UNIT I: MATRIX CHARACTERISTIC EQUATION

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UNIT I: MATRIX

CHARACTERISTIC EQUATION

Let ‘A’ be a given matrix. Let be a scalar. The equation det is called the

characteristic equation of the matrix A.

1. Find the Characteristic equation of A =

Solution: The Characteristic equation of A is =0 ie. Where = Trace of A

& = Therefore = 4 & = -5 implies that

2. Find the Characteristic equation of A =

Solution: The Characteristic equation of A is =0 ie., Where =

Trace of A , =Sum of the minors of the major diagonal elements & = = 3 & = -1 &

implies that

EIGEN VALUE

The values of obtained from the characteristic equation are called the Eigen values of A.

EIGEN VECTOR

Let A be a square matrix of order ‘n’ and be a scalar, X be a nonzero column vector such that AX = X.

The non-zero column vector which satisfies is called eigen vector or latent

vector.

LINEARLY DEPENDENT AND INDEPENDENT EIGEN VECTOR

Let ‘A’ be the matrix whose columns are eigen vectors.

(i) If then the eigen vectors are linearly dependent.

(ii) If then the eigen vectors are linearly independent.

1. Find the eigen values and eigen vectors of

Solution: The Characteristic equation of A is =0 ie.

Where = Trace of A , =Sum of the minors of the major diagonal elements & =

www.Vidyarthiplus.com


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= 18 & = 45 & implies that

To find eigen vector : By the definition we have ie., (

CASE (I) : When λ= -3 , Substituting in (1) we get

; ;

Solving using cross multiplication rule If then

CASE (ii) : When λ= 5 , Substituting in (1) we get

; ;


The eigen vectors are

Since the eigen values are repeated the eigen vectors are linearly dependent.

2. Find the eigen values and eigen vectors of

Solution: The Characteristic equation of A is =0 ie. . Where

= Trace of A , =Sum of the minors of the major diagonal elements & =

= 0 & & implies that


CASE (I) : When λ= -1 , Substituting in (1) we get

All the three equations reduce to one and the same equation

Two of the unknowns, say and are to be treated as free variables. Taking and ,

we get and taking and , we get .



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CASE (ii) : When λ= 2 , Substituting in (1) we get

; ;

Solving using cross multiplication rule

The eigen vectors are

Though two of the eigen values are equal, the eigen vectors are are linearly independent.

NOTE:

(i) The eigen vector corresponding to an eigen value is not unique.

(ii) If all the eigen values of a matrix are distinct, then the corresponding eigen vectors are linearly

independent.

(iii) If two or more eigen values are equal, then the eigen vectors may be linearly indenpent or linearly

dependent.

PROPERTIES OF EIGEN VALUES AND EIGEN VECTORS:

Property 1: (I) The sum of the Eigen values of a matrix is equal to the sum of the elements of the principal

diagonal (trace of the matrix). i.e.,

(ii)The product of the Eigen values of a matrix is equal to the determinant of the matrix.

i.e.,

Property 2: A square matrix A and its transpose have the same Eigen values.

Property 3: The characteristic roots of a triangular matrix are just the diagonal elements of the matrix.

Property 4: If is an Eigen value of a matrix A, then λ

λ is the Eigen value of

Property 5: If is an Eigen value of an orthogonal matrix A, then λ

λ is also its Eigen value.

Property 6: If are the Eigen values of a matrix A, then has the Eigen

values (m being a positive integer)

Property7: If are the Eigen values of a matrix A, then are the Eigen

values of the matrix KA.



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Property 8: Property7: If are the Eigen values of a matrix A and if K is a scalar then

are the Eigen values of the matrix A-KI.

Property 9: The Eigen values of a real symmetric matrix are real numbers.

Property 10: The Eigen vectors corresponding to distinct Eigen values of a real symmetric matrix are

orthogonal.

Property 11: The similar matrices have same Eigen values.

Property 12: Eigen vectors of a symmetric matrix corresponding to different Eigen values are orthogonal.

Property 13: If A and B are matrices and B is a non singular matrix then A and have same

Eigen values.

Property 14: Two Eigen vectors and are called orthogonal vectors if

Property 15: If be distinct Eigen values of a matrix then corresponding Eigen

vectors form a linearly independent set.

Note: The absolute value of a determinant (|detA|) is the product of the absolute values of the eigen values

of matrix A.

c = 0 is an eigen value of A if A is a singular (noninvertible) matrix

· If A is a n x n triangular matrix (upper triangular, lower triangular) or diagonal matrix, the eigen

values of A are the diagonal entries of A.

· A and its transpose matrix have same eigen values.

· Eigen values of a symmetric matrix are all real.

· Eigen vectors of a symmetric matrix are orthogonal, but only for distinct eigen values.

· The dominant or principal eigen vector of a matrix is an eigen vector corresponding to the eigen

value of largest magnitude (for real numbers, largest absolute value) of that matrix.

· For a transition matrix, the dominant eigen value is always 1.

· The smallest eigen value of matrix A is the same as the inverse (reciprocal) of the largest eigen value

of A-1

; i.e. of the inverse of A.

1. Find the Sum and the product of the Eigen values of A =

Solution: From the property of Eigen values i.e. &

Therefore Sum of the Eigen values =1+5+1= 7 & Product of the Eigen values =

i.e., 1(5-1)-1(1-3) +5(1-15) = 4+2-70 =-64. Therefore &



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2. If A = write down the sum and product of the Eigen values of A.


Therefore Sum of the Eigen values =1+2+3= 6 & Product of the Eigen values =

Ie, 1(6-4)-1(3-2) +1(2-2) = 2-1 =1. Therefore &

3. Find the Sum and the product of the Eigen values of A =


Therefore Sum of the Eigen values =6 & Product of the Eigen values = =6

4. Prove that the Eigen values of are the same as those of A =

Solution: The Characteristic equation of A is =0

ie. Where = Trace of A & =

Therefore = 2 & = -3 implies that . Eigen values of A are 3 and -1.

(By the property of Eigen values we know that if are eigen values A then & are eigen

values of also if are eigen values -kA)

Since Eigen values of A are 3 and -1. Eigen values of are

Therefore Eigen values of are -1 and 3.

5. If the Sum of the two eigen values and trace of matrix A are equal. Find the value of

Solution: Let be the eigen values of A. From the property of Eigen values we know that

& Given that is

implies that

Therefore Product of the Eigen values =

6. Prove that if X is an eigen vector of A corresponding to the eigen value λ. Then for any nonzero

scalar multiple of A, X is an eigen vector.

Solution: By definition of eigen values → (1) Pre multiplying by k on both sides of (1)

is the Eigen values of (kA) & is the Eigen vector of (kA).



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7. Two eigen values of a matrix A = are equal to 1 each. Find the eigen values of A & .

Solution: From the property of Eigen values we know that .

Given . Therefore the Eigen values of A are 1, 1, 5.

(By the property of Eigen values we know that if are eigen values A then & are eigen

values of ) Therefore the Eigen values of are 1, 1, .

8. Find the eigen value of A = corresponding to the eigen vector .

Solution: By the definition we have ie. (

2- =0 2 = .BY property = 6 . Therefore eigen

values of A are 2 and 6.

9. If A is an orthogonal matrix. Show that is also orthogonal matrix.

Solution: Since A is an orthogonal matrix .Implies that (A = = ) Let B = , to prove

B is orthogonal we have to check = .

To prove: = = = , Since =A.

10. Find the constants a and b such that the matrix has 3 and -2 as its eigen values.

Solution: BY property & &

Given = 1 & -6 Therefore

a(1-a)=-2 a = 2 & a = -1. b = -1 & b = 2. Therefore when a=-1 then b=2

and when a = 2 then b = -1.

11. Given that verify that eigen values of are the squares of those of A.


i.e., Where = Trace of A & = Therefore = 7 & = 6 implies that

Eigen values are 1 & 6.

The Characteristic equation of is =0

Eigen values of are 1 and 36, that are the squares of the eigen values of A,

namely 1 and 6.



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12. The product of two eigen values of the matrix is 14. Find the third eigen

value.

Solution: By the property . Given that .

13. Find the eigen values of .


i.e., Where = Trace of A & = Therefore = 6 & = 5 implies that

Eigen values are 1 & 5.

By the property .

(i.e.) 1 & 25 are the eigen values of .

The eigen values of are 2(1) & 2(25) = 2, 50.

14. Find the sum of the squares of the eigen values of .

Solution: By the property “The eigen values of a upper or lower triangular matrix are the main diagonal

elements”.

Eigen values of A = 3, 2, 5.

Sum of the squares of the eigen values of A =9+4+25 =38.

15. Find the sum of the eigen values of the inverse of .

Solution: By the property “The eigen values of a upper or lower triangular matrix are the main diagonal

elements”.

Eigen values of A = 3, 4, 5.

By the property .

Sum of the eigen values of .

CAYLEY HAMILTON THEOREM:

Every square matrix satisfies its own characteristic equation.



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This means that, if is the characteristic equation of a

square matrix A of order n, , where I is the unit matrix of

order n.

1. If A is non singular matrix then we can get , using this theorem

2. Higher positive integral powers of A can be computed

1. Verify the Cayley Hamilton theorem for the matrix A =

Solution: Every square matrix satisfies its own characteristic equation is the statement of Cayley

Hamilton theorem. The Characteristic equation of A is =0

ie., Where = Trace of A & =

Therefore = 8 & = 14 implies that We have to check

A2

= = & 8A =

L.H.S =

= = =R.H.S Cayley Hamilton theorem is

verified.

2. Using Cayley Hamilton theorem find given A =

Solution: The Characteristic equation of A is =0 i.e., Where = Trace

of A & = Therefore = 4 & = -5 implies that By Cayley Hamilton

theorem we have …..(1) , Premultiplying by on both sides of (1) we get

= =

3. Using Cayley Hamilton theorem find for A =

Solution: The Characteristic equation of A is =0 ie., Where

= Trace of A , =Sum of the minors of the major diagonal elements & = = 3 & = -1

& implies that 3



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(Every square matrix satisfies its own characteristic equation is the statement of Cayley Hamilton

theorem.) By Cayley Hamilton theorem we have …..(1) Premultiplying

by on both sides of (1) we get =

=

=


SOLUTION : The Characteristic equation of A is =0 ie., Where

= Trace of A , =Sum of the minors of the major diagonal elements & = = 5 & =9

& implies that 5 (Every square matrix satisfies its own characteristic

equation is the statement of Cayley Hamilton theorem.) By Cayley Hamilton theorem we have

5 …..(1) , Premultiplying by on both sides of (1)

we get =

=

=


SOLUTION : The Characteristic equation of A is =0 ie., Where

= Trace of A , =Sum of the minors of the major diagonal elements & = = 8 & =8

& implies that 8 (Every square matrix satisfies its own characteristic

equation is the statement of Cayley Hamilton theorem.) By Cayley Hamilton theorem we have

…..(1)

Premultiplying by on both sides of (1) we get =

=



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=

6. Verify Cayley Hamilton theorem and also find interms of , A & I of A =

SOLUTION : The Characteristic equation of A is =0 ie.,

Where = Trace of A , =Sum of the minors of the major diagonal elements & = = 5 &

=7 & implies that 5


theorem.)

To verify C.H.T we have check : 5

Consider L.H.S of (I) : 5

= -

= = R.H.S of (i)

Therefore C.H.T is verified.

By Cayley Hamilton theorem we have 5 …..(1) ,

=

=

=

Premultiplying by on both sides of (1) we get

= ………(2)

=

=

Premultiplying by on both sides of (2) we get

=



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=

=

7. Verify Cayley Hamilton theorem and also find interms of , A & I of A =

SOLUTION : The Characteristic equation of A is =0 ie.,

Where = Trace of A , =Sum of the minors of the major diagonal elements & = = 6 &

=8 & implies that 6


theorem.)

To verify C.H.T we have check : 6

Consider L.H.S of (I) : 6

= -

= = R.H.S of (i)

Therefore C.H.T is verified.

By Cayley Hamilton theorem we have 6 …..(1) ,

6 .

=

(

=

=

=

8. Verify the Cayley Hamilton Theorem and hence find

Ans: : The Characteristic equation of A is =0 ie.,



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Where = Trace of A , =Sum of the minors of the major diagonal elements & = = 6

& = -9 & Þ 6

By Cayley Hamilton theorem we have 6 …..(1)

Premultiplying by on both sides of (1) we get =

= =

DIAGONALISATION OF A MATRIX

The process of finding a matrix M such that , where D is a diagonal matrix, if called

diagonalisation of the Matrix A

Note:

DIAGONALISATION BY ORTHOGONAL

TRANSFORMATION OR ORTHOGONAL REDUCTION

If A is a real symmetric matrix, then the eigen vectors of A will be not only linearly independent but also

pair wise orthogonal. If we normalize each eigen vector i.e. divide each element of by the square

root of the sum of the square\s of all the elements of and use the normalized eigen vectors of A to form

the normalised modal matrix N, then it can be proved that N is an orthogonal matrix. By a property of

orthogonal matrix, .

The similarity transformation takes the form .

Transforming A into D by means of the transformation is known as orthogonal transformation or

orthogonal reduction.

NOTE:- Diagonalisation by orthogonal transformation is possible only for a real symmetric matrix.

1. Diagonalise the matrix by an orthogonal transformation.

SOLUTION: Given A =

The Characteristic equation of A is =0

i.e.


= 12 & =36 & implies that

igen values of the matrix A are 2 , 2 & 8.



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CASE (I) : When λ=8 , Substituting in (2) we get


CASE(ii) : When λ=2 , Substituting in (2) we get

We have only one equation with three unknowns, let

if If then

CASE(iii) : When λ=2 , Let From orthogonal transformation we know that

must be mutually perpendicular to each other. , &


Modal matrix M=

Normalised modal matrix

= =D



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Solution: Given A =


i.e.




To find eigen vector: By the definition we have ie., (




We have only one equation with three unknowns, let if

If then






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Modal matrix M=


= =D


Solution: Given A =


i.e.










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CASE(iii) : When λ=14 , Substituting in (2) we get


Modal matrix M=

Normalized modal matrix

= =D

QUADRATIC FORMS

A homogeneous polynomial of the second degree in any number of variables is called a quadratic form.

For example, is a quadratic form in three variables.

The symmetric matrix is called the matrix of the quadratic form Q.

NOTE:

To find the symmetric matrix A of a quadratic form, the coefficient of is placed in the position and

is placed in each of the and positions.

CANONICAL FORM OF A QUADRATIC FORM: Let be a quadratic form in n variables

.

In the linear transformation X = PY, if P is chosen such that is a diagonal matrix of the form

, then the quadratic form Q gets reduced as Q



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This form of Q is called the sum of the squares form of Q or the canonical form of Q.

NATURE OF QUADRATIC FORMS:

Rank: When the quadratic form is reduced to the canonical form it contains only r terms which is the

rank of A.

Index: The number of positive terms in the canonical form is called the index (p) of the quadratic

form.

Signature: The difference between the number of positive and negative terms is called signature (s) of

the quadratic form [s = 2p-r].

The quadratic form in n variables is said to be

(i) Positive definite: If r = n and p = n or if all the eigen values of A are positive.

(ii) Positive semi definite: If r < n and p = r or if all the eigen values of and atleast one eigen value is

zero.

(iii) Negative definite: If r = n and p = 0 or if all the eigen values of A are negative.

(iv) Negative semi definite: If r < n and p = 0 or if all the eigen values of and atleast one eigen

value is zero.

(v) Indefinite: In all other cases or if A has positive as well as negative eigen values.

RULES FOR FINDING NATURE OF QUADRATIC FORM USING PRINCIPAL SUB-

DETERMINANTS:

In this method we can determine the nature of the quadratic form without reducing it to the canonical form.

Le A be a square matrix of order n.

…………………………………

Here are called the principal subdeterminants of A. From ,the

nature of the quadratic form can be determined.



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1. A Q.F is positive definite if are all positive i.e., for all n.

2. A Q.F is negative definite if are all negative and are all positive

i.e., for all n.

3. A Q.F is positive semi- definite if and atleast one .

4. A Q.F is negative semi- definite if and atleast one .

5. A Q.F is indefinite in all other cases.

1. Without reducing to canonical form find the nature of the Quadratic form

Solution: Matrix of the Quadratic form is A =

, = 0 & = 0 -2-2 =-4

Since 0 & Nature of the Quadratic form is indefinite.

2. Reduce the quadratic form to canonical form using

orthogonal transformation also find its nature, rank , index & signature.

Solution: Quadratic form =

Matrix form of Quadratic form = where X = & A =


i.e.


= 3 & =0 & implies that 3

igen values of the matrix A are -1, 2 & 2.

To find Eigen vector : By the definition we have ie., (

CASE (I) : When λ=-1 , Substituting in (2) we get

2



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if If then

CASE (iii) : When λ=2 , Let From orthogonal transformation we know that



Modal matrix M=


= =D

Let X = NY be an orthogonal transformation which changes the quadratic form to canonical form.

where Substituting X = NY in (1) we get

Q.F =

=

=

=



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=

Q.F = which is the canonical form of the quadratic form

Nature of the Q.F = Indefinite

Rank of the Q.F (r) = 3

Index of the Q.F (p) = 2

Signature of the Q.F (s) =2p-r = 1.


orthogonal transformation also find its nature, rank, index & signature.




i.e.



igen values of the matrix A are 0, 3 & 15.



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, & are mutually perpendicular to each

other.

Modal matrix M=


= =D



Q.F =

=

=

=

=


Nature of the Q.F = Positive semi definite





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i.e.


= 12 & =36 & implies that 12







if If then





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Modal matrix M=


= =D



Q.F =

=

=

=

=

Q.F = 8 which is the canonical form of the quadratic form

Nature of the Q.F = Positive definite




5. Reduce the quadratic form to canonical form using orthogonal

transformation also find its nature, rank , index & signature.






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i.e.






2







other.

Modal matrix M=




25

= =D



Q.F =

=

=

=

=


Nature of the Q.F = Positive semi definite









i.e.








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other.

Modal matrix M=


= =D



Q.F =



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=

=

=

=











i.e.

Where = Trace of A, =Sum of the minors of the major diagonal elements & =









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other.

Modal matrix M=


= =D



Q.F =

=

=

=

=





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i.e.

Where = Trace of A, =Sum of the minors of the major diagonal elements & =


igen values of the matrix A are -2, 6 & 6.


CASE (I) : When λ=-2 , Substituting in (2) we get

4

0



We have only one equation with three unknowns, let ,

if If then



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Modal matrix M=


= =D



Q.F =

=

=

=

=

Q.F = -2 which is the canonical form of the quadratic form

Nature of the Q.F = In definite








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i.e.



igen values of the matrix A are 2, -1 & -1.



-2


CASE(ii) : When λ=-1 , Substituting in (2) we get


if If then

CASE(iii) : When λ=-1 , Let From orthogonal transformation we know that





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Modal matrix M=


= =D



Q.F =

=

=

=

=


Nature of the Q.F = In definite



Signature of the Q.F (s) =2p-r =- 1.



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UNIT – II: THREE DIMENSIONAL ANALYTICAL GEOMETRY

FORMULAE:

1. The equation of the straight line through the point 1 1 1( , , )p x y z and having direction cosines

( , , )l m n is 1 1 1x x y y z z

l m n

- - -= =

2. The equation of the straight line through the point 2 2 2( , , )B x y z and having direction ratios

( , , )a b c is 1 1 1x x y y z z

a b c

- - -= =

3. The equation of the straight line passing through the points 1 1 1( , , )A x y z and 2 2 2( , , )B x y z is

1 1 1

2 1 2 1 2 1

x x y y z z

x x y y z z

- - -= =

- - -

4. Angle between the straight lines:

(I) 1 2 1 2 1 2cos l l m m n nq = + +

d. r of 1 1 1( , , )oA l m n=

d. r of 2 2 2( , , )oB l m n=

(ii) if the lines are perpendicular θ=90⁰ ,cos 90=0

1 2 1 2 1 2l l m m n n+ + =0

(iii)If the lines are parallel , θ=0, cos0=1 1 2 1 2 1 2 1l l m m n n+ + = or 1 1 1

2 2 2

l m n

l m n= =

(iv) d.r of OA 1 1 1( , , )a b c= d.r of OB 2 2 2( , , )a b c=

1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

cosa a b b c c

a b c a b cq

+ +=

+ + + +

(v) If the lines are perpendicular, cos90 0=

1 2 1 2 1 2 0a a b b c cÞ + + =

(vi) If the lines are parallel, cos0 1=

2 2 2 2 2 2

1 2 1 2 1 2 1 1 1 2 2 2a a b b c c a b c a b c+ + = + + + +

1 1 1

2 2 2

a b cor

a b c= =

(vii) The co-ordinate of any point on the straight line is



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1 1 1

1 1 1

,

, ,

x x y y z zk

a b c

x x ak y y bk z z ck

- - -= = =

= + = + = +

5. The conditions for the lines are coplanar is

2 1 2 1 2 1

1 1 1

2 2 2

0

x x y y z z

l m n

l m n

- - -

=

6. The equation of the coplanar plane is

1 1 1

1 1 1

2 2 2

0

x x y y z z

l m n

l m n

- - -

=

7. The sphere having centre at the origin and radius r is 2 2 2 2x y z r+ + =

8. The sphere having centre at the point ( , , )a b c and radius r is

2 2 2 2( ) ( ) ( )x a y b z c r- + - + - =

9. The general equation of the sphere is with centre ( , , )u v w- - - and radius

2 2 2r u v w d= + + -

10. The equation of the sphere having the points 1 1 1( , , )x y z and 2 2 2( , , )x y z as the extremities of the

diameter is 1 2 1 2 1 2( )( ) ( )( ) ( )( ) 0x x x x y y y y z z z z- - + - - + - - =

CONE

DEFINITION: A cone is defined as a surface generated by a straight line which passes through a fixed

point and satisfies one or more conditioni.e.ie, it may intersect a fixed curve.

Note:

1. The fixed point is said to be the vertex of the cone

2. The fixed curve is said to be the guiding curve of the cone

3. The straight line in any position is called the generator of the cone.

FORMULA:

The equation of the cone with vertex 1 1 1( , , )x y z and whose generators intersect the guiding curve

2 22 2 2 0, 0ax hxy by gx fy c z+ + + + + = = is

1. Find the equation of the cone with vertex at (1,1,1) and which passes through the curve

given by x2+ y

2=4 ,z=2

2 2 21 1 1 1 1 1 1 1 1 1 1 1 1 1 1( ) 2 ( )( ) ( ) 2 ( )( ) 2 ( )( ) ( ) 0a zx z x h zx z x zy z y b zy z y g zx z x z z f zy z y z z c z z- + - - + - + - - + - - + - =



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Solution: let the equation of the generator be

(1)

Its point of intersection with the plane z=2 is

(1) Þ (2)

This point will lie on x2+ y

2=4 Þ

(2

+ (2=4 (3)

Eliminating l,m,n from (1) & (3) gives

Þ X2 + y

2 -2z

2 +2xz+2zy-4x-4y+4 = 0

2. Find the equation of the cone whose vertex is the origin and the guiding curve is

x2 + y

2 +z

2 +4x+2y-6z+ 5 =0, 2x+y+2z+5 = 0

Solution: Equation of the line passing through the origin is

- (1)

Any point on this line (1) is (x, y, z) = (lr, mr,nr)

This point (x, y, z) lies on the guiding curve

x2 + y

2 +z

2 +4x+2y-6z+ 5 =0 , 2x+y+2z+5 = 0

This gives (l2+m

2+n

2) r

2+ (4l+2m-6n) r+ 5 = 0 (2)

r (2l+m+2n) +5 = 0 (3)

Eliminating r from (2) & (3) gives

l2+4m

2 2ln

2 – 4lm+6mn+ 2ln = 0 (4)

Eliminating (l, m, n) from (1) & (4) gives

x2+ 4y

2 +21z

2 -4xy + 6yz +12xz= 0

3. Find the equation of the cone whose vertex at (1,2,3) and which passes through the curve

x2+y

2 +z

2=4 , x+y+z=1 (AU MAY/JUNE 2009)

Solution: Equation of any line through (1, 2, 3) is

=r (1)

Any point on this line is (x,y,z) = (lr+1,mr+2,nr+3) (2)

But this point (x,y,z) lies on the curve x2+y

2 +z

2=4 , x+y+z=1

Þ(Lr+1)2+ (mr+2)

2 + (nr+3)

2 = 4 (3)

(Lr+1)+ (mr+2) + (nr+3) =1 (4)

Solve (4) gives r =



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Using in (3) 5l2 +3m

2 +n

2-2ml-4nl-6nm=0 (5)

Eliminating (l, m, n) from (1) & (5), we get

5x2+3y

2+z

2-2xy-6yz-4xz+6x+8y+10z-26=0

4. Find the Equation of the cone whose vertex is the origin and guiding curve is

Solution: The required cone is homogeneous equation of second degree with vertex at origin

and passes through the given curves hence we have

From (1) & (2) We have

5. Find the equation of the cone whose vertex is the point (1,1,0) and whose base is the curve

Solution: The Guiding curve is the intersection of

Any Generator through (1,1,0) is

Since the point lies on (1) we get

Since this is homogeneous in l, m, n Hence we substitute



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6. Find the equation of the cone formed by rotating line about the y-axis

Solution: If the curve x=f(y) in the xy plane is rotated about y-axis , The equation of the

surface of revolution thus generated is

This is the required equation of cone.

RIGHT-CIRCULAR CONE

DEFINITION: A right circular cone is a surface generated by a straight line which passes through a

fixed point and makes a constant angle with a fixed line through the fixed point.

The equation of right circular cone vertex is 1 1 1( , , )x y z ,the semi vertical angle a and axis the line

1 1 1x x y y z z

l m n

- - -= = is

2 2 2 2 2 2 2 21 1 1 1 1 1( )(( ) ( ) ( ) )cos ( ( ) ( ) ( ))l m n x x y y z z l x x m y y n z za+ + - + - + - = - + - + -

1. Find the equation of the right circular cone whose vertex is the point (2,1,-3) and semi

vertical angle is and the axis is parallel to the straight line

Solution: Vertex O (2, 1,-3), semi vertical angle α=450

d. r. for the axis OC parallel to SD d. r. for OC = 1.3.-4

axis equation is

P(x, y, z) any point on the cone . d. r. for OP= x-2,y-1,z+3 =a1,b1c1

d. r. for OC=1,3,-4 =a2,,b2 ,c2

cos = cos 45 =

= 12x2+4y

2-3z

2 -6xy+24yz+8xz+30x

+100y-130z-117=0

2. Find the equation of the right circular cone having its vertex at the origin and passing

through the circle y2

+z2=25, x =4, also find its semi vertical angle.



38

Solution: The Equation of the cone with vertex at the origin and

guiding curve y2

+z2=25, x =4is obtained by making

y2

+z2=25, with the help of x =4

y2

+z2=25(

2 = 16(y

2 +z

2) =25x

2 and the semi vertical angle is

=510 21’

3. Find the equation of the right circular cone which

passes through the point (2,1,3)with vertex at (1,1,2) and axis

parallel to the line

Solution: d. r. for VM=2,-4, 3

d. r. for VA=2-1,1-1,3-2=1,0,1

cos = --------------------------------------- (1)

Let Q(x, y, z) be any point on the cone

d. r. for VQ=x-1,y-1,z-2

d. r. for VM= 2,-4,3

cos = -------------- (2)

From (1) & (2) equating the R.H.S

Þ 17x2

-7y2+7z

2+32xy+48yz-24zx-18x-114y-52z+118=0

4. Find the equation of the right circular cone generated when the straight line 2y+3z=6, x=0

revaluated about z-axis.

Solution: If the curve y= f (z) in the yz plane is rotated about the z-axis

The equation of the surface of resolution thus generated is

X2+y

2= (f (z))

2

Here Y= (f (z)) = Hence X2+y

2=

2

Þ 4(X2+y

2)-9z

2+36z-36=0

5. Find the equation of the right circular cone which contains the three coordinate axes as

generators. Obtain the semi vertical angle and the equations of the axis of the cone.

Solution: Let l, m, n, be the d. c. of the axis.

Let be the angle.The axes of the coordinates are generators of the cone and each of the them

is inclined at angle axes.

Since 1, 0, 0, are d.c of the axis

cos =l.1+m.0+n.0=l



39

Similarly cos =m=n

Since l2+m

2+n

2=1

Þ3cos2

=1 cos =

=cos-1

( is the semi vertical angle of the cone

ertex is at the origin, the equations of axis are

x=

Now ertex of the cone is at the origin, the d.c of axis

Þ cos =

x2+y

2+z

2= (x

2

xy

6. Find the equation of the right circular cone whose vertex is at the origin and base is the

circle x=a, y2

+z2=b

2.

Solution: Here axis is along x-axis

If is the semi-vertical angle, than Cos =

d. c of axis are (1,0,0) the equation of the cone is

cos =

=

2(y

2+z

2) =b

2x

2

CYLINDER

DEFINITION: A cylinder is a surface generated by a straight line which is parallel to a fixed line and

it has to intersect a given fixed curve. The straight line is any position called a generator and the fixed

point is called the guiding curve of the cylinder.

The equation of cylinders whose generators are parallel to the line x y z

l m n= = and intersect the curve

2 22 2 2 0, 0ax hxy by gx fy c z+ + + + + = = is

2 2 2( ) 2 ( )( ) ( ) 2 ( ) 2 ( ) 0a nx lz h nx lz ny mz b ny mz gn nx lz fn ny mz n c- + - - + - + - + - + =

1. Find the equation of the cylinder whose generators are parallel to the line x = , and

whose guiding curve is the ellipse x2+2y

2=1,z=3



40

Solution: Let p(x1,y1,z1) be any point on the cylinder. Then the equation of the generator

through p is: =

Since Generators are parallel to the line: , then equation of the generator is

= ------------- (1)

The line (1) meets at Z=3 Þ from(1)

This point will lie on the ellipse: , if:

----------- (2)

The Locus of P(x1,y1,z1) is: =9

2. Find the equation of the cylinder whose generating lines have the d. c’s (l, m ,n) and which

passes through the circumference of the fixed circle in the ZOX plane.

Solution: Let p(x1,y1,z1) be any point on the cylinder. The given circle is , y=0

Then the equation of the generator through p is = ---- (1)

The point where the line (1) meets y=0 is

------------------------------------------------- (2)

Point (2) lies on

Þ

Locus of the point (x1,y1,z1) is:

3. Find the equation of the cylinder whose generators are parallel to z-axis and which passes

through the curve of intersection of , x+y+z=1

Solution: Let p(x1,y1,z1) be any point on the cylinder.

The d. c of z axis = (0, 0, 1)

Then the equation of the generator through p is:

= ------------------------------------- (1)

If the point (x1,y1,z1) lies on x+y+z=1







43

Solution:

The Equation of the Right circular cylinder is

THE SPHERE

DEFINITION: A sphere is the locus of a point moving at a constant distance form a fixed point. The

constant distance is the radius and the fixed point is the centre of the sphere.

PLANE SECTION OF A SPHERE:

A plane section of a sphere is a circle sphere S: 2 2 2 2 2 2 0x y z ux vy wz d+ + + + + + = plane U:

1 0ax by cz d+ + + = the combined equation (S,U) is a circle.

The equation of the sphere through the circle a ( , )S U is 1S S KU= +

Ie, 2 2 2

1 1(2 ) (2 ) (2 ) 0S x y z x u ak y v bk z w ck d kd= + + + + + + + + + + =

EQUATION OF THE TANGENT PLANE

The sphere is 2 2 2 2 2 2 0x y z ux vy wz d+ + + + + + = and the point of contact is 1 1 1( , , )x y z then

Equation of the Tangent plane is 1 1 1 1 1 1( ) ( ) ( ) 0xx yy zz u x x v y y w z z d+ + + + + + + + + =

CONDITION FOR TANGENCY:

Condition for tangency is perpendicular from centre to the plane = radius

12 2 2

2 2 2

au bv cw du v w d

a b c

- - - += + + -

+ +

CONDITION FOR ORTHOGONALITY OF TWO SPHERES:

The condition for orthogonality of two spheres 2 221 1 1 1 1: 2 2 2 0S x y z u x v y w z d+ + + + + + = and

2 2 2

2 2 2 2 2: 2 2 2 0S x y z u x v y w z d+ + + + + + = is 1 2 1 2 1 2 1 22( )u u v v w w d d+ + = +

1. Find the centre and radius of the sphere

Solution: The General equation

Centre=(-u,-v,-w) =(-3,3,-4) d=9

Radius= r =



44

2. A sphere of constant radius K passes through the origin O and meets the axes in A, B, C.

Prove that the locus of the centroid of the triangle ABC is the sphere

Solution: S:

As S passes through O (0, 0, 0): d=0

S: ------------------- (1)

= -------------------- (2)

As A is in x axis (y=0, z=0)

In (1)Þ

X=-2u & x=0

A (-2u, 0, 0), B (0,-2v, 0), c (0, 0,-2w)

Let G(x,y,z) be the centroid of triangle ABC.

X= y= z=

U= v= w=

(2) Þ Locus of G(x,y,z) is:

=

3. Find the equation of the concentric sphere with

and passing through the origin.

Solution: Given sphere is:

------------------ (1)

Equation of concentric sphere is

------------------- (2)

As sphere (2) passes through (0, 0, 0) K=0

Put in (2)

4. Find the equation of the sphere concentric with and

passing through the point (1,2,3)

Solution: The equation of the sphere concentric with

------------------ (i)

(I) passes through the point (1,2,3), k= -2

The required equation of the sphere is



45

5. