| = r (R + r cos(φ)).
The area of the torus is thus:
A =
∫02π∫02π r (R + r cos(φ)) dφ dθ
= [2π r (Rφ + r sin(φ))]2πφ=0
= 4π2rR.
We temporarily denote the fixed sectional radius
by r 0, so that the solid torus has
0 ≤ r ≤ r 0. Then the volume is:
V = ∫0r0∫02π∫02π r (R + r cos(φ)) dφ dθ dr
=
∫0r0
A(r) dr
=
∫0r0
4π2rR dr
=
2π2r02R .
Then replacing r 0 with r, we get
V = 2π 2r 2R .
Note that A(r) = d⁄ drV(r).
This is clear from the ∫ A(r) dr
in the calculation, which comes about because
of the unit orthogonality of the radial coordinate
∂T⁄ ∂r above. Geometrically, this means
that an increment of Δr produces a
layer of uniform thickness Δr
over the toric surface, so that the rate
of change of volume is the area.
4a. We compute: a×b = (1,2,−1) × (1,0,1) = (2,−2,−2).
4b. For v = (x,y,z), we have:
det(a,b,v)
= (1)(0)(z) + (x)(2)(1) + (1)(y)(−1)
− (1)(2)(z) − (1)(y)(1) − (x)(0)(−1)
= 2x − 2y − 2z
= (a×b)•v.
4c,d. All three vectors a, b, a× b have dot products equal to zero.
In particular, the angle between a and b is
θ = π⁄ 2, so:
|a||b| sin(θ) = (√6)(√2)(1)
= 2√3,
|a×b| = 2√3.
It is easy to see from elementary geometry that
the area of a parallelogram is the product of the edge-lengths times the sine of the angle between them.
11/14 Lect 28.
⊞
Directional curl. Curl Thm in R3 (Stokes' Thm): ∬S (curl F)•dS =
∮ F(c)•dc.
- Class notes
- Directional curl
- Recall curl of 3D vector field F = (p,q,r):
curl F
=
∇×F
=
(∂r⁄∂y − ∂q⁄∂z ,
∂p⁄∂z − ∂r⁄∂x ,
∂q⁄∂x − ∂p⁄∂y).
For a plane near a point a
spanned by basis vectors u,v,
define directional curl to mean the rate of circulation
of F around a small curve c(t)
in the given plane, per unit area enclosed:
curluv F =
limc→a
1⁄area(c)
∮ F(c) • dc .
- Theorem:
The directional curl is equal to the component of the vector curl perpendicular to the plane:
curluv F =
(curl F) • n⁄|n| ,
where n = u×v is the normal vector of the plane,
and n⁄|n| is the unit normal vector.
Proof:
For h > 0, consider the parallelogram curve
ch(t) defined by the four linear segments:
a + tu ,
a + u + tv ,
a + (h−t)u + v ,
a + (h−t)v ,
each running over t ∈ [0,h].
The area enclosed is the length of the cross product
|hu × hv| = h2|u×v|,
and we compute:
curluv F =
limh→0
1⁄h2|u×v|
(∫01 F(a+tu) • u dt
+
∫01 F(a+u+tv) • v dt
+
∫01 F(a+(h−t)u+v) • (−u) dt
+
∫01 F(a+(h−t)v) • (−v) dt).
Substituting F(a + r(t)) = F(a) + DFa(r(t)) + o(h), and writing DF(u) • v in terms of matrix products as
uT DFT v, where u, v are considered as column vectors, we compute the limit:
1⁄|u×v|
(uT DFT v − v DFT u)
=
1⁄|u×v|
uT(DFT−DF)v
=
1⁄|u×v|
det(curl F, u, v)
=
(curl F) • u×v⁄|u×v| .
The above applies to a parallelogram curve,
but the Theorem holds for arbitrary c(t)
converging to a in the plane of u,v.
- Curl Theorem
- Motivation for Curl Theorem. As in two dimensions,
the total circulation of a vector field F
around a curve c should be equal to
the integral of the rate of circulation
over any surface S enclosed by c.
- What do we mean by the "integral of the rate of circulation"?
- Surface S = P(S*) ⊂ R3 parametrized by
P(u,v) = (x(u,v), y(u,v), z(u,v))
over parameter region S* ⊂ R2,
with c(t) going around the boundary or edge of S.
- Tangent plane of S has basis
∂P⁄∂u ,
∂P⁄∂v .
Normal vector n(u,v) = ∂P⁄∂u ×
∂P⁄∂v perpendicular to S
at each point P(u,v)
- Integral of rate of circulation is:
∬S*
(
curl F in tangent
| plane of S | )
()
du dv
=
∬S*
(curl F) •
n⁄|n| |n| du dv
=
∬S* (curl F) • n du dv
=
∬S (curl F) • dS,
where we use the "vector surface area element"
dS = n du dv.
-
Curl Theorem (Stokes' Theorem): The integral of the rate of circulation of F over a surface S is equal to the total circulation around the boundary curve of S. That is, if a surface S = P(S*) has boundary curve c, then
∬S (curl F) • dS
=
∮ F(c) • dc,
or more explicitly:
∬S* curl F(P(u,v)) •
(∂P⁄∂u×∂P⁄∂v) du dv
=
∫01 F(c(t))
• c'(t) dt.
- Justification:
Line integral around the boundary of S
is equal to the sum of line integrals
around small parameter boxes making up S:
line integrals along opposite internal boundaries cancel,
leaving only external boundary contributions.
The circulation around each parameter box
is approximately directional curl times area (by definition),
and this is given by (curl F)•n,
as described above.
- Universal stretching factor: Gramian determinant
- Let S ⊂ Rn be a k-dimensional "surface" inside an n-dimensional space,
parametrized by a function P : Rk → Rn
with variables P(u1, . . . , uk)
= (x1, . . . , xn),
so that S = P(S*) over a parameter region S* ⊂ Rk.
We wish to write a substitution formula for the k-dimensional "area" of S as an integral over S*:
area(S) = ∫···∫D* d(P)
du1···duk
for a suitable stretching factor d(P)(u1, . . . , uk).
- The Jacobian matrix DP = [∂P⁄∂u1
| ··· |
∂P⁄∂uk ] is the n × k matrix
whose columnns are the k tangent vectors of P along u-axis directions.
Define the Gramian matrix
of Gram(P) as the k × k matrix of dot products of these tangent vectors:
Gram(P) = [ ∂P⁄∂ui • ∂P⁄∂uj ], where i,j = 1, . . . , k.
- The stretching factor is the Gramian determinant, the square root
of the determinant of the Gramian matrix:
d(P) = √det G(P)
= √det
[ ∂P⁄∂ui • ∂P⁄∂uj ].
In fact, the determinant of the Gramian matrix and the square root are always positive real numbers.
- The simplest example is a curve P = c : R → Rn.
There is only k = 1 tangent vector, the Gramian is the scalar c'(t) • c'(t),
and the stretching factor is √(c'(t) • c'(t)) = |c'(t)|,
which is the correct integrand in the length formula L = ∫ |c'(t)| dt.
- Proof: We show that the
k-dimensional volume of the parallelopiped spanned
by vectors v1, . . . , vk ∈ Rn is equal to the Gramian determinant.
- Assume v1, . . . , vk are
linearly independent vectors spanning a k-dimensional subspace
V ⊂ Rn.
Let L : Rk → Rn be the linear mapping with matrix [L] =
[ v1 | ··· | vk ],
so that L(ei) = vi,
where ei is a standard basis vector.
The image of L is L(Rk) = V ⊂ Rn,
and the image of the unit cube in Rk
is the parallelopiped spanned by v1, . . . , vk .
- Let ũ = {u1, . . . , uk}
be an orthonormal basis of V.
Restricting the codomain of L,
define the linear mapping L̃ : Rk → V,
represented by a k × k matrix [L̃],
with inputs in standard coordinates of Rk
and outputs in ũ-coordinates of V,
both orthonormal coordinate systems.
Then the stretching factor of L and L̃
(and the k-dimensional volume of the above parallelopiped) is
det[L̃]. We wish to compute this in terms of [L].
- Extend ũ to an orthonormal basis
u = {u1, . . . , uk, uk+1, . . . , un} of Rn,
and let [L]u denote the matrix
of L with inputs in standard coordinates, and outputs in u-coordinates.
The top k × k submatrix of [L]u is [L̃],
while the bottom (n−k) × k submatrix is identically zero.
Letting U =
[u1 | ··· | un ]
be the orthogonal
change-of-basis matrix from u-coordinates to
e(n)-coordinates, with U−1 = UT,
we have:
- Recall the Gramian G = [L]T [L] = [vi • vj],
the k × k matrix of dot products of the vi
column vectors.
We may compute its determinant as:
det G = det [L]T [L]
=
det [L]T U UT [L]
=
det [L]uT
[L]u
=
det [L̃]T [L̃]
=
(det[L̃])2.
Therefore, the k-volume of the parallelopiped is: det[L̃] = √det(G).
Reading: [MT] Ch 8.2 Stokes' Theorem pp. 439-446.
- Consider the vector field F(x,y,z) = (0,0,x+y).
- Compute the vector curl F
- Compute the directional curl around the coordinate axes
i, j, k
(i.e. along the planes perpendicular to these vectors).
- Around what direction is the directional curl maximal?
- Sketch F and verify the above computations geometrically.
Hint: Draw the arrows F(x,y,0) from the xy coordinate plane.
- Consider the upper unit hemisphere surface S defined by
x2+y2+z2 = 1, x ≥ 0,
and take F(x,y,z) = (0, 0, y(z+1)).
- Parametrize S by P(u,v) with P : S* → S,
and find its normal vector n(u,v).
- Compute curl F.
- Compute the flux integral of curl F through S:
that is ∬S curl F • dn = ∬S* F(P(u,v)) •
(∂P⁄∂u×∂P⁄∂v) du dv.
- Parametrize the circular boundary of S by c(t),
and compute its tangent vector c'(t).
- Compute the circulation of F around c,
that is ∮ F(c) • dc,
and verify that it gives the same answer as the double integral (c).
- Consider the tilted ellipse surface S
along the plane z = x+1 and inside the cylinder
x2 + y2 ≤ 1.
Also let F(x,y,z) = (−y,x,z).
- Parametrize S and its boundary curve c.
Hint: First write the x- and y-coordinates of P(u,v) = (x(u,v), y(u,v), z(u,v)), then find z.
- Find the normal vector n(u,v) for S,
and the tangent vector c'(t).
- Compute the total circulation of F around c.
- Compute the flux of curl F across S.
By the Curl Theorem, this should be the same as the
circulation of F around the boundary curve in part (c).
- Extra Credit: Give an analog of the Conservative Vector Field Theorem, starting from: (i) a given G has a vector potential F with G = curl F;
and ending with: (iv) div F = 0.
Using the integral theorems, we can now prove all the equivalences, analogously to the original case, except for (ii) ⇒ (i): to do this, find an integral
formula that computes a vector potential F
for a given G.
- Gramian:
For each of our previous area and volume substitution formulas, namely for
F : R2 → R2,
R2 → R3,
R3 → R3,
show that the known stretching factor is equal to the square
root Gramian determinant (after simplification).
⊞
Solutions
1a. The determinant formula gives: curl F
= (1,−1,0).
1b. The directional curls around the x,y,z axes are the
components of the vector curl:
curl(1,0,0) = 1, curl(0,1,0) = −1,
curl(0,0,1) = 0.
1c. For a direction v with |v| = 1,
the directional curl around the axis v is:
curlv F = (curl F) • v = |curl F| |v| cos(θ)
This is maximal when the angle θ = 0, so that v =
curl F⁄ |curl F|,
i.e. in the direction of the vector curl.
1d.
The vector field, being independent of z, has the same vectors
on each horizontal plane.
At every point, the main rotation is around the direction
(1,−1,0). This is evident on the line y = x itself,
but is also true on lines like y = x + 1, since the arrows
get longer at points farther from y = x.
2a. The spherical coordinate parametrization of the vertical half-sphere is:
P(φ,θ) = (sin(φ) cos(θ), sin(φ) sin(θ), cos(φ))
for 0 ≤ φ ≤ π , −π⁄2 ≤ θ ≤ π⁄2 .
Normal vector: n
=
∂P⁄ ∂φ× ∂P⁄ ∂θ
=
(sin 2(φ)cos(θ), sin 2(φ)sin(θ), sin(φ)cos(φ))
2b. For F(x,y,z) = (0, 0, y(z+1)), we have:
curl F(x,y,z)
=
(∂⁄∂y(y(z+1)) − ∂⁄∂z(0),
−∂⁄∂x(y(z+1)) + ∂⁄∂z(0),
∂⁄∂x(0) − ∂⁄∂y(0))
=
(z+1, 0, 0).
2c. The integral of the curl over the surface is:
∬S curl F • dn
=
∫
−π/2
π/2
∫ 0
π
curl F(sin(φ)cos(θ), sin(φ)sin(θ), cos(φ)) • n(φ,θ) dφ dθ
=
∫
−π/2
π/2
∫ 0
π
(cos(φ)+1, 0, 0)
• (sin2(φ)cos(θ), sin2(φ)sin(θ), sin(φ)cos(φ))
dφ dθ
=
∫
−π/2
π/2
∫ 0
π
(cos(φ)+1) sin2(φ) cos(θ)
dφ dθ
=
(∫
−π/2
π/2
cos(θ)
dθ)
(∫ 0
π
(cos(φ)+1) sin2(φ)
dφ)
= (2)(π⁄2)
= π.
2d. The vertical boundary circle ∂S,
with equations x = 0 and y2+z2 = 1,
is given parametrically by
c(t) = (0, cos(t), sin(t)) for 0 ≤ t ≤ 2π.
Note that in curling the fingers of my right hand
around c,
my thumb points along the positive x-axis, which is
the positive direction for S.
The tangent vector is c'(t) = (0, −sin(t), cos(t)).
2e. The circulation around the boundary is:
∮ F(c) • dc
=
∫02π F(0, cos(t), sin(t))
• (0, cos(t), sin(t))' dt
=
∫02π (0, 0, cos(t)(sin(t)+1))
• (0, −sin(t), cos(t)) dt
=
∫02π cos2(t)(sin(t)+1) dt
=
π.
This agrees with part (c), as it must by the Curl Theorem.
3a. Cylindrical coorinates: P(r,θ) = (r cos(θ), r sin(θ), r cos(θ) + 1) for 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π.
The boundary is:
c(t) = (cos(t), sin(t), cos(t)+1) for 0 ≤ t ≤ 2π.
3b. n(r,θ) =
∂P⁄∂r
×
∂P⁄∂r
=
(cos(θ), sin(θ), cos(θ)) ×
(−r sin(θ), r cos(θ), −r sin(θ))
= (−r, 0, r).
The normal vector function
is so simple because S lies in a plane
orthogonal to (−1, 0, 1). The length |n| is the stretching factor of the parametrization: the grid boxes of P have size proportional to the radius r.
The tangent vector of the curve is:
c'(t) = (−sin(t), cos(t), −sin(t)).
3c. ∮ F(c) • dc
=
∫02π
F(cos(t), sin(t), cos(t)+1)
• (cos(t), sin(t), cos(t)+1)' dt
=
∫02π
(−sin(t), cos(t), cos(t)+1) •
(−sin(t), cos(t), −sin(t)) dt
= 2π.
