# Appendix-A

### Geometry

Circle of radius r: circumference $$2\pi r$$; area $$\pi r^2$$.

Sphere of radius r: area $$4\pi r^2$$ ; $$volume =\frac {4}{3}\pi r^3$$ Right circular cylinder of radius r and height h: area = $$2\pi r^2 + 2\pi r h; volume = \pi r^2 h$$ Triangle of base a and altitude h: area = $$\frac{1}{2}ah$$

If $$ax^2+bx+c =0$$, the roots are $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

### Trigonometric Functions of Angle $$\theta$$

$$\sin\theta=\frac{y}{r}$$ $$~~~\cos\theta=\frac{x}{r}$$

$$\tan\theta=\frac{y}{x}$$ $$~~~\cot\theta=\frac{x}{y}$$

$$\sec\theta=\frac{r}{x}$$ $$~~~\csc\theta=\frac{r}{y}$$ ### Pythagorean Theorem

In this right triangle,

$$a^2+ b^2 = c^2$$ ### Triangles

Angles are A, B, C Opposite sides are a, b, c Angles $$A + B + C = 180^0$$

$$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$$

$$c^2 = a^2+b^2-2ab\cos c$$

$$Exterior angle: D =A+C$$ ### Mathematical Signs and Symbols

Here are some commonly used signs and symbols in mathematics and statistics:

x equals y: $$x = y$$

x is less than y: $$x < y$$

x is greater than y: $$x > y$$

x is less than or equal to y: $$x \leq y$$

x is greater than or equal to y: $$x \geq y$$

x is not equal to y: $$x \neq y$$

x is approximately equal to y: $$x\approx y$$

x is proportional to y: $$x\propto y$$

x to the power n: $$x^{n}$$

x subscript n: $$x_{n}$$

x bar: $$\bar{x}$$

x hat: $$\hat{x}$$

a division b: $$\frac{a}{b}$$

x is a proper subset of y: $$x\subset y$$

x is not a proper subset of y: $$x\not\subset y$$

x is a subset of y: $$x\subseteq y$$

x is not a subset of y: $$x\nsubseteq y$$

X∼Normal(μ,σ) $$X \sim \text{Normal}(\mu, \sigma)$$

### Trigonometric Identities

$$\sin\theta=1/\csc\theta$$

$$\cos\theta=1/\sec\theta$$

$$\tan\theta=1/\cot\theta$$

$$\sin(90^{\circ}-\theta)=\cos\theta$$

$$\cos(90^{\circ}-\theta)=\sin\theta$$

$$\tan(90^{\circ}-\theta)=\cot\theta$$

$$\sin^2\theta+\cos^2\theta=1$$

$$\sec^2\theta-\tan^2\theta=1$$

$$\tan\theta=\sin\theta/\cos\theta$$

$$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$$

$$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$$

$$\tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}$$

$$\sin2\theta=2\sin\theta\cos\theta$$

$$\cos2\theta=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1$$

$$\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$$

$$\sin\alpha+\sin\beta=2\sin\frac{1}{2}(\alpha+\beta)\cos\frac{1}{2}(\alpha-\beta)$$

$$\cos\alpha+\cos\beta=2\cos\frac{1}{2}(\alpha+\beta)\cos\frac{1}{2}(\alpha-\beta)$$

### Derivatives

$$\frac{d}{dx}[af(x)]=a\frac{d}{dx}f(x)$$

$$\frac{d}{dx}[f(x)+g(x)]=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)$$

