Past exam of the mathematical methods II course of the University of Oxford

i

Solution

The real logarithm restricts the approach to

x \to 0^{+}

. The Taylor expansion

sin x - x = - x^{3} /6 + O (x^{5})

gives

\frac{1}{x} + \frac{l n x}{s i n x - x} = \frac{- 6 l n x}{x ^{3}} (1 + O (x^{2})) + \frac{1}{x} .

The positive first term dominates, since

x^{2} /∣ ln x ∣ \to 0

. Thus the requested extended limit is

+ \infty

. There is no two-sided real limit across zero because the real logarithm is undefined for

x < 0

ii

Solution

For

x > 2

, write

\frac{1}{x} + \frac{l n x}{s i n x - x} = \frac{1 - l n x / ( 1 - s i n x / x )}{x} .

Since

∣ sin x ∣ \leq 1

, the denominator factor tends to one, and

ln x / x \to 0

. Therefore the limit is

0

. The function is eventually negative, so it approaches zero from below. Oscillation of

sin x

does not obstruct this limit, because it is bounded while

x

grows.

2

Solution

Use logarithmic differentiation on the positive function:

ln f = x^{- 1} ln (x^{2} + 2)

. The product rule and chain rule give

f^{'} (x) = (x^{2} + 2)^{1/ x} [\frac{2}{x ^{2} + 2} - \frac{ln ( x ^{2} + 2 )}{x ^{2}}], x \neq = 0.

The positive base permits this real definition on both components of the indicated domain. There is no need to treat a variable exponent as a constant.

3

Solution

For an exact differential

P d x + Q d y

, equality of mixed partial derivatives requires

P_{y} = Q_{x}

. The matching pair is

P = f_{1}

Q = f_{3}

, since

\partial_{y} f_{1} = \frac{c o s y}{x} - e^{x} sin 2 y = \partial_{x} f_{3} .

The other pairings do not give the same partial derivatives on an open region. Integrating

f_{1}

with respect to

x

gives

ln x sin y + e^{x} cos^{2} y + C (y)

. Its

y

derivative agrees with

f_{3}

precisely when

C^{'} (y) = 0

. Hence

f (x, y) = ln x sin y + e^{x} cos^{2} y + C, x > 0.

This also proves existence, rather than relying only on the necessary mixed-partial derivative test.

4

Solution

Put

h = x - 1/ 2

and

k = y - 1/ (22)

. At the expansion point the exponent is one and the gradient is

(2 e, 2 e)

. The first-order multivariable Taylor expansion is

f = e + 2 e (h + k) + O (h^{2} + k^{2}) .

The directional error of a first-order Taylor approximation is determined by the omitted terms, not just the equal first partial derivatives. The Hessian matrix at the expansion point is

e (4222)

, so

f - e - 2 e (h + k) = e (2 h^{2} + 2 hk + k^{2}) + O ((h^{2} + k^{2})^{3/2}) .

For equal small steps along the coordinate axes, the leading error is

2 e h^{2}

along

x

and

e k^{2}

along

y

. The error grows faster along the $x$ axis, although the first-order change in the function is equal along the two axes.

5

Solution

Choose the normal pointing away from the origin, equivalently the outward normal of the first-octant tetrahedron. Parametrize the face by

r = (x, y, 2 - x /2 - y /2)

D = {x, y \geq 0 : x + y \leq 4}

. Its oriented area element is

r_{x} \times r_{y} d x d y = (1/2, 1/2, 1) d x d y .

Thus the surface integral is

\iint_{D} (- \frac{x}{2} + \frac{y}{2} + 2 - \frac{x}{2} - \frac{y}{2}) d x d y = \int_{0}^{4} (2 - x) (4 - x) d x = \frac{16}{3} .

The divergence theorem checks this result:

\nabla \cdot F = 1

, the tetrahedron's volume is

4 \cdot 4 \cdot 2/6 = 16/3

, and each coordinate-plane face has zero flux. The opposite orientation gives

- 16/3

; a signed flux requires an orientation.

6

Solution

The triangular support can be written

0 \leq x \leq π /2

π /2 - x \leq y \leq π /2

. The density is nonnegative there, and the normalization of a probability distribution requires

1 = A \int_{0}^{π /2} sin x (1 - cos x) d x = \frac{A}{2} .

Hence

A = 2

. Multiplying the joint probability density by the required function cancels

sin x

on the interior. Its expectation is

E [\frac{1}{sin X}] = 2 \int_{0}^{π /2} (1 - cos x) d x = π - 2.