3d. We have curl F = (0,0,2), meaning that near every point, F primarily rotates around a vertical axis. The flux of the curl through S is:
∬S F × dn
=
∫01 ∫02π (0,0,2) ×
(−r, 0, r) dθ dr
=
∫01 4πr dr
=
2π, which agrees with part (c), confirming
the Curl Theorem in this case.
4. If the Jacobian [DP] is a square matrix, this follows from
properties of the determinant:
det([DP]T [DP]) =
det[DP]T det[DP] =
(det[DP])2.
For DP : R2 → R3,
it follows from a basic formula relating cross product
and dot product:
|u × v|2
= |u|2 |v|2
− (u • v)2.
(This is a special case of the fact that the
Hodge ☆ operator on differentials is an isometry.)
11/16 Lect 29.
⊞
Flux, Divergence Thm (Gauss' Thm): ∯S F•dS
= ∫∫∫R (div F) dR.
- Class notes
- Parametric surface S = P(S*) for P(u,v) = (x(u,v),y(u,v),z(u,v))
Normal vector n =
∂P⁄∂u ×
∂P⁄∂v
-
Orientation of S means a choice of positive normal direction n
- Flux of vector field F across oriented surface S:
Net flow of F from negative to positive side
of S
∬S F • dS
=
∬S* F(P(u,v)) •
n(u,v) du dv
Justification: Flux = ∬S F •
n⁄|n| |n| du dv, where:
F •
n⁄|n| = component of
flow of F perpendicular to S (i.e. across surface S)
|n| = |∂P⁄∂u×∂P⁄∂v| =
area stretching factor of parameterization P(u,v)
- Ex: S = unit hemisphere x2 + y2 + z2 = 1, x ≥ 0 (right of yz-plane)
Parametrize: P(φ,θ) = (sin(φ)cos(θ), sin(φ)sin(θ), cos(φ))
for 0 ≤ φ ≤ π , −π⁄2 ≤ θ ≤ π⁄2
Normal vector: n
=
∂P⁄∂φ×∂P⁄∂θ
=
(sin2(φ)cos(θ), sin2(φ)sin(θ), sin(φ)cos(φ)),
points outward
Vector field F(x,y,z) = (0, 0, z+1), pointing upward on S, getting longer as you go up
Net flow is positive: weak inflow of F through bottom of S,
stronger outflow through top
Compute: net flow of F through S
= ∬S F • dn
= ∬S* F(P(φ,θ)) • n(φ,θ) dφ dθ
= ∬S* F(sin(φ)cos(θ), sin(φ)sin(θ), cos(φ)) •
(sin2(φ)cos(θ), sin2(φ)sin(θ), sin(φ)cos(φ)) dφ dθ
= ∬S*
(0, 0, cos(φ)+1) •
(sin2(φ)cos(θ), sin2(φ)sin(θ), sin(φ)cos(φ)) dφ dθ
=
∫
−π/2
π/2
∫ 0
π
(cos(φ)+1) sin(φ) cos(φ)
dφ dθ
= 2⁄3
- Divergence of F means
rate of flux of F across a small closed surface S
near a, per unit area enclosed.
div F = limS→a
∬S F•dS / vol(S).
In coordinates: F(x,y,z) = (p(x,y,z), q(x,y,z), r(x,y,z))
div F =
∂p⁄∂x
+ ∂q⁄∂y
+ ∂r⁄∂z
=
(∂⁄∂x ,
∂⁄∂y ,
∂⁄∂z)
• (p,q,r)
=
∇ • F.
- Divergence Theorem (Gauss' Theorem):
The outward flux of a vector field F
across a closed surface S is equal to
the integral of the rate of flux of F
in the solid region R enclosed by S.
That is, if a solid region R = Q(R*)
has boundary surface S = P(S*),
then
∬S F • dS
=
∫∫∫R (div F) dR,
or more explicitly:
∬S* F(P(u,v)) •
(∂P⁄∂u×∂P⁄∂v) du dv =
∫∫∫R* (div F)(Q(u,v,w)) det[DQ] du dv dw.
- Justification:
Flux integral across the boundary surface S
is equal to the sum of flux integrals
out of small parameter boxes making up R:
flux integrals across common internal boundaries cancel,
leaving only external boundary contributions.
The flux out of each parameter box
is approximately div F times volume (by definition),
that is, (div F) det[DQ] du dv dw.
Reading: [MT] Ch 5.5 Triple Integrals, p. 294.
Ch 7.6 Surface Integrals of Vector Fields, p. 400.
Ch 8.4 Gauss' Theorem, p. 461.
HW: Finish WHW 7 below.
- Divergence Theorem.
Consider the solid region R under the above plane
z = x+1, above the xy-plane, and inside the cylinder
x2 + y2 ≤ 1,
the shape of a wedge cut out of a tree-trunk:
R = {(x,y,z) | x2 + y2 ≤ 1 , 0 ≤ z ≤ x+1}.
Again let F(x,y,z) = (−y,x,z).
- Parametrize R and its three
boundary surfaces, the top S1, side S2, and bottom S3.
(The top is the same ellipse as in #1.)
- Compute the stretching factor det[DQ(u,v,w)]
of the parametrization Q(u,v,w) of region R.
- Compute the normal vectors n(u,v) for S1, S2, S3.
You might have to take negative of a normal
to make sure it points out from R.
- Compute the total flux of F across the
three surfaces.
- Compute the integral of div F on R.
By the Divergence Theorem, this should the same as the
flux of F across the boundary surfaces in part (d).
⊞
Solutions
1a. R is parametrized in cylindrical coordinates
by Q(r,θ,z) = (r cos(θ), r sin(θ), z)
over the parameter region:
R* = {(r,θ,z) | 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2π , 0 ≤ z ≤ r cos(θ) + 1}.
Here the z-range is defined by floor and ceiling functions
depending on the previous variables r, θ.
The top surface S1 is the same as S from the previous HW,
P1(r,θ)
= (r cos(θ), r sin(θ), r cos(θ) + 1)
for 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π..
The side surface S2 is defined by P2(θ,z)
= (cos(θ), sin(θ), z) over
S2* defined by 0 ≤ θ ≤ 2π
and 0 ≤ z ≤ cos(θ)+1.
The bottom surface S3 is the unit disk
P3(r,θ) = (r cos(θ), r sin(θ), 0)
over S3* defined by 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π.
1b. The stretching factor of Q(r,θ,z) is the determinant of its Jacobian matrix, i.e. the volume of the parallelopiped spanned by the gridline tangent vectors
∂Q⁄∂r,
∂Q⁄∂θ,
∂Q⁄∂z.
This works out to det[DQ] = r, the same as for polar coordinates, since the z direction is not stretched at all.
1c. The normal vector of S1 is dn1 = (−r, 0, r), which correctly points upward (outward from R).
The normal vector of S2
is dn2 = (cos(θ), sin(θ), 0),
which points outward.
The normal vector of S3 seems to be
(cos(θ), sin(θ), 0) × (−r sin(θ), r cos(θ), 0) = (0, 0, r),
but this points upward into R,
so we must use its outward-pointing negative:
n3 = (0, 0, −r).
1d. The total flux of F over the top surface
is: ∬S1 F • dn2 =
∬S1* F(P1(r, θ)) • n1(r,θ) dr dθ
= ∫01 ∫02π (−r sin(θ), r cos(θ), r cos(θ) + 1) • (−r, 0, r) dr dθ
= π.
From drawing R and F,
we can see that F is parallel to the surface along S2 and S3, so it has no flux through these surfaces. This can be confirmed by calculating
∬S2 F • dn2 =
∬S3 F • dn3 = 0.
Therefore the total flux over the surface ∂R = S1 ∪ S2 ∪ S3
is π + 0 + 0 = π.
1e. We have div F =
∂⁄∂x(−y)
+
∂⁄∂y(x)
+
∂⁄∂z(z)
= 1.
That is, F has a constant rate of flux from
every point, due to the smaller inflow at the bottom of
a small box than the outflow at the top.
Thus ∭R div F dx dy dz
= ∭R* 1 det[DQ] dz dθ dr
= ∫01 ∫02π
∫0r cos(θ)+1 r dz dθ dr = π, which agrees with (d), confirming
the Divergence Theorem in this case.
11/17 Recitation 12. WHW 8: Planetary orbits.
11/18 Lect 30. Review Integral Thms & Conservative Vect Fld Thm.
Postponed to 11/23:
⊞
Proof of Integral Theorems
- Class notes
- Consider a vector field F in the plane R2,
a simple plane region R, and the curve c
which is the counterclockwise boundary of R. (For more general R, possibly with holes, the boundary curves should
be oriented so that the region is on the
left as you traverse each curve.)
- In coordinates, we write
F(x,y) = (p(x,y), q(x,y)) and c(t) = (x(t), y(t)).
We can write a line integral as:
∮ F • dc
= ∮c (p dx + q dy)
= ∫ p(x(t), y(t)) x'(t) dt + ∫ q(x(t), y(t)) y'(t) dt.
- Curl Theorem (Green's Theorem):
In R2,
the integral of the rate of rotation over a region R
is equal to the total rotation around the boundary of R:
∬R curl(F) dy dx
= ∮ F(c) • dc.
In coordinates:
∬R (∂q⁄∂x
− ∂p⁄∂y) dy dx
= ∮c (p dx + q dy).
- Divergence Theorem (Green's Theorem):
In R2,
the integral of the rate of outflow over a region R
is equal to the total net outflow across the boundary of R:
∬R div(F) dy dx
= ∮ F(c) • dn,
where n is the outward-pointing normal to c,
meaning dn = (−c')⊥ dt = (y'(t), −x'(t)) dt.
In coordinates:
∬R (∂p⁄∂x
+ ∂q⁄∂y) dy dx
= ∮c (p dy − q dx).
- The two coordinate equations are equivalent to each other.
In the first equation, we replace F = (p, q)
by F⊥ = (−q, p) to obtain
the second equation.
Reading: [MT] Ch 8.1 Green's Theorem, Ch 8.2 Stokes' Theorem, Ch 8.4 Gauss' Theorem
HW:
- Go through our proof of the Curl Theorem for the specific
case of the vector field F(x,y) = (x+1, xy) and the y-simple
region R = {(x,y) with −1 ≤ x ≤ 1,
0 ≤ y ≤ 1+x2}.
- First, compute the two sides of the Curl Theorem
as directly as possible, and check equality.
- Starting our proof requires regions
which are both x-simple and y-simple (see M&T p. 283).
Split R into two such regions
R = R1 ∪ R2, joined along a common boundary line.
Hint:
As an inverse for y = x2 + 1, use the function
b(y) = √(y−1) for y ≥ 1, b(y) = 0 for y ≤ 1.
- Simplify ∬Ri (∂q⁄∂x
− ∂p⁄∂y) dy dx
for each simple region R1, R2.
Rather than computing out all the integrals and derivatives,
consider the terms ∬Ri ∂q⁄∂x dx dy
and −∬Ri ∂p⁄∂y dy dx
separately, and use the one-variable
Fundamental Theorem of Calculus.
- Rewrite the above terms of the line integrals
∮ci F • dci
=
∮ci p dx + ∮ci q dy,
where c1, c2
are the boundaries of the two regions.
- Explain why the sum of the line integrals for the
two boundaries is equal to the line integral for the
boundary of R. (What happens to the integral over the
extra line between the regions?)
- Give a proof of the Divergence Theorem
by mimicking our proof of the Curl Theorem.
(As discussed above, the Curl Theorem implies
the Divergence Theorem by simply substituting
F⊥ for F.
Nevertheless, prove the second theorem without assuming the first.)
- If F is a vector field with curl(F) = 1 for
all (x,y), then the Curl Theorem says that the area of the region
R is equal to the line integral of F over the boundary
curve of R.
- Let F = ½(−y, x),
and let R be the region given in polar coordinates
by r ≤ f(θ), for some function f.
Use the Curl Theorem to give a simple formula
for the area of R.
Obtain the same formula directly using the polar change
of variables.
- M&T p. 437 #13. Find the area
under one arch of the cycloid curve (x,y) = (t−sin(t), 1−cos(t)) by computing the line integral of:
F(x,y) = (0, x);
or (−y,0); or ½(−y, x).
Which vector field gives the simplest computation?
- Go over the proofs of Stokes' and Gauss' Theorems in R3 on [MT] p. 441, 464-465, and summarize the arguments in words, carefully noting the previous results used.
⊞
Solutions
1a. From the picture of F(x,y) = (x+1, xy), we can see that curl(F) > 0 on the whole region. Also, F • c' is positive on the left, bottom, and right boundaries of R, while it is moderately negative on the top boundary (oriented right-to-left); thus the total rotation around the boundary should be positive.
Computing the left side of the Curl Theorem, with curl(F) = y − 0:
∬R curl(F) dy dx
=
∫−11
∫01+x2 y dy dx
=
∫−11
½(1+x2)2 dx
=
28⁄15 .
Computing the right side of the Curl Theorem for the four
counter-clockwise boundary curves of R:
∮ F • dc
| =
| ∫−11
F(t,0) • (t,0)' dt +
∫02
F(1,t) • (1,t)' dt
| |
| + ∫1−1
F(t,1+t2) • (t,1+t2)' dt +
∫20
F(−1,t) • (−1,t)' dt
|
| =
| ∫−11
(t+1,0) • (1,0) dt +
∫02
(2,t) • (0,1) dt
| |
| + ∫1−1
(t+1,t+t3) • (1,2t) dt +
∫20
(0,−t) • (0,1) dt
|
| = | 2 + 2 − 62⁄15
+ 2 =
28⁄15.
|
Notice that we parametrized the left boundary curve
as c4(t) = (−1, t) from the top corner
t = 2 to the bottom corner t = 0, giving reversed limits of integration
∫ 20. (Similarly with the top curve.)
1b. Split the region along the y-axis so that
each half is doubly simple, with no vertical or horizontal
bays or dents:
R1 = {(x,y) with 0 ≤ x ≤ 1, 0 ≤ y ≤ 1+x2}
= {(x,y) with 0 ≤ y ≤ 2, b(y) ≤ x ≤ 1},
R2 = {(x,y) with −1 ≤ x ≤ 0, 0 ≤ y ≤ 1+x2}
= {(x,y) with 0 ≤ y ≤ 2, −1 ≤ x ≤ −b(y)},
where b(y) = √(y−1) for y ≥ 1 and b(y) = 0 for y ≤ 1.
1c. For the region R1, and F = (p,q) = (x+1, xy),
the curl integral is:
∬R1 curl(F) dy dx
| =
| ∬R1 (∂q⁄∂x
− ∂p⁄∂y) dy dx
|
| =
| ∫
y=0
2
∫
x=b(y)
1
∂⁄∂x(xy) dx dy
−
∫
x=0
1
∫
y=0
x2+1
∂⁄∂y(x+1) dy dx
| =
| ∫
y=0
2
[xy]x=b(y)
x=1
dy
−
∫
x=0
1
[x+1]y=0
x2+1
dx
| =
| ∫
y=0
2
1y − b(y)y
dy
−
∫
x=0
1
(x+1) − (x+1) dx.
| =
| ∫
y=0
2
q(1,y) − q(b(y),y)
dy
−
∫
x=0
1
p(x,x2+1) − p(x,0) dx.
| | | | |
We have a similar computation for R 2.
1d. The line integral over the boundary curve
c1 = (x(t), y(t)) is:
∮ F • dc1
=
∮ (p(c1),0) • dc1+
∮ (0,q(c)) • dc1
=
∫ p(x(t),y(t)) x'(t) dt +
∫ q(x(t),y(t)) y'(t).
