Chaitanya's Random Pages

June 30, 2013

The quarter final streaks of Connors and Federer

Filed under: sport — ckrao @ 10:36 am

Roger Federer‘s quarter final streak win in grand slam tennis singles tournaments ended at 36 recently, the magnitude of which indicates his consistency, longevity and fitness. His loss in the second round at Wimbledon 2013 was interestingly his 1111th career match and it was the first time he had ever lost a second round match in a grand slam match (he was 49-0 prior to this).

Jimmy Connors had the second longest streak of 27 ending in the 4th round of Wimbledon 1983 to Kevin Curren. Due to his association with World Team Tennis, he was not allowed to participate in the French Open during his peak years, and he also chose to skip the Australian Open after 1975. This is why the streak lasted a period of 10 years (age 20 to 30). (Connors was able to remain a top ten player for 15 years!)

Federer won 141 matches in his streak of first four round wins (there were four withdrawals from opponents) while Connors won 109 (his two Australian Opens had only three rounds before the quarter final stage). Both had 22 6-0 wins.

Here are the scores of their 5-set matches prior to quarter finals in grand slam singles matches. They each had one occasion of being pushed to 10-8 in the fifth set. (Note that from 1975 to 1979 the first three rounds of US Opens were best of 3 sets.)

5-set matches of Connors:

• 1981 US Open, 3rd round vs Andres Gomez: 6-7, 6-3, 6-1, 4-6, 7-6
• 1980 French Open, 2nd round vs Jean-Francois Caujolle (FRA): 3-6, 2-6, 7-5, 6-1, 6-1
• 1978 US Open, 4th round vs Adriano Panatta: 4-6, 6-4, 6-1, 1-6, 7-5
• 1977 Wimbledon, 4th round vs Stan Smith: 7-9, 6-2, 3-6, 6-3, 6-3
• 1974 Wimbledon, 2nd round vs Phil Dent: 5-7, 6-3, 3-6, 6-3, 10-8
• 1973 Australian Open, 3rd round (R16) vs Syd Ball:  6-4, 5-7, 6-7, 6-3, 6-4

5-set matches of Federer:

• 2013 French Open, 4th round vs Gilles Simon: 6-1, 4-6, 2-6, 6-2, 6-3
• 2012 Wimbledon, 3rd round vs Julien Benneteau: 4-6, 6-7(3), 6-2, 7-6(6), 6-1
• 2011 Australian Open, 2nd round vs Gilles Simon: 6-2, 6-3, 4-6, 4-6, 6-3
• 2010 Wimbledon, 1st r0und vs Alejandro Falla: 5-7, 4-6, 6-4, 7-6(1), 6-0
• 2009 French Open, 4th round vs Tommy Haas: 6-7(4), 5-7, 6-4, 6-0, 6-2
• 2009 Australian Open, 4th round vs Tomas Berdych: 4-6, 6-7(4), 6-4, 6-4, 6-2
• 2008 US Open, 4th round vs Igor Andreev: 6-7(5), 7-6(5), 6-3, 3-6, 6-3
• 2008 Australian Open, 3rd round vs Janko Tipsarevic: 6-7(5), 7-6(1), 5-7, 6-1, 10-8
• 2006 Australian Open, 4th round vs Tommy Haas: 6-4, 6-0, 3-6, 4-6, 6-2

Hence Federer came back from 2 sets down on 4 occasions while Connors only had to do so once.

You can see their full listings of the grand slam matches of these two great players via their ATP pages here (for Connors) and here (for Federer). More statistics on Federer’s streak can be found here and here (both Wall Street Journal Daily Fix blog entries).

June 29, 2013

Line, surface and volume integrals of different types

Filed under: mathematics — ckrao @ 1:45 am

In this post we list a number of types of line, surface and volume integrals in three dimensional space which differ according to whether the integrand or differential components are scalars or vectors. Some worked examples can be found at [3] and [4].

Line Integrals

We list five types of which the first and fourth are the most commonly encountered.

