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This is also an instance of divergence of a curl is zero, since the potential is smooth off the axis. The classical field is undefined on the axis, so that identity must not be applied there without specifying an extension.

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 16 / a / i / Solution by

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The logarithm is defined away from the

z

axis, where

x^{2} + y^{2} > 0

. For the vector potential

ψ e_{z} = (0, 0, ψ)

, its curl is

(ψ_{y}, - ψ_{x}, 0)

. Since

ψ_{x} = \frac{a x}{x ^{2} + y ^{2}}, ψ_{y} = \frac{a y}{x ^{2} + y ^{2}},

(1)

we obtain

u = (\frac{a y}{x ^{2} + y ^{2}}, - \frac{a x}{x ^{2} + y ^{2}}, 0), x^{2} + y^{2} > 0.

(2)

For

a > 0

this is clockwise circulation about the axis, the velocity structure of a point vortex in each transverse plane.

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 15 / c / ii / Solution by

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For the general solution in part (i),

y (0) = A + B = 1, y^{'} (0) = - \frac{3}{4} - 2 B = 0.

(1)

Thus

B = - 3/8

and

A = 11/8

, giving the solution of the initial value problem

y (x) = \frac{3}{4} x^{2} - \frac{3}{4} x + \frac{11}{8} - \frac{3}{8} e^{- 2 x} .

(2)

Direct substitution checks both the differential equation and the two initial conditions.

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The dependent variable itself does not occur, so reduce order by putting

p = y^{'}

. Then

p^{'} + 2 p = 3 x

, a first-order linear differential equation. Its integrating factor is

e^{2 x}

(e^{2 x} p)^{'} = 3 x e^{2 x}, p = \frac{3}{2} x - \frac{3}{4} + C e^{- 2 x} .

(1)

Integrating once more and renaming the constants gives

y (x) = \frac{3}{4} x^{2} - \frac{3}{4} x + A + B e^{- 2 x} .

(2)

The two homogeneous terms are the constant solution and the exponential associated with the roots

0, - 2

of the characteristic equation.

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 15 / b / iii / Solution by

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The equation

u = v^{'}

fixes

u

from the same two constants:

u (t) = 2 A e^{2 t} - 2 B e^{- 2 t} .

(1)

Indeed

u^{'} = 4 A e^{2 t} + 4 B e^{- 2 t} = 4 v + 2

, so both original equations hold. No further independent integration constant is available in this two-dimensional linear system of ordinary differential equations.

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The characteristic equation of the homogeneous equation is

λ^{2} - 4 = 0

, giving exponential solutions

e^{2 t}

and

e^{- 2 t}

. A constant particular solution is

v_{p} = - 1/2

. Thus the general solution is

v (t) = A e^{2 t} + B e^{- 2 t} - \frac{1}{2}, A, B \in R .

(1)

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Differentiate

v^{'} = u

once, then substitute the other equation. The required second-order linear differential equation is

v^{''} = u^{'} = 4 v + 2, v^{''} - 4 v = 2.

(1)

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 14 / a / iv / Solution by

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Let

T

count tests up to and including the first failure. By independence of random variables, the event

T = j

requires

j - 1

passes and then a failure. With pass probability

p = 1/2

and failure probability

q = 1/2

, this is the geometric distribution

P (T = j) = q p^{j - 1}, j = 1, 2, \dots .

(1)

Using part (iii), its expected value is

E [T] = q j = 1 \sum \infty j p^{j - 1} = \frac{q}{( 1 - p ) ^{2}} = \frac{1}{q} = 2.

(2)

This counts the test that discovers the fault. The expected number of earlier passing tests alone would be one.

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 14 / a / iii / Solution by

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The absolutely convergent geometric series has sum

\sum_{k = 0}^{\infty} r^{k} = (1 - r)^{- 1}

for

0 < r < 1

. Its Cauchy product with itself is

(\sum_{k = 0}^{\infty} r^{k})^{2} = \sum_{m = 0}^{\infty} (\sum_{k = 0}^{m} 1) r^{m} = \sum_{m = 0}^{\infty} (m + 1) r^{m} .

(1)

There are exactly

m + 1

pairs of nonnegative integers whose sum is

m

. Relabeling

j = m + 1

proves

j = 1 \sum \infty j r^{j - 1} = \frac{1}{( 1 - r ) ^{2}} .

(2)

Absolute convergence justifies the multiplication and regrouping; no differentiation of an infinite series is needed.