For the p-term, we parametrize the
boundary of R 1 by four segments consistent with
its y-simple description:
(x(t),y(t)) = (t,0), (1,t), (t,t2+1), (0,t),
the last two traveled in reverse.
For the q-term, we parametrize the same
boundary consistently with the x-simple description:
(x(t),y(t)) = (t,0), (1,t), (t,b(t)),
with the last traveled in reverse.
Then we may compute:
∮ (p(c1),0) • dc1+
∮ (0,q(c)) • dc1
| =
| ∫01 p(t,0) (t)'
+ ∫02 p(1,t) (1)'
+ ∫10 p(t,t2+1) (t)' dt
+ ∫10 p(0,t) (0)' dt
| |
| + ∫01 q(t,0) (0)'
+ ∫02 q(1,t) (t)'
+ ∫20 q(t,b(t)) (t)' dt.
|
The vertical and horizontal terms containing (1)' and (0)' all vanish, and the remaining four terms match those
at the end of 1(c).
This proves the Curl Theorem for region R 1,
and the proof for R 2 is similar.
1e. Assume we know the Curl Theorem for the two regions R 1 and R 2.
Adding the double integrals on the left
side of the Theorem for R 1 and R 2,
we clearly get the double integral over R.
Adding the line integrals along the two boundary curves,
we get the line integral over the boundary of R,
plus the line integral along the boundary line separating the subregions, taken twice in opposite directions, which
cancel out. The left side equals the right side
over the whole region R.
3a. For F = ½(−y,x), we easily calculate curl F = 1,
so the left side of the Curl Theorem gives the area of the region.
The right side is the line integral of F along
the boundary curve given by (r,θ) = (f(t),t),
namely c(t) = (f(t)cos(t), f(t)sin(t)):
∮ F(c) • dc
=
∫t=02π ½(−f(t)sin(t), f(t)cos(t)) • (f(t)cos(t), f(t)sin(t))' dt
=
∫t=02π ½ f(t)2 dt.
This can also be obtained from parametrizing R by
the polar coordinate map
T(r,θ) = (r cos(θ), r sin(θ))
for (r,θ) in the parameter region defined
by 0 ≤ θ ≤ 2π, 0 ≤ r ≤ f(θ);
then using the Change of Variables formula.
4. To prove Stokes' Theorem ∮c F•dc
= ∬S (curl F)•dS, write the surface in space as a graph (or more generally a parametrized surface) over a parameter region in the plane.
Explicitly compute both sides of the equation in coordinates,
making use of the Chain Rule.
Reduce the LHS to a line integral
around the parameter region boundary,
and the RHS to the integral of a certain
function over the parameter region.
Then notice that these are equal by Green's Theorem.
To prove Gauss' Theorem ∬S F•dS = ∭R div(F) dR,
assume R is a symple solid region, meaning it can be
parametrized as lying between floor and ceiling graph surfaces
in the z-direction, and above a parameter region in the xy-plane;
and symmetrically for the y- and x- directions.
Then the LHS and RHS of the equation each split into three terms
involving the three components of the field.
Computing each of these with respect to the corresponding
parametrization of R, a straightforward calculation
and an application of the single-variable Fundamental Theorem of Calculus shows the two sides are equal.
11/21 Lect 31.
⊞
Newton's Laws, Gravity field.
- Class notes
- Newton's Law of Gravitation: A point mass M at the orgin produces a gravitational field:
G(v) = −Mv⁄|v|3.
Magnitude = inverse square of radius, direction = toward origin
- Superposition: Several masses Mi
at positions ci
produce a field
G(v) =
−∑i Mi(v−ci)⁄|v−ci|3.
- Newton's Force Law (2nd Law of Motion): A particle of mass m in the presence of a field G is subject to a force F = mG,
and moves along a trajectory c(t) satisfying
F(c(t)) = m c''(t), i.e. F = ma,
which reduces to:
G(c(t)) = c''(t).
In other words, the particle accelerates
in the direction of G.
For a point-mass field, this produces trajectories which
are conic sections: ellipse, parabola, hyperbola.
- Does such a gravitational field G have a potential function g(x,y,z) with G = ∇g?
- By the Conservative Vector Field Theorem, this is equivalent to curl G = (0,0,0). This is easily checked from the formulas, and is evident geometrically.
- The potential of G(v) = −Mv⁄|v|3
is g(v) = M⁄|v|.
This can be found by a line integral over a radial line c = tv going inward from t = ∞ to t = 1.
- Physicists define potential energy φ = −g,
and φ(v) = −M⁄|v| can be pictured as a "gravity well" near the central mass.
- The field G(v) = −Mv⁄|v|3 also satisfies div G = 0 everywhere except the origin. By the Divergence Theorem, this means:
∬S G • dS
=
∭R div(G) dR = 0
for any surface S enclosing a region R which does not contain the origin.
- Nevertheless, ∬S G • dS
= −4πM
for any surface enclosing the origin, so
we may say that div(G) is infinitely concentrated
at the origin, the location of the point-mass.
In fact, this is the only radial vector field
with zero divergence outside the origin.
We get similar results with several point-masses.
- Gauss' Divergence Law:
A continuous distribution of mass with density function μ(x,y,z), giving g/cm3 at each point, induces a gravitational field with
div(Gμ) = 4πμ.
(Here 4π is the surface area of a unit sphere.)
This is an equivalent formulation of Newton's inverse-square law. (See exercises below.)
Reading: [MT] Ch 4.3 pp. 238-241,
Kepler's Laws, Newton's Equation, Euler's Method
Feynman Lectures on Physics (Vol I, Ch 9).
HW:
WHW 8 below.
- Newton's Gravity Law
- Compute directly that G(v)
=
Cv⁄|v|3 has curl(G) = 0 and div(G) = 0.
- Show that any radial vector field G(v) = f(|v|) v⁄|v|3 has curl(G) = 0. Here f(t) is any single-variable function.
Hint: Chain Rule
and Product Rule for Curl:
for scalar function g(x,y,z)
and vector field F(x,y,z):
curl(g F) =
∇×(g F) =
∇g × F
+ g ∇×F.
where we use the notation ∇×F = curl F.
- Show that if G(v)
= f(|v|) v⁄|v|3 with
div(G) = 0 for v ≠ 0,
then G(v) = Cv⁄|v|3.
This shows that Gauss' Law implies Newton's inverse-square law for a point-mass.
Hint: Chain Rule and Product Rule for Divergence:
div(g F) =
∇ • (g F) =
∇g • F
+ g ∇•F,
where we use the notation ∇•F = div F.
- Let G(v) =
−∑i Mi(v−ci)⁄|v−ci|3
be the gravity field of several point-masses.
- Show curl G = 0 and
div G = 0 except at the mass points ci.
- Find the potential function of G.
Hint: No long computations needed: use the previous problem, and linearity of the derivative operations.
- Consider the gravitational field Gμ
of a continuous mass distribution with density μ(x,y,z) within a region R.
- Explain intuitively why:
Gμ(v) =
∭t∈R
−μ(t)(v−t)⁄|v−t|3 dR,
where we integrate over t = (t1, t2, t3) ∈ R,
and the spatial integral of a vector field is defined
as the limit of appropriate Riemann sums
(equivalent to integrating each component as a scalar function).
- Show that the above field has div(Gμ) = 4πμ,
proving that Newton's Gravitation Law implies Gauss' Law.
Hint: Move the derivatives with respect to the components of v = (x,y,z) inside the integral.
- Using the above formula,
consider the gravitational field Gr
of a sphere with radius r centered at the origin,
and having uniform mass density μ = 1.
Show that for a point v outside the sphere, this field is Gr(v) = −Mv⁄|v|3,
the same as the field of a point-mass at the origin
having the same total mass M as the sphere.
Hint: Integrate using an appropriate coordinate system.
⊞
Solutions
1a. Compute in coordinates:
G(x,y,z) =
( Cx⁄ (x2+y2+z2)3/2 ,
Cy⁄ (x2+y2+z2)3/2 ,
Cz⁄ (x2+y2+z2)3/2 ).
1b. Use ∇|v|
= v⁄|v| ,
and ∇f(|v|)
= f '(|v|) v⁄|v| . Then the terms in the Product Rule both vanish.
1c. Write the vector field as f(|v|) v⁄|v|3, and use the Product Rule
to get f '(|v|) = 0, so that f(|v|) = C a constant.
2a. Each term Gi coming from a point-mass has zero curl and divergence, using the first problem (with a shift v → v−ci). Then
div(∑i Gi)
= ∑i div(Gi).
2b. The potential of each Gi is
Mi ⁄|v−ci| ,
and the sum of the potentials is the potential of the sum.
3a. A small cube has mass
μ(t) Δt1Δt2Δt3,
approximately concentrated at the point t = (t1,t2,t3),
so its gravity field at v = (x,y,z) is −μ(t)(v−t)⁄|v−t|3.
Summing these fields over all small cubes gives a
Riemann sum for the spatial integral,
and taking the limit as Δt1, Δt2, Δt3 → 0
gives the integral itself.
11/23 Extra Topic
11/24-27 Thanksgiving Break, no class
11/28-12/2: Lect 32
⊞
Maxwell's Equations, Electromagnetism.
- Combined class notes
- Electric field E = E(x,y,z;t)
= (Ex, Ey, Ez)
Magnetic field B = B(x,y,z;t)
= (Bx, By, Bz)
- Lorentz force law: particle with charge q, velocity v has F = q(E + v×B)
E pulls or repels charges, B deflects moving charges perpendicular to
their velocity
- Maxwell Equations: charge density δ, current density J
| Name | Differential | Integral | Description
| (A) | Gauss Law | div E = δ
| ∯S E•dn = ∫∫∫R δ dV, S=∂R
| Electric flux out of closed surface
equals charge enclosed
| (B) | No Magnetic Monopoles | div B = 0
| ∯S B•dn = 0
| No magnetic flux out of closed surface
| (C) | Faraday Law | curl E = −dB⁄dt
| ∮C E•dc = −d⁄dt ∫∫S B•dn, C=∂S
| Voltage induced around closed loop equals negative rate of change of enclosed magnetic flux
| (D) | Ampere Law | curl B = J + dE⁄dt
| ∮C B•dc = ∫∫S J•dn + d⁄dt ∫∫S E•dn, C=∂S
| Magnetic field around closed loop equals enclosed current plus displacement current (rate of change of enclosed electric flux)
|
- Physical examples
- Gauss Law: Static charge attracts objects,
makes hair repel & stand on end
- Faraday Law: Two rings with same axis, AC current in one ring induces voltage E around other ring
Changing E around one ring
⇒ changing axial B ⇒
E curling around other ring
- Ampere Law: Current-bearing wires attract (if parallel currents), repel (if opposite currents).
Current in one wire ⇒ B through other wire
⇒ moving charge in other wire deflected
- Ampere Law: Current J equivalent to displacement current dE⁄dt
Parallel capacitor plates discharging through a connecting wire
Effect on B is same for flux of J
and of decreasing E between plates
- Static solutions: dE⁄dt = dB⁄dt = 0
- curl E = 0 ⇒ E = −grad φ for electric potential energy φ
div B = 0 ⇒ B = curl A for
magnetic vector potential A
- Point charge: E(x)
=
x⁄|x|3
=
x̂⁄|x|2 ,
φ(p) =
1⁄|x| , B = 0
div E = 0, but delta function at (0,0,0)
Notation: x̂ = x⁄|x| , the unit vector of x
- Line charge, unit density along ℓ:
E(x)
=
x⊥ℓ⁄2π|x⊥ℓ|2
=
x̂⊥ℓ⁄2π|x⊥ℓ| ,
φ(x) =
−1⁄2π log |x⊥ℓ|, B = 0
div E = 0, but unit linear density delta function on z-axis.
Notation: x⊥ℓ
is the projection of x perpendicular to ℓ
- Line current, unit along ℓ = ẑ:
B(x) =
ℓ×x⁄2π|ℓ×x|2 ,
B(x,y,z) =
(−y,x,0)⁄2π(x2+y2) ,
A(x,y,z) = (0, 0, −1⁄4π log(x2+y2)),
E = 0
curl B = 0, but ẑ-valued delta function on z-axis
- Thin solenoid (current rotating around z-axis cylinder): E = 0
B = 0, but (0,0,1)-valued delta function on z-axis,
A(x,y,z) =
(−y,x,0)⁄2π(x2+y2)
- Electric dipole with axis ẑ: B = 0
- E =
limε→0
(Eεẑ − E−εẑ)⁄2ε , scaled limit of converging point-charges
-
E(x) =
(3(ẑ•x)x − |x|2ẑ)⁄|x|5 =
(3(ẑ•x̂)x̂
−
ẑ)⁄|x|3
E(x,y,z) =
(3xz, 3yz, 2z2−x2−y2)⁄(x2+y2+z2)5/2
- φ(x) = ẑ•x⁄|x|2
= ẑ•x̂⁄|x|
- Magnetic dipole with axis ẑ: E = 0
- B =
limC→0
(B of xy-loop C)⁄(xy-area enclosed), scaled limit of unit current around small xy-loop C
- B(x,y,z) =
(3xz, 3yz, 2z2−x2−y2)⁄(x2+y2+z2)5/2
- A(x) = ẑ×x⁄|x|2
= ẑ×x̂⁄|x|
- Maxwell eqns reduce to Poisson eqn: ∇2φ = −δ , ∇2A = −J
- General static solutions: Poisson integral of fundamental point-charge solution
- Electrostatic:
E(x) =
1⁄4π
∫∫∫u δ(u) (x−u)⁄|x−u|3 dVu
φ(x) =
1⁄4π
∫∫∫u δ(u)⁄|x−u| dVu
,
B = 0
- Magnetostatic:
B(x) =
1⁄4π
∫∫∫u J(u) × (x−u)⁄|x−u|3 dVu
A(x) =
1⁄4π
∫∫∫u J(u)⁄|x−u| dVu
,
E = 0
- Biot-Savart: steady current I along loop C = u(t)
B(x) =
−I⁄4π
∮C (x−u)⁄|x−u|3 × du
A(x) =
I⁄4π
∮C 1⁄|x−u| du
,
- Electric field at edge of a charged plate.
Conformal mapping f(z) = z2,
f(v+iu) = (v2−u2) + (2vu)i,
has inverse function:
g(w) = √w
= (r cos(θ) + i r sin(θ))½
= √r cos(θ⁄2)
+ i √r sin(θ⁄2),
g(x+iy) = v(x,y) + i u(x,y)
=
1⁄2sgn(y)(x+(x2+y2)½)½
+ 1⁄2(−x+(x2+y2)½)½.
Take the ray with u(x,y) = 0 to be a side view of
a half-plane of uniform negative charge;
then u(x,y) is the electric potential energy,
and E = −∇u(x,y) = −∇v(x,y)⊥ is the electric field.
-
⊞
Vector potential: reversing curl F
- Smooth manifold M with basepoint po
- Deformation retraction r : M×[0,1] → M, (p,t) ↦ r(p,t)
with:
r|M×0 = id,
r|M×1 = po,
r|po×[0,1] = po
- Chain homotopy K : Ωi(M) → Ωi−1(M), K(ω) = η
ηp(v1,...,vi−1)
=
∫10
ωr(p,t)(∂r⁄∂t(p,t),
Dr(p,t)(v1),...,Dr(p,t)(vi−1))
dt
with d K − K d = id on Ωi for i > 0
- Poincare Lemma: closed ⇔ exact, Hi(M) = 0 for i > 0.