The singularity at

x = 0

lies at a support vertex of zero probability; the displayed integral establishes integrability. Assigning any value to the random function at that single vertex does not change the expectation.

7

Solution

The two planes bound

z - 1 \leq y \leq 1 - z

, and the elliptical cross-section has

- 1 \leq z \leq 1

. Under

x = 2 r cos θ

z = r sin θ

, the absolute Jacobian determinant for

(x, z)

2 r

. Therefore

V = \int_{0}^{2 π} \int_{0}^{1} \int_{r s i n θ - 1}^{1 - r s i n θ} 2 r d y d r d θ = \int_{0}^{2 π} \int_{0}^{1} 4 r (1 - r sin θ) d r d θ = 4 π .

As a check, the cross-sectional ellipse has area

2 π

, while the part of the thickness proportional to

z

integrates to zero by reflection. The mean thickness is therefore two.

8

Solution

For positive wavenumber, the condition gives

ω = C k^{- n /2}

on the positive-frequency branch. The group velocity and phase velocity are consequently related by

v_{g} = \frac{d ω}{d k} = - \frac{n}{2} \frac{ω}{k} = - \frac{n}{2} v_{p} .

The required positive ratio

v_{g} / v_{p} = 2

gives

n = - 4

. Isotropy makes these velocities parallel to the wave vector when this coefficient is positive. Identifying group velocity with energy transport assumes the usual lossless narrowband wave packet regime.

9

a

Solution

Since

a

is constant, direct differentiation gives

(r \cdot \nabla) a = 0

(a \cdot \nabla) r = a

\nabla \cdot r = 3

and

\nabla \cdot a = 0

. The stated right-hand side is therefore

2 a

. Independently, index notation gives

[\nabla \times (a \times r)]_{i} = ϵ_{ijk} ϵ_{k l m} \partial_{j} (a_{l} x_{m}) = 2 a_{i} .

Thus

\nabla \times (a \times r) = 2 a

. A conservative vector field has zero curl, so the field is not conservative unless $a = 0$ . For a closed planar circuit, its circulation is

2 a \cdot A

by Stokes theorem; the zero-vector exception is conservative with a constant potential.

b

i

Solution

For a sufficiently smooth vector field on an oriented surface

S

with compatible boundary orientation, Stokes theorem states

\oint_{\partial S} F \cdot d r = \iint_{S} (\nabla \times F) \cdot n d S .

Apply it to

F = (P, Q, 0)

in the plane. The closed line integral is the area integral of

Q_{x} - P_{y}

. Thus

P_{y} = Q_{x}

makes it vanish whenever a spanning surface lies within the smooth domain. In a simply connected domain this implies path independence: join one path from

A

B

to the reverse of another, and their difference is a zero closed line integral.

The domain qualification is essential. On the punctured plane,

P = - y / (x^{2} + y^{2})

Q = x / (x^{2} + y^{2})

satisfy

P_{y} = Q_{x}

, but the line integral around the unit circle is

2 π

. A disk spanning that circle passes through the singularity. Hence the printed implication requires an appropriate domain assumption; zero curl alone is not a global path independence theorem.

ii

Solution

Let

F^{'} = f

and

G^{'} = g

on the relevant intervals. The chain rule gives

d {F (x y) + G (y / x)} = [y f (x y) - \frac{y}{x ^{2}} g (y / x)] d x + [x f (x y) + \frac{1}{x} g (y / x)] d y .

Thus the one-form is explicitly an exact differential. Fixing its potential to vanish at

(1, 1)

determines it uniquely on the connected domain:

ϕ (x, y) = \int_{1}^{x y} f (s) d s + \int_{1}^{y / x} g (t) d t .

Continuity of

f, g

on the integration ranges suffices. For arbitrary smooth data defined throughout

x > 0

, that half-plane also supplies the simply connected domain in the previous argument.

For the specialization

f (s) = s^{- 2}

, the potential based at

(1, 1)

is defined in the positive quadrant; it cannot be continued through

x y = 0

using this formula. Taking

a, b > 0

keeps both endpoints in that component. Their ratios

y / x

are both

b / a

, so the

G

term cancels and

ϕ (ab, b^{2}) - ϕ (a, b) = \int_{ab}^{a b^{3}} s^{- 2} d s = \frac{b ^{2} - 1}{a b ^{3}} .