1. $\displaystyle \int_C f(x,y,z) \ {\text d} \ell$

(both integrand and differential are scalars)
This is an integral with respect to arc length. An example might be to find the weight of a curve (e.g. a wire) with density given by $f(x,y,z)$. If $C$ is parametrised by $\boldsymbol{\ell}(t) = (x(t), y(t), z(t))$ for $a \leq t \leq b$, then

$\displaystyle \int_C f(x,y,z) \ {\text d} \ell = \int_a^b f(x(t),y(t),z(t)) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}\ {\text d} t.$

2. $\displaystyle \int_C \boldsymbol{f}(x,y,z) \ {\text d} \ell$

(the integrand is a vector, the differential a scalar)
An example might be finding the centre of mass of a curve (e.g. a wire). If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, then

$\displaystyle \int_C \boldsymbol{f}(x,y,z) \ {\text d} \ell = \boldsymbol{\hat{x}} \int_C f_1(x,y,z) \ {\text d} \ell + \boldsymbol{\hat{y}} \int_C f_2(x,y,z) \ {\text d} \ell + \boldsymbol{\hat{z}} \int_C f_3(x,y,z) \ {\text d} \ell.$

Hence it is the sum involving three integrals of type 1.

For finding the centre of mass, $\boldsymbol{f}(x,y,z) = \left(x \boldsymbol{\hat{x}} + y \boldsymbol{\hat{y}} + z \boldsymbol{\hat{z}}\right)\rho(x,y,z)/m$, where $\rho(x,y,z)$ is the density of the curve at $(x,y,z)$ and $m$ is its mass.

3. $\displaystyle \int_C f(x,y,z) \ {\text d} \boldsymbol{\ell}$

(the integrand is a scalar, the differential a vector)
This represents a vector sum of a curve weighted by the integrand, where in the special case $f(x,y,z) = 1$, the integral is simply the vector joining the endpoints of the curve. If $C$ is parametrised by $\boldsymbol{\ell}(t) = (x(t), y(t), z(t))$ for $a \leq t \leq b$, then

\begin{aligned} \int_C f(x,y,z) \ {\text d} \boldsymbol{\ell} &= \boldsymbol{\hat{x}} \int_a^b f(x(t),y(t),z(t)) \ \frac{{\text d} x}{{\text d} t}\ {\text d}t \ + \boldsymbol{\hat{y}} \int_a^b f(x(t),y(t),z(t)) \ \frac{{\text d} y}{{\text d} t}\ {\text d}t \\ & \quad \quad + \boldsymbol{\hat{z}} \int_a^b f(x(t),y(t),z(t)) \ \frac{{\text d} z}{{\text d} t}\ {\text d}t. \end{aligned}

4. $\displaystyle \int_C \boldsymbol{f}(x,y,z) . {\text d} \boldsymbol{\ell}$

(the integrand and differential are vectors)
This is equivalent to a scalar line integral of the tangential component of $\boldsymbol{f}$ along the curve $C$. One common use is for calculating the work done by a force field on a particle along a curve. If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, then

$\displaystyle \int_C \boldsymbol{f}(x,y,z) . {\text d} \boldsymbol{\ell} = \int_C f_1(x,y,z)\ {\text d}x + \int_C f_2(x,y,z)\ {\text d}y + \int_C f_3(x,y,z)\ {\text d}z,$

where each of the integrals can be worked out through parametrisation of the curve as in 3.

5. $\displaystyle \int_C \boldsymbol{f}(x,y,z) \times {\text d} \boldsymbol{\ell}$

(the integrand and differential are vectors)
Examples of such an integral are calculating the force on a loop carrying an electric current ($\displaystyle F = I \oint_C d\boldsymbol{\ell} \times \boldsymbol{B}$, where $I$ is the current and $\boldsymbol{B}$ the magnetic field) or simply calculating the area enclosed by a loop $\displaystyle \left( \frac{1}{2} \oint_C \boldsymbol{\ell} \times d\boldsymbol{\ell} \right)$. If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, then

$\displaystyle \int_C \boldsymbol{f} \times {\text d} \boldsymbol{\ell} = \boldsymbol{\hat{x}}\int_C f_2\ {\text d}z - f_3\ {\text d}y\ + \boldsymbol{\hat{y}} \int_C f_3\ {\text d}x - f_1\ {\text d}z \ + \boldsymbol{\hat{z}}\int_C f_1\ {\text d}y - f_2\ {\text d}x,$

where each of the integrals can be worked out through parametrisation of the curve as in 3.