Proof: Given dω = 0,
get ω = d(Kω) − K(dω) =
dη.
- K is adjoint to the cone operator K^ on singular chains:
(i−1)-simplex σ goes to i-simplex K^(σ)
by cone-ing σ to po via r(p,t).
- M = Rn, r(p,t) = tp
- K(f dxj1···dxji)p
=
∑ik=1 (−1)k−1
(∫10 ti−1 f(tp) dt)
dxj1···
dxjk···dxji
- n = 3, p = (x,y,z)
- Vect field B = (Bx, By, Bz) = Bz dx dy − By dx dz
+ Bx dy dz ∈ Ω2
- Vect field A = (Ax, Ay, Az) =
Ax dx + Ay dy + Az dz
∈ Ω1
- Given B with dB = div B = 0,
find A with curl A = dA = B :
A = K(B) =
| |
[∫10 (By(tp) tz
− Bz(tp) ty) dt] dx
| − |
[∫10 (Bx(tp) tz
− Bz(tp) tx) dt] dy
| + |
[∫10 (Bx(tp) ty
− By(tp) tx) dt] dz
|
|
- Can also take r(p,t) shrinking one dimension at a time (coordinate axis paths to origin), cf. [MT] Ch 8.4 Ex #20, p 461.
- Summary for R3
- f −grad→ F −curl→ G −div→ g
- grad f = F, K : Ω1 → Ω0,
f(p) = (∫10 F(tp) dt) • p
- curl F = G, K : Ω2 → Ω1,
F(p) = (∫10 t G(tp) dt) × p
- div G = g,
K : Ω3 → Ω2,
G(p) = (∫10 t2 g(tp) dt) p☆
p☆ = (x,y,z)☆
= (z,−y, x)
- Key examples
- Curl of rotational field = axial field:
curl (−y,x,0) = (0,0,2)
curl 1⁄x2+y2 (−y,x,0)
= 0 except for (0,0,1) times delta-function of z-axis
- Curl of concentrated axial field = rotational field:
curl (0,0,1⁄x2+y2) = −2⁄x2+y2 (−y,x,0)
-
⊞
Electromagnetic wave equations
- Renormalization: Replace A by A + ∇ψ ⇒
Coulomb gauge div A = 0
Poisson eqn ∇2ψ = −div A,
Poisson integral ψ(p) = −∫∫∫ div A⁄4π|p−u|2 dVu
- Reduce Maxwell equations to wave equations
- (B) ∇•B = 0 ⇔
B = ∇×A for some A
- (C) 0
=
∇×E + ∂B⁄∂t
=
∇×(E + ∂A⁄∂t)
⇔
E + ∂A⁄∂t
=
−∇φ for some φ
- (A) ρ =
∇•E =
∇•(−∇φ −
∂A⁄∂t)
=
−∇2φ −
∂⁄∂t(∇•A)
⇔
∇2φ −
∂2φ⁄∂t2
=
−ρ
− ∂⁄∂t(∇•A +
∂φ⁄∂t)
- (D) J =
∇×B −
∂E⁄∂t
=
∇×(∇×A) −
∂⁄∂t(∂A⁄∂t + ∇φ)
=
∇(∇•A) −
∇2A
+ ∂2A⁄∂t2
+ ∂⁄∂t∇φ
⇔
∇2A −
∂2A⁄∂t2
= −J
+ ∇(∇•A + ∂φ⁄∂t)
- Gauge shift: A = Ao + ∇ψ,
φ = φo − ∂ψ⁄∂t
Then:
∇×A = B,
−∇φ = E +
∂A⁄∂t
⇔ ∇×Ao = B,
−∇φo = E +
∂Ao⁄∂t
Lorenz gauge: solve inhomogeneous wave eq
∇2ψ −
∂2ψ⁄∂t2
=
−∇Ao −
∂φo⁄∂t
⇔
∇•A +
∂φ⁄∂t
= 0
- Conclusion: Maxwell eqns reduce to inhomogeneous wave eqns
∇2φ − ∂2φ⁄∂t2
= −ρ ,
∇2A −
∂2A⁄∂t2
= −J
- Free space solutions: Feynmann II-20, ρ = 0 , J = 0
-
∇2A = ∂A⁄∂t ⇒
∇2B = ∂B⁄∂t
∇2φ = ∂φ⁄∂t ⇒
∇2E = ∂E⁄∂t
- Linear wave:
E = E(x,t) = (0, f1(x−t) + g1(x+t), f2(x−t) + g2(x+t)) ,
B = B(x,t) = (0,−f2(x−t) + g2(x+t), f1(x−t) − g1(x+t))
- General solution is superposition of this type
along x,y,z directions
- Spherical waves: E = E(r,t) = 1⁄r (F(r−t) + G(r+t))
for r = √x²+y²+z²
Physical solutions move outward only, so G = 0
- Poisson integrals for dynamic potentials
- φ(p) =
1⁄4π
∫∫∫u ρ(u,t')⁄|p−u| dVu
- A(p) =
1⁄4π
∫∫∫u J(u,t')⁄|p−u| dVu
- Retarded time variable: t' = t − |p−u|
Meaning: a charge in position u at time t
only affects position p after time |p−u|
-
We have taken speed of light c = 1; in general
t' = t − 1⁄c |p−u|
∇2φ − 1⁄c2 ∂2φ⁄∂t2
= −ρ ,
∇2A −
1⁄c2 ∂2A⁄∂t2
= −1⁄c2 J
Reading: Wikipedia, [MT] Ch 8.4,
Feynmann Lectures (Vol 2)
⊞
HW
- Point-charge electric field, inverse-square radial:
E(v) =
v⁄|v|3
defined for v ≠ 0.
- Write E in coordinates,
and compute that div E = 0 whenever it is defined.
- Compute the flux of E out of a sphere of radius r centered at the origin. Use spherical coordinates.
- For any spatial region R, show the flux of E out of R is equal to zero if R does not contain the origin, and to 4π if it does contain the origin. Hint: Use the Divergence Theorem along with the previous results.
- Give a physical interpretation of (c) using Gauss's Law.
What electric charge δ would cause this field?
- Line-current magnetic field along z-axis, inverse linear cylindrical:
B(x,y,0) =
(−y, x, 0)⁄(x2+y2)
defined for (x,y,z) ≠ (0,0,z).
- Show that B can be written in the vector form
B(v) =
ℓ×v⁄|ℓ×v|2 ,
where ℓ = k = (0,0,1) is the current direction.
The same formula would hold for any unit axis vector
ℓ.
- Using the coordinate form,
compute that curl B = 0 wherever it is defined.
- Compute the circulation of B around a circle of radius r centered at the origin in the xy-plane.
- For any surface S, show the circulation of B around the boundary of S is equal to zero if S is not punctured by the z-axis, and to 2π if it is punctured. Hint: Use the Curl Theorem along with the previous results.
- Give a physical interpretation of (c) using Ampere's Law.
What electric current J would cause this field?
- Consider the magnetic field B induced by
a steady current around a unit circle c in the xy-plane.
The Biot-Savart Law computes the magnetic field as:
B(v) =
∫ (v−c(t))⁄|v−c(t)|3 × c'(t) dt.
Compute the field on the z-axis, B(0, 0, z).
Hint: In the integral, v = (0, 0, z) is constant, and only t is a variable. The field should always point upward along the axis, consistent with the right-hand rule.
⊞
Soln
1a. Writing
E(x,y,z) =
(x,y,z)⁄ (x2+y2+z2)3/2,
apply the definition of div E,
then compute the partial derivatives
using the Quotient and Chain Rules, until everything cancels.
1b. Parametrize the sphere using P(θ,φ) = Sph(ρ,θ,φ) with constant radius ρ = r,
and normal vector
∂P⁄∂θ × ∂P⁄∂φ
= r2 sin(φ) ̂n, where ̂n
is the unit normal radial vector.
The electric field is equal to ̂n⁄r2,
and the integral reduces to
∫02π
∫0π sin(φ) dφ dθ = 4π.
1c. If R does not contain the origin, then
the Divergence Theorem says the flux of E out of R
is equal to the integral of the div E = 0 inside R.
If R does contain the origin, split R into a ball of small radius
at the origin, and the remaining region R' which does not contain the origin. Then the flux of E out of R is the sum of the
fluxes out of the small ball and out of R', since they join along a small sphere boundary, with opposite normals.
By (a) and (b), this gives 4π + 0.
1d. Gauss' Law says the flux of the electric field out of a region R is equal to the total charge inside the region. The above results mean that there is a point charge of 4π at the origin, and none elsewhere.
2a. Recall the definition of parallel and perpendicular projections from Lect 1.
2b. Apply the definition of curl B,
then compute the partial derivatives
using the Quotient and Chain Rules, until everything cancels.
2c. Parametrize the circle by cylindrical coordinates.
The integrand becomes constant,
since the rotational field is parallel to the tangent vector.
2d. If S does not meet the z-axis, then
the Curl Theorem says the circulation of B around S
is equal to the flux integral of curl B = 0 through S.
If S does meet the z-axis, split S into a disk of small radius
at the origin, and the remaining region S' which does not meet the z-axis. Then the circulation of B around S is the sum of the
circulations around the small disk and around S', since they join along a small circle boundary, with opposite tangents.
By (b) and (c), this gives 2π + 0.
2e. Ampere's Law for static fields says the curl of the magnetic field around a region S is equal to flux of the current through the surface. The above results mean that there is a line current of 2π along the z-axis, and none elsewhere.
3. Compute the vector integrand in coordinates, reducing
to an integral of the form
∫ (f1(t), f2(t), f3(t)) dt
=
(∫ f1(t) dt, ∫ f2(t) dt, ∫ f3(t) dt).
Compute the three ordinary integrals to get
B(0, 0, z) =
(0 , 0, −2π⁄(1+z2)3/2).
12/1 Recitation 13. WHW 9: Vector potential theorem.
12/5-9 Review material.
⊞
Syllabus.
Class notes
- Geometric objects
- Vectors v = (x1, . . . , xn) ∈ Rn,
with addition, subtraction, scalar multiplication
Dot product: u•v = |u| |v| cos θuv
Cross product for v = (x,y,z):
(u×v) • w = det(u,v,w),
|u×v| = |u| |v| sin θuv
- Curve C ⊂ Rn
parametrized by c : R → Rn,
c(t) = (x1(t), . . . , xn(t)) over t ∈ [a,b]
- Region R ⊂ Rn
(plane or solid) with boundary ∂R,
points (x1, . . . , xn) ∈ R defined by:
- Simple region (successive floor/ceiling):
Plane: a ≤ x ≤ b , c(x) ≤ y ≤ d(x)
Space:
a1 ≤ x ≤ b1 ,
a2(x) ≤ y ≤
b2(x) ,
a3(x,y) ≤ z ≤
b3(x,y)
General: a1 ≤ x1 ≤ b1 ,
a2(x1) ≤ x2 ≤
b2(x1) ,
a3(x1,x2) ≤ x3 ≤
b3(x1,x2) , . . .
- Coordinate parameter function P : Rn → Rn
over simple parameter region u ∈ R*
P(u1, . . . ,un) =
(x1(u1, . . . ,un), . . . ,
xn(u1, . . . ,un))
- Inequality defining region: f(x1, . . . , xn) ≤ c,
with boundary contour f(x1, . . . , xn) = c
- Surface S ⊂ R3
with boundary curve ∂S, defined by:
- Parameter function
P : R2 → R3,
P(u,v) = (x(u,v), y(u,v), z(u,v))
over simple parameter region (u,v) ∈ S*
- Contour equation f(x,y,z) = c, cut off by inequalities g(x,y,z) ≤ a.
- Parametrized k-dimensional "surface" S ⊂ Rn
- P : Rk → Rn,
P(u1, . . . , uk) = (x1, . . . , xn)
- S = P(S*) for parameter region S* ⊂ Rk
- Functions
- Scalar function (scalar field, 0-form) f : Rn → R
at each point gives a number f(x)
= height, temperature, potential energy . . .
- Mapping: general parameter function
P : Rk → Rn
takes points to points
- General field Φ on Rn:
at each point gives a local object
Φx
- Vector field F : Rn
→ Rn, at each point an arrow
F(x)
= force, flow velocity . . .
- Extra: Co-vector field, 1-form φ,
at each point a linear function
φx :
Rn → R
- Extra: k-form ω, at each point a function
ωx : (Rn)k → R
ωx(h1, . . . , hk) is multilinear and alternating
in h1, . . . , hk
- Derivative operations
- Nabla ∇ = (∂⁄∂x1 ,
∂⁄∂x2 , . . . )
- Partial derivative ∂f⁄∂xi , grad f(x1, . . . ,xn) = ∇f
= ∂f⁄∂x1 +
· · · + ∂f⁄∂xn
Linear approximation: f(a+h) = f(a) + ∇f(a)•h + o(h)
Directional derivative ∂f⁄∂v
= ∇f • v for |v| = 1
- Total derivative of P(u1, . . . , uk)
= (x1, . . . , xn)
- Linear function
DPa : Rk → Rn,
P(a+h) =
P(a) + DPa(h) + o(h)
- Jacobian matrix (m rows) × (n columns):
[DPa]
=
[∂P⁄∂u1 | . . . | ∂P⁄∂uk ]
=
- Extra: Hessian second-derivative matrix
- Hessian n×n matrix H(a) = [ ∂2f⁄∂xi∂xj(a) ]
- Quadratic Taylor series approximation
f(a+h) = f(a) + ∇f(a)•h
+ 1⁄2hT•H(a)•h + o(|h|2)
for n×1 column vector h and 1×n row vector hT
- Divergence div F = ∇ • F
= ∂f1⁄∂x1 +
· · · + ∂fn⁄∂xn
for F = (f1, . . . , fn)
- Planar curl of F = (p,q): curl F = curl(p,q) = ∂q⁄∂x
−
∂p⁄∂y
Spatial curl of F = (p,q,r):
curl F = ∇ × F
= (curlyz(q,r), −curlxz(p,r), curlxy(p,q))
- Tangent vectors from parameter functions: c'(t) or
∂P⁄∂u1 , . . . ,
∂P⁄∂uk
- Derivative of combination: Product Rules, Chain Rule
Obtained from product or composition of linear approx
- Extra: Exterior derivative of function
f : Rn → R is 1-form
φ = df
df =
∂f⁄∂x1 dx1
+ ··· +
∂f⁄∂xn dxk
dual to gradient
- Exterior derivative takes k-form ω to
(k+1)-form dω
d(f dxi1∧···∧ dxik)
= df ∧ (dxi1∧···∧ dxik)
- Integral operations
- Arclength ∫C dC = ∫ab |c'(t)| dt
- Line integral ∫C F • dC
= ∫ F(c(t)) • c'(t) dt, work done by F pulling along c;
For closed loop C, this is called circulation: ∮C F • dC
- Size of a region
- Area of a parametrized region R in the plane: ∫∫R dA
= ∫∫R* |det[DP]| du dv
- Volume of a parametrized solid region R in space: ∫∫∫R dV
=
∫∫∫R* |det[DP]| du dv dw.