No condition on

g

beyond existence on the traversed ratios enters this difference.

10

a

Solution

Write

A (s) = s (s^{2} - 4)

, so

z = A (x) A (y)

and

z_{x} = (3 x^{2} - 4) y (y^{2} - 4), z_{y} = (3 y^{2} - 4) x (x^{2} - 4) .

Solving these simultaneously gives thirteen stationary points: the nine grid points with

x, y \in {- 2, 0, 2}

and the four points

x, y \in {- 2/ 3, 2/ 3}

. There are no mixed extra cases, because

A

and

A^{'}

have no common root.

At a grid point the Hessian matrix has zero diagonal and off-diagonal entry

A^{'} (x) A^{'} (y) \neq = 0

, so its determinant is negative: all nine are saddle points. At the other four points the Hessian matrix is

- 16 x y I

. Thus the equal-sign points are local maxima with value

256/27

, and the opposite-sign points are local minima with value

- 256/27

In the requested nonnegative quadrant, the stationary points are the four boundary saddle points

(0, 0), (0, 2), (2, 0), (2, 2)

and the interior local maximum

(2/ 3, 2/ 3)

. The zero contours are

x = 0, 2

and

y = 0, 2

. The gradient is perpendicular to regular contours and points toward increasing

z

. A boundary point's classification above refers to the function on the whole plane; its behavior restricted to the quadrant can be one-sided. The interior local maximum is not a global maximum on the unbounded quadrant, since

z \to + \infty

along

x = y \to \infty

Figure 1.
Contours, zeros, stationary points and gradient directions for the polynomial surface in the nonnegative quadrant
. **Details**.

b

Solution

The divergence of

F = \nabla z

is the Laplacian

\nabla \cdot F = 6 x y (x^{2} + y^{2} - 8) .

Its area integral over the square is

\int_{0}^{2} \int_{0}^{2} 6 x y (x^{2} + y^{2} - 8) d x d y = 6 (8 + 8 - 32) = - 96 .

For the boundary flux,

\int_{0}^{2} A (s) d s = - 4

. On

x = 0

F \cdot n = 4 A (y)

- 16

; on

x = 2

it is

8 A (y)

- 32

. The corresponding horizontal sides give

- 16

and

- 32

. The total is again

- 96

, verifying the divergence theorem with outward normals. With a counterclockwise boundary, this is the same orientation as

M d y - N d x

11

a

Solution

First solve the supplied equations as a linear system, with

ω_{0}^{2} = g / l

and

α > 0

. A normal mode

x = a e^{iω t}

obeys

\frac{ω ^{2}}{ω _{0}^{2}} a = D a, D = (1 + 2 α - α - α α) .

Its characteristic polynomial is

λ^{2} - (1 + 3 α) λ + α (1 + α)

. Hence

ω_{\pm}^{2} = \frac{g}{2 l} [1 + 3 α \pm 1 + 2 α + 5 α^{2}] .

Both are positive, since the trace and determinant of

D

are positive.

There is a physical discrepancy in the source. The horizontal-coordinate mass matrix of an unequal-mass double pendulum follows from the stated horizontal displacements, which give

T_{2} = \frac{m}{2} \overset{x}{˙}_{1}^{2} + \frac{α m}{2} \overset{x}{˙}_{2}^{2}, V_{2} = \frac{m g}{2 l} [x_{1}^{2} + α x_{1}^{2} + α (x_{2} - x_{1})^{2}] .

The Euler-Lagrange equations reproduce the supplied upper equation, but give

\overset{x}{¨}_{2} = - (g / l) (x_{2} - x_{1})

, without its extra

α

. The true normal modes of a double pendulum consequently have

ω_{\pm}^{2} = (g / l) [1 + α \pm α (1 + α)]

. The boxed answer and the remaining parts below concern the explicitly supplied linear system. The two models agree at

α = 1

, but not for a general mass ratio.

b

Solution

α = 1

, the supplied and physical systems agree. Their squared angular frequencies are

(g / l) (2 \pm 2)

. The upper row gives

a_{2} / a_{1} = 3 - λ

, so

\frac{a _{2}}{a _{1}} = 1 + 2 for the slower mode, \frac{a _{2}}{a _{1}} = 1 - 2 for the faster mode .

In the slower normal mode, both masses move horizontally in phase and the lower has the larger amplitude. In the faster normal mode, the lower mass moves in the opposite direction, with magnitude

(2 - 1) ∣ a_{1} ∣

. These are horizontal displacement ratios. The rod angles instead satisfy

θ_{2} / θ_{1} = \pm 2

, because

x_{1} = l θ_{1}

x_{2} = l (θ_{1} + θ_{2})

; confusing these ratios changes the sketch.

Figure 1.
Equal-mass normal-mode snapshots with horizontal displacements enlarged for readability; arrow lengths show the relative displacement amplitudes
. **Details**.

c

Solution

For the supplied system, the small-

α

Taylor expansion gives

1 + 2 α + 5 α^{2} = 1 + α + O (α^{2})

, hence

ω_{-}^{2} = αg / l + O (α^{2} g / l)

. Setting

ω_{-} = (1/4) g / l

at this order yields

α ≃ \frac{1}{16} .

This is an approximation in the squared frequency. The exact positive solution of

λ_{-} = 1/16

α = (229 - 13) /32 ≃ 0.06665

, showing the size of the neglected correction. The physical unequal-mass double pendulum has a different small-mass limit, so the source's requested tuning does not follow for that model.

d

Solution

For the supplied equations,

λ_{+} \to 1

and

a_{2} / a_{1} = α / (α - λ_{+}) \sim - α

. Thus the higher-frequency normal mode becomes an upper-mass oscillation at

g / l

with a vanishing, opposite-sign lower horizontal displacement. At

α = 0

the supplied lower equation is

\overset{x}{¨}_{2} = 0

and its other mode has zero frequency.

This limiting picture belongs to the given matrix. For the physical unequal-mass double pendulum, both normal mode angular frequencies tend to

g / l

as the lower mass tends to zero; at exactly zero mass its independent equation needs a limiting test-body interpretation. The source's extra

α

changes precisely this conclusion.

12

a

Solution

Use separation of variables,

y = X (x) T (t)

. A nontrivial separated solution satisfies

X^{''} / X = T^{''} / (c^{2} T) = - k^{2}

. Fixed endpoints force

X (0) = X (L) = 0

, so

X = sin (nπ x / L)

n = 1, 2, \dots

. Therefore

ω_{n} = \frac{nπ c}{L}, f_{n} = \frac{n c}{2 L} .

Here

c > 0

is the wave speed. Zero or negative

k^{2}

gives no nonzero fixed-end spatial solution. Each time coefficient is a simple harmonic oscillator.

b

Solution

Expand the initial velocity in the Fourier sine basis,

V (x) = \sum_{n \geq 1} V_{n} sin (nπ x / L)

, where

V_{n} = (2/ L) \int_{0}^{L} V (x) sin (nπ x / L) d x

. Zero initial displacement removes the cosine-in-time coefficients. Thus

y (x, t) = n \geq 1 \sum \frac{V _{n}}{ω _{n}} sin \frac{nπ x}{L} sin (ω_{n} t) .

The trigonometric identity

2 sin a sin b = cos (a - b) - cos (a + b)

rewrites each summand as

\frac{V _{n}}{2 ω _{n}} [cos (\frac{nπ x}{L} - ω_{n} t) - cos (\frac{nπ x}{L} + ω_{n} t)] .

It is therefore a superposition of travelling waves of speeds

+ c

and

- c

, whose interference enforces the fixed endpoints. For general square-integrable initial velocity, the series is interpreted in the standard finite-energy sense.

c

Solution

The product-to-sum identity gives

V_{1} = V_{0}

V_{2} = V_{0} /2

and

V_{3} = 2 V_{0} /3

, with all other modal coefficients zero. If

μ

is the string's linear mass density, the initial potential energy vanishes and orthogonality gives the modal energy of a string

E_{n} = \frac{μ}{2} \int_{0}^{L} V_{n}^{2} sin^{2} \frac{nπ x}{L} d x = \frac{μL}{4} V_{n}^{2} .

Consequently

E_{1} : E_{2} : E_{3} = 36 : 9 : 16

for

V_{0} \neq = 0

. The fundamental mode carries the most energy, with fraction

36/61

. Comparing displacement coefficients without their frequency factors would give the wrong energy test.

d

Solution

Substitute the three nonzero initial-velocity coefficients into the general Fourier sine series, with

ω_{1} = π c / L

y (x, t) = \frac{V _{0}}{ω _{1}} [sin \frac{π x}{L} sin ω_{1} t + \frac{1}{4} sin \frac{2 π x}{L} sin 2 ω_{1} t + \frac{2}{9} sin \frac{3 π x}{L} sin 3 ω_{1} t] .