Surface Integrals

We list five types analogous to the line integrals above.

1. $\displaystyle \iint_S f(x,y,z) \ {\text d} S$

(both integrand and differential are scalars)
An example might be to find the weight of a surface (e.g. a thin shell) with density given by $f(x,y,z)$. Given $S$ is parametrised by $\boldsymbol{r}(u,v)$ for $a \leq u \leq b, c \leq v \leq d$, we have

$\displaystyle \iint_S f(x,y,z)\ {\text d} S = \int_c^d \int_a^b f(x(u,v), y(u,v), z(u,v))\ \left|\frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v} \right|\ {\text d}u {\text d}v.$

2. $\displaystyle \iint_S \boldsymbol{f}(x,y,z) \ {\text d} S$

(the integrand is a vector, the differential a scalar)
An example might be finding the centre of mass of a surface (e.g. a shell). If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, then

$\displaystyle \iint_S \boldsymbol{f}(x,y,z) \ {\text d} S = \boldsymbol{\hat{x}} \iint_S f_1(x,y,z) \ {\text d} S + \boldsymbol{\hat{y}} \iint_S f_2(x,y,z) \ {\text d} S + \boldsymbol{\hat{z}} \iint_S f_3(x,y,z) \ {\text d} S.$

Hence it is the sum involving three scalar integrals of type 1.

For finding the centre of mass, $\boldsymbol{f}(x,y,z) = \left(x \boldsymbol{\hat{x}} + y \boldsymbol{\hat{y}} + z \boldsymbol{\hat{z}}\right)\rho(x,y,z)/m$, where $\rho(x,y,z)$ is the density of the surface at $(x,y,z)$ and $m$ is its mass.

3. $\displaystyle \iint_S f(x,y,z) \ {\text d} \boldsymbol{S}$

(the integrand is a scalar, the differential a vector)
The vector differential $\boldsymbol{S}$ is a vector area element of surface $S$ in a direction normal to the surface. The integral represents a vector area of a curve weighted by the integrand, where in the special case $f(x,y,z) = 1$, the components of the integral are the signed areas of the projection of the surface onto the $yz, xz$ and $xy$ planes.

If $S$ is parametrised by $\boldsymbol{r}(u,v)$ for $a \leq u \leq b, c \leq v \leq d$, and if $\boldsymbol{J}(u,v) = J_x(u,v) \boldsymbol{\hat{x}} + J_y(u,v)\boldsymbol{\hat{y}} + J_z(u,v) \boldsymbol{\hat{z}}$ is the vector Jacobian $\frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v}$, then

\begin{aligned}\iint_S f(x,y,z) \ {\text d} \boldsymbol{S} &= \int_c^d \int_a^b f(x(u,v),y(u,v),z(u,v)) \ \boldsymbol{J}(u,v)\ {\text d}u {\text d}v \\ &= \boldsymbol{\hat{x}} \int_c^d \int_a^b f(x(u,v),y(u,v),z(u,v)) J_x(u,v) \ {\text d}u {\text d}v \ + \boldsymbol{\hat{y}} \int_c^d \int_a^b f(x(u,v),y(u,v),z(u,v)) J_y(u,v) \ {\text d}u {\text d}v \\ & \ + \boldsymbol{\hat{z}} \int_c^d \int_a^b f(x(u,v),y(u,v),z(u,v)) J_z(u,v) \ {\text d}u {\text d}v.\end{aligned}

4. $\displaystyle \iint_S \boldsymbol{f}(x,y,z) . {\text d} \boldsymbol{S}$

(the integrand and differential are vectors)
This is equivalent to the flux of the vector field $\boldsymbol{f}$ through the surface $S$. One use is for calculating the rate of fluid flow through a surface given its velocity field.

