Math. 311 (Fulling) 23 April 2004

Test C – Solutions

Name:

Calculators may be used for simple arithmetic operations only!

1. (20 pts.) Let v⃗1 = (1, 1, 1, 1) , v⃗2 = (1, −1, 2, −1) .

(a) Find an orthonormal basis, {û1, û2}, for the two-dimensional subspace of R4 spanned

by v⃗1 and v⃗2 .

∥v⃗_{1}∥^{2} = 1 + 1 + 1 + 1 = 4 , so a normalized vector is

û_{1} = \frac{v⃗_{1}}{∥v⃗_{1}∥} = \frac{1}{2} (1, 1, 1, 1).

Now ⟨û_{1}, v⃗_{2}⟩ = \frac{1}{2} (1 − 1 + 2 − 1) = \frac{1}{2} , so the part of v⃗_{2} parallel to v⃗_{1} is

v⃗_{2∥} = û_{1}⟨û_{1}, v⃗_{2}⟩ = \frac{1}{4} (1, 1, 1, 1),

and hence the perpendicular part is

v⃗_{2⊥} = v⃗_{2} − v⃗_{2∥} = (1, −1, 2, −1) − ( \frac{1}{4} , \frac{1}{4} , \frac{1}{4} , \frac{1}{4} ) = \frac{1}{4} (3, −5, 7, −5).

Next we have

∥v⃗_{2⊥}∥^{2} = \frac{1}{16} (9 + 25 + 49 + 25) = \frac{108}{16} .

(Actually, I could have dropped the factor \frac{1}{4} , since it cancels out at the next step.) Therefore, the normalized vector orthogonal to û_{1} is

û_{2} = \frac{v⃗_{2⊥}}{∥v⃗_{2⊥}∥} = \frac{1}{\sqrt{108}} (3, −5, 7, −5) = \frac{1}{6\sqrt{3}} (3, −5, 7, −5).

(b) Give a formula for P , the orthogonal projection operator onto that subspace. (That

is, for any v⃗ in R4 , P (v⃗) is the part of v⃗ “parallel” to the plane span{v⃗1, v⃗2}.)

This is just the next step in a Gram–Schmidt construction:

P (v⃗) = v⃗_{∥} = û_{1}⟨û_{1}, v⃗⟩ + û_{2}⟨û_{2}, v⃗⟩,

where û_{1} and û_{2} were found in (a).

2. (30 pts.) In the (x, y) plane define new coordinates (u, v) by x = \frac{u}{2} , y = \frac{u2}{4} + v.

(a) Find the tangent vectors to the coordinate curves (as functions of u and v).

\frac{∂r⃗}{∂u} = ( \frac{1}{2} ; \frac{u}{2} ) , \frac{∂r⃗}{∂v} = ( \begin{matrix}0 \\ 1 \end{matrix} ) .

Note: Since these vectors are not orthogonal, there is no point in normalizing them to unit length. The same remark applies to the vectors in (b).

311C-S04 Page 2

(b) Find the normal vectors to the coordinate “surfaces” (which are actually curves in this

two-dimensional case), as functions of u and v.

From (a), the Jacobian of the coordinate transformation and its inverse (by the 2 × 2 Cramer’s rule) are

J = ( \begin{matrix}\frac{1}{2} & 0 \\ \frac{u}{2} & 1 \end{matrix} ) , J^{−1} = \frac{1}{1/2} ( \begin{matrix}1 & 0 \\ − \frac{u}{2} & \frac{1}{2} \end{matrix} ) = ( \begin{matrix}2 & 0 \\ −u & 1 \end{matrix} ) .

The standard normal vectors are the rows of J^{−1} :

∇u = (2, 0), ∇v = (−u, 1).

(c) Evaluate ∫∫ v2 dx dy over the region bounded by the curves v = 0 , u = 2 , v = 1 ,

and u = 1 .

From J found in (b), we have det J = \frac{1}{2} . So the integral is

∫ ∫ v^{2} det J du dv = ∫_{1}^{2} du ∫_{0}^{1} \frac{1}{2} v^{2} dv = \frac{1}{2} \frac{v^{3}}{3} |_{0}^{1} = \frac{1}{6} .

(d) Sketch the curves v = 0 and u = 2 , the region in (c), and the two sets of basis vectors in (a) and (b) evaluated (and drawn) at the point (u, v) = (2, 0) . Clearly label the vectors as ∇u, \frac{∂r⃗}{∂u} , etc.

The Cartesian coordinates of the point are (x, y) = (1, 1) . The u = constant curves are vertical lines; the v = constant curves are parabolas, y = x^{2}+v .

\frac{∂r⃗}{∂u} = ( 1^{\frac{1}{2}} ) , \frac{∂r⃗}{∂v} = ( \begin{matrix}0 \\ 1 \end{matrix} ) .

∇u = (2, 0), ∇v = (−2, 1).

Remark: ∇u is orthogonal to \frac{∂r⃗}{∂v} and ∇v is orthogonal to \frac{∂r⃗}{∂u} , although the elements of each basis are not orthogonal to each other.

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3. (28 pts.) Find a quadrature rule (approximate integration formula) of the form

∫{∞ \atop 0} f(t) e−t dt ≈ a1f(0) + a2f(1) + a3f(10)

by requiring that the rule gives the exact answer for all f in P2 (the quadratic polynomials). Use Cramer’s rule to solve for the coefficients, showing intermediate steps. Useful information: ∫ ∞ 0 tne−t dt = n! .