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 14 / a / ii / Solution by

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Passing corresponds to

- π /6 \leq X \leq π /6

in the normalized variable from part (i). Integrating its probability density,

p_{pass} = \frac{1}{2} \int_{- π /6}^{π /6} cos x d x = sin \frac{π}{6} = \frac{1}{2} .

(1)

The complementary probability of detecting the fault in one test is also

1/2

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The probability density is symmetric about

300 mL

, vanishes at

270

and

330 mL

, and has its single maximum

π / (120 mL)

300 mL

. It is zero outside that interval. Its area is one: with the dimensionless change of variables

X = \frac{π ( v - 300 mL )}{60 mL}, p_{X} (x) = \frac{1}{2} cos x, - \frac{π}{2} \leq x \leq \frac{π}{2},

(1)

normalization is

\frac{1}{2} \int_{- π /2}^{π /2} cos x d x = 1

Figure 1.
Coffee-volume density and the two-cup tail region
. Original graphs. Left: the density with the passing interval shaded. Right: transformed independent cup volumes $X, Y$ and the triangular region $X + Y \geq π /2$ used for the probability in part (b). **Details**.

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 11 / d / iii / Solution by

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The square

d (t)^{2} = (t + 1)^{2} + 1/2

is minimized when its first term vanishes. Therefore

t_{m i n} = - 1, d_{m i n} = \frac{1}{2} = s .

(1)

This agrees with the distance between skew lines found in part (b). The corresponding parameter on the first line is its orthogonal projection value

λ_{m i n} = \frac{( r _{2} ( - 1 ) - a _{1} ) \cdot n _{1}}{∥ n _{1} ∥ ^{2}} = - \frac{3}{2} .

(2)

The closest pair is

(- 1/2, - 3/2, 0)

and

(- 1, - 1, 0)

; their connecting vector

(- 1/2, 1/2, 0)

is perpendicular to both direction vectors.

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The graph is the upper branch of the hyperbola

d^{2} - (t + 1)^{2} = 1/2

. It is symmetric about

t = - 1

, positive everywhere, and has its lowest point at

(- 1, 1/ 2)

. Its derivatives are

d^{'} (t) = \frac{t + 1}{d ( t )}, d^{''} (t) = \frac{1}{2 d ( t ) ^{3}} > 0,

(1)

so it is strictly convex. Since

d (t) - ∣ t + 1∣ = \frac{1/2}{( t + 1 ) ^{2} + 1/2 + ∣ t + 1∣} ⟶ 0,

(2)

its two oblique asymptotes are

d = t + 1 (t \to + \infty), d = - t - 1 (t \to - \infty) .

(3)

Figure 1.
Distance from a moving point to a skew line
. Original plot of $d (t) = (t + 1)^{2} + 1/2$ , its minimum and its two asymptotes. **Details**.

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For a point

p = r_{2} (t)

, the magnitude of

(p - a_{1}) \times n_{1}

is the area of a parallelogram. Dividing by its base length

∥ n_{1} ∥

gives the perpendicular Euclidean distance to the first line. Explicitly,

p - a_{1} = (t - 1, t, t + 1), (p - a_{1}) \times n_{1} = (- (t + 1), t + 1, - 1) .

(1)

Hence

d (t) = \frac{2 ( t + 1 ) ^{2} + 1}{2} = (t + 1)^{2} + \frac{1}{2} .

(2)

This use of the cross product removes any component parallel to the first line automatically.

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Here

a_{2} \cdot a_{2} = 1

a_{2} \cdot n_{2} = 1

and

n_{2} \cdot n_{2} = 3

. Therefore

d_{0} = 1 - \frac{1}{3} = \frac{2}{3} .

(1)

The closest point has parameter

t_{0} = - 1/3

and position

(- 1/3, - 1/3, 2/3)

, whose norm gives the same result.

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Use only the dot product to minimize the squared Euclidean distance from the origin:

D (t)^{2} = (a_{2} + t n_{2}) \cdot (a_{2} + t n_{2}) = ∥ a_{2} ∥^{2} + 2 t (a_{2} \cdot n_{2}) + t^{2} ∥ n_{2} ∥^{2} .

(1)

Completing the square gives

D (t)^{2} = ∥ n_{2} ∥^{2} (t + \frac{a _{2} \cdot n _{2}}{∥ n _{2} ∥ ^{2}})^{2} + ∥ a_{2} ∥^{2} - \frac{( a _{2} \cdot n _{2} ) ^{2}}{∥ n _{2} ∥ ^{2}} .