- Area of a parametrized surface S in space: ∫∫S dA
= ∫∫S* |dS|
= ∫∫S* |∂P⁄∂u×∂P⁄∂v| du dv
- Flux integral: net flow of F across boundary,
toward right-hand or outward normal n
- Planar
∫C F • dn
=
∫ab F(c(t)) •
(−c'(t)⊥) dt
- Spatial ∫∫S F • dS
= ∫∫S* F(P(u,v)) • (∂P⁄∂u×∂P⁄∂v) du dv
- Extra: Integral of k-form ω on parametrized
k-dim surface S = P(S*),
P(u1, . . . , uk) = (x1, . . . , xn)
∫S ω =
∫S* ω(∂P⁄∂u1 ,
. . . , ∂P⁄∂uk)
du1··· duk
- Fundamental Theorems of Calculus
- Gradient Theorem: The integral of the rate of change of f along C is equal to the total change of f between the endpoints:
∫C ∇f • dC =
f(b) − f(a),
for C from a to b.
- Curl Theorem (Stokes):
The integral of the rate of circulation of F
over a surface
is equal to the total circulation of F around the boundary curve C.
- Planar ∫∫R curl F dA = ∮C F • dC
- Spatial ∫∫S curl F • dn = ∮C F • dC
- Divergence Theorem (Gauss): The integral of the rate of flux of F over a region is equal to the total flux of F out across the boundary.
- Planar ∫∫R div F dA = ∮C F • dn, C is boundary curve of R
- Spatial ∫∫∫R div F dV = ∫∫S F • dn, S is boundary surface of R
- Extra: Stokes Theorem for Differential Forms: For k-form ω, (k+1)-dimensional surface S ⊂ Rn
with k-dimensional boundary surface ∂S:
∫∂S ω
= ∫S dω
- Conservative Vector Field Theorem: The following conditions are equivalent for a vector field F:
- Conservative: There is a potential function f with ∇f = F.
In this case,
f(a) = f(0) + ∫C F • dC for any curve C from 0 to a.
- Path independent:
∫C1 F • dC1 =
∫C2 F • dC2
for any curves C1, C2 between the same endpoints.
- Zero circulation: Around any closed loop
C, ∮C F • dc = 0.
- Irrotational: curl F(v) = 0 at all v
Poincare Lemma: For k-form ω on Rn,
there exists (k−1)-form η with ω = dη
if and only if dω = 0.
- Max/min of f : Rn → R over region R
- Critical point inside R: solve ∇f(v) = 0,
for v ∈ R
- Parametrize boundary S = P(S*), critical points ∇f(P(u)) = 0
- Also consider endpoints of parametrized curve
- Choose absolute max & min values among f(v)
for all critical & end points v
- Constrained critical point on boundary ∂R:
Define ∂R as contour g(v) = c:
solve ∇f(v) = λ ∇g(v) for Lagrange multiplier λ ∈ R
and v ∈ ∂R,
Parametrize ∂R by P(u),
solve ∇f(P(u)) = 0.
- Miscellaneous points where ∇f or ∇g is undefined
- Extra: Second-derivative test for
critical point ∇f(v) = 0,
using Hessian H
- if H is positive-definite
(all upper-left det > 0), then
v is a local minimum
- if H is negative-definite
(−H is positive-definite), then
v is a local maximum
- if H is not pos/neg definite, then v is a saddle or stationary point
⊞
Problems I: Vector algebra, vector fields, potential, Jacobian derivative
- Vector algebra. Consider the plane containing
(endpoints of) the vectors
a = (0,1,1), b = (1,1,0), c = (2,2,2).
- Find a parametric equation P(u,v) = (x(u,v), y(u,v), z(u,v)) for the plane.
- Find an equation for the plane as a level surface: ax + by + cz = d.
(You will need a normal direction vector n, orthogonal to
the directions along the plane.)
- Find the area of the triangle △abc,
which is half the area of an appropriate parallelogram.
- Find the distance from the origin to the above plane.
(Take a vector from the origin to the plane,
and project it to the normal direction n.)
- Find the volume of the tetrahedron with vertices
0,a,b,c as 1⁄3
the area of the base triangle in part (c), times the height from part (d).
- Find the above tetrahedron volume another way, as
1⁄6 the volume of the parallelopiped
spanned by edges a,b,c.
- Line integrals. Consider the vector field F(x,y) = (2x, 1).
- Sketch the field F, plotting sample vectors at grid points.
- Does it look like F is the gradient field (uphill vectors) of a
potential function, i.e. F = ∇f?
Sketch the contour lines and graph of
this hypothetical f(x,y).
- Review the arguments from WHW 5 that F is conservative (having a potential function) if and only if it is irrotational (zero curl).
- Gradient Theorem:
F = ∇f
⇒
f(c(1)) = f(c(0))
+ ∮F(c)•dc
is path independent (depends only on endpoints).
- ∮F path independent ⇔
F has zero circulation (∮F = 0 on any closed loop)
- Curl Theorem:
F has zero circulatiion ⇐ curl F = 0 everywhere.
- Recall the pictorial meaning of curl F,
why curl(0,x) = 1, curl(y,0) = −1, and
curl(p,q) = ∂q⁄∂x
− ∂p⁄∂y .
- Compute curl F.
- Compute f(a,b) = ∮ F(c)•dc
= ∫10 F(c(t))•c'(t) dt
for the straight line c(t) = (ta,tb) from (0,0) to (a,b).
Compare to (b) above.
- Linear mappings and matrices
- Let R0 :
R2 → R2
be the 1⁄4 counterclockwise rotation.
Write its 2×2 matrix [ R0 ]
= [ R0(i)
| R0(j) ].
- In R3, compute the 3×3 matrix of a 1⁄4
rotation Rz around the z-axis, and similarly
Rx around the x-axis.
- Multiply matrices to calculate the composite mapping: [Rz∘ Rx]
= [ Rz ] • [ Rx ].
- Picture Rz∘ Rx by its effect on the unit vectors
i, j, k, and describe it as a rotation.
- Consider the mapping f :
R2 → R3
defined by f(v) = (|v|, |v−i|,
|v−j|), recording the distances from v
to the origin and the two coordinate vectors.
- Find the Jacobian matrix of f at c = (½, 0),
and write the affine linear approximation for f(x,y) near this point.
- Explain the form of the Jacobian geometrically: how could you have
approximately predicted [ Df(c) ] without computing?
- For g(x,y,z) = x+y+z, use the Chain Rule compute the Jacobian matrix of
g∘f, and check that ∇(g∘f)(v) =
v⁄|v| +
v−i⁄|v−i| +
v−j⁄|v−j|.
Extra Credit: Show the sum of distances
g(f(v)) is minimized when the three vectors
v, v−i, v−j
are all 120° from each other.
- Let F : R3 → R3
be a differentiable vector field, and f : R3 → R a scalar function. Compute and simplify the following
second derivatives.
- curl grad f = ∇ × (∇f)
- div curl F = ∇ • (∇ × F)
- div grad f = ∇ • (∇f).
This is called the Laplacian of f, usually denoted ∇2f.
⊞
Solutions
1a. Subtracting any two points on the plane P gives a direction
along (parallel to) the plane,
for example c− a = (2,1,1)
and c− b = (1,1,2).
To get to any point on the plane,
move in these two directions from a base point c:
p(s,t) = (2,2,2) + s (2,1,1) + t (1,1,2)
= (2+2s+t, 2+s+t, 2+s+2t).
1b. A cross product u×w has direction orthogonal
to u and w, and length
|u| |w| sin θuw.
Thus, we get a normal vector, orthogonal to P,
as: n = (2,1,1) × (1,1,2) = (1,−3,1).
Now, for any point v = (x,y,z) on P, the direction
from c to v is orthogonal to n:
(v−c) • n = 0
⇔
v • n = c • n
⇔
x−3y+z = 2
1c. The area of a triangle with side lengths u,w enclosing angle θ is ½uw sin θ. Our triangle has side vectors u = c−a and w = c−b, so the area is half the length of the cross product n = u×w from in part (b):
area(△abc) = ½|n| =
|(1,−3,1)| = ½√11.
1d. To get a vector m from the origin directly to the plane P, take any
vector from the origin to P, such as c, and project it
to the normal direction n orthogonal to P.
The length of this projection is c dotted with
the unit normal vector n⁄|n|:
|m| =
| c • n⁄|n| |
=
|c•n|⁄|n|
=
2⁄√11
1e. We can think of the tetrahedron as a cone (or pyramid) whose base is △abc and whose altitude is the vector m.
The volume of any cone is ⅓ the base area times the height, which
we know from parts (c) and (d), so
the volume of our tetrahedron is: 1⁄3(√11⁄2)(2⁄√11) = 1⁄3 .
1f. The signed volume of the parallelpiped with edge vectors a, b, c is the determinant of the matrix of these three vectors (written as columns or rows: det(0,1,1; 1,1,0; 2,2,2) = −2,
so the absolute volume is 2.
The volume of a tetrahedron is 1⁄6 of this,
namely 1⁄3 as before.
2a. The vector F(x,y) = (2x,1) depends only on x, and gives the same vector everywhere
on a vertical line y = c: a vector pointing upward and away from the y-axis.
Wolfram Alpha gives:
2b. F looks like the gradient field of a function which slopes upward along
the y-axis, and gets steeper away from the y-axis. The contour curves f(x,y) = c
cut orthogonally to the vectors of F:
Lifting these contours to their proper heights makes the graph z = f(x,y):
2c. We recall some of the reasoning for the equivalence: F has a potential function if and only if curl F = 0.
First step. Assume F = ∇f for some potential function f.
Gradient Theorem:
The line integral of the rate of increase of f along a curve
is the total increase along the curve:
∮ ∇f(c) • dc = f(c(1)) − f(c(0)).
We can then compute f(a,b) by taking the line integral of
∇f(x,y) from a base point (0,0) to (a,b).
This line integral depends only on the endpoints (0,0) and (a,b),
and is independent of how c(t) travels between them.
Second step. Suppose F is path indepenent.
Given two paths c1, c2
between two points, we can follow the first path out and
the second one back to get a closed curve c.
Since the line integrals along c1 and
c2 are equal, their difference,
the line integral around c, is zero.
That is, the vector field is irrotational, having zero line integral around
any closed curve.
Third step. Curl Theorem:
The double integral of the rate of rotation of F over a region R is
equal to the total rotation of F around the boundary of R.
If F is irrotational, having zero total rotation around any
closed curve,
this means the rate of rotation at every point must be zero:
curl F(x,y) = 0 everywhere.
2d. For F(x,y) = (p(x,y), q(x,y)),
the curl F(x,y) =
∂q⁄∂x
−
∂p⁄∂y
measures how strongly a small paddlewheel at (x,y)
would be turned by a fluid flowing with velocity F.
A left-to-right increase in the vertical component of F
leads to a positive (couterclockwise) curl, hence the term
∂q⁄∂x;
a bottom-to-top increase in the horizontal component of F
leads to a negative (clockwise) curl, hence the term
−∂p⁄∂y.
2e. curl F =
∂⁄∂x(1)
−
∂⁄∂y(2x) = 0.
Thus F is irrotational, and must have a potential function f.
2f. We compute the potential function f(x,y) at (x,y) = (a,b)
as the line integral of F(x,y) = (2x, 1)
along the straight-line path c(t) = (at, bt)
for 0 ≤ t ≤ 1:
f(a,b) =
∮ F(c) • dc
=
∮ F(c(t)) • c'(t) dt
=
∮01 F(at, bt) • (at, bt)' dt
=
∫01 (2at,1) • (a,b) dt
=
∫01 2a2t + b dt
=
a2 + b.
Thus, f(x,y) = x 2 + y, and we can verify that ∇f = F.
3a. The quarter-turn ℓ0
takes i to j = (0,1),
and j to −i = (−1,0).
The corresponding matrix of column-vectors is:
3b. The z-axis quarter-rotation takes i to j, j to
−i, and k to itself. These output vectors are the column vectors of the matrix; and similarly for the x-axis quarter-rotation.
(Note: we are using the right-handed rotation, in which the thumb points
along the positive axis of rotation, and the fingers curl in the rotation direction.)
[ ℓz ]
| =
|
|
| [ ℓx ]
| =
|
|
3c. Multiplying the above gives the matrix of composite linear mapping
ℓz(ℓx(v)):
3d. The columns of the above matrix are the outputs of
the coordinate vectors:
the mapping takes i → j → k
→ i. That is, the equilateral
triangle with corners at i,j,k
is rotated around its center
⅓(i+j+k)
= (⅓,⅓,⅓). This means the linear mapping is a
120° rotation around the axis (⅓,⅓,⅓),
or equivalently the axis (1,1,1).
Note that composing in the other order,
ℓx ∘ ℓz,
gives a 120° rotation
around a different axis.
4a. The Jacobian at (x,y) = (½, 0) of:
f(x,y) = ( (x2+y2)1/2,
((x−1)2+y2)1/2,
(x2+(y−1)2)1/2 )
is:
[ Df(½, 0) ]
| =
|
x⁄(x2+y2)1/2
| y⁄(x2+y2)1/2
|
(x−1)⁄((x−1)2+y2)1/2
| y⁄((x−1)2+y2)1/2
|
x⁄(x2+(y−1)2)1/2
| (y−1)⁄(x2+(y−1)2)1/2
|
| =
|
|
4b. The point (x,y) = (½, 0) is on the line between 0
and i. We can predict the first column of the Jacobian
by seeing that moving along the x-axis will have exactly opposite
effects on the distances to 0 and i,
which accounts for the +1 and −1;
while it increases the distance to j at a positive
rate, accounting for the bottom entry.
As for the second column, moving vertically will affect
the distances to 0 and i only tangentially
(quadratically), so the first two rows are zero;
while the last row is negative, since the vertical motion
decreases the distance to j.
4c. The derivative of g(x,y,z) = x+y+z is the row-vector gradient:
∇g(x,y,z) = [1 1 1]. Multiplying a 1×3 matrix
by a 3×2 matrix gives a 1×2 matrix, whch is the gradient
of g(f(x,y)).
5. See the identities on [MT] p. 255.
5a. curl grad f = (0,0,0) for any function, which you can check from the definitions, coordinate-by-coordinate.
5b. div curl F = 0.
5c. div grad f = ∂2f⁄∂x2
+
∂2f⁄∂y2
+
∂2f⁄∂z2 .
⊞
Problems II: Parametrization, max/min, divergence and flux
These problems refer to the solid region R inside the
tilted cone
(x−z)2 + y2 ≤ z2, above z = 0, and below the plane
z = 2 − ½x.
The horizontal slice of the cone at z = c is a circle
(x−c)2 + y2 = c2
with radius c and center (c,0,c). These centers make
an axis along (1,0,1), which intersects the plane
z = 2 − ½x at (4⁄3, 0, 4⁄3).
The plane cuts the cone in an
ellipse with center (4⁄3, 0, 4⁄3),
and with extreme points
(4⁄3, 4⁄3, 4⁄3)
and (0,0,2).
-
Consider the region R inside the cone
(x−z)2 + y2 ≤ z2, above z = 0, and below the plane
z = 2 − ½x.
- Describe the horizontal slices of the bottom
surface
(x−z)2 + y2 = z2, holding z constant.
Sketch these slices in 3 dimensions,
forming the surface of a tilted cone.
Hint: The slices are circles.
- The edge curve of R is
given by the intersection of the lower surface
(x−z)2 + y2 = z2 and the upper surface
z = 2 − ½x.
Describe the xy-shadow under this curve.
