Easy to understand maths

Thursday, 7 March 2019

Posted by Parth sharma on March 07, 2019 with 1 comment

Asymptotes

A straight line L is called an asymptotes of an infinite branches of a curve if the perpendicular distance of a point P on that branch from the straight line L tend to zero as P moves to infinity along the branch. An asymptote of a curve is a line to which the curve converge. In other words, the curve and its asymptotes get infinity close but they never meet. They are use in graphing rational equation.

In most cases, the asymptotes of a curve can be found by taking the limit of a value where the function is not defined. Typical examples would be and or the point where the denominator of a rational function equals zero. Asymptotes are broadly classified into three categories
1. Horizontal asymptote
2. Vertical asymptote
3. Oblique asymptote

1. Horizontal asymptote

Horizontal asymptote is a line which is parallel to x-axis. When x goes to infinity then curve approaches some constant value.

2. Vertical asymptote

Vertical asymptote is a line which is parallel to y-axis. When x goes to infinity then curve approaches some constant value.

3. Oblique asymptote

An asymptote, which is neither parallel to x-axis nor y-axis is called an oblique asymptote. When x goes to infinity then curve goes toward a line y=mx+c.

☆Rule to find asymptotes parallel to x-axis

Equate to zero the real linear factor in the coefficient of higher power of x in the equation of the given curve.
It should be noted properly that if coefficient of higher power of x in the equation of the given curve is a constant or has no real linear factor, then the curve has no asymptote parallel to x-axis.

☆Rule to find asymptotes parallel to y-axis

Equate to zero the real linear factor in the coefficient of higher power of y in the equation of the given curve.
It should be noted properly that if coefficient of higher power of y in the equation of the given curve is a constant or has no real linear factor, then the curve has no asymptote parallel to y-axis.

Example

Q1. Find the vertical and horizontal asymptotes of the curve $3xy+5x-4y-3=0$
Sol.
The equation of the given curve is $3xy+5x-4y-3=0$ (1)
The coefficient of highest power of x in (1) is $3y+5$

$3y+5=0$ is the only asymptote parallel to x-axis
Horizontal asymptote of given curve is

$3y+5=0$

The coefficient of highest power of y in (1) is

$3x-4$

$3x-4=0$ is the only asymptote parallel to y-axis
Vertical asymptote of the given curve is $3x-4=0$ .

Oblique asymptotes

Rule to find oblique asymptotes

(1). Find $\lim_{x\rightarrow \infty }\frac{y}{x}$ is the equation of the curve and denote it by m.

(2). Find $\lim_{x\rightarrow \infty } \left ( y-mx \right )$ in the equation of the curve and denote it by c.

Then $y=mx +c$ is an asymptote of the curve $f\left ( x,y \right )=0$ .

Example

Q2. Find the asymptotes of the curve $y=\frac{x^{2}+2x-1}{x}$ .
Sol.
Here equation of the curve is $y=\frac{x^{2}+2x-1}{x}$
Or $xy=x^{2}+2x-1$
Or $x^{2}-xy+2x-1=0$
The coefficient of highest power of x is 1, which is constant.
Given curve has no asymptote parallel to x-axis

The coefficient of highest power of y is -x
The asymptote of given curve parallel to y-axis is -x=0 or x=0
Let us now find the oblique asymptote $y=mx +c$ .
To determine m and c :
$m =\lim_{x\rightarrow \infty }\frac{y}{x}=\lim_{x\rightarrow \infty }\frac{x^{2}+2x-1}{x}=\lim_{x\rightarrow \infty }\left ( 1+\frac{2}{x} -\frac{1}{x^{2}}\right )=1$
$c=\lim_{x\rightarrow \infty }\left ( y-mx \right )=\lim_{x\rightarrow \infty }\left ( \frac{x^{2}+2x-1}{x}-1.x \right )$
$=\lim_{x\rightarrow \infty }\frac{x^{2}+2x-1-x^{2}}{x}=\lim_{x\rightarrow \infty }\frac{2x-1}{x}$
$=\lim_{x\rightarrow \infty }\left ( 2-\frac{1}{x} \right )=2-0=2$
Therefore oblique asymptote is given by

$y=x+2$
Hence the given curve has two asymptotes given by $x=0$ and $y=x+2$ .

☆Rule to find the asymptotes of a rational algebraic curve :

Step1. Find   $\phi _{n}\left ( m \right ),\phi _{n-1}\left ( m \right )$ by putting x=1 and y=m in the nth degree terms and in the (n-1)th degree respectively of the given curve $f\left ( x,y \right )=0$ .

Step2. Find all real roots of   $\phi _{n}\left ( m \right )=0$ .

Step3. If m₁ is non-repeated root of   $\phi _{n}\left ( m \right )=0$ , then the corresponding to the value of c is given by $c\, \phi _{n}'\left ( m \right )+\phi _{n-1}\left ( m \right )=0$ , provided   $\phi _{n}'\left ( m_{1} \right )\neq 0$
If   $\phi _{n}'\left ( m_{1} \right )\neq 0$ , then there is no asymptote to the curve corresponding to the value m₁ of m.

Step4. If m₁ is a repeated root occurring twice, then corresponding values of c are given by

$\frac{c^{2}}{2!}\: \phi ''_{n}\left ( m \right )+c\: \phi '_{n-1}\left ( m \right )+\phi _{n-2}\left ( m \right )=0$ provided

$\phi ''_{n}\left ( m _{1}\right )\neq 0.$
In this case there are two parallel asymptotes to the curve.
Similarly we can proceed when m₁ is repeated three or more times.

Example

Q3. Find the asymptotes of the curve

$x^{3}+2x^{2}y-xy^{2}-2y^{3}+xy-y^{2}-1=0$ .
Sol.
The equation of given curve is $x^{3}+2x^{2}y-xy^{2}-2y^{3}+xy-y^{2}-1=0$ (1)
The coefficient of highest power of x in (1) is 1, which is constant.
There is no asymptotes parallel to x-axis
The coefficient of highest power of y in (1) is -2, which is constant
There is no asymptotes parallel to y-axis

For oblique asymptotes

[Putting x=1 and y=m in degree 3 term]
$\phi _{3}\left ( m \right )=1+2m-m^{2}-2m^{3}$
$\phi _{2}\left ( m \right )=m-m^{2}$
$\phi _{1}\left ( m \right )=0$
$\phi _{0}\left ( m \right )=-1$
Now $\phi _{3}\left ( m \right )=0$

$\therefore \: \: \, \: 1+2m-m^{2}-2m^{3}=0$
$\Rightarrow 1\left ( 1+2m \right )-m^{2}\left ( 1+2m \right )=0$
$\Rightarrow \left ( 1-m^{2} \right )\left ( 1+2m \right )=0$
$\Rightarrow \left ( 1-m \right )\left ( 1+m \right )\left ( 1+2m \right )=0$
$\therefore\, \, \, m=1,-1,-\frac{1}{2}$
$\phi' _{3}\left ( m \right )=2-2m-6m^{2}$
When m=1, $c=-\frac{\phi _{2}\left ( m \right )}{\phi' _{3}\left ( m \right )}=\frac{\phi _{2}\left ( 1 \right )}{\phi '_{3}\left ( 1 \right )}=-\frac{1-1}{2-2-6}=0$
Corresponding asymptotes is y=x, or x-y=0

When m=-1,

$c=-\frac{\phi _{2}\left ( m \right )}{\phi' _{3}\left ( m \right )}=\frac{\phi _{2}\left ( -1 \right )}{\phi '_{3}\left ( -1 \right )}=-\frac{-1-1}{2+2-6}=\frac{-2}{-2}=-1$
Corresponding asymptotes is y=-1x-1 i.e., x+y+1=0

When $m=-\frac{1}{2},$

$c=-\frac{\phi _{2}\left ( m \right )}{\phi' _{3}\left ( m \right )}=\frac{\phi _{2}\left ( -\frac{1}{2} \right )}{\phi '_{3}\left ( -\frac{1}{2} \right )}=-\frac{-\frac{1}{2}-\frac{1}{4}}{2+1-\frac{3}{2}}=\frac{\frac{3}{4}}{\frac{3}{2}}=\frac{1}{2}$
Corresponding asymptotes is $y=-\frac{1}{2}x+\frac{1}{2}\: \, \: i.e,\: \, x+2y=1$

Hence asymptotes of the given curve are

$x-y=0,\: \: x+y+1=0,\, \: x+2y=1$

Friday, 1 March 2019

Posted by Parth sharma on March 01, 2019 with 3 comments

Calculus

Multiple point

It refers to a point on a curve through which two or more branches of the curve pass.

Singular point

A point on the curve at which the curve behaves in an extraordinary manner is called a singular point.
There are two types of singular points:
1. Points of inflection
2. Multiple point

Point of inflection:-A curve's Inflection point is the point at which the curve's concavity changes. A point that separates the convex part of the curve from part of the curve is called a point of inflection.

Multiple point:- A point on the curve through which more than one branches of the curve pass is called a multiple point.

Double point:- A point on the curve through which two branches of the curve pass is called a double point.

Classification of double points.

There are three kinds of double points.

1. Node

A node is a point in the curve through which two real branches of the curve pass and two tangents at which are real and distinct. Thus P is a node.

2. Cusp

A double point on the curve through which two real branches of the curve pass and thw tangent at which are real and coincident is called cusp. Thus P is a cusp.

3. Conjugate point

A conjugate point on a curve is ap point in tbe neighbourhood of which there are no other real point of the curve.
The two tangent at a conjugate point are in general imaginary but sometimes they may be real.

Note 1. Conjugate point is also called isolated points.

Note 2. The determination of the nature of double point depends basically on the nature of the two branches of the curve passing through it and not on the tangents to the curve at that point. Generally when the tangent at a double point are real, the branches are also real. But there are cases, when the tangent may be real, yet the branches may be imaginary.

☆Working rule for finding the nature of origin which is a double point

Find the tangent at the origin by equating to zero the lowest degree terms in x and y of the equation of the curve. If the origin is a double point, then we shall get two tangents which may be real or imaginary.

1. If two tangent are imaginary, then origin is a conjugate point.

2. If two tangent are real and coincident, then origin is a cusp or a conjugate point.

3. If two tangent are real and distinct, then origin is a node or a conjugate point.

To be sure, examine the nature of curve in the neighbourhood of origin. If the curve has real branches through the origin, then it is a node, otherwise a conjugate point.

Example

Q1. Examine the nature of origin for the curve : $x^{4}-ax^{2}y +axy^{2}+a^{2}y^{2}=0,a>0$
Sol.
The equation of curve is

$x^{4}-ax^{2}y +axy^{2}+a^{2}y^{2}=0,a>0$

Equation to zero, the lowest degree terms, the tangent at the origin are given by

$a^{2}y^{2}=0\:\: i.e.\: \: y^{2}=0$ $i.e. \: \, y=0,\: \: y=0$
There are two real and coincident tangent at the origin
Origin is either a cusp or a conjugate point

To be sure, we study the nature of given curve near the origin.

Now: $x^{4}-ax^{2}y +axy^{2}+a^{2}y^{2}=0,a>0$
$\therefore y=\frac{ax^{2}\pm \sqrt{a^{2}x^{4}-4ax^{4}(x+a)}}{2a(x+a)}$

$\therefore y=\frac{ax^{2}\pm x^{2}\sqrt{-4ax-3a^{2}}}{2a(x+a)}$
Now, for small value of $x\neq 0,\: \: -4ax-3a^{2}$ is negative and so y is imaginary in the neighbourhood of origin.

Origin is a conjugate point.

☆Working rule for finding the position and nature of double points of the curve

Step1. Find $\frac{\partial f}{\partial x},\:\frac{\partial f}{\partial y},\: \frac{\partial^2 f}{\partial x^2},\: \frac{\partial^2f }{\partial x \partial y},\: \frac{\partial^2 f}{\partial y^2}$

Step2. Solve the equation $\frac{\partial f}{\partial x}=0\: \: and\: \: \frac{\partial f}{\partial y}=0$ to get possible double points
Reject those points which do not satisfied the equation f (x,y)=0 of the curve remaining are the double points.

Step3. At each double point, calculate $D=\left ( \frac{\partial^2 f}{\partial x^2 \partial y } \right )^2-\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2}$
(a) If D is positive, double point is a node or conjugate point.
(b) If D=0, double point is a cusp or conjugate point.

In these cases (a) and (b), find the nature by shifting the origin to the double points and then testing the nature of tangent and existence of the curve in the neighbourhood of new origin.

(c) If D is negative, double point is a conjugate point.

Example

Q2. Find the position and nature of the double points on the curve $\left ( x-2 \right )^{2}=y\left ( y-1 \right )^{2}$
Sol.
The equation of curve is $\left ( x-2 \right )^{2}=y\left ( y-1 \right )^{2}$ (1)
$\frac{\partial f}{\partial x}=2\left ( x-2 \right )$

   $\frac{\partial f}{\partial x}=-2y\left ( y-1 \right )-\left ( y-1 \right )^{2}=-\left ( y-1 \right )\left ( 3y-1 \right )$
For the double points    $\frac{\partial f}{\partial x}=0, \: \: \frac{\partial f}{\partial y},\: \: f\left ( x,y \right ) =0$
Now   $\frac{\partial f}{\partial x}=0\: \Rightarrow 2\left ( x-2 \right )=0\, \, \Rightarrow x=2$
And $\frac{\partial f}{\partial y}= 0\: \: \Rightarrow -\left ( y-1 \right )\left ( 3y-1 \right )=0\, \, \Rightarrow y=1,\frac{1}{3}$
The possible double points are $\left ( 2,1 \right )\left ( 2,\frac{1}{3} \right )$
Now   $\left ( 2,\frac{1}{3} \right )$ does not satisfy (1)

(2,1) is the only double point
$\frac{\partial^2 f}{\partial x^2}=2,\: \, \frac{\partial^2 f}{\partial y^2}=-3\left ( y-1 \right )-\left ( 3y-1 \right ),\frac{\partial^2 f}{\partial x \partial y }=0$

$\frac{\partial^2 f}{\partial x^2}=2,\: \, \frac{\partial^2 f}{\partial y^2}=-3\left ( 1-1 \right )-\left ( 3-1 \right )=-2,\frac{\partial^2 f}{\partial x \partial y }=0$
$\left ( \frac{\partial^2f }{\partial x \partial y } \right )^{2}-\frac{\partial^2 f}{\partial x^2}.\frac{\partial^2 f}{\partial x y^2}=\left ( 0 \right )^{2}-\left ( 2 \right )\left ( -2 \right )=4>0$
There is a node at the point (2,1).

Thursday, 7 March 2019

Asymptotes

1. Horizontal asymptote

2. Vertical asymptote

3. Oblique asymptote

☆Rule to find asymptotes parallel to x-axis

☆Rule to find asymptotes parallel to y-axis

Example

Oblique asymptotes

Rule to find oblique asymptotes

Example

☆Rule to find the asymptotes of a rational algebraic curve :

Example

Friday, 1 March 2019

Multiple point

Singular point

Classification of double points.

1. Node

2. Cusp

3. Conjugate point

Example

☆Working rule for finding the position and nature of double points of the curve

Example

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