Every term satisfies the wave equation and the endpoint conditions. Differentiating in time at zero recovers the three coefficients

V_{0}, V_{0} /2, 2 V_{0} /3

, providing a direct initial-data check.

e

Solution

Assume

V_{0} \neq = 0

. Linear independence of the Fourier sine basis makes a flat string equivalent to vanishing of all three temporal coefficients. The fundamental requires

sin (ω_{1} t) = 0

, which automatically zeros the other two. Thus the earliest positive return is

t_{flat} = \frac{π}{ω _{1}} = \frac{L}{c} .

The velocity need not have returned to its initial value at that time; full recurrence of the displacement-and-velocity data occurs at

2 L / c

A nonzero pure fundamental profile would require

sin 2 ω_{1} t = sin 3 ω_{1} t = 0

. The common zeros have

ω_{1} t

an integer multiple of

π

, so the fundamental vanishes too. A pure second or third harmonic would already require the fundamental to vanish, again making the whole string flat. The string never has a nonzero pure sine profile. If

V_{0} = 0

, it remains identically zero and there is no earliest positive isolated return.

2021 resit (A10289W1)

Source paper: www.whatdotheyknow.com/request/physics_preliminary_past_papers/response/2680788/attach/3/2018%202021%20Physics%20prelims.zip?cookie_passthrough=1

The examination date is 9 September 2021. The source archive labels its member 2020/L/a10289w1.pdf; the date on the paper establishes the year. Solutions cover all questions, including every option in Section B.

e^{x^{2}} = 1 + x^{2} + \frac{x ^{4}}{2} + \frac{x ^{6}}{6} + O (x^{8}), x sin x = x^{2} - \frac{x ^{4}}{6} + \frac{x ^{6}}{120} + O (x^{8}) .

Both the numerator and denominator first become nonzero at order

x^{4}

. Dividing by that common power gives the limit

x \to 0 lim \frac{1 + x ^{2} - e ^{x^{2}}}{x sin x - x ^{2}} = x \to 0 lim \frac{- 1/2 + O ( x ^{2} )}{- 1/6 + O ( x ^{2} )} = 3.

b

Solution

The fundamental theorem of calculus differentiates the variable upper endpoint:

I^{'} (x) = \frac{2}{π} e^{- x^{2}}, I^{''} (x) = - \frac{4 x}{π} e^{- x^{2}}, I^{'''} (x) = \frac{- 4 + 8 x ^{2}}{π} e^{- x^{2}} .

At zero,

I (0) = 0

I^{'} (0) = 2/ π

I^{''} (0) = 0

and

I^{'''} (0) = - 4/ π

. The integrand is even, so its integral from zero is an odd function; all even coefficients in the Maclaurin series vanish. Hence

I (x) = \frac{2}{π} (x - \frac{x ^{3}}{3}) + O (x^{5}) .

This is the local expansion of the error function.

2

Solution

First method: substitute the trajectory. Along the parametric curve, the temperature is

T (t) = 20 e^{- c o s^{2} t} - (1 + sin t)^{3}

. The chain rule gives

\frac{d T}{d t} = 40 sin t cos t e^{- c o s^{2} t} - 3 (1 + sin t)^{2} cos t .

Second method: use the spatial derivatives. The partial derivatives of the scalar field are

T_{x} = - 40 x e^{- x^{2}}

and

T_{z} = - 3 (z + 1)^{2}

, while the velocity components are

\overset{x}{˙} = - sin t

and

\overset{z}{˙} = cos t

. The multivariable chain rule therefore gives

\frac{d T}{d t} = T_{x} \overset{x}{˙} + T_{z} \overset{z}{˙} = (- 40 cos t e^{- c o s^{2} t}) (- sin t) - 3 (1 + sin t)^{2} cos t,

which agrees with the first calculation. This is the gradient dotted with the velocity, rather than a partial derivative at a fixed position.

3

Solution

Apply the product-to-sum formula

2 cos u cos v = cos (u + v) + cos (u - v)

to the amplitude modulation:

A = a cos (ω t - k x) + \frac{b}{2} cos ((ω + ω_{m}) t - k x) + \frac{b}{2} cos ((ω - ω_{m}) t - k x) .