Hint: It is an ellipse.
- Describe R as a z-simple region above its
xy-shadow.
- Use the above description of R
to set up integrals for
the volume and surface area of R.
- Parametrize R by "conical coordinates".
- Start by parametrizing the top ellipse
using polar coordinates with respect to
its major and minor axis vectors
(from the center to the extreme points),
then scale each point (x,y,z) on the ellipse
by a parameter s, getting (sx, sy, sz) for 0 ≤ s ≤ 1.
- Use the above parametrization to compute the volume and surface area of R.
(Add the areas of ellipse and cone.)
-
Let f(x,y,z) = (x-1)2 + y2 + z2, the square distance of (x,y,z) from (1,0,0).
- From the picture, estimate the
critical points of f on R, its boundary surfaces, and its corner boundary circle.
- Find the critical points
by solving ∇f(v) = 0 for v ∈ R.
- Find the constrained critical points by
solving ∇f(v) = λ ∇g(v), for a scalar λ and v ∈ ∂R, where g(x,y,z) = (x−z)2 + y2 − z2 (lower surface contour) or g(x,y,z) = 2 − ½x − z (upper surface countour).
- Find the critical points on the cone and disk surfaces of R by parametrizing them.
- Find the critical points on the corner edge circle (where the surfaces join)
by parametrizing it as c(t) and solving
f(c(t))' = 0.
Also consider the vertex point of the cone.
- Determine the max and min points and values of f on R.
- For the above f, let F = ∇f.
- Sketch this vector field.
- Find div F and curl F, and
explain why the answers are expected.
- Determine the integral of div F over R.
- Determine the flux of F through the boundary surfaces of R.
Explain why the result is the same as the previous integral.
⊞
Solutions.
1a. The axis vectors of the ellipse are:
a = (4⁄3,4⁄3,4⁄3) −
(4⁄3,0,4⁄3)
=
(0,4⁄3,0),
b = (2,0,0) −
(4⁄3,0,4⁄3)
= (2⁄3,0,−4⁄3).
Then:
P(r,t,s) = s ( (4⁄3,0,4⁄3) + r cos(t) a + r sin(t) b )
=
(4⁄3 s + 2⁄3 s r sin(t),
4⁄3 s r cos(t),
4⁄3 − 2⁄3 s r sin(t)).
⊞
Problems III: Interesting solids.
Class notes
- (p. 270 #7) A vertical cylinder of radius r (representing a tree trunk)
has a wedge W lying above the xy-plane
and below the plane through the y-axis at angle θ from horizontal.
- Find the volume of the wedge as the double integral ∬ f(x,y) dA.
Interpret the inner integral as an area.
Hint: Compute the integral in the most advantageous order,
dA = dy dx or dx dy.
- Write the above volume a triple integral ∫∫∫ dz dy dx,
specifying W as the points (x,y,z) with a1 ≤ x ≤ a2, b1(x) ≤ y ≤ b2(x),
c1(x,y) ≤ z ≤ c2(x,y).
- Write the above triple integral in five other ways corresponding
to dV = dz dx dy = dy dz dx = dy dx dz = dx dz dy = dx dy dz.
- Define the solid S as the region above the surface z = x2 − 2
and under z = 2 − y2.
- Describe the boundary curve where the two surfaces meet, first by equations in x,y,z; then give it as a parametrized
curve c(t).
- Find the most coordinate convenient plane where we can project the
solid S, and describe its shadow.
Set up and evaluate the triple integral giving the volume of S.
- For each solid below, determine:
- Volume using rectangular coordinates (taking an appropriate order of integration)
- Volume using a cylindrical (or elliptic cylindrical) parametrization
- Surface area using any convenient parametrization.
The solids are constructed from specified
slices at a given position.
- Pillow solid: For −1 ≤ z ≤ 1,
the horizontal slice at height z is the ellipse
(x⁄(z2−1))2 + y2 ≤ 1
with width 2 in the y-direction, width 2|z2−1| in the x-direction.
- Trilobite: Base is unit circle x2 + y2 ≤ 1 in xy-plane,
vertical slices above x = a are squares defined by
|y| ≤ √(1−a2),
0 ≤ z ≤ 2√(1−a2).
Hint: This solid is an intersection of a vertical circular
cylinder and a horizontal parabolic cylinder:
its surface can be constructed out of paper.
- Extra Credit:
Consider the circular cylinder
C1 consisting of the points within distance 1 from the x-axis;
and similarly C2 for the y-axis.
Now define the solid S = C1 ∩ C2,
the region inside both cylinders.
- Give equations for the two cylinders, and describe S by inequalities of the
form:
a1 ≤ z ≤ a2
b1(z) ≤ y ≤ b2(z)
c1(z,y) ≤ x ≤ c2(z,y)
Hint: The xy-shadow of S is the largest horizontal cross-section.
Find the level z = c of this cross-section, and plug z = c into the equations.
- Find the volume of S by evaluating a triple integral.
⊞
Solutions
1a. See picture on p. 270 #7.
The wedge is defined by x2 + y2 ≤ r
and 0 ≤ z ≤ mx, where m = tan θ.
Writing the xy-shadow in y-simple form: -r ≤ x ≤ r, √−(r2−x2) ≤ y ≤ (r2−x2); and also 0 ≤ z ≤ mx.
The volume integral becomes:
∫r0 ∫√(r2−x2)
−√(r2−x2) mx dy dx
= 2⁄3 mr3.
The inner integral is the area of the rectangular vertical cross-section
of the solid in the plane y = c, where c is a fixed value of x.
2 & 3a. See class notes.
⊞
Problems IV: Line, double and triple integral practice.
HW: Work out book examples before looking at solutions
- [MT] 7.2 Examples 1, 2, 8, p 360.
- [MT] 7.3 Example 3, p 380.
- [MT] 7.5 Examples 1, 2, 3, p 394.
- [MT] 7.6 Examples 1, 4, p 401; Exercise 1, p 411.
Problem sheets
16-9 #12. Consider the region R bounded by the four circles:
x2 + y2 = 2x,
x2 + y2 = 6x,
x2 + y2 = 2y,
x2 + y2 = 8y.
This region is adapted to a coordinate system called
the Apollonian or inversion coordinates
( parabolic pencil of circles),
whose coordinate grid curves
are circles centered on the x-axis or the y-axis,
all of which pass through the origin.
This is given by the inversion transformation:
(x,y) = T(u,v) =
(2u⁄(u2+v2) ,
2v⁄(u2+v2)).
This is a kind of "reflection" across the circle
of radius √2 ;
T takes a point (u,v) with radius r to a point
(x,y) with the same polar angle and
radius 2⁄ r. That is, T switches the
inside and outside of the circle, with the origin going to infinity.
- Write the boundary circles of R in the form
(x−a)2 + (y−b)2 = r2
to show they are indeed Apollonian grid circles.
(Hint: Complete the square.)
- Show that the line u = c in the (u,v)-plane
corresponds to the circle (x−a)2 + y2 = a2 in the (x,y)-plane, where a = 1⁄c;
and similarly, the line v = c corresponds to
the circle x2 + (y−a)2 = a2.
In the picture above, identify the blue circles
as u = c and the red circles as v = c, for appropriate constants.
- Parametrize the region R as the points T(u,v) for
a1 ≤ u ≤ a2 ,
b1 ≤ v ≤ b2 , for appropriate constants.
- Compute the Jacobian matrix [DT(u,v)]
and its determinant |det[DT(u,v)]|
= |∂(x,y)⁄∂(u,v)|,
as it is written in the Change of Variables Formula.
- Use Change of Variables to rewrite the
following integral in uv coordinates:
∬R 1⁄(x2+y2)2 dx dy.
- Verify that the inversion mapping T is its own inverse:
that is, if (x,y) = T(u,v), then the variables u,v satisfy
the equation (u,v) = T(x,y).
⊞
Solutions
16-9 #12(a). The equation x2 + y2 = 2ax
is equivalent to (x−a)2 + y2 = a2, a circle centered at (a,0) with radius a,
so that the circle contains the origin.
The first two circles have centers (1,0) and (3,0),
and similarly the second two have centers (0,1) and (0,4).
12(b). The line u = c is parametrized by (c,t) for t ∈ R, which corresponds to (x,y) = T(c,t).
It is easy to verify that
x = 2c⁄(c2+t2)
and y = 2t⁄(c2+t2)
satisfy x2 + y2 = 2⁄c x, which is equivalent to the circle equation. Similarly for v = c.
12(c). From the picture, we see
R = { (x,y) = T(u,v) for 1⁄3 ≤ u ≤ 1 ,
1⁄4 ≤ v ≤ 1 }
12(d). You can do this with Wolfram Alpha by pasting partial results
into a scratch file, then pasting back into W|A.
Result: the Jacobian determinant (streching factor) is
|−4⁄(u2+v2)2|
= 4⁄(u2+v2)2 .
12(e). The integral reduces
to ∫1⅓ ∫1¼
(u2+v2)2
1⁄4(u2+v2)2
dv du
= 1⁄8.
12(f). Use W|A to solve the system of equations
x = 2u⁄(u2+v2)
and y = 2v⁄(u2+v2)
for the variables u,v. Keep in mind that the equations
are symmetric upon switching u,v and simultaneously switching x,y,
so a solution for u can be switched into a solution for v.
Note: This is also evident from the polar description
of T as an inversion across a circle of radius √2.
Furthermore, we can write T as a conjugate inversion
in the plane of complex numbers.
That is, we let (x,y) correspond to
the complex variable z = x + iy
with conjugate
z = x − iy;
then T(x,y) corresponds to the mapping
z ↦ 2⁄z
=
2z⁄zz.
16-6 #13. The solid bounded by planes y + z = 4, y = 0, z = 0,
and the parabolic cylinder y = 4−x2.
First examine the boundary surfaces involving only x,y: the
back wall y = 0, and the curving front wall y = 4−x2.
Next consider the roughly horizontal surfaces involving z:
the floor z = 0 and the slanted ceiling z = 4−x.
This solid is clearly z-simple: a single floor
and ceiling in the z-direction,
above an xy-shadow contained between curves
y = 0 and y = 4−x 2.
Answer: S = { (x,y,z) with −2 ≤ x ≤ 2, 0 ≤ y ≤ 4−x 2, 0 ≤ z ≤ 4−x }.
The volume is 128⁄ 5 .
16-6 #19. The solid bounded by planes x + y + z = 2 and x = 0,
and the parabolic cylinders y = z2 and z = y2.
First examine the boundary surfaces involving only y,z: the
parabolic cylinders cutting each other in a squashed cylinder.
Next consider the surfaces involving x:
x = 0 and x = 2 − y − z, which cut off the ends of
the squashed cylinder.
This solid is clearly x-simple: a single floor
and ceiling in the x-direction,
above a yz-shadow contained between the parabolas
z = y 2 and y = z 2.
Answer: S = { (x,y,z) with 0 ≤ z ≤ 1, √z ≤ y ≤ z 2, 0 ≤ x ≤ 2 − y − z }.
The volume is 11⁄ 30 .
12/8 Recitation 14. WHW 9 Electric & magnetic fields.
12/14 Final Exam
Wed Dec 14, 2022 at 5:45-7:45pm in our usual classroom EB 3400.
Extra: Introduction to differential forms
⊞
Differential 1-form = covector field
- Class notes
- For a vector space V = Rn, a
covector or dual vector
is a linear function φ : V → R.
Vector v corresponds to
covector φ = v* defined by dot product
v*(x) = v • x.
Dual vector space
V* = {covectors φ} = {v* for v ∈ V}.
Add covectors, mult by scalar.
- Linear approximation of
f : Rn → R
is f(a+h)
≈ f(a) + Dfa(h)
for linear Dfa : Rn → R
Jacobian matrix [Dfa] = ∇f(a) gradient.
Dfa(h) = ∇f(a) • h, so Dfa = ∇f(a)*
- Derivative Df gives a covector at each point a ;
it is a covector field, denoted df = Df
analogous to vector field F giving
a vector at each point
- Differential form (of degree 1)
means any covector field φ, not necessarily df.
- Example of R2:
Constant vector fields x̂ = (1,0),
ŷ = (0,1)
are basis for general F = (p,q)
= p(x,y) x̂ + q(x,y) ŷ
Constant differential forms
dx(h) = dx(hx , hy) = hx ,
dy(h) = dy(hx , hy) = hy
are basis for general φ =
p(x,y) dx + q(x,y) dy = F*
F is conservative if F = ∇f;
analogously, φ is exact if φ = df
Not every F is conservative,
so not every φ is exact
- Define line integral of a diff form φ = p dx + q dy
over curve
c(t), t ∈ [0,1]:
∫c φ =
∫10 φc(t)(c'(t)) dt
=
∫10 p(x(t),y(t)) x'(t)
+ q(x(t),y(t)) y'(t) dt.
This is another notation for
line integral of F = (p,q),
since φc(t)(c'(t))
= F(c(t)) • c'(t).
Reading: [MT] Ch 8.5 pp. 476-8.
HW: Finish WHW 8 due 4/17 below.
- In R2,
consider the 1-form (covector field)
φ(x,y) = dx + x dy.
Here we use the basis 1-forms
dx(x,y)(hx , hy)
= hx ,
dy(x,y)(hx , hy) = hy.
- Give the dual vector field F
such that φx(h)
= F(x) • h.
- Sketch φ as a field of little parallel-lines icons,
along with the vector field F.
- In R3,
let f(x,y,z) = x2 − yz.
- Compute the gradient vector field ∇f
and its covector field df = (∇f)*.
- For the curve c(t) =
(1, t, t2) for t ∈ [0,1], compute the line integral ∫c ∇f • dc,
and the equivalent ∫c φ.
⊞
Solutions
1. The dual of dx is (1,0), of dy is (0,1), so
the dual of φ = dx + x dy is F(x,y) = (1, x).
Since the parallel lines show (scaled) level curves
φ (x,y)(b>h) = F • h = 0 and 1, longer F corresponds to closer lines in φ.
2a. We have ∇f(x,y,z) = (2x, −z, y) and
its dual φ = df = 2x dx − z dy + y dz.
2b. The line integral of φ = df is another notation for that of ∇f:
∫c φ
=
∫1t=0
φc(t)(c'(t)) dt
=
∫1t=0
φ(1,t,t2)(0,1,2t) dt
=
∫1t=0
2(1)(0) − (t2)(1) + (t)(2t) dt
= ··· =
f(1,1,1) − f(0,0,0).
⊞
Exterior derivative, 2-forms, wedge product, Stokes Thm.
- Class notes
- Recall differential 1-forms
- Natural linear approximation of function
f : Rn → R
near point x, for small h:
f(x+h) ≈ f(x) + dfx(h)
using 1-form φ = df = (∇f)*
- Field of covectors: at each x = (x1 , . . . , xn) ∈ Rn,
linear fun φx = dfx : Rn → R
- In coordinates df = ∂f⁄∂x1dx1 + · · · +
∂f⁄∂xndxn ,
where dxi(h) =
dxi(h1 , . . . , hn) = hi
- Reverse derivative with line integral ∫c φ
= ∫c df
= ∫ ∇f • dc
= f(c(1)) − f(c(0))
- Arbitrary 1-form φ = p1 dx1
+ · · · + p1 dxn corresponds to vector field F = (p1 , . . . , pn)
by covector duality operation φ = F*,
φx(h) = F(x) • h
- Arbitrary 1-form φ is a derivative φ = df
⇔
φ = F* with F
conservative, F = ∇f.
- For concreteness, we work with n = 3,
x = (x, y, z) ∈ R3,
but everything works for Rn
-
Repeat above constructions:
define exterior derivative of
arbitrary 1-form φ
= p dx + q dy + r dz
- Natural measure of 1-form φ is circulation:
line integral over loop
∮c φ
-
Derivative of φ is rate of circulation near each point x:
ηx =
dφx =
limc→x
∮c φ ⁄(area enclosed by c)
- To describe this derivative of a 1-form,
need a new kind of field called a 2-form.
- Rate of circulation near x depends on
the choice of c → x
For vectors h1 , h2
∈ Rn and small Δx, Δy → 0,
define parallelogram boundary loop c:
x
to x + Δx h1
to x + Δx h1
+ Δy h2
to x + Δy h2
back to x
-
Define 2-form dφx(h1,h2) as rate of circulation for this loop c
- Geometric properties of 2-form
ηx(h1,h2) = dφx(h1,h2)
- Similar to determinant det(h1,h2,h3)
- Alternating: ηx(h2,h1)
= −ηx(h1,h2)
since switching vectors gives opposite circulation
- Bilinear: ηx(h1,h2)
is linear function of each parallelogram vector variable
h1, h2
since in circulation integral,
p,q,r components of φ may be replaced by linear approximations
- A general 2-form η is any field of alternating bilinear functions, not necessarily dφ
- Since φ = df has zero circulation by Gradient Theorem,
get derivative η = dφ = 0
d(df) = 0
- Coordinate basis of 2-forms
- Combine 1-forms φ, ψ using
wedge product (exterior product)
2-form η = φ ∧ ψ :
(φ ∧ ψ)(h1,h2)
=
φ(h1)
ψ(h2)
−
φ(h2)
ψ(h1)
Ordinary product of linear functions
gives bilinear function,
switching variables gives alternating
- Hence φ ∧ φ = 0
- Wedges of coordinate dx, dy, dz give only 3 basis 2-forms
dx∧dy, dx∧dz, dy∧dz
since dy∧dx = −dx∧dy and
dx∧dx = 0
- Standard formula for arbitrary 2-form:
η =
p dy∧dz − q dx∧dz + r dx∧dy
For R3, this corresponds
to vector field F = (p, q, r)
by Hodge star operation:
η = F☆
- Compute exterior derivative η = dφ in coordinates
- Lemma: For any function
f : R3 → R and any 1-form ψ with dψ = 0,
we have d(f ψ) = (df) ∧ ψ.
- Corollary: If φ = F* for F = (p,q,r), then
dφ = G☆ for
G = curl F:
d(F*) = (curl F)☆
In this sense, exterior derivative of a 1-form is equivalent
to curl of a vector field.
Proof: Compute
dφ = d(p dx + q dy + r dz)
= dp ∧ dx + dq ∧ dy + dr ∧ dz.
- Stokes Theorem
- Natural measure of a 2-form η is its integral over a surface S ⊂ R3
S defined by parameter function
P : S* → S, P(u,v) = (x(u,v), y(u,v), z(u,v))
- Define surface integral:
∬S η =
∫∫(u,v)∈S* ηP(u,v)(∂P⁄∂u ,
∂P⁄∂u) du dv
- The surface integral of a 2-form is equal to the flux integral of the corresponding vector field
If η = F☆,
then ∬S φ =
∬S F • dS
- Denote boundary curve of a surface S as
c = ∂S
- Stokes Theorem: The circulation of a 1-form φ
around a loop c
is equal to the integral of the rate of circulation dφ
over any surface S enclosed by the loop:
∫∂S φ =
∬S dφ .
- By the above correspondences, this is equivalent
to the Curl Theorem
∮ F • dc =
∬S (curl F) • dS.
The differential forms version is more geometrically natural,
not using cross-product tricks.
It is also much more general, working on arbitrary Rn,
and even on an n-manifold.
- The notation ∂S
is meant to suggest that the boundary operation
c = ∂S is "dual" to the exterior derivative dφ. This is the beginning of deRham cohomology theory.
- Since η = dφ is flux-free,
we have dη = 0:
d(dφ) = 0
Reading: [MT] Ch 8.5, including odd-numbered exercises (answers in back).
HW: Finish WHW 8, due 4/17 below.
- Forms on R2
- Show that the only bilinear alternating functions on R2 are multiples of
the determinant:
c det
= c dx ∧ dy.
Thus any 2-form can be written η = f(x,y) dx ∧ dy
for a general function
f : R2 → R.
- Adapt the above definition of ∬S η
to the case where the surface S is a parametrized region of R2, and compare it with our previous integral of a function f(x,y) over S:
∬S f(x,y) dx ∧ dy =
∬S f(x,y) dx dy.
- Using the above definitions,
show that d(p dx + q dy) = curl(p,q) dx ∧ dy.
- Recall that we sketch 1-forms using little parallel-lines icons, and we can similarly draw 2-forms by transverse
pairs of parallel lines with orientation:
Note that close-together contour lines indicate a rapidly-increasing function.
Problem: On one set of axes, sketch the 1-form
φ = x2 dy;
on another sketch d(x2) together with dy;
and on a third sketch the derivative
dφ = 2x dx∧dy.
- For R3,
carefully unravel the coordinate
definitions in the above notes
to verify the following formulas.
- If φ = p dx + q dy + r dz = F* = (p,q,r)*, then:
dφ
= (curlyz F) dy∧dz
+ (curlxz F) dx∧dz
+ (curlxy F) dx∧dy
= (curl F)☆,
where the coefficients denote 2-dimensional curl
of F projected to a coordinate plane.
- If φ = F*, then
∫c φ =
∫ F • dc.
- If η = G☆,
then
∬S η = ∬S G • dS.
⊞
Solutions
1a. Consider an alternating bilinear function:
d(h1,h2)
=
d( h1xi + h1yj ,
h2xi + h2yj ),
and expand using the properties to get
d(h1 , h2)
=
d(i,j) (h1xh2y
− h2xh1y)
=
c (dx(h1) dy(h2)
−
dx(h2) dy(h1))
=
c (dx ∧ dy)(h1,h2)
for c = d( i, j).
1b. By definition, for a parametrization
P : S* → S,
P(u,v) = (x(u,v), y(u,v)):
∬S f(x,y) dx ∧ dy
=
∬S* f(P(u,v))
(dx∧dy)(∂P⁄∂u ,
∂P⁄∂u)
du dv
=
∬S* f(P(u,v))
det(∂P⁄∂u ,
∂P⁄∂u)
du dv
=
∬S* f(x,y) dx dy.
The last equality is by the Substitution Formula.
1c. Expand d(p dx + q dy) =
dp ∧ dx + dq ∧ dy
=
(∂p⁄∂x dx +
∂p⁄∂y dy) ∧ dx
+ · · ·
2. The formulas are all analogs of the 2-dimensional computations in #1, using the definition of the derivative dφ, the line integral ∫c φ from the previous lecture, the Hodge star G☆, and the integral of a 2-form ∬S η.
In (c), the computation of η(∂P⁄∂u , ∂P⁄∂v)
ends up equal to
the triple product
G • (∂P⁄∂u
× ∂P⁄∂v),
because cross product is also alternating and bilinear,
like η.
⊞
Stokes Theorem for
k-form on Rn:
∫∂S ω = ∫S dω.
- Class notes
- Points x = (x1 , . . . , xn) ∈
Rn,
coordinate 1-forms
dxi(h) = hi
for h = (h1 , . . . , hn)
∈ Rn
- k-form on Rn:
field of alternating k-linear functions ωx(h1 , . . . , hk)
for hj ∈ Rn
- Basis of coordinate k-forms
dxi1 ∧ · · ·
∧ dxik
corresponds to multi-indexes
1 ≤ i1 < · · ·
< ik ≤ n:
(dxi1 ∧ · · ·
∧ dxik)(h1 , . . . , hk)
=
∑ ±hj1,i1 · · ·
hjk,ik ,
where hj
= (hj,1 , . . . , hj,k),
and (j1 , . . . , jk)
runs over all permutations of {1, . . . , k},
with sign ±1 depending on whether the permutation
performs an even or odd number of transpositions.
General k-form is:
ω =
∑ fi1···ik
dxi1 ∧ · · ·
∧ dxik
where the sum runs over all
1 ≤ i1 < · · ·
< ik ≤ n.
- Integrate a k-form ω over k-dim surface S ⊂ Rn, parametrized
by P : S* → S,
P(u1 , . . . , uk)
= (x1 , . . . , xn)
∫S ω
= ∫···∫u∈S*
ωP(u)(∂P⁄∂u1 , . . . ,
∂P⁄∂uk) du1··· duk
- Wedge product of 1-form φ and k-form ω
is (k+1)-form υ = φ ∧ ω:
υ(h1 , . . . , hk+1)
=
(φ ∧ ω)(h1 , . . . , hk+1)
=
∑j (−1)j−1 φ(hj) ω(h1 , . . . ,
ĥj ,
. . . , hk+1),
where j = 1, . . . , k+1, and ĥj indicates a deleted input vector.
- Exterior derivative of k-form ω is (k+1)-form υ = dω
Geometrically, dω(h1 , . . . , hk+1) measures rate near each point of:
(
| integral of ω out of
| k-dim boundary surface of parallelopiped
| spanned by
h1 , . . . , hk+1
|
| )
| /
| (
| (k+1)-dimensional
| volume of parallelopiped)
|
| )
|
Algebraically, for
ω = ∑ fi1···ik
dxi1 ∧ · · ·
∧ dxik , we compute:
dω =
∑ dfi1···ik ∧
dxi1 ∧ · · ·
∧ dxik
- Stokes Theorem: For a k-form ω and a
(k+1)-dimensional surface S in Rn,
with k-dimensional boundary ∂S, we have:
∫∂S ω =
∫S dω .
- Proposition: For any k-form ω, we have:
d(dω) = 0.
Proof:
Let R be a (k+2)-dimensional parallelopiped;
then its boundary S = ∂R is a closed
(k+1)-dimensional surface
without boundary: ∂S = ∂(∂R) = ∅,
the empty set.
Applying Stokes twice:
∫R d(dω)
=
∫∂R dω
=
∫∂(∂R) ω
=
∫∅ ω
= 0
Since this holds for any R, we must have d(dω) = 0.
Alternatively, the geometric definition
of d(dω) is a limit of such
integrals ∫∂R dω = 0,
so d(dω) = 0.
- Poincare Lemma: We say a form
ω over Rn is exact if ω = dη for some η.
Then ω is exact if and only if dω = 0.
This means that if ω has zero derivative,
then it has a "potential" form η, generalizing
the usual scalar potential and the vector potential of a vector field.
- Volume forms. Consider a k-manifold S ⊂ Rn.
The inner product induces a volume n-form ω
defined by:
ω(v1, . . . , vn)
=
det[v1, . . . , vn],
where [v1, . . . , vn] is
an n × n matrix of column vectors
(with respect to the standard orthonormal basis).
This also induces a volume k-form η on S, defined by
the Gramian determinant of the k × k dot-product matrix:
η(v1, . . . , vk)
=
G(v1, . . . , vk)
=
±√det[vi • vj]i,j .
Another volume form is obtained by a parametrization
P : S* → S:
∂P =
(∂P⁄∂u1)*
∧ · · · ∧
(∂P⁄∂uk)*
=
G[DP] η,
where G[DP] is the Gramian of the tangent vectors.
(Proof: Evaluate both sides on the basis of
tangent vectors DP.)
Finally, the Hodge star operator takes a 1-form
φ = F* to an (n−1)-form ☆φ
by the rule:
γ ∧ ☆φ =
G* ∧ ☆(F*)
= (G • F) ω.
Now suppose k = n−1, and S is a level
curve of a function g : Rn → R.
The we have the volume form:
☆dg = ☆(∇g)*
=
|∇g| η.
(Proof: Evaluate left and right sides of
|∇g|2 ω = dg ∧ ☆dg
?=? |∇g| dg ∧ η
on a basis of Rn
consisting of ∇g and an orthonormal basis of
tangent vectors of S.)
Example: For a surface
S ⊂ R3 defined as a graph z = f(x,y),
we have P(x,y) = (x, y, f(x,y)) ∈ S,
as well as g(x,y,z) = z − f(x,y) = 0 on S.
Then ∂P⁄∂x ×
∂P⁄∂y
= ∇g = (−∂f⁄∂x ,
−∂f⁄∂y , 1),
and
∂P
=
(∂P⁄∂x)* ∧
(∂P⁄∂y)*
=
☆(∂P⁄∂x ×
∂P⁄∂y)*
=
☆(∇g)*
= ☆dg
= √(1 + (∂f⁄∂x)2 + (∂f⁄∂y)2) η.
Reading: [MT] Ch 8.5, including odd-numbered exercises (answers in back).
HW:
- In R3, for any 2-form η = G☆,
work out the above definitions to show that:
dη = (div G) dx∧dy∧dz
= (div G) det.
- In R3, for any 3-form ω = f(x,y,z) dx∧dy∧dz
and 3-dimensional region R,
work out the definitions to show that:
∫R ω = ∫R f(x,y,z)
dx dy dz.
- Using the above notes, show without computation
that div(curl F) = 0 for any vector field
F on R3.
- In R4, for a vector field
F(x,y,z,w) = (p, q, r, s),
use differential forms to compute an analog of curl for F.
Hint: Compute the 2-form d(F*).
This is in no way dual to a vector field, but
the coefficients give the data needed to compute
the rate of circulation.
⊞
Solutions
1. As for the R2 case in the previous HW,
expand
d( p dy∧dz − q dx∧dz + r dx∧dy )
=
dp ∧ (dy∧dz) − dq ∧ (dx∧dz)
+ dr ∧ (dx∧dy)
using
dp = ∂p⁄ ∂x dx + ∂p⁄ ∂y dy + ∂p⁄ ∂z dz, etc.,
until it is in terms of dx∧dy∧dz.
The coefficient is div G.
2. As for the R2 case, this reduces to the Substitution Formula.
3. This follows from d(d(F*)) = 0.
4. For φ = F* = p dx + q dy + r dz + s dw,
a 1-form on R4,
expand out the exteior deriative
dφ =
dp ∧ dx + dq ∧ dy + dr ∧ dz + ds ∧ dw,
remembering that dp = ∂p⁄ ∂x dx + · · · + ∂p⁄ ∂w dw , and dy∧dx = − dx∧dy, etc.
The 6 coefficients of the basis
dx∧dy , dx∧dz , dx∧dw , dy∧dz , dy∧dw , dz∧dw
form a vector in R6 which is the analog of curl, but it is much more natural to think of this as a 2-form on R4.
Weekly Homework to hand in
-
⊞
WHW 1 due Wed 9/7: Geometry theorems via vector algebra
Choose ONE of the following theorems from Euclidean plane geometry,
and give an algebraic proof using vector operations.
(Turn in only ONE: the first is straightforward, the second more challenging.)
-
Thales' Theorem: For a triangle inscribed in a half-circle with one side equal to the diameter, the angle opposite the diameter is a right angle. That is, if points A,B,C lie on a circle with diameter AB, then angle ∠ACB = 90°.
- Angle Bisector Ratio Theorem: In a triangle, an angle bisector cuts the opposite edge into segments whose ratio is equal to the ratio of the two sides of the angle. That is, if triangle △ABC has angle bisector AD for a point D on the edge BC, then DB⁄DC = AB⁄AC .
