Class 12 Chapter 10 (Vector Algebra) Class Notes Part I

 Vector Algebra

 

Introduction

In this chapter we will come to know about some interesting facts of vectors and how they are used in our day-to-day life.

Following are some examples where vectors are used.

• Motion of aeroplane is based on the concept of vectors. When the motion of the aeroplane is considered both speed and in which direction it flies in the sky is also taken into account.

• The landing instructions which a pilot receives  from air traffic controller.
• Suppose when  a  ball  is  thrown  let it be thrown with some speed (for e.g.:-5m/s) and the direction in which it is thrown is let north-east .This shows we are considering a vector as the ball has both speed and direction. 

• Vectors are to derive results in engineering and science. Fluid mechanics, static, Electrical Engineering etc.
• In game  of  cricket  the  bowler  bowls  the ball with some  speed  and  in  some direction  and  also  the  batsman  hits  the  ball with  some  speed  and  in particular direction. So the  concept  of  vectors  is involved

• A motor can be either used for pumping water or in a car. Motors work on the principle of torque which is a vector quantity.

• In the  game  of  carom board  the striker is  hit with  some  speed  and  in some direction.

 

Vectors

Vectors are the  physical quantities  which  have  both direction as well as magnitude. Vectors are represented by an arrow. The point  A (as shown in the figure)   from  where   the vector  \(|\overrightarrow {AB} |\)  starts is called its  initial point. The point B where  it  ends  is  called  its  terminal point.

The distance between  initial  and  terminal points  of a  vector  is  called  the  magnitude (or length) of the vector, denoted by   \(|\overrightarrow {AB} |\)   , or   \(|\overrightarrow {a} |\)    or |a|.

The arrow indicates the direction of the vector.

 

 

Scalar & Vector

Consider the following measures and to classify them as scalar and vectors.

Right handed rectangular coordinate system

Right hand thumb rule  is  applied  to understand  the  orientation  conventions  for  vectors  in three dimensions. Suppose if we fold our 4 fingers as shown in the diagram and move from X to Y axis then the thumb gives the direction of Z-axis.

Position Vector

Consider a  point  P in  space,  having  coordinates (x, y, z)  with  respect  to  the  origin O (0, 0, 0). Then, the vector  \(|\overrightarrow {AB} |\)  having  O and P as its initial and  terminal points, respectively, is called position vector of the point P with respect to O.

The value of  \(|\overrightarrow {OP} | = \sqrt {{x^2} + {y^2} + {z^2}} \).

 

Direction Angle & Direction Cosines

Consider a  position vector \(\overrightarrow {OP} \) (or \(\overrightarrow r \)) of a point P(x, y,  z) as  shown  in the figure. The angles α, β and γ made by the vector \(\overrightarrow r \)  with the (+)ve  directions of  x,y and z-axes. The cosine values  of  these  angles, i.e.,  cos α, cos β and cos γ  are  known as direction  cosines of the vector \(\overrightarrow r \). They are denoted by (l, m and n). Therefore, cos α = l, cosβ = m and cos γ = n.

Consider the right angled triangle OAP,

\( \Rightarrow \cos \alpha  = \left( {\dfrac{x}{r}} \right)\) where (\(r = \,|\overrightarrow r |\)). Similarly from right angled triangles(OBP) and (OCP) we get  \(\cos \beta  = \left( {\dfrac{y}{r}} \right)\)  and \(\cos \gamma  = \left( {\dfrac{z}{r}} \right)\) . The coordinates of the point P can also be written as (lr, mr, nr). The lr, mr, nr  are  called as  direction  ratios of vector \(\overrightarrow r \). The direction ratios are denoted as a ,b and c respectively. 

In general l² + m² + n² = 1.

 

Problem: - 

Write two different vectors having same direction.

Answer: -

Consider \(\overrightarrow a  = \left( {\widehat i + \widehat j + \widehat k} \right)\) and \(\overrightarrow b  = \left( {2\widehat i + 2\widehat j + 2\widehat k} \right)\). The direction cosines of \(\overrightarrow a \) are given by,

\(l = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \dfrac{1}{{\sqrt 3 }}\)

\(m = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \dfrac{1}{{\sqrt 3 }}\)

\(n = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }} = \dfrac{1}{{\sqrt 3 }}\)

The direction cosines of \(\overrightarrow b \) are given by,

\(l = \dfrac{2}{{\sqrt {{2^2} + {2^2} + {2^2}} }} = \dfrac{1}{{\sqrt 3 }}\)

\(m = \dfrac{2}{{\sqrt {{2^2} + {2^2} + {2^2}} }} = \dfrac{1}{{\sqrt 3 }}\)

\(n = \dfrac{2}{{\sqrt {{2^2} + {2^2} + {2^2}} }} = \dfrac{1}{{\sqrt 3 }}\)

The direction cosines of \(\overrightarrow a \) and \(\overrightarrow b \) are the same. Hence, the two vectors have the same direction.

Problem:-

Find the direction cosines of the vector î +2ĵ + 3k̂

Answer:-

Let \(\overrightarrow a \) = î +2ĵ + 3k̂.

Therefore \(|\overrightarrow a |\,\, = \dfrac{1}{{\sqrt {{1^2} + {2^2} + {3^2}} }} = \dfrac{1}{{\sqrt {1 + 4 + 9} }} = \dfrac{1}{{\sqrt {14} }}.\)

Hence, the direction cosines of \(\overrightarrow a \) are \(\dfrac{1}{{\sqrt {14} }},\,\,\dfrac{2}{{\sqrt {14} }},\,\,\dfrac{3}{{\sqrt {14} }}.\)

 

Types of Vectors

• Zero Vector
• Unit Vector
• Coinitial Vector
• Collinear Vector
• Equal Vectors
• Negative Vectors
• Free Vectors

Zero Vector:-

A vector, whose initial and terminal points coincide, is called a zero vector or (null vector). It is denoted by \(\overrightarrow 0 \). Zero vector cannot be assigned a definite direction as it has zero magnitude.

Unit Vector:-

A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. The unit vector in the direction of a given vector \(\overrightarrow a \)  is denoted by \({\widehat a}\).

Coinitial Vectors:-

Two or more vectors having same  initial point are called Coinitial vectors.

 

Collinear Vectors:-

Two or more vectors are said  to be  collinear  if they are parallel to the same line, irrespective of their magnitudes and directions.

For example: - Consider 3 vectors as shown in the figure, they all are parallel to each other but their magnitudes are different as well as the directions. But they are said to be collinear vectors because they are parallel to each other.

 

Equal Vectors:-

Two vectors are said to be equal, if they have the same magnitude and direction regardless of the positions of their initial points.

For example: - Consider 2 vectors whose magnitudes and their directions are same irrespective of origin, then they are known as equal vectors.

 

Negative of a Vector:-

A vector whose magnitude is same as that of a given vector but direction is opposite to that of it is called negative of the given vector.

 

Free Vectors:-

Vectors that don’t change even if it is displaced  in  parallel direction without changing  its  magnitude and direction are  called  free vectors.