The amplitude modulation sidebands have angular frequencies

ω + ω_{m}

and

ω - ω_{m}

, each with amplitude

b /2

in this real cosine convention. The carrier wave has angular frequency

ω

and amplitude

a

This identity is exact for the supplied space-time expression. A physical qualification is needed if every term is meant to be a freely propagating electromagnetic wave with speed of light

c

: the dispersion relation requires

k_{\pm} = (ω \pm ω_{m}) / c

, whereas the prescribed modulation at every position leaves all three wavenumbers equal to

k = ω / c

. A freely propagating modulated wave instead has envelope

a + b cos (ω_{m} (t - x / c))

; its analogous expansion has

k_{\pm}

. For example, at

x = t = 0

the prescribed expression has

A_{tt} - c^{2} A_{xx} = - b ω_{m}^{2}

, which is nonzero for nontrivial modulation. The given formula can describe modulation imposed throughout space or the temporal signal at one position.

4

Solution

Differentiate the parametric curve:

\overset{x}{˙} = a t cos t, \overset{y}{˙} = a t sin t .

The arc length element is

d s = \overset{x}{˙}^{2} + \overset{y}{˙}^{2} d t = ∣ a ∣ t d t

0 \leq t \leq 2

. Thus

L = \int_{0}^{2} ∣ a ∣ t d t = 2∣ a ∣.

For a positive geometric scale

a

, this is

2 a

. The absolute value is necessary when the unspecified real constant

a

is negative;

a = 0

gives zero arc length.

5

Solution

The gradient of the scalar field is

\nabla f = (2 x + 9 y - 2 z^{2}, 9 x, - 4 x z), \nabla f (2, 4, 8) = (- 88, 18, - 64) .

For a unit vector

\hat{n}

, the directional derivative is

D_{\hat{n}} f = \nabla f \cdot \hat{n}

. The Cauchy-Schwarz inequality bounds its magnitude by

∥\nabla f ∥ = 23041

, with equality precisely in the two directions parallel to the gradient. Therefore

\hat{n} = \pm \frac{( - 44 , 9 , - 32 )}{3041}, ∥ \hat{n} ∥ = 1 max ∣ D_{\hat{n}} f ∣ = 23041 .

The plus choice gives the greatest positive directional derivative, and the minus choice gives the most negative directional derivative. Both maximize its magnitude.

6

Solution

For the scalar surface integral over an obliquely cut cylinder, split the scalar surface integral into the curved cylinder and its two planar caps. Write

s = z - x

, so the solid has

x^{2} + y^{2} \leq 4

and

- 3 \leq s \leq 2

. On the curved cylinder, use the parametrization

R (θ, s) = (2 cos θ, 2 sin θ, 2 cos θ + s), 0 \leq θ \leq 2 π, - 3 \leq s \leq 2.

The surface area element follows from

R_{θ} \times R_{s} = (2 cos θ, 2 sin θ, 0), d S = 2 d θ d s, x - z = - s .

Consequently the curved contribution is

I_{side} = \int_{0}^{2 π} \int_{- 3}^{2} (- s) 2 d s d θ = 10 π .

On either cap

z = x + s

, the surface area element is

1 + z_{x}^{2} + z_{y}^{2} d x d y = 2 d x d y

over a radius-two disk. The lower cap has

x - z = 3

and the upper cap has

x - z = - 2

. Thus

I_{lower} = 32 (4 π) = 122 π, I_{upper} = - 22 (4 π) = - 82 π,

(4)

and the complete scalar surface integral is

\iint_{S} (x - z) d S = π (10 + 42) .

(5)

The surface area element is unsigned: opposite outward normals on the caps do not introduce extra minus signs.

Figure 1.
Original sketch of the solid inside the cylinder between two parallel sloping caps
. **Details**.

0.04 = 1/25

, its normalization of a probability distribution is

\int_{- \infty}^{\infty} f_{T} (t) d t = \int_{0}^{\infty} \frac{1}{25} e^{- t /25} d t = [- e^{- t /25}]_{0}^{\infty} = 1.

It is an exponential distribution with rate

1/25

per month.

P (T < 8) = \int_{0}^{8} \frac{1}{25} e^{- t /25} d t = 1 - e^{- 8/25} = 0.274 (three significant figures) .