Guide
- Here is a model proof for HW 1 #3 above.
- In the statement of the Theorem, clearly identify the hypothesis (the setup or "if" part which is assumed), and the conclusion (the payoff or "then" part to be deduced).
- Start the proof by translating the geometric terms into vector algebra, so that
the assumed hypothesis and the desired conclusion become equations.
Introduce vectors u,v pointing along the relevant sides of the triangle.
Translate statements about angles into vector equations by applying
u•v = |u||v|cos(θ).
No need to use coordinates:
the proofs are cleaner if you write each vector as a single letter, not (x,y).
Keep pictorial geometric arguments and trigonometry to a minimum:
this is an exercise in algebra.
- Carefully expand and simplify the left side of the desired equation using known properties of the dot product such as commutative and distributive laws,
as well as the hypothesis equality.
Once you have transformed the left side into the right side,
you have proved the theorem.
- In your final draft, start by stating the Theorem.
You may include a suggestive drawing, but your argument should be readable without it.
For a valid logical structure, do not simplify the desired equation into 0 = 0.
(Deducing a true statement does not verify the assumption you started from!)
Rather, organize your scratchwork into a chain of equalities from the left side to the right side,
where each equality in the chain is clearly true, not a tentative or desired equality.
- You may turn in hand-written papers, but the best tool to
present mathematics is TeX, which you can download free.
- WHW 2 due Mon 9/12:
Plane linear mappings and their matrices
(tex source file)
-
⊞
WHW 3 due Mon 9/19: Graph landscape
Invent a function f : R2 → R whose graph z = f(x,y) makes an interesting landscape
- To start exploring, graph the one-variable bump functions:
g(r) =
e−r2, 1⁄(1+r2) , 1⁄(1+|r|) ,
max(0, 1−|r|).
Example: WolframAlpha: plot e^(-r^2) for r=-2.5 to 2.5).
Turn each of these curves into a 3D mountain z = f(x,y) by substituting r = √(x2+y2), obtaining
the surface of revolution f(x,y) = g(√(x2+y2))
(W|A: plot3d 1/(1+sqrt(x^2+y^2)) for x = -2 to 2 and y = -2 to 2).
In GeoGebra, you can also draw a rectangular plot, or polar coordinate plot,
such as z = 2e−r2 = 2e−(x2 + y2) over the unit circle.
Experiment with other g(r).
- Be artistic. More techniques:
- A long ridge with a given cross-section: z = g(x) or g(y).
What is going on with: f(x,y) = exp(−(y−sin(x))2)?
- Experiment with the effect of taking
f(x−a, y−b) and f(ax, bx)
for some constants a,b.
- Superimpose several mountains and valleys
by adding and subtracting such functions.
- Try multiplying functions:
exp(−x2−y2) ∗ sin(x)
scales the waves of f(x,y) = sin(x) by the bump function
exp(−x2−y2).
- Taking the square root of a positive function produces
cusps at the zero points: egg-crate function
f(x,y) = −√(2−cos(x)−cos(y)).
- Functions of the form f(x,y) = g(x) + h(y) can be visualized by
drawing the backbone z = g(x) above the x-axis, then hanging identical rib curves z = h(y)
perpendicularly to the backbone, with the rib above a given line x = c being raised by g(c).
- Find the (x,y) window which best shows the
interesting features of your function.
In addition to the graph, find the contour plot and gradient vector field ∇f.
Hint: It is probably best to draw the gradient vectors by hand onto the contour plot:
remember that the closer the contours, the longer the gradient (since its length is the uphill slope).
- Once you have your function, locate its critical points,
which are the flat points on its graph.
These are the values (x,y) where ∇f(x,y) = (df⁄dx,df⁄dy) = (0,0). (W|A will solve the equations for you most of the time.)
Classify each as a min (valley like z = x2+y2),
max (hill like −x2−y2)
or saddle (like x2−y2),
or possibly a ditch or ridge (like ±x2).
Note how each of these points appears on the countour plot and gradient field.
- Software tips.
Turn in:
- For your most interesting-looking function f(x,y), a printout of the graph, countour plot and gradient vector field
(copy and paste the W|A images).
- A clear description of how you built up the
formula for your function f(x,y) from its basic elements.
Explain how each part of the formula controls some feature of the picture, not just "I tried this formula at random, and the computer drew that picture, I don't know why."
- Location and classification of the critical points.
(At least give approximate location if solving the equations is too difficult.)
- WHW 4 due Mon 9/26: Optimizing a sawmill (tex)
-
⊞
WHW 5 extended to Fri 10/7: Conservative Vector Field Theorem
All of the following conditions are equivalent for a vector field F(x,y):
- F is conservative: there is a potential function f with F = ∇f.
- F is path independent:
the line integral between two fixed points
is the same for any choice of path between them;
for any paths c1, c2
with c1(0) = c2(0) = (a,b)
and c1(1) = c2(1) = (c,d),
we have
∫ F(c1)•dc1
= ∫ F(c2)•dc2
.
- F has zero circulation:
the line integral around any loop is zero; if c(t) for 0 ≤ t ≤ 1 is a closed curve with c(0) = c(1),
then ∮ F(c)•dc = 0.
- F is irrotational: curl F(x,y) = 0 at all points (x,y).
Problem: Write up the proof of this Theorem,
showing the three forward implications
(i) ⇒ (ii) ⇒ (iii) ⇒ (iv)
and the three reverse implications
(iv) ⇒ (iii) ⇒ (ii) ⇒ (i).
You may quote and use the theorems discussed in the daily notes.
Hints: Model solution ( revised; tex)
- We discussed the Conservative Vector Field Theorem in Lect 11.
- As in all proofs, start by stating the Theorem at the beginning, briefly: you don't need to include definitions of your terms.
- In each section of the proof, start by stating the hypothesis (the setup assumed) and finish by stating the conclusion deduced.
- You may use any theorems we have covered,
explicitly quoting the results you use.
- (i) ⇒ (ii): Use the Gradient Theorem ∫ ∇f(c)•dc = f(c(1)) − f(c(0)).
- (ii) ⇒ (iii): For a curve c(t) with
0 ≤ t ≤ 1, and its reverse curve
c−(t) = c(1−t),
show that ∫ F(c−)•dc−
= −∫ F(c)•dc
using the single-variable Substitution Theorem:
for u = g(t) = 1−t, we have:
Then, for a given loop, cut it into two paths from the beginning point to the half-way point.
- (iii) ⇒ (iv): Use the Circulation limit definiton of curl from Lect 11: curl F(x,y) =
limc→(x,y)
∫ F(c)•dc⁄area(c) .
- (iv) ⇒ (iii): Use the Curl Theorem (Fund Thm for Curl)
∬D curl F(x,y) dx dy
=
∮ F(c)•dc,
where c is the counterclockwise boundary of a region D.
Note: This proves (iii) only for curves which bound a region D,
but you may ignore more complicated curves (such as self-intersecting ones).
- (iii) ⇒ (ii): Easy converse of (ii) ⇒ (iii).
- (ii) ⇒ (i): Here you need a real argument.
Assume that f(x,y) = ∮ F(c)•dc, the integral of F = (p,q)
along c from (0,0) to (x,y),
is a function depending only on (x,y), not on the path chosen;
then you must show ∂f⁄∂x = p
and ∂f⁄∂y = q.
For the first, compute f(x,y) along the rectangular path from (0,0) to (0,y) to (x,y),
and use the single-variable First Fundamental Theorem of Calculus,
applied to the variable x as a bound of integration.
That is, if you parametrize the path correctly, f(x,y) = ∫0x p(t,y) dt + (term without x),
and ∂f⁄∂x(x,y) = p(x,y),
as desired. So far, this will work for any F.
But now, assuming path independence, you get the same f(x,y)
from the other rectangular path,
and show the other formula
∂f⁄∂y = q(x,y).
- WHW 6 due Wed 11/2:
Substution Formula and areas
(tex).
⊞
Solution
16-9 #12. Consider the region R bounded by the four circles:
x2 + y2 = 2x,
x2 + y2 = 6x,
x2 + y2 = 2y,
x2 + y2 = 8y.
This region is adapted to a coordinate system called
the Apollonian or inversion coordinates
( parabolic pencil of circles),
whose coordinate grid curves
are circles centered on the x-axis or the y-axis,
all of which pass through the origin.
This is given by the inversion transformation:
(x,y) = T(u,v) =
(2u⁄(u2+v2) ,
2v⁄(u2+v2)).
This is a kind of "reflection" across the circle
of radius √2 ;
T takes a point (u,v) with radius r to a point
(x,y) with the same polar angle and
radius 2⁄ r. That is, T switches the
inside and outside of the circle, with the origin going to infinity.
- Write the boundary circles of R in the form
(x−a)2 + (y−b)2 = r2
to show they are indeed Apollonian grid circles.
(Hint: Complete the square.)
- Show that the line u = c in the (u,v)-plane
corresponds to the circle (x−a)2 + y2 = a2 in the (x,y)-plane, where a = 1⁄c;
and similarly, the line v = c corresponds to
the circle x2 + (y−a)2 = a2.
In the picture above, identify the blue circles
as u = c and the red circles as v = c, for appropriate constants.
- Parametrize the region R as the points T(u,v) for
a1 ≤ u ≤ a2 ,
b1 ≤ v ≤ b2 , for appropriate constants.
- Compute the Jacobian matrix [DT(u,v)]
and its determinant |det[DT(u,v)]|
= |∂(x,y)⁄∂(u,v)|,
as it is written in the Change of Variables Formula.
- Use Change of Variables to rewrite the
following integral in uv coordinates:
∬R 1⁄(x2+y2)2 dx dy.
- Verify that the inversion mapping T is its own inverse:
that is, if (x,y) = T(u,v), then the variables u,v satisfy
the equation (u,v) = T(x,y).
Solution
16-9 #12(a). The equation x2 + y2 = 2ax
is equivalent to (x−a)2 + y2 = a2, a circle centered at (a,0) with radius a,
so that the circle contains the origin.
The first two circles have centers (1,0) and (3,0),
and similarly the second two have centers (0,1) and (0,4).
12(b). The line u = c is parametrized by (c,t) for t ∈ R, which corresponds to (x,y) = T(c,t).
It is easy to verify that
x = 2c⁄(c2+t2)
and y = 2t⁄(c2+t2)
satisfy x2 + y2 = 2⁄c x, which is equivalent to the circle equation. Similarly for v = c.
12(c). From the picture, we see
R = { (x,y) = T(u,v) for 1⁄3 ≤ u ≤ 1 ,
1⁄4 ≤ v ≤ 1 }
12(d). You can do this with Wolfram Alpha by pasting partial results
into a scratch file, then pasting back into W|A.
Result: the Jacobian determinant (streching factor) is
|−4⁄(u2+v2)2|
= 4⁄(u2+v2)2 .
12(e). The integral reduces
to ∫1⅓ ∫1¼
1⁄16(u2+v2)2
4⁄(u2+v2)2
dv du
= 1⁄8 .
12(f). Use W|A to solve the system of equations
x = 2u⁄(u2+v2)
and y = 2v⁄(u2+v2)
for the variables u,v. Keep in mind that the equations
are symmetric upon switching u,v and simultaneously switching x,y,
so a solution for u can be switched into a solution for v.
Note: This is also evident from the polar description
of T as an inversion across a circle of radius √2.
Furthermore, we can write T as a conjugate inversion
in the plane of complex numbers.
That is, we let (x,y) correspond to
the complex variable z = x + iy
with conjugate
z = x − iy;
then T(x,y) corresponds to the mapping
z ↦ 2/ z
=
2z⁄zz
=
2z⁄|z|2.
-
⊞
WHW 7 due Fri 11/18: Parametrized curves and surfaces in space.
Solutions (Mathematica).
For each surface:
- Write a parametric formula
P(t,s) = (x(t,s), y(t,s), z(t,s))
over a parameter region S* in the (t,s)-plane.
Use whatever parameter variables you like:
P(θ,φ), etc.
- Print out a picture of the surface.
Use any plotting software you can find:
Mathematica
(example),
GeoGebra
(surface),
Desmos,
W|A,
Zweig,
URI.
- Helix curve (corkscrew, spring) coiling upward around the z-axis and above the xy unit circle (radius 1). Write and draw
c(t) = (x(t), y(t), z(t)).
- Helical staircase surface, consisting of horizontal rays from z-axis to helix
- Tube surface at distance r from the helix. Hint: Use the Frenet frame from Lect 27.
- Ram's horn
- Elliptical surface on a tilted plane, with major axis
from a = (1,2,3) to −a = (−1,−2,−3), and minor axis from
b = (2,−1,0) to −b = (−2,1,0). Note a ⊥ b.
Hint: Use polar coordinates with respect to the axes
a, b instead of i, j.
- Spherical coordinates on the globe
- Draw a globe surface with some mountains on it,
adapting some of the topographic features from WHW 3.
In that project, we regarded the globe from up close,
treating longitude, latitude, altitude as rectangular coordinates.
Now use Sph to transfer those pictures to a sphere.
- On a parameter plane with (latitude, longitude) coordinates,
draw line segments between the following consecutive points,
making a polygon:
(-124, 49), (-90, 49), (-83, 42), (-70, 47), (-67, 45), (-82, 31), (-81, 25), (-84, 31), (-94, 31), (-98, 26), (-108, 32), (-117, 32), (-124, 40), (-124, 49)
Mathematica: Graphics[Polygon[{{x1,y1}, {x2,y2}, . . . }]];
GeoGebra: Polygon((x1,y1), (x2,y2), . . .), Settings > Zoom to fit.
Q: What does this represent? Why does it look too wide?
Extra Credit: Use the appropriate linear mapping
to convert (latitude, longitude) into (θ,φ),
and use Sph(4000, θ, φ) to draw the corresponding polygon on the globe (radius 4000 mi).
- On a map, find a latitude-longitude rectangle D*
that roughly approximates the map of the continental United States.
For the corresponding region D = Sph(D*) on the Earth,
compute its area using the area substitution formula.
That is, fixing the Earth's radius at ρ = 4000 mi,
parametrize the globe as P(θ,φ) = Sph(4000,θ,φ) and compute:
Area = ∫∫D* |∂P⁄∂θ×∂P⁄∂φ| dθ dφ .
How well does this match the actual area of the continental
U.S.?
- WHW 8
due Mon 11/28: Euler's Method & Newtonian trajectories (tex).
References: Kepler's Laws, Newton's Equation, Euler's Method,
Feynman Ch I-9-6.
- WHW 9 due Tue 12/13. Vector potential and electromagnetic fields (tex).
Notes & Tools
- Review notes on Calculus I and Calculus II.
- Marsden & Tromba Theoretical Supplement
- Calculus explainer:
3BLUE1BROWN
- Vector Calculus Identities (Wikipedia)
-
⊞
Software
- Wolfram Alpha plotting
- parametric curve c(t) = (x(t), y(t))
- graph surface z = f(x,y),
scaled vertically,
contour plot f(x,y) = c
- vector field F(x,y) = (p(x,y), q(x,y)),
arrow plot,
stream plot,
kludge if F(x,y) indep of x or y
- multiply matrices
- Get Wolfram Mathematica with a free MSU license. Plotting examples: 2D, 3D.
- Surface plotters:
GeoGebra
(surface),
Zweig,
Desmos,
URI
- Old Math 254H:
Spring 2017,
|