Problem:- Represent graphically a displacement of 40 km, 30° west of south.

Answer: - 

The vector \(\overrightarrow {OP} \) represents the required displacement.

 

Addition of Vectors

Consider a scenario when a man moves from A to B and then from B to C (as shown in the fig).

Then the  net displacement made by a man from point A to point C, is given by the vector \(\overrightarrow {AC} \) and expressed as:-

                   \(\overrightarrow {AC}  = \overrightarrow {AB}  + \overrightarrow {BC} \) 

This is  known as  the  triangle law of addition.

Head to tail vector addition

To add two vectors, position them so that the initial point of one coincides with the terminal of the other.

Consider two vectors \(\overrightarrow a \) and \(\overrightarrow b \) and move them parallel to each other such that tail of \(\overrightarrow b \)  touches  the  head  of \(\overrightarrow a \). Then the  resultant vector will be given as vector (\(\overrightarrow a  + \overrightarrow b \)). Also \(\overrightarrow {AC}  =  - \,\overrightarrow {CA} \)  from equation (\(\overrightarrow {AC}  = \overrightarrow {AB}  + \overrightarrow {BC} \)).

\(\overrightarrow {AA}  = \overrightarrow {AB}  + \overrightarrow {BC}  + \overrightarrow {CA}  = 0\). This implies when we consider the sides of the triangle in order, it leads to zero resultant as both initial and terminal points coincides with each other. This is shown in the figure (ii).

Constructing a vector as shown in the figure (iii) \(\overrightarrow {BC'} \)  and  the  magnitude  of \(\overrightarrow {BC'} \)  is  same as \(\overrightarrow {BC} \)  but the direction is opposite  to  the vector \(\overrightarrow {BC} \), i.e.

\(\overrightarrow {BC'}  =  - \,\overrightarrow {BC} \)

By applying triangle law of addition we have, \(\overrightarrow {AC'}  = \overrightarrow {AB}  + \overrightarrow {BC} \).

\[ = \overrightarrow {AB}  + ( - \,\overrightarrow {BC} )\]

\[\overrightarrow {AC'}  = \overrightarrow a  - \,\overrightarrow b \]

Therefore vector  \(\overrightarrow {AC'} \) is  said to represent the  difference  of  \(\overrightarrow a \) and \(\overrightarrow b \).

 

 

Parallelogram law of vector addition:-

If we have two vectors (\(\overrightarrow a \) and \(\overrightarrow a \)) represented by the two adjacent sides of a parallelogram in magnitude and direction, then their sum (\(\overrightarrow a  + \overrightarrow b \))  is represented in magnitude and direction by the diagonal of the parallelogram through common point.

By using parallelogram law of vector, it will be  given as: \(\overrightarrow {OC}  = \overrightarrow {OA}  + \overrightarrow {AC} \)     (Fig)

In triangle  OAC  applying triangle law of addition,

 

\(\overrightarrow {OC}  = \overrightarrow {OA}  + \overrightarrow {AC} \)

Therefore, \(\overrightarrow {OC}  = \overrightarrow {a}  + \overrightarrow {b} \)

Again applying triangle law of addition in triangle OCB,

\(\overrightarrow {OC}  = \overrightarrow {OB}  + \overrightarrow {CB} \)

\(\overrightarrow {OC}  = \overrightarrow {a}  + \overrightarrow {b} \)

Hence proved.

Problem:- Find the sum of the vectors \(\overrightarrow a  = \left( {\widehat i - 2\widehat j + \widehat k} \right)\) and \(\overrightarrow b  = \left( { - 2\widehat i - 4\widehat j + 5\widehat k} \right)\) and \(\overrightarrow c  = \left( {\widehat i - 6\widehat j - 7\widehat k} \right)\).

Answer:-

The given vectors are \(\overrightarrow a  = \left( {\widehat i - 2\widehat j + \widehat k} \right)\) and \(\overrightarrow b  = \left( { - 2\widehat i - 4\widehat j + 5\widehat k} \right)\) and \(\overrightarrow c  = \left( {\widehat i - 6\widehat j - 7\widehat k} \right)\).

Therefore \(\overrightarrow a  + \overrightarrow b  + \overrightarrow c  = (1 - 2 + 1)\widehat i + ( - 2 + 4 - 6)\widehat j + (1 + 5 - 7)\widehat k\)

 \( = \left( {0.\widehat i - 4\widehat j - 1\widehat k} \right)\)

\({ =  - 4\widehat j - \widehat k}\) .

 

Properties of vector addition

Commutative property: - For any two vectors \(\overrightarrow a \) and \(\overrightarrow b \)  if we add (\(\overrightarrow a  + \overrightarrow b \))  or (\(\overrightarrow b  + \overrightarrow b \) ) ; the result will be same i.e. (\(\overrightarrow a  + \overrightarrow b \)) = (\(\overrightarrow a  + \overrightarrow b \)).

 

Associative property:-

For any three vectors \(\overrightarrow a \),  \(\overrightarrow b \)  and  \(\overrightarrow c \)  then: 

\((\overrightarrow a  + \overrightarrow b ) + \overrightarrow c  = \overrightarrow a  + (\overrightarrow b  + \overrightarrow c )\)

If we  add  first \(\overrightarrow a \) and  \(\overrightarrow b \)  and  then add  the result to \(\overrightarrow c \) it  will  be same  if we first add  (\(\overrightarrow b \)  and \(\overrightarrow c \))  and  then  adding  the  result  the vector \(\overrightarrow a \).

 

Additive Identity:- Suppose if  add  \(\overrightarrow 0 \)  to any vector \(\overrightarrow a \) then  we  will get  the  vector \(\overrightarrow a \).

(\(\overrightarrow a \) + \(\overrightarrow 0 \)) = (\(\overrightarrow 0 \) + \(\overrightarrow a \)) = \(\overrightarrow a \).

Therefore \(\overrightarrow 0 \)  is  the  additive  identity  for  the  vector addition.

Multiplication of a Vector by a Scalar:-

Let \(\overrightarrow a \) be any vector and  λ  be a scalar. Multiplication of vector \(\overrightarrow a \) by  a scalar  λ  is  denoted as : \(\lambda \overrightarrow a \). \(\lambda \overrightarrow a \) is also a vector which  has  the  direction same or opposite to that of vector \(\overrightarrow a \).

The magnitude of vector  (\(\lambda \overrightarrow a \)  is given as : \(|\lambda \overrightarrow a |\, = |\lambda |\,\,|\overrightarrow a |\) 

Geometrically can be shown as:-

Additive Inverse: -

If we add a vector \(\overrightarrow a \)  with  itself  but in opposite direction then the resultant is \(\overrightarrow a \). If λ (scalar) = – 1, then \(\lambda \overrightarrow a  =  - \overrightarrow a \), the magnitude of negative  vector  is same as  \(\overrightarrow a \)  but  direction  is  opposite  to \(\overrightarrow a \). The vector \(\left( { - \overrightarrow a } \right)\) is  called  the negative (or additive inverse)  of vector \(\overrightarrow a \). It is given  as \(\overrightarrow a \) + (\( - \overrightarrow a \)) = (\( - \overrightarrow a \)) + \(\overrightarrow a \) = \(\overrightarrow 0 \).