If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, $S$ is parametrised by $\boldsymbol{r}(u,v)$ for $a \leq u \leq b, c \leq v \leq d$, and if $\boldsymbol{J}(u,v) = J_x(u,v) \boldsymbol{\hat{x}} + J_y(u,v)\boldsymbol{\hat{y}} + J_z(u,v) \boldsymbol{\hat{z}}$ is the vector Jacobian $\frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v}$, then we use $\boldsymbol{S} = \boldsymbol{J}(u,v) \ {\text d}u {\text d}v$ to find

\begin{aligned} \iint_S \boldsymbol{f}(x,y,z). {\text d} \boldsymbol{S} &= \int_c^d \int_a^b f_1(x(u,v), y(u,v), z(u,v)) J_x(u,v) + f_2(x(u,v), y(u,v), z(u,v)) J_y(u,v)\\ & \quad \quad +f_3(x(u,v), y(u,v), z(u,v)) J_z(u,v) \ {\text d}u {\text d}v.\end{aligned}

5. $\displaystyle \iint_S \boldsymbol{f}(x,y,z) \times {\text d} \boldsymbol{S}$

(the integrand and differential are vectors)
This is not commonly encountered, but an example of such an integral is calculating the angular velocity of a three-dimensional object with volume $V$ given a velocity field $\boldsymbol{u}$ on its surface $S$ $\displaystyle \left( \frac{1}{2V} \oint_S \boldsymbol{u} \times d\boldsymbol{S} \right)$.

If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, $S$ is parametrised by $\boldsymbol{r}(u,v)$ for $a \leq u \leq b, c \leq v \leq d$, and if $\boldsymbol{J}(u,v) = J_x(u,v) \boldsymbol{\hat{x}} + J_y(u,v)\boldsymbol{\hat{y}} + J_z(u,v) \boldsymbol{\hat{z}}$ is the vector Jacobian $\frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v}$, then we use $\boldsymbol{S} = \boldsymbol{J}(u,v) \ {\text d}u {\text d}v$ to find

\begin{aligned} \iint_S \boldsymbol{f}(x,y,z) \times {\text d} \boldsymbol{S} &= \boldsymbol{\hat{x}}\int_c^d \int_a^b f_2 J_z- f_3 J_y\ {\text d}u {\text d}v\ + \boldsymbol{\hat{y}} \int_c^d \int_a^b f_3 J_x- f_1 J_z\ {\text d}u {\text d}v\\ & \quad \quad + \boldsymbol{\hat{z}}\int_c^d \int_a^b f_1 J_y- f_2 J_x\ {\text d}u {\text d}v.\end{aligned}

Volume Integrals

In  three dimensions, the volume differential ${\text d} V$ can only be a scalar, so we have two types of integrals.

1. $\displaystyle \iiint_V f(x,y,z) \ {\text d} V$

(integrand is a scalar)
Such an integral may arise when $f(x,y,z)$ represents a density and we integrate over the volume to find the mass. In Cartesian coordinates we would simply have ${\text d}V = {\text d}x {\text d}y {\text d}z$.

2. $\displaystyle \iiint_V \boldsymbol{f}(x,y,z) \ {\text d} V$

(integrand is a vector)
An example of such an integral may be finding the centre of mass of a body. If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, then

$\displaystyle \iiint_V \boldsymbol{f}(x,y,z) \ {\text d} V = \boldsymbol{\hat{x}} \iiint_V f_1(x,y,z) \ {\text d} V + \boldsymbol{\hat{y}} \iiint_V f_2(x,y,z) \ {\text d} V + \boldsymbol{\hat{z}} \iiint_V f_3(x,y,z) \ {\text d} V.$

For finding the centre of mass, $\boldsymbol{f}(x,y,z) = \left(x \boldsymbol{\hat{x}} + y \boldsymbol{\hat{y}} + z \boldsymbol{\hat{z}}\right)\rho(x,y,z)/m$, where $\rho(x,y,z)$ is the density of the body at $(x,y,z)$ and $m$ is its mass.

References

[1] C. Tai, Generalized vector and dyadic analysis: applied mathematics in field theory, IEEE Press, 1987.

[2] K. F. Riley, M. P. Hobson and S. J. Bence, Mathematical methods for Physics and Engineering, Cambridge University Press, 2006.

[3] Physical Applications of Line Integrals – Math24.net

[4] Physical Applications of Surface Integrals – Math24.net

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