(2)

The square is minimized at

t_{0} = - (a_{2} \cdot n_{2}) /∥ n_{2} ∥^{2}

, yielding

d_{0} = a_{2} \cdot a_{2} - \frac{( a _{2} \cdot n _{2} ) ^{2}}{n _{2} \cdot n _{2}} .

(3)

Geometrically this subtracts the squared length of the orthogonal projection parallel to the line. At the minimizing point,

(a_{2} + t_{0} n_{2}) \cdot n_{2} = 0

, as required for perpendicular distance.

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 19 / a / iii / Solution by

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For every positive integer

n

, the bound

∣ sin (ωn) ∣ \leq 1

gives

∣ f (n) ∣ = A n^{p} ∣ sin (ωn) ∣ \leq A n^{p} .

(1)

For

p < - 1

, the positive p-series

\sum n^{p}

converges. The comparison test therefore proves

n = 1 \sum \infty f (n) converges absolutely, and hence converges .

(2)

No special arithmetic condition on

ω

is required for this sufficient condition.

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Use the derivative definition and the specified value

f (0) = 0

\frac{f ( h ) - f ( 0 )}{h} = A ω ∣ h ∣^{p} \frac{s i n ( ωh )}{ωh} .

(1)

The last factor tends to one. Therefore the limit is zero when

p > 0

, equals

A ω

when

p = 0

, and diverges to

+ \infty

when

p < 0

. A finite derivative exists exactly when

p \geq 0, f^{'} (0) = {0, A ω, p > 0, p = 0.

(2)

In particular

- 1 < p < 0

gives continuity without differentiability. The threshold is an example of regularity of a power times a function with a simple zero.

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 Pinned article: Introduction to the OurBigBook Project

Welcome to the OurBigBook Project! Our goal is to create the perfect publishing platform for STEM subjects, and get university-level students to write the best free STEM tutorials ever.

Everyone is welcome to create an account and play with the site: ourbigbook.com/go/register. We belive that students themselves can write amazing tutorials, but teachers are welcome too. You can write about anything you want, it doesn't have to be STEM or even educational. Silly test content is very welcome and you won't be penalized in any way. Just keep it legal!

Video 1.

Intro to OurBigBook

. Source.

We have two killer features:

topics: topics group articles by different users with the same title, e.g. here is the topic for the "Fundamental Theorem of Calculus" ourbigbook.com/go/topic/fundamental-theorem-of-calculus
Articles of different users are sorted by upvote within each article page. This feature is a bit like:
- a Wikipedia where each user can have their own version of each article
- a Q&A website like Stack Overflow, where multiple people can give their views on a given topic, and the best ones are sorted by upvote. Except you don't need to wait for someone to ask first, and any topic goes, no matter how narrow or broad
This feature makes it possible for readers to find better explanations of any topic created by other writers. And it allows writers to create an explanation in a place that readers might actually find it.
Figure 1.
Screenshot of the "Derivative" topic page
. View it live at: ourbigbook.com/go/topic/derivative
Video 2.
OurBigBook Web topics demo
. Source.
local editing: you can store all your personal knowledge base content locally in a plaintext markup format that can be edited locally and published either:
- to OurBigBook.com to get awesome multi-user features like topics and likes
- as HTML files to a static website, which you can host yourself for free on many external providers like GitHub Pages, and remain in full control
This way you can be sure that even if OurBigBook.com were to go down one day (which we have no plans to do as it is quite cheap to host!), your content will still be perfectly readable as a static site.
Figure 2.
You can publish local OurBigBook lightweight markup files to either OurBigBook.com or as a static website
.
Figure 3.
Visual Studio Code extension installation
.
Figure 4.
Visual Studio Code extension tree navigation
.
Figure 5.
Web editor
. You can also edit articles on the Web editor without installing anything locally.
Video 3.
Edit locally and publish demo
. Source. This shows editing OurBigBook Markup and publishing it using the Visual Studio Code extension.
Video 4.
OurBigBook Visual Studio Code extension editing and navigation demo
. Source.
Internal cross file references done right:
Infinitely deep tables of contents:
Figure 6.
Dynamic article tree with infinitely deep table of contents
.
Live URL: ourbigbook.com/cirosantilli/chordate
Descendant pages can also show up as toplevel e.g.: ourbigbook.com/cirosantilli/chordate-subclade

All our software is open source and hosted at: github.com/ourbigbook/ourbigbook

Further documentation can be found at: docs.ourbigbook.com

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