311C-S04 Page 3 Requiring that the rule gives the right answer on the standard basis for P_{2} yields three equations,

a_{1} + a_{2} + a_{3} = ∫_{0}^{∞} e^{−t} dt = 1,
a_{2} + 10a_{3} = ∫_{0}^{∞} te^{−t} dt = 1,
a_{2} + 100a_{3} = ∫_{0}^{∞} t^{2}e^{−t} dt = 2.

The determinant of the system is

Δ = | \begin{matrix}1 & 1 & 1 \\ 0 & 1 & 10 \\ 0 & 1 & 100 \end{matrix} | = | \begin{matrix}1 & 10 \\ 1 & 100 \end{matrix} | = 90.

Therefore,

a_{1} = \frac{1}{90} | \begin{matrix}1 & 1 & 1 \\ 1 & 1 & 10 \\ 2 & 1 & 100 \end{matrix} | = \frac{1}{90} [| \begin{matrix}1 & 10 \\ 1 & 100 \end{matrix} | − | \begin{matrix}1 & 10 \\ 2 & 100 \end{matrix} | + | \begin{matrix}1 & 1 \\ 2 & 1 \end{matrix} |]

= \frac{1}{90} (90 − 80 − 1) = \frac{9}{90} = \frac{1}{10} ,

a_{2} = \frac{1}{90} | \begin{matrix}1 & 1 & 1 \\ 0 & 1 & 10 \\ 0 & 2 & 100 \end{matrix} | = \frac{1}{90} | \begin{matrix}1 & 10 \\ 2 & 100 \end{matrix} | = \frac{80}{90} = \frac{8}{9} ,

a_{3} = \frac{1}{90} | \begin{matrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 2 \end{matrix} | = \frac{1}{90} | \begin{matrix}1 & 1 \\ 1 & 2 \end{matrix} | = \frac{1}{90} .

Thus, finally,

∫_{0}^{∞} f(t) e^{−t} dt ≈ \frac{1}{10} f(0) + \frac{8}{9} f(1) + \frac{1}{90} f(10).

4. (22 pts.) Let B⃗(x, y, z) = \frac{x}{(x2 + y2)n} î + \frac{y}{(x2 + y2)n} ĵ, where n is an arbitrary, fixed number. (Note that Bz = 0 .)

(a) Calculate ∫∫S B⃗ · dS⃗ when S is the piece of cylindrical surface defined in standard

cylindrical coordinates by

r = 2, 0 < θ < π, 0 < z < 3.

(The result will be a function of n.)

Easy way: Note that B⃗ = \frac{x}{r^{2n}} î + \frac{y}{r^{2n}} ĵ is perpendicular to the surface. The unit normal vector is n̂ = \frac{x}{r} î + \frac{y}{r} ĵ. Thus

B⃗ · n̂ = \frac{x^{2} + y^{2}}{r^{2n+1}} = \frac{1}{r^{2n−1}} ,

which is constant on the cylinder. So we merely need to multiply by the area of S :

∫ ∫_{S} B⃗ · dS⃗ = \frac{1}{2^{2n−1}} × 6π = \frac{3π}{4^{n−1}} .

Hard way: Since x = 2 cos θ , y = 2 sin θ , z = z , we have

∫ ∫_{S} B⃗ · dS⃗ = ∫ ∫ [B_{x} dy dz + B_{y} dz dx + B_{z} dx dy]

= ∫ ∫ [\frac{2 cos θ}{4^{n}} (2 cos θdθ) dz + \frac{2 sin θ}{4^{n}} dz (−2 sin θ) dθ]

= \frac{1}{4^{n−1}} ∫ ∫ [cos^{2} θ dθ dz + sin^{2} θ dθ dz]

= \frac{1}{4^{n−1}} ∫_{0}^{π} dθ ∫_{0}^{3} dz = \frac{3π}{4^{n−1}} .

(b) For what value(s) of n does there exist a vector potential A⃗(x, y, z) such that B⃗ = ∇×A⃗ (everywhere except possibly on the axis, x = y = 0 )?

We need

0 = ∇ · B⃗ = \frac{∂}{∂x} (\frac{x}{(x^{2} + y^{2})^{n}} ) + \frac{∂}{∂y} (\frac{y}{(x^{2} + y^{2})^{n}} )

= \frac{1}{(x^{2} + y^{2})^{n}} − \frac{2nx^{2}}{(x^{2} + y^{2})^{n+1}} + \frac{1}{(x^{2} + y^{2})^{n}} − \frac{2ny^{2}}{(x^{2} + y^{2})^{n+1}}

= \frac{2x^{2} + 2y^{2} − 2n(x^{2} + y^{2})}{(x^{2} + y^{2})^{n+1}}

= \frac{2(1 − n)}{(x^{2} + y^{2})^{n}} .

(This calculation would be easier if we knew the formula for the divergence operator in cylindrical coordinates, but we haven’t studied that.) So the needed condition is

n = 1.

(This means that B⃗ = \frac{n̂}{r} . On the axis, B⃗ , n̂, and ∇ · B⃗ are all undefined. Since n̂ points outward, it is geometrically obvious that ∇ · B⃗ should be regarded as infinite on the axis, much as the divergence of the electric field of a point charge is infinite at the origin. In this case we have “magnetic monopole charge” concentrated along the whole axis.)