Unit Vector in a direction

Unit vector is a vector of magnitude 1. It is denoted by  \({\widehat a}\).

Let λ(scalar) \( = \left( {\dfrac{1}{{|\overrightarrow a |}}} \right)\), then \(|\lambda \overrightarrow a |\, = |\lambda |\,\,|\overrightarrow a |\). \( \Rightarrow \left( {\dfrac{1}{{|\overrightarrow a |}}} \right)|\overrightarrow a |\, = 1\). Therefore  \(\lambda \overrightarrow a \) represents the unit vector in the direction of \(\overrightarrow a \).

Unit vector can be written as:  \(\widehat a = \left( {\dfrac{1}{{|\overrightarrow a |}}} \right) \times \overrightarrow a \).

 

Components of a vector

Unit Vector

 

Consider points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on x-axis y-axis and  z-axis, respectively. \(|\overrightarrow {OA} |\, = 1,\,\,|\overrightarrow {OB} |\, = 1\) and \(|\overrightarrow {OC} |\, = 1\) , this shows all the 3 vectors having  the same magnitude 1.

The vectors \(\overrightarrow {OA} ,\,\,\,\overrightarrow {OB} \,\,\,and\,\,\overrightarrow {OC} \)  are known as unit vectors along the axes OX, OY and OZ  respectively  and  they are denoted by î , ĵ  and  k̂ respectively.

Problem:-

Find the unit vector in the direction of the vector \({\overrightarrow a }\) = (î + ĵ +2 k̂).

Answer:

The unit vector n̂ in the direction of vector \({\overrightarrow a }\) = (î + ĵ +2 k̂).

\(|\overrightarrow a | = \sqrt {{1^2} + {1^2} + {2^2}}  = \sqrt 6 \).

Therefore \(\widehat a = \left( {\dfrac{{\overrightarrow a }}{{|\overrightarrow a |}}} \right)\)

\( \Rightarrow \widehat a = \dfrac{{\widehat i + \widehat j + 2\widehat k}}{{\sqrt 6 }} = \dfrac{1}{{\sqrt 6 }}\widehat i + \dfrac{1}{{\sqrt 6 }}\widehat j + \dfrac{2}{{\sqrt 6 }}\widehat k\)

Problem:-

Find the unit vector in the direction of vector a + b, \(\overrightarrow a  = (2\widehat i - \widehat j + 2\widehat k)\) and \(\overrightarrow b  = (- \widehat i + \widehat j - \widehat k)\).

Answer:-

The given vectors are \(\overrightarrow a  = (2\widehat i - \widehat j + 2\widehat k)\) and \(\overrightarrow b  = (- \widehat i + \widehat j - \widehat k)\).

Therefore \(\overrightarrow a  + \overrightarrow b  = (2 - 1)\widehat i + ( - 1 + 1)\widehat j + (2 - 1)\widehat k\)

\( = (1)\widehat i + (0)\widehat j + (1)\widehat k\)

= î + k̂

\(|\overrightarrow a  + \overrightarrow b | = \sqrt {{1^2} + {1^2}}  = \sqrt 2 \)

Hence, the unit vector in the direction of is \(\overrightarrow a  + \overrightarrow b \) is

\(\dfrac{{\overrightarrow a  + \overrightarrow b }}{{|\overrightarrow a  + \overrightarrow b |}} = \dfrac{{\widehat i + \widehat k}}{{\sqrt 2 }} = \dfrac{{\widehat i}}{{\sqrt 2 }} + \dfrac{{\widehat k}}{{\sqrt 2 }}\)

Position Vector:-

Consider a position vector \(\overrightarrow {OP} \) of a point P(x, y, z) as shown in the figure. Let P₁ = perpendicular from P onto the plane XOY .This shows  P₁P  is  parallel  to the z-axis. Also from the figure we can see that the î , ĵ and k̂  are the unit vectors along  x, y and z-axes respectively.

 

By the definition of the coordinates of P,  \(\overrightarrow {{P_1}P}  = \overrightarrow {OR}  = z\widehat k\) . Similarly, \(\overrightarrow {Q{P_1}}  = \overrightarrow {OS}  = y\widehat j\) and \(\overrightarrow {OQ}  = x\widehat i\) .

Therefore \(\overrightarrow {O{P_1}}  = \overrightarrow {OQ}  + \overrightarrow {Q{P_1}}  = x\widehat i + y\widehat j\)  and \(\overrightarrow {OP}  = \overrightarrow {O{P_1}}  + \overrightarrow {{P_1}P}  = x\widehat i + y\widehat j + z\widehat k\).

Hence, the position vector of P with reference to O is given as \(\overrightarrow {OP} \) or \(\overrightarrow r  = x\widehat i + y\widehat j + z\widehat k\). Where r = position vector OP. This form of any vector is known as component form. x, y and z are called scalar components of vector \(\overrightarrow r \). x î + yĵ  + z k̂  are called vector components of vector \(\overrightarrow r \) along the 3 respective axes.

The length of any vector r⃗  = x î  + yĵ  + z k̂  is given as | \(\overrightarrow r \)| = \(\sqrt {{x^2} + {y^2} + {z^2}} \).

Problem: - Compute the magnitude of the following vectors: \(\overrightarrow a  = \widehat i + \widehat j + \widehat k\) and \(\overrightarrow b  = 2 \widehat i - 7 \widehat j - 3 \widehat k\).

Answer:

The given vectors are: \(\overrightarrow a  = \widehat i + \widehat j + \widehat k\) and \(\overrightarrow b  = 2 \widehat i - 7 \widehat j - 3 \widehat k\).

\(|\overrightarrow a | = \sqrt {{1^2} + {1^2} + {1^2}}  = \sqrt 3 \)

\(|\overrightarrow b | = \sqrt {{{(2)}^2} + {{( - 7)}^2} + {{( - 3)}^2}}  = \sqrt {62} \)

 

Problem: -

Write two different vectors having same magnitude.

Answer: -

Consider \(\overrightarrow a  = \widehat i - 2\widehat j + 3\widehat k\) and \(\overrightarrow b  = 2\widehat i + \widehat j - 3\widehat k\).

It can be observed that:

\(|\overrightarrow a | = \sqrt {{{(1)}^2} + {{( - 2)}^2} + {{(3)}^2}}  = \sqrt {1 + 4 + 9}  = \sqrt {14} \) and

 \(|\overrightarrow b | = \sqrt {{{(2)}^2} + {{(1)}^2} + {{( - 3)}^2}}  = \sqrt {4 + 1 + 9}  = \sqrt {14} \).

Hence, \(\overrightarrow a \) and \(\overrightarrow b \) are two different vectors having the same magnitude. The vectors are different because they have different directions.