$$\frac{d}{dx}[f(x)g(x)]=f(x)\frac{d}{dx}g(x)+g(x)\frac{d}{dx}f(x)$$

$$\frac{d}{dx}f(u)=\left[\frac{d}{du}f(u)\right]\frac{du}{dx}$$

$$\frac{d}{dx}x^m=mx^{m-1}$$

$$\frac{d}{dx}\sin x=\cos x$$

$$\frac{d}{dx}\cos x=-\sin x$$

$$\frac{d}{dx}\tan x=\sec^2x$$

$$\frac{d}{dx}\cot x=-\csc^2x$$

$$\frac{d}{dx}\sec x=\tan x\sec x$$

$$\frac{d}{dx}\csc x=-\cot x\csc x$$

$$\frac{d}{dx}e^x=e^x$$

$$\frac{d}{dx}\ln x=\frac{1}{x}$$

$$\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$$

$$\frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}}$$

$$\frac{d}{dx}\tan^{-1}x=-\frac{1}{1+x^2}$$

### Integrals

$$\int af(x)\,dx=a\int f(x)\,dx$$

$$\int[f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx$$

$$\int x^m\,dx=\frac{x^{m+1}}{m+1}\ (m\neq1) ~~~~~~=\ln x\ (m=-1)$$

$$\int\sin{x}\,dx=-\cos x$$

$$\int\cos{x}\,dx=\sin x$$

$$\int\tan{x}\,dx=\ln|\sec x|$$

$$\int\sin^2{ax}\,dx=\frac{x}{2}-\frac{\sin{2ax}}{4a}$$

$$\int\cos^2{ax}\,dx=\frac{x}{2}+\frac{\sin{2ax}}{4a}$$

$$\int\sin{ax}\cos{ax}\,dx=-\frac{\cos2ax}{4a}$$

$$\int e^{ax}\,dx=\frac{1}{a}e^{ax}$$

$$\int xe^{ax}\,dx=\frac{e^{ax}}{a^2}(ax-1)$$

$$\int\ln{ax}\,dx=x\ln{ax}-x$$

$$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}$$

$$\int\frac{dx}{a^2-x^2}-\frac{1}{2a}\ln\left|\frac{x+a}{x-a}\right|$$

$$\int\frac{dx}{\sqrt{a^2+x^2}}=\sinh^{-1}\frac{x}{a}$$

$$\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}$$

$$\int\sqrt{a^2+x^2}\,dx=\frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\sinh^{-1}\frac{x}{a}$$

$$\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\int{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}$$

### Binomial Theorem

$$(1+x)^n=1 + \frac{nx}{1!}+\frac{n(n-1)x^2}{2!}+........(x^2 <1)$$

### Exponential Expansion

$$e^x = 1 + x + \frac{x^2}{2!} +\frac{x^3}{3!}+.........$$

### Logarithmic Expansion

$$ln(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - .....(|x| <1)$$

### Trigonometric Expansions $$(\theta ~in ~radians)$$

$$\sin\theta = \theta -\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}+......$$

$$\cos\theta = \theta -\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+......$$

$$\tan\theta = \theta + \frac{\theta^3}{3}+\frac{2\theta^5}{15}+\frac{17\theta^7}{315}+.....$$+$$\frac{2^{2n}(2^n-1)B_n\theta^{2n-1}}{(2n)!}+......$$

$$\sec\theta = 1 + \frac{\theta^2}{2}+\frac{5\theta^4}{24}+\frac{61\theta^6}{720}+.....$$$$+ \frac{E_n\theta^{2n}}{(2n)!}+......$$

where $$B_n$$ and $$E_n$$ are Bernoulli and Euler Numbers, respectively.

### Scalar Products

The dot product or scalar product of two vectors $$\vec a$$ and $$\vec b$$:

$\begin{equation} {\vec a} \cdot {\vec b} \equiv a_x b_x + a_y b_y + a_z b_z = {\rm scalar~number}. \end{equation}$

Let us rotate the coordinate axes though $$\theta$$ degrees about $$Oz$$. Then, $${\vec a} \cdot {\vec b}$$ takes the following form in the new coordinate system.

$$\vec a \cdot \vec b$$ $$=(a_x\cos\theta+a_y\sin\theta)(b_x\cos\theta + b_y\sin\theta)$$

$$(-a_x\sin\theta + a_y\cos\theta)(-b_x\sin \theta + b_y\cos\theta) +a_z b_z$$
$$=a_x b_x + a_y b_y + a_z b_z$$

Thus, $${\vec a} \cdot {\vec b}$$ is invariant under rotation about $$Oz$$. Similarly, it can easily be shown that it is also invariant under rotation about $$Ox$$ and $$Oy$$. Therefore, $${\vec a} \cdot {\vec b}$$ is a scalar, and that the above definition is well established.

It cab be easily shown that the dot product follows the commutative and distributive laws of algebra:

$${\vec a} \cdot {\vec b}$$ $${=\vec b} \cdot {\vec a},$$
$${\vec a}\cdot({\vec b}+{\vec c})$$ $${=\vec a} \cdot {\vec b} + {\vec a}\cdot {\vec c}.$$

We have shown that the dot product $${\vec a} \cdot {\vec b}$$ is coordinate independent. But what is the geometric significance of this? In the special case where $${\vec a} = {\vec b}$$, we get

$\begin{equation} {\vec a} \cdot {\vec b} = a_x^{~2}+a_y^{~2} + a_z^{~2} = \vert{\vec a}\vert^2=a^2. \end{equation}$

So, the invariance of $${\vec a} \cdot {\vec a}$$ is equivalent to the invariance of the magnitude of vector $${\bf a}$$ under transformation.

Let us now investigate the general case. The length squared of $$AB$$ in the vector triangle as shown in Figure A-4 is $\begin{equation} ({\vec b} - {\vec a} ) \cdot ({\vec b} - {\vec a} ) = \vert{\vec a}\vert^2 + \vert{\vec b}\vert^2 - 2\,{\vec a} \cdot {\vec b}. \end{equation}$

Applying the “cosine rule” of trigonometry, we get

$\begin{equation} (AB)^2 = (OA)^2 + (OB)^2 - 2 \,(OA)\,(OB)\,\cos\theta, \end{equation}$

where $$(AB)$$ denotes the length of side $$AB$$. It follows that

$\begin{equation} {\vec a} \cdot {\vec b} = {|\vec a|} {|\vec b|}\cos\theta. \end{equation}$

In this case, the invariance of $${\vec a} \cdot {\vec b}$$ under transformation is equivalent to the invariance of the angle subtended between the two vectors. Note that if $${\vec a} \cdot {\vec b} =0$$ then either $${|\vec a|}=0$$, $${|\vec b|}=0$$, or the vectors $$\vec a$$ and $$\vec b$$ are mutually perpendicular. The angle subtended between two vectors can easily be obtained from the dot product as follows:

$\begin{equation} \cos\theta = \frac{{\vec a} \cdot {\vec b}}{{|\vec a|}{|\vec b|}}. \end{equation}$

The work $$W$$ performed by a constant force $$\vec F$$ which moves an object through a displacement $$\vec r$$ is the product of the magnitude of $$\vec F$$ times the displacement in the direction of $$\vec F$$. If the angle subtended between $$\vec F$$ and $$\vec r$$ is $$\theta$$ then

$\begin{equation} W = {|\vec F|}({|\vec r|}\cos\theta) = {\vec F}\cdot {\vec r}. \end{equation}$

The work $$dW$$ performed by a non-constant force $${\vec F}$$ which moves an object through an infinitesimal displacement $$d{\vec r}$$ in a time interval $$dt$$ is $$dW={\vec F}\cdot d{\vec r}$$. Thus, the rate at which the force does work on the object, which is usually referred to as the power, is $$P = dW/dt = {\vec F}\cdot d{\vec r}/dt$$, or $$P={\vec F}\cdot {\vec v}$$, where $${\vec v} = d{\vec r}/dt$$ is the object’s instantaneous velocity.

### Vector Product

Let’s, consider, the cross product or vector product:

$\begin{equation} {\vec a}\times{\vec b} \equiv (a_y \, b_z-a_z\, b_y,\, a_z\, b_x - a_x\, b_z,\, a_x\, b_y - a_y\, b_x) ={\vec c}. \end{equation}$

Let’s rotate the coordinate axes through an angle $$\theta$$ about $$Oz$$. In the new coordinate system,

$$c_{x'}$$ $$=$$ $$(-a_x\, \sin\theta + a_y\,\cos\theta)\,b_z - a_z\,(-b_x\, \sin\theta + b_y\,\cos\theta)$$
$$=$$ $$(a_y\, b_z - a_z\, b_y)\, \cos\theta + (a_z\, b_x-a_x\, b_z)\,\sin\theta$$
$$=$$ $$c_x\,\cos\theta +c_y\,\sin\theta.$$

Thus, the $$x$$-component of $${\vec a}\times{\vec b}$$ transforms correctly. It can easily be shown that the other components transform correctly as well, and that all components also transform correctly under rotation about $$Ox$$ and $$Oy$$. Thus, $${\vec a}\times{\vec b}$$ is a proper vector. Incidentally, $${\vec a}\times{\vec b}$$ is the only simple combination of the components of two vectors that transforms like a vector (which is non-coplanar with $${\vec a}$$ and $${\vec b}$$). Unlike the dot or scalar product of vectors, the cross product does not hold commutative law,

$\begin{equation} {\vec a}\times{\vec b} = - {\vec b} \times{\vec a}, \end{equation}$

however it holds distributive law,

$\begin{equation} {\vec a}\times({\vec b} +{\vec c})= {\vec a} \times{\vec b}+{\vec a}\times{\vec c}, \end{equation}$

but it does not hold associative law,

$\begin{equation} {\vec a}\times({\vec b} \times{\vec c})\neq ({\vec a}\times{\vec b}) \times{\vec c}. \end{equation}$

The cross product transforms like a vector, which means that it must have a well-defined direction and magnitude. We can show that $${\vec a}\times{\vec b}$$ is perpendicular to both $${\vec a}$$ and $${\vec b}$$. Consider $${\vec a}\cdot {\vec a}\times{\vec b}$$. If this is zero then the cross product must be perpendicular to $${\vec a}$$. Now,

$${\vec a}\cdot {\vec a}\times{\vec b}$$ $$=$$ $$a_x\,(a_y\, b_z-a_z\, b_y) + a_y\, (a_z\, b_x- a_x \,b_z) +a_z\,(a_x \,b_y - a_y\, b_x)$$
$$=$$ $$0.$$

Therefore, $${\vec a}\times{\vec b}$$ is perpendicular to $${\vec a}$$. Likewise, it can be demonstrated that $${\bf a}\times{\vec b}$$ is perpendicular to $${\vec b}$$. The vectors $$\vec a$$, $$\vec b$$, and $${\vec a}\times{\vec b}$$ form a right-handed set, like the unit vectors $${\vec e}_x$$, $${\vec e}_y$$, and $${\vec e}_z$$. In fact, $${\vec e}_x\times {\vec e}_y={\vec e}_z$$. This defines a unique direction for $${\vec a}\times{\vec b}$$, which is obtained from a right-hand rule as shown in Figure A-5 below: Let us now calculate the magnitude of $${\vec a}\times{\vec b}$$. We have

$$({\vec a}\times{\vec b})^2$$ = $$(a_y \,b_z-a_z\, b_y)^2 +(a_z \,b_x - a_x\, b_z)^2 +(a_x \,b_y-a_y \,b_x)^2$$
$$=(a_x^{~2}+a_y^{~2}+a_z^{~2})\,(b_x^{~2}+b_y^{~2}+b_z^{~2})-(a_x\, b_x + a_y \,b_y + a_z\, b_z)^2$$
$$=\displaystyle \vert{\vec a}\vert^2 \,\vert{\vec b}\vert^2 - ({\vec a}\cdot {\vec b})^2$$

$$={|\vec a|^2}{|\vec b|^2}-({|\vec a|}\cdot{|\vec b|})^2 = {|\vec a|^2}{|\vec b|^2}-{|\vec a|^2}{|\vec b|^2}\cos^2\theta$$ $$= {|\vec a|^2}{|\vec b|^2}\sin^2\theta$$

Thus,

$\begin{equation} \vert{\vec a}\times{\vec b}\vert = \vert{\vec a}\vert\,\vert{\vec b}\vert\,\sin\theta, \end{equation}$

where $$\theta$$ is the angle subtended between $${\vec a}$$ and $${\vec b}$$. Clearly, $${\vec a}\times{\vec a} = {\vec0}$$ for any vector, since $$\theta$$ is always zero in this case. Also, if $${\vec a}\times{\vec b} = {\vec0}$$ then either $$\vert{\vec a}\vert=0$$, $$\vert{\vec b}\vert=0$$, or $${\bf b}$$ is parallel (or antiparallel) to $${\bf a}$$.

Consider the parallelogram defined by the vectors $${\vec a}$$ and $${\vec b}$$ as shown in Figure A-6 below. The scalar area of the parallelogram is $$a\,b \sin\theta$$. By convention, the vector area has the magnitude of the scalar area, and is normal to the plane of the parallelogram, in the sense obtained from a right-hand circulation rule by rotating $${\vec a}$$ on to $${\vec b}$$ (through an acute angle): i.e., if the fingers of the right-hand circulate in the direction of rotation then the thumb of the right-hand indicates the direction of the vector area. So, the vector area is coming out of the page in Figure A-6. It follows that

$\begin{equation} {\vec S} = {\vec a}\times {\vec b}, \end{equation}$

Suppose that a force $${\vec F}$$ is applied at position $${\vec r}$$ Figure A-7. The torque about the origin $$O$$ is the product of the magnitude of the force and the length of the lever arm $$OQ$$. Thus, the magnitude of the torque is $$\vert{\vec F}\vert\,\vert{\vec r}\vert\,\sin\theta$$. The direction of the torque is conventionally defined as the direction of the axis through $$O$$ about which the force tries to rotate objects, in the sense determined by a right-hand circulation rule. Hence, the torque is out of the page in Figure A-7. It follows that the vector torque is given by

$\begin{equation} \vec\tau= {\vec r}\times{\vec F}. \end{equation}$

The angular momentum, $${\vec l}$$, of a particle of linear momentum $${\vec p}$$ and position vector $${\vec r}$$ is simply defined as the moment of its momentum about the origin: i.e.,

$\begin{equation} {\vec l}={\vec r}\times {\vec p}. \end{equation}$

### Products of Vectors: Unit Vector Notation

$$\vec a \times \vec b = -\vec b \times \vec a = \begin{vmatrix} \hat i & \hat j & \hat k \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}$$

$$=\hat i\begin{vmatrix} a_y & a_z \\ b_y & b_z \end{vmatrix} -\hat j\begin{vmatrix} a_x & a_z \\ b_x & b_z \end{vmatrix}+\hat k\begin{vmatrix} a_x & a_y \\ b_x & b_y \end{vmatrix}$$

$$=\left(a_y b_z-b_y a_z\right)\hat i+\left(a_z b_x-b_z a_x\right)\hat j$$ $$+\left(a_x b_y-b_x a_y\right)\hat k$$

$$\left|\vec a \times \vec b\right|=ab\sin\theta$$

$$\vec a.(\vec b \times \vec c)=\vec b.(\vec c \times \vec a)=\vec c.(\vec a \times \vec b)$$ $$\vec a \times (\vec b \times \vec c) = (\vec a.\vec c)\vec b - (\vec a.\vec b)\vec c$$

### Cramer’s Rule

Two simultaneous equations in unknowns x and y, $$a_1x+b_1y=c_1~~~and~~~a_2x+b_2y=c_2$$

have the solutions

$$x=\frac{\begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix}}{{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}}}=\frac{c_1b_2-c_2b_2}{a_1b_2-a_2b_1}$$

and

$$y=\frac{\begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}}{{\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}}}=\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}$$