The strict endpoint does not affect the probability for this continuous probability distribution.

ii

Solution

The survival function of this exponential distribution is

P (T > t) = e^{- t /25}

for

t \geq 0

. Therefore

P (T > 20) = \int_{20}^{\infty} \frac{1}{25} e^{- t /25} d t = e^{- 20/25} = 0.449 (three significant figures) .

iii

Solution

Subtract the two values of the survival function, or integrate the probability density function directly:

P (14 < T < 30) = \int_{14}^{30} \frac{1}{25} e^{- t /25} d t = e^{- 14/25} - e^{- 30/25} = 0.270 (three significant figures) .

The final zero records the requested third significant figure.

c

Solution

The expectation follows by integration by parts, with

u = t

and

d v = e^{- t /25} d t /25

E [T] = \int_{0}^{\infty} t \frac{1}{25} e^{- t /25} d t = [- t e^{- t /25}]_{0}^{\infty} + \int_{0}^{\infty} e^{- t /25} d t .

The endpoint term vanishes because the exponential function decays faster than

t

grows. Hence the mean lifetime is

E [T] = 25 months .

F

on a neighborhood of an oriented, piecewise smooth surface

S

, Stokes theorem states

\oint_{\partial S} F \cdot d r = \iint_{S} (\nabla \times F) \cdot \hat{n} d S .

Here

\partial S

is the boundary curve,

d r

is its directed tangent element,

\hat{n}

is the chosen unit normal vector,

d S

is the surface area element, and

\nabla \times F

is the curl. The right-hand rule relates the boundary direction to

\hat{n}

. Thus circulation around the edge equals the flux of curl through the spanning surface. Reversing both orientations changes both signs. One can replace a difficult line integral by an easier surface integral, or replace one spanning surface by another within the region of regularity.

For a bounded volume

V

with piecewise smooth closed boundary and a continuously differentiable vector field on a neighborhood of its closure, the divergence theorem states

\iint_{\partial V} F \cdot \hat{n}_{out} d S = ∭_{V} \nabla \cdot F d V .

Here

\hat{n}_{out}

points out of

V

, including into any holes,

d V

is the volume element, and

\nabla \cdot F

is the divergence. It equates total outward flux to integrated source strength. It converts a closed surface integral into a volume integral, and explains why internal fluxes cancel in a subdivided region. Singularities cannot be included without checking the hypotheses.

For a conservative force, the work around every closed path is zero. Stokes theorem gives zero flux of curl through every sufficiently small patch. Dividing by its area and shrinking the patch shows that every component of the continuous curl vanishes:

\oint_{C} F \cdot d r = 0, \nabla \times F = 0, F = - \nabla U .

Here

U

is potential energy; conversely, zero curl implies a conservative vector field on a simply connected domain. The divergence theorem then gives the compatible relation

\iint_{\partial V} F \cdot \hat{n} d S = - ∭_{V} \nabla^{2} U d V

. There is no general zero-divergence condition on a conservative force:

F = - r = - \nabla (r^{2} /2)

is a concrete counterexample, with

\nabla \cdot F = - 3

b

Solution

Take

0 < r_{1} < r_{2}

, so the shell excludes the possible singularity at the origin. The divergence and curl of a radial vector field formula in spherical coordinates gives

\nabla \cdot F = \frac{1}{r ^{2}} \frac{d}{d r} (r^{2} r^{- β}) = (2 - β) r^{- β - 1} .

With volume element

d V = r^{2} sin θ d r d θ d ϕ

, direct integration gives

∭_{V} \nabla \cdot F d V = \int_{r_{1}}^{r_{2}} \int_{0}^{π} \int_{0}^{2 π} (2 - β) r^{1 - β} sin θ d ϕ d θ d r = 4 π (r_{2}^{2 - β} - r_{1}^{2 - β}) .

For the independent boundary calculation, the outer sphere has outward unit normal vector

+ \hat{r}

and the inner sphere has outward unit normal vector

- \hat{r}

. Their fluxes are

Φ_{outer} = 4 π r_{2}^{2} r_{2}^{- β}, Φ_{inner} = - 4 π r_{1}^{2} r_{1}^{- β} .

Their sum equals the direct volume integral, verifying the divergence theorem. The restriction

β \neq = 2

permits the displayed power-antiderivative step. At

β = 2

, treated separately, both the shell volume integral and net boundary flux are zero; the excluded origin must still not be filled in.

c

Solution

Away from the origin, write

F = r^{- β - 1} r

. The curl product rule for a scalar field and the gradient of a radial function give

\nabla \times F = \nabla (r^{- β - 1}) \times r + r^{- β - 1} \nabla \times r = - (β + 1) r^{- β - 3} r \times r = 0.

Consequently the surface integral of curl over the upper hemisphere is zero. Its outward unit normal vector induces a counterclockwise equator when viewed from above. Parametrize that boundary by

r (ϕ) = r_{2} (cos ϕ, sin ϕ, 0), d r = r_{2} (- sin ϕ, cos ϕ, 0) d ϕ .

The radial vector field is perpendicular to this tangent, so

\oint_{\partial S} F \cdot d r = 0 = \iint_{S} (\nabla \times F) \cdot \hat{n} d S .

This verifies Stokes theorem directly without passing through the singular origin.

(A (x, y), B (x, y), 0)

, with no dependence on

z

. Its curl is

\nabla \times V = i \partial_{x} A j \partial_{y} B k \partial_{z} 0 = (\frac{\partial B}{\partial x} - \frac{\partial A}{\partial y}) k .

Thus, for continuously differentiable

A

and

B

, the necessary and sufficient pointwise condition is

\frac{\partial B}{\partial x} = \frac{\partial A}{\partial y} .

To deduce global path independence from this local condition also requires a suitable domain, for example a simply connected domain.

b

Solution

Set

P = 10 x^{4} - 2 x y^{3}

and

Q = - 3 x^{2} y^{2}

. Their partial derivatives satisfy

P_{y} = - 6 x y^{2} = Q_{x}

throughout

R^{2}

. The vector field is smooth on the entire simply connected plane, so zero curl proves path independence.

Choose the broken straight path

(0, 0) \to (2, 0) \to (2, 1)

. Along its horizontal segment

y = 0

and

d y = 0

I_{1} = \int_{0}^{2} 10 x^{4} d x = 64.

Along its vertical segment

x = 2

and

d x = 0

I_{2} = \int_{0}^{1} - 12 y^{2} d y = - 4.

Therefore the line integral is

I = I_{1} + I_{2} = 60.

Figure 1.
Original sketch of the chosen oriented path through the corner at (2,0)
. **Details**.

c

Solution

An exact differential is an expression

P d x + Q d y

that equals

df = f_{x} d x + f_{y} d y

for a single-valued differentiable function

f

on the domain. Its line integral is an endpoint difference by the fundamental theorem for line integrals.

Integrating

f_{x} = P

with respect to

x

gives

f = 2 x^{5} - x^{2} y^{3} + g (y)

. Matching

f_{y} = - 3 x^{2} y^{2} + g^{'} (y)

Q

requires

g^{'} (y) = 0

. Hence

df = d (2 x^{5} - x^{2} y^{3}), I = f (2, 1) - f (0, 0) = 64 - 4 = 60.

The arbitrary additive constant cancels. This constructs the vector-field potential globally and provides a second proof of path independence.

d

Solution

Assume a simple, piecewise smooth closed curve oriented counterclockwise around its enclosed region

D

. Apply Green's theorem with

P = - y

and

Q = x

\oint_{C} (x d y - y d x) = \iint_{D} (Q_{x} - P_{y}) d x d y = \iint_{D} 2 d x d y = 2 area (D) .

The orientation assumption is essential. A clockwise unit circle,

x = cos θ

y = - sin θ

for

0 \leq θ \leq 2 π

, gives

- 2 π

although its enclosed area is

π

. For a self-intersecting curve the same line integral measures twice the signed area, with multiplicity, rather than necessarily twice an ordinary enclosed area.

For positive semiaxes

a, b

, the ellipse has the counterclockwise parametrization

x = a cos θ

y = b sin θ

. Then

x d y - y d x = ab (cos^{2} θ + sin^{2} θ) d θ = ab d θ .

Therefore

area (ellipse) = \frac{1}{2} \int_{0}^{2 π} ab d θ = πab .

If nonzero signed constants are allowed in place of positive semiaxes, the geometric area is

π ∣ ab ∣

ξ = x + v t

and

F (ξ) = A e^{- ξ^{2} / x_{0}^{2}}

, with

x_{0} \neq = 0

. The chain rule gives

y_{xx} = (\frac{4 ξ ^{2}}{x _{0}^{4}} - \frac{2}{x _{0}^{2}}) y, y_{tt} = v^{2} (\frac{4 ξ ^{2}}{x _{0}^{4}} - \frac{2}{x _{0}^{2}}) y .

For a nonzero pulse the wave equation on a string therefore requires

v^{2} = \frac{T}{μ}, ∣ v ∣ = \frac{T}{μ} .