 

Components of a vector: Vectors Operations

Let there be two vectors given by \(\overrightarrow a  = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\) and \(\overrightarrow b  = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\)

• Sum of vectors a and b is given by: \(\overrightarrow a  + \overrightarrow b  = ({a_1} + {a_2})\widehat i + ({b_1} + {b_2})\widehat j + ({a_3} + {b_3})\widehat k\)
• Difference of vectors a and b is given by: \(\overrightarrow a  - \overrightarrow b  = ({a_1} - {a_2})\widehat i + ({b_1} - {b_2})\widehat j + ({a_3} - {b_3})\widehat k\)
• Vectors a  & b are equal if and only if a₁ = b₁, a₂ = b₂ and a₃ = b₃.
• Multiplication of vector by Scalar \(\lambda \overrightarrow a  = (\lambda {a_1})\widehat i + (\lambda {a_2})\widehat j + (\lambda {a_3})\widehat k\)

Distributive law

Let \(\overrightarrow a \) and \(\overrightarrow b \)  be any two vectors, and k and m be any scalars. Then

\[k\overrightarrow a  + m\overrightarrow a  = (k + m)\overrightarrow a \]

\[km\overrightarrow a  = (km)\overrightarrow a \]

\[k(\overrightarrow a  + \overrightarrow b ) = k\overrightarrow a  + k\overrightarrow b \]

Collinear Vectors

For any value of λ, the vector (λa) is always collinear to the vector \(\overrightarrow a \).

Two vectors \(\overrightarrow a \) and \(\overrightarrow a \) are collinear if and only if there exists a nonzero scalar λ such that  \(\overrightarrow b  = \lambda \overrightarrow a \).

\( \Rightarrow {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k = \lambda ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k)\)  where \(\overrightarrow b  = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\) and \(\overrightarrow a  = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\)

\( \Rightarrow {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k = (\lambda {a_1})\widehat i + (\lambda {a_2})\widehat j + (\lambda {a_3})\widehat k\)

\( \Rightarrow {b_1} = \lambda {a_1},{b_2} = \lambda {a_2},{b_3} = \lambda {a_3}\)

Therefore, \(\left( {\dfrac{{{b_1}}}{{{a_1}}}} \right) = \left( {\dfrac{{{b_2}}}{{{a_2}}}} \right) = \left( {\dfrac{{{b_3}}}{{{a_3}}}} \right) = \lambda \)

Note:-

• If \(\overrightarrow a  = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\), then a₁, a₂ and a₃  are  also  called as direction ratios of \(\overrightarrow a \).
• If l, m, n are direction cosines of a vector, then  \(l\widehat i + m\widehat j + n\widehat k = (\cos \alpha )\widehat i + (\cos \beta )\widehat j + (\cos \gamma )\widehat k\) is the unit vector in the direction of that vector, where α, β and γ are the angles which the vector makes with x, y and z axes respectively.

 

Problem:-

Show that the vectors \(2\widehat i - 3\widehat j + 4\widehat k\) and \( - 4\widehat i + 6\widehat j - 8\widehat k\) are collinear.

Answer:-

Let \(\overrightarrow a  = 2\widehat i - 3\widehat j + 4\widehat k\) and \(\overrightarrow b  =  - 4\widehat i + 6\widehat j - 8\widehat k\).

It is observed  that \(\overrightarrow b  =  - 4\widehat i + 6\widehat j - 8\widehat k =  - 2(2\widehat i - 3\widehat j + 4\widehat k) =  - 2\overrightarrow a \)

Therefore \(\overrightarrow b  = \lambda \overrightarrow a \) where λ = - 2

Hence, the given vectors  are  collinear.

Problem:-

Find a vector in the direction of vector \(5\widehat i - \widehat j + 2\widehat k\) which has magnitude 8 units.

Answer:-

Let \(\overrightarrow a  = 5\widehat i - \widehat j + 2\widehat k\), So, \(|\overrightarrow a |\,\, = \sqrt {{5^2}{{( - 1)}^2} + {2^2}}  = \sqrt {25 + 1 + 4}  = \sqrt {30} \)

Therefore \(\widehat a = \left( {\dfrac{{\overrightarrow a }}{{|\overrightarrow a |}}} \right) = \dfrac{{5\widehat i - \widehat j + 2\widehat k}}{{\sqrt {30} }}\)

Hence, the vector in the direction of vector \({5\widehat i - \widehat j + 2\widehat k}\) which has magnitude 8 units is given by,

\(8\widehat a = 8\left( {\dfrac{{5\widehat i - \widehat j + 2\widehat k}}{{\sqrt {30} }}} \right) = \dfrac{{40}}{{\sqrt {30} }}\widehat i - \dfrac{8}{{\sqrt {30} }}\widehat j + \dfrac{{16}}{{\sqrt {30} }}\widehat k\).

Problem:-

Show that the vector \(\widehat i + \widehat j + \widehat k\) is equally inclined to the axes OX, OY, and OZ.

Answer:-

Let \(\overrightarrow a  = \widehat i + \widehat j + \widehat k\). Then, \(|\overrightarrow a |\,\, = \sqrt {{1^2} + {1^2} + {1^2}}  = \sqrt 3 \).

Therefore, the direction cosines of \(\overrightarrow a \) are \(\dfrac{1}{{\sqrt 3 }}\),  \(\dfrac{1}{{\sqrt 3 }}\),  \(\dfrac{1}{{\sqrt 3 }}\).

Now, let α, β and γ be the angles formed by with the positive directions of x, y, and z axes.

\(\cos \alpha  = \dfrac{1}{{\sqrt 3 }},\,\,\cos \beta  = \dfrac{1}{{\sqrt 3 }},\,\,\cos \gamma  = \dfrac{1}{{\sqrt 3 }}\)

Hence, the given vector is equally inclined to axes OX, OY and OZ.

 

Vectors joining two points

Let P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂)  are  initial point and terminal point respectively, then the vector joining P₁ and P₂ is given by vector \(\overrightarrow {{P_1}{P_2}} \).

Joining the points P₁ and P₂ with the origin O, and applying triangle law of addition. Then from triangle OP₁P₂(figure) : \(\overrightarrow {O{P_1}}  + \overrightarrow {{P_1}{P_2}}  = \overrightarrow {O{P_2}} \).

The vector will be  always specified as:  (Terminal point – Initial point).

This implies \(\overrightarrow {{P_1}{P_2}}  = \overrightarrow {O{P_2}}  - \overrightarrow {O{P_1}} \).

Therefore \(\overrightarrow {{P_1}{P_2}}  = \left( {{x_2}\widehat i + {y_2}\widehat j + {z_2}\widehat k} \right) - \left( {{x_1}\widehat i + {y_1}\widehat j + {z_1}\widehat k} \right) = ({x_2} - {x_1})\widehat i + ({y_2} - {y_1})\widehat j + ({z_2} - {z_1})\)

The magnitude of \(\overrightarrow {{P_1}{P_2}} \) is given as: \(|\overrightarrow {{P_1}{P_2}} | = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \)

 

Problem:-

Find the vector with initial point (2, 1) and terminal point (-5, 7).

Answer:-

 \(\overrightarrow {AB}  = ( - 5 - 2)\widehat i + (7 - 1)\widehat j =  - 7\widehat i + 6\widehat j\)

Problem:-

Find the direction cosines of the vector joining the points A (1, 2, – 3) and B (– 1, – 2, 1).

Answer:-

The given points are A (1, 2, – 3) and B (– 1, – 2, 1).

Therefore \(\overrightarrow {AB}  = ( - 1 - 1)\widehat i + ( - 2 - 2)\widehat j + (1 + 3)\widehat k =  - 2\widehat i - 4\widehat j + 4\widehat k\)

Therefore \(|\overrightarrow {AB} | = \sqrt {{{( - 2)}^2} + ( - 4) + {4^2}}  = \sqrt {4 + 16 + 16}  = \sqrt {36}  = 6\)

Hence, the direction cosines of \(\overrightarrow {AB} \) are \(\dfrac{{ - 2}}{6}\),  \(\dfrac{{ - 4}}{6}\),  \(\dfrac{{4}}{6}\) = \(\dfrac{{ - 1}}{3}\), \(\dfrac{{ - 2}}{3}\), \(\dfrac{{2}}{3}\).

Problem:-

Find  the  unit vector  in the direction of vector \(\overrightarrow {PQ} \), where P and Q are the points (1, 2, 3) and  (4, 5, 6), respectively.

Answer:-

The given points are P (1, 2, 3 ) and Q (4, 5, 6 ).

Therefore \(\overrightarrow {PQ}  = (4 - 1)\widehat i + (5 - 2)\widehat j + (6 - 3)\widehat k = 3\widehat i + 3\widehat j + 3\widehat k\)

\(|\overrightarrow {PQ} | = \sqrt {{3^2} + {3^2} + {3^2}}  = \sqrt {27}  = 3\sqrt 3 \)

Hence, the unit vector in the direction of \(\overrightarrow {PQ} \) is

\(\dfrac{{\overrightarrow {PQ} }}{{|\overrightarrow {PQ} |}} = \dfrac{{3\widehat i + 3\widehat j + 3\widehat k}}{{3\sqrt 3 }} = \dfrac{3}{{3\sqrt 3 }}\widehat i + \dfrac{3}{{3\sqrt 3 }}\widehat j + \dfrac{3}{{3\sqrt 3 }}\widehat k\)

\( = \dfrac{1}{{\sqrt 3 }}\widehat i + \dfrac{1}{{\sqrt 3 }}\widehat j + \dfrac{1}{{\sqrt 3 }}\widehat k\)

Problem:-

Show that the points A, B and C with position vectors, \(\overrightarrow a  = 3\widehat i - 4\widehat j - 4\widehat k,\,\,\,\overrightarrow b  = 2\widehat i - \widehat j + \widehat k\) and \(\overrightarrow c  = \widehat i - 3\widehat j - 5\widehat k\) respectively form the vertices of a right angled triangle.

Answer:-

Position vectors of points A, B, and C are respectively given as:

\(\overrightarrow a  = 3\widehat i - 4\widehat j - 4\widehat k,\,\,\,\overrightarrow b  = 2\widehat i - \widehat j + \widehat k,\,\,\,\overrightarrow c  = \widehat i - 3\widehat j - 5\widehat k\)

Therefore \(\overrightarrow {AB}  = (\overrightarrow b  - \overrightarrow a ) = (2 - 3)\widehat i + ( - 1 + 4)\widehat j + (1 + 4)\widehat k =  - \widehat i + 3\widehat j + 5\widehat k\)

\(\overrightarrow {BC}  = (\overrightarrow c  - \overrightarrow b ) = (1 - 2)\widehat i + ( - 3 + 1)\widehat j + ( - 5 - 1)\widehat k =  - \widehat i - 2\widehat j - 6\widehat k\)

\(\overrightarrow {CA}  = (\overrightarrow a  - \overrightarrow c ) = (3 - 1)\widehat i + ( - 4 + 3)\widehat j + ( - 4 + 5)\widehat k = 2\widehat i - \widehat j + \widehat k\)

Therefore

\(|\overrightarrow {AB} {|^2}\) = (-1)²+ (3)²+ (5)² = 35

\(|\overrightarrow {BC} {|^2}\) = (-1)²+ (-2)²+ (-6)² = 41

\(|\overrightarrow {CA} {|^2}\) = (2)²+ (-1)²+ (1)² = 6

Therefore \(|\overrightarrow {AB} {|^2}\) + \(|\overrightarrow {CA} {|^2}\) = 36 + 6 = \(|\overrightarrow {BC} {|^2}\)

Hence, ABC is a right-angled triangle.

 

 Section Formula

Let P and Q be two points represented by the position vectors  \(\overrightarrow {OP} \)  and \[\overrightarrow {OQ} \) respectively w.r.t. to origin.

Then the line segment which joins the points P and Q can be divided by the third point R in two ways :- Internally and Externally.

Case 1:- When R divides PQ  internally ,

 

If R divides \(\overrightarrow {PQ} \) such that m (\(\overrightarrow {QR} \)) = n (\[\overrightarrow {PR} \)),

Where m and n  are positive scalar quantities.

Suppose R divides  \(\overrightarrow {PQ} \)  internally in the ratio m : n.

Consider triangles ORQ  and  OPR (from the figure):

\(\overrightarrow {RQ}  = \overrightarrow {OQ}  - \overrightarrow {OR}  = \overrightarrow b  - \overrightarrow r \)  (Equation(1))

\(\overrightarrow {PR}  = \overrightarrow {OR}  - \overrightarrow {OP}  = \overrightarrow r  - \overrightarrow a \)  (Equation (2))

Therefore from (1) and (2) after simplification

\(m(\overrightarrow b  - \overrightarrow r ) = n(\overrightarrow r  - \overrightarrow a ) \Rightarrow \overrightarrow r  = \dfrac{{m\overrightarrow b  + n\overrightarrow a }}{{m + n}}\)

Hence the  position  vector of the  point R which divides P and Q  internally in the ratio  (m:n) is given by:  \(\overrightarrow {OR}  = \dfrac{{m\overrightarrow b  + n\overrightarrow a }}{{m + n}}\) 

Case 2:-When R divides PQ externally,

 

Let we have 2 vectors P and R and the point  R is present outside the \(\overrightarrow {PQ} \).

The \(\overrightarrow {PQ} \) is  divided  externally in the ratio (m : n).

Therefore \(\dfrac{{\overrightarrow {PQ} }}{{\overrightarrow {QR} }} = \dfrac{m}{n}\)  (equation (1))

From the figure we have : \(\overrightarrow a  + \overrightarrow {PR}  = \overrightarrow r \)

\( \Rightarrow \overrightarrow {PR}  = \overrightarrow r  - \overrightarrow a \).

Also \(\overrightarrow b  + \overrightarrow {QR}  = \overrightarrow r \)

\( \Rightarrow \overrightarrow {QR}  = \overrightarrow r  - \overrightarrow b \)

From equation(1) :  \(\dfrac{{\overrightarrow r  - \overrightarrow a }}{{\overrightarrow r  - \overrightarrow b }} = \dfrac{m}{n}\)

On simplification we will get \(\overrightarrow r  = \dfrac{{m\overrightarrow b  - n\overrightarrow a }}{{m - n}}\)

Hence the position vector of the point R which divides P and Q externally in the ratio (m:n) is given by : \(\overrightarrow OR  = \dfrac{{m\overrightarrow b  - n\overrightarrow a }}{{m - n}}\) 

Problem:-

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are  (î + 2ĵ - k̂)  and (-î +  ĵ +   k̂) respectively, in the ration 2 : 1.

(i) Internally   (ii) Externally

Answer:

The position vector of point R dividing the line segment joining two points, P and Q  in the ratio m: n is given by:

Internally: \(\dfrac{{m\overrightarrow b  + n\overrightarrow a }}{{m + n}}\)

Externally: \(\dfrac{{m\overrightarrow b  - n\overrightarrow a }}{{m - n}}\) 

Position vectors  of P and Q are given as:

\(\overrightarrow {OP}  = \widehat i + 2\widehat j - \widehat k\) and \[\overrightarrow {OQ}  = - \widehat i + \widehat j + \widehat k\)

(i) The position vector of point R  which divides the line joining two  points P and Q internally in the ratio 2 : 1 is given by,

\(\overrightarrow {OR}  = \dfrac{{2( - \widehat i + \widehat j + \widehat k) + 1(\widehat i + 2\widehat j - \widehat k)}}{{2 + 1}} = \dfrac{{ - 2\widehat i + 2\widehat j + 2\widehat k + \widehat i + 2\widehat j - \widehat k}}{3}\)

\(\overrightarrow {OR}  = \dfrac{{ - \widehat i + 4\widehat j + \widehat k}}{3} =  - \dfrac{1}{3}\widehat i + \dfrac{4}{3}\widehat j + \dfrac{4}{3}\widehat k\) 

(ii) The  position vector of point R which divides  the line  joining two points P and Q externally in the ratio 2:1 is given by,

\(\overrightarrow {OR}  = \dfrac{{2( - \widehat i + \widehat j + \widehat k) - 1(\widehat i + 2\widehat j - \widehat k)}}{{2 - 1}} = \dfrac{{ - 2\widehat i + 2\widehat j + 2\widehat k - \widehat i - 2\widehat j + \widehat k}}{1}\)

\(\overrightarrow {OR}  =  - 3\widehat i + 3\widehat k\)

 

Problem:-

Find the position vector of the midpoint of the vector joining the points P (2, 3, 4) and Q (4, 1, – 2).

Answer:-

The position vector of mid-point R of the vector joining points

 P (2, 3, 4) and Q (4, 1, –2) is given by,

\(\overrightarrow {OR}  = \frac{{(2\widehat i + 3\widehat j + 4\widehat k) + (4\widehat i + \widehat j - 2\widehat k)}}{2} = \frac{{6\widehat i + 4\widehat j + 2\widehat k}}{2}\)

\(\overrightarrow {OR}  = 3\widehat i + 2\widehat j + \widehat k\)

 

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Class 12 chapter 9 (differential Equations) Class Notes

Differential Equations

Introduction

Differential equation plays an important role in the modern science.

In Physics, “Newton’s Law of cooling”- This states that the rate of change of the temperature of an object is directly proportional to the difference between its own temperature and the ambient temperature”.

\[\dfrac{{dT}}{{dt}} =  - K\left[ {T(t) - {T_a}(t)} \right]\]

This is differential equation

Solving a RC, RLC or RL circuit in physics involves differential equation.

\({V_{emf}} = {K_e}\dfrac{{d\theta }}{{dt}}\)   -- A differential Equation

Hence, an in-depth study of differential equations has assumed prime importance in all modern scientific investigations.

 

What is differential equation?

An equation that involves a relationship between a function and one or more derivatives is called a differential equation.

Example:  \(a\dfrac{{dy}}{{dx}} + x = y\)  A Derivative form \(\dfrac{{dy}}{{dx}} = \cos x\)

Above are the examples of differential equation since it involve a derivative form and the function in form of x and y.

But,
ax² + bx + c = 0 or \(x\sin x - x = 1\). These equations have variable x but do not include their derivatives hence these functions are not differential equation

Hyperbola equation: \(\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\), Hyperbola Equation is not a differential equation

  

Similarly y = sinx not a differential Equation  

Notation used: 

\(y' = \frac{{dy}}{{dx}}\) and \(y'' = \frac{{{d^2}y}}{{d{x^2}}}\), For n^th order \({y_n} = \frac{{{d^n}y}}{{d{x^n}}}\)

Order and Degree of Differential Equation

Order of differential Equation

The highest derivative in a differential equation is said to be a order of differential equation.

Example:

Now let’s find out order of the function : \(\dfrac{{{d^3}y}}{{d{x^3}}} + x{\left( {\dfrac{{dy}}{{dx}}} \right)^4} = 0\)

Order here is 3 because of the highest order of derivative i.e. d³y/dx³ 

Note: That there is difference in order and degree as shown in our next section.

Degree of differential Equation-
It is the highest power of the highest order derivative involved in the differential function.

Note:- The conditions to find the degree of differential equation is that the function should only be polynomial function if the differential equation contains log, exponential and trigonometric function of the derivative then degree is not defined i.e. the equation has to be polynomial function to define degree of a differential equation.

Example: \(\dfrac{{{d^3}y}}{{d{x^3}}} + x{\left( {\dfrac{{dy}}{{dx}}} \right)^4} = 0\)

Here the order is 3 and degree is 1 i.e the highest power of highest derivative. 

y''' + y' + y = 0          Here order is 3 and degree = 1.

\(\dfrac{{dy}}{{dx}} + \sin \left( {\dfrac{{dy}}{{dx}}} \right) = 0\)             – Here the order is 1 but degree is not defined.

Note: Order and degree (if defined) of a differential equation are always positive integers.

 

General and Particular Solution of Differential Equation

But for differential equation like y’-sinx=0. Solution of y is in function in terms of variables not in numbers.

We classify solution of differential equation into two parts-

1. General Solution
2. Particular Solution

Let us understand General solution of differential equation with an example.
 

Example: \(\frac{{dy}}{{dx}} - k\sin x = 0\)  .........{ A differential equation

Integrating both side \(\int {dy}  = \int {k\sin xdx} \)

We get solution as  y = - Kcosx + C 

This is called a “general solution” of a differential equation as we get the solutions in constant form of K and C (K and C can be any value from - ∞ to + ∞ )

Particular solution:
Now the solution we get is a General solution but if we assign some value to it by giving some conditions Let’s say we get the value of k and c be - 4 and 7 resp. then the equation will became y = 4cosx + 7    {This equation is a particular solution of a given differential equation}.

Formation of differential equation whose general solution is given

Now let’s learn how to form the differential equation of a given function. Please note that we need to eliminated the parameter like a,b  etc. to form a differential equation

Procedure to find a differential equation

1. If the given family F of curves depends on only one parameter then we can represent the equation of the form

F (x, y, a) = 0 ..(1)

Example: Parabola Equation y² = 4ax (as shown in fig.) this can be represented in an equation of the form  F(x, y, a) : y² = 4ax…..(1)

Now to find out differential equation we differentiate equation (1)
with respect to x, we get an equation involving  y^′, y, x, and a,

i.e.,   \(y\frac{{dy}}{{dx}} = 2a\) ……..(2)      

We get a differential equation of a family of a parabola by putting the value of parameter “a” from (2) in equation (1)

We get         \(y = 2x\frac{{dy}}{{dx}}\)   ………..this a differential equation.

 i.e  F(x,y,y’)=0

2. If two parameters are given say “a and b” in the function, then we represent the equation of the form F (x, y, a ,b) = 0

Example:  y = asin(x + b) this can be represented by an equation of the  form F(x, y, a, b) : y = asin(x + b)……….(1)

Now since it involve two parameter (a, b) we need to differentiate twice since it is not possible to eliminate two parameters a and b from the two equations only and so, we required a third equation.

\(\dfrac{{dy}}{{dx}} = a\cos (x + b)\)                       --(2)

\(\frac{{{d^2}y}}{{d{x^2}}} =  - a\sin (x + b)\)       --(3)

and since  we put the value in 3rd in the equation, we get \(\dfrac{{{d^2}y}}{{d{x^2}}} =  - y\)  or  \(\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0\)

i.e F(x,y,y’,y”) = 0

Note:- Similarly we follow the same steps for 3 or more parameters if involved.

 

Problem:  Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Sol. The equation of the family of hyperbolas with the centre at origin and foci along the x-axis is: \(\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\)     ..(1)

           

Differentiating both sides of equation (1) with respect to x, we get:

\(\dfrac{{2x}}{{{a^2}}} + \dfrac{{2yy'}}{{{b^2}}} = 0\) or \(\dfrac{{x}}{{{a^2}}} + \dfrac{{yy'}}{{{b^2}}} = 0\)   ----(2)

Again Differentiating equation (2) with respect to x, we get: \(\dfrac{1}{{{a^2}}} - \dfrac{{y'y' + yy}}{{{b^2}}} = 0\) or \(\dfrac{1}{{{a^2}}} = \dfrac{{{{(y')}^2} + yy}}{{{b^2}}}\)     ....(3)

Now substituting value of \(\dfrac{1}{{{a^2}}}\) from equation in (3) in the equation (2) we get,

Solving the equation we get,

x(y’)²+xyy’’-yy’=0 or xyy’’+x(y’)²-yy’=0

This is a required differential equation of Hyperbola.

Differential Equations with Variables Separable

A first order-first degree differential equation is of the form

dy/dx = F(x,y)   --(1)

If F (x, y) can be expressed as a product g (x) h(y), where, g(x) is a function of x and h(y) is a function of y, then the differential equation (1) is said to be of “variable separable” type. Then differential equation (1) has the form             

dy/dx=  h(y).g(x)

d(y)/h(y) = g(x). dx   ... (3)   where h(y) ≠ 0

Integrating both sides of equation (3), we get

∫ (1/h(y)).dy =      ∫   g(x). dx                                                   =  ... (4)

Thus, (4) provides the solutions of given differential equation in the form

H(y) = G(x) + C

Here, H (y) and G (x) are the anti-derivatives of 1/h(y) and g(x) respectively and C is the arbitrary constant.

Problem: find the general solution of the following of question :  sec² x tan y dx + sec² y tan x dy = 0

Solution: Sec²y tanx dy= - sec² x tan y dx

Integrating both side 

Let tany=t and sec²ydy = dt

log(tany)= -log(tanx)+logC

tany.tanx=C

This is required general solution of differential equation.

 

Homogeneous Function

A function F(x, y) is said to be homogeneous function of degree n if

F(λx, λy) = λⁿ F(x, y) for any nonzero constant λ.

Eg:    F(x, y) = y² + 2xy    

F(λx, λy) = (λy)² + 2λxλy

F(λx, λy) = λ² (y² + 2xy )  

Thus, F(x, y) = y² + 2xy      is  a homogeneous function.

 

But,  F(x, y) = sin x + cos y is not a homogeneous function since it cannot be expressed as F(λx, λy) = λⁿ F(x, y)

Homogeneous Differential Equation

A differential equation of the form dy/dx = F (x, y) is said to be homogenous if F(x, y) is a homogenous function of degree zero

Example: Below Equation is Homogeneous Differential Equation

          

Since,

F(λx, λy) = λ⁰F(x,y)  

i.e  F(λx, λy) = F(x,y)

 

Method to solve a given homogeneous differential equation

     ………(1)

We make the substitution of    y = v . x  ……….(2)

Differentiating equation (2) with respect to x, we get

      ……..(3)

 

Substituting the value of  dy/dx  from equation (3) in equation (1), we get

   ………(4)

 

Arranging the variables,                                          ………(5)

 

Equation (5) gives general solution (primitive) of the differential equation (1) when we replace v by y/x

Note: If the homogeneous differential equation is in the form dy/dx= F(x, y) where, F (x, y) is homogenous function of degree zero, then we make substitution of x/y = v   i.e., x = vy and we proceed further to find the general solution as discussed above by writing dy/dx = (x,y) = g(y/x)

First order Linear differential equations

A differential equation of the from  dy/dx + Py = Q

where, P and Q are constants or functions of x only, is known as a first order linear differential equation.

Example:

Another form of first order linear differential equation can be dx/dy+ Px =Q

where, P and Q are constants or functions of y only, is known as a first order linear differential equation

Steps involved to solve first order linear differential equation:

(i) Write the given differential equation in the form dy/dx + Py = Q  ,  where P, Q are constants or functions of x only.

(ii) Find the Integrating Factor (I.F)

(iii) Write the solution of the given differential equation as

y (IF) =∫Q + IF. dx  + C

 

Problem:  Solve equation:     dy/dx + y/x = x² 

Solution. The given differential equation is in the form of dy/dx + Py = Q

(Here P=1/x and Q=x²)

 

This is the required general solution of the given differential equation.

 

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Class 12 Chapter 8 (Application of Integrals) class notes

  Application of Integrals

Introduction

Calculus is widely seen as functions, differentiation, and integration. It has a lot of applications in the fields of Engineering, Science, and Mathematics. We have already studied to find the area bounded by the curve y = f(x), the ordinates x = a, x = b and x-axis, while calculating definite integral as the limit of a sum.

                         

 

In this chapter, we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses. We will learn how to use integrals to find out areas under different conditions.

Area under Simple Curves

Let we want to find the area bounded be the curve y = f(x), x-axis and the ordinates x = a and x = b.

From the figure, we assume that area under the curve is composed of large number of very thin vertical strips. Let us take an arbitrary strip of height y and width dx, then dA (area of the elementary strip) = y dx, where, y = f(x). This area is called the elementary area which is located at an arbitrary position within the region which is specified by some value of x between a and b.

Total area A of the region between x-axis, ordinates x = a, x = b and the curve y = f(x) is calculated by adding up the elementary areas of thin strips across the region PQRSP.

The area A under the curve f(x) bounded by x = a and x = b is given by

A = _aꭍ^b dA = _aꭍ^b y dx = _aꭍ^b f(x) dx

The area A of the region bounded by the curve x = g (y), y-axis 

and the lines y = c, y = d as shown in the figure, is given by

A = _cꭍ^d x dy = _cꭍ^d g(y) dy

Example: Find the area of the region bounded by the curve y² = x and the lines x = 1, x = 4 and the x-axis.

Solution:

The area of the region bounded by the curve, y² = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.

So, area of ABCD = ʃ₁⁴ y dx

                               = ʃ₁⁴ √x dx

                               = [x^3/2/(3/2)]₁⁴ 

                               = (2/3)[4^3/2 - 1^3/2]

                               = (2/3)[8 – 1]

                               = (2/3) * 7

                               = 14/3 units

Example: Find the area of the region bounded by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

Solution:

The area of the region bounded by the curve, x² = 4y, y = 2, and y = 4, and the y-axis is the area ABCD.

So, area of ABCD = ʃ₂⁴ x dy

                               = ʃ₂⁴ 2√y dy

                               = 2[y^3/2/(3/2)]₂⁴ 

                               = (4/3)[4^3/2 - 2^3/2]

                               = (4/3)[8 – 2√2]

     p;                         = (32 – 8√2)/3 units

The area of the region bounded by a curve and a line

In this section, we will find the area of the region bounded by a line and a circle, a line and a parabola, a line and an ellipse when their standard forms are given.

Example: Find the area of the region in the first quadrant enclosed by x-axis, line x = √3y and the circle  x² + y² = 4

Solution:

The area of the region bounded by the circle x² + y² = 4, x = √3y, and the x-axis is the  area OAB.

The point of intersection of the line and the circle in the first quadrant is (√3, 1).

Area OAB = Area ∆OCA + Area ACB

Area ∆OCA = (1/2) * OC * AC = (1/2) * √3 * 1 = √3/2

Area ACB = ʃ_√3² y dx

                  = ʃ_√3² √(4 – x²) dx

                  = [x√(4 – x²)/2 + (4/2) * sin^-1 x/2]_√3² 

                  = [2 * π/2 - (√3/2) * √(4 - 3) - 2sin^-1 √3/2]

                  = [π - √3/2 – 2(π/3)]

                  = [π/3 - √3/2]

Therefore, area enclosed by x-axis, x = √3y and the circle x² + y² = 4 in the first quadrant

= √3/2 + [π/3 - √3/2] = π/3 units

Example: Find the area of the smaller part of the circle x² + y² = a² cut off by the line x = a/√2

Solution:

The area of the smaller part of the circle, x² + y² = a² cut off by the line x = a/√2 is the area ABCDA.

It can be observed that the area ABCD is symmetrical about x-axis.

                                                        

So, Area ABCD = 2 * Area ABC

Now, Area of ABC = ʃ_a/√2^a y dx

                                 = ʃ_a/√2^a √(a² – x²) dx

                                 = [x√( a² – x²)/2 + (a²/2) * sin^-1 x/a]_a/√2^a 

                                 = [a²/2 * π/2 - (a/2√2) * √( a² - a²/2) - a²/2 * sin^-1 1/√2]

                                 = a²/2 * π/2 - (a/2√2) * (a/√2) - a²/2 * (π/4)

                                  = a²π/4 - a²/4 - a²π/8

                                  = (a²/4)(π - 1 - π/2)

                                  = (a²/4)(π/2 - 1)          

So, area of ABCD = 2[(a²/4)(π/2 - 1)] = (a²/2)(π/2 - 1)

Therefore, The area of the smaller part of the circle, x² + y² = a² cut off by the line x = a/√2 is

(a²/2)(π/2 - 1) units.

 

Area between Two Curves

The most general class of problems in calculating the area of bounded regions is the one involving the regions between two curves. Let we are given two curves represented by y = f(x), y = g(x), where f(x) ≥ g(x) in [a, b] as shown in figure.

                                                

Here the points of intersection of these two curves are given by x = a and x = b obtained by taking common values of y from the given equation of two curves.

Elementary strip has height f(x) – g(x) and width dx so that the elementary area

dA = [f(x) – g(x)] dx

Total area A can be calculated as A = _aꭍ^b [f(x) – g(x)]

Alternatively,

A = [area bounded by y = f (x), x-axis and the lines x = a, x = b]

– [area bounded by y = g (x), x-axis and the lines x = a, x = b]

A = _aꭍ^b f(x) = _aꭍ^b g(x) = _aꭍ^b [f(x) – g(x)], where f(x) ≥ g(x) in [a, b]

If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the figure, then the area of the regions bounded by curves can be calculated as

Total Area = Area of the region ACBDA + Area of the region BPRQB.

                   = _aꭍ^c [f(x) – g(x)] + _cꭍ^b [g(x) – f(x)]  

                            

Example:

Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y.

Solution:

The required area is represented by the shaded area OBCDO.

                                                  

Solving the given equation of circle, 4x² + 4y² = 9, and parabola, x² = 4y, we get the point of intersection as B(√2, 1/2) and D(-√2, 1/2).

It can be observed that the required area is symmetrical about y-axis.

So, Area OBCDO = 2 * Area OBCO

Now, draw BM perpendicular to OA.

Therefore, the coordinates of M is (√2, 0).

Therefore, Area OBCO = Area OMBCO – Area OMBO

                                         = ʃ₀^√2  {√(9 – 4x²)/4} dx - ʃ₀^√2 x²/4 dx

                                         = (1/2)ʃ₀^√2 √(9 – 4x²) dx – (1/4)ʃ₀^√2 x² dx

                                        = (1/4)[x√(9 – 4x²) + (9/2) * sin^-1 2x/3]₀^√2  - (1/4)[x³/3]₀^√2            

                                        = (1/4)[√2 * √(9 – 8) + (9/2) * sin^-1 2√2/3] - (1/12)(√2)³

                                        = (√2/4) + (9/8) * sin^-1 2√2/3 - √2/6

                                        = (√2/12) + (9/8) * sin^-1 2√2/3

                                        = (1/2)[√2/6 + (9/4) * sin^-1 2√2/3]

Therefore, the required area OBCDO = 2 * (1/2)[√2/6 + (9/4) * sin^-1 2√2/3]

                                                                   = [√2/6 + (9/4) * sin^-1 2√2/3] units

Example:

Find the area between the curves y = x and y = x²

Solution:

The required area is represented by the shaded area OBAO as

The points of intersection of the curves, y = x and y = x² is A (1, 1).

Now, draw AC perpendicular to x-axis.

So, Area (OBAO) = Area (∆OCA) – Area (OCABO)

                              = ʃ₀¹ x dx - ʃ₀¹ x² dx

                              = [x²/2]₀¹ – [x³/3]₀¹

                              = 1/2 - 1/3

                              = 1/6 units

 

 

 

 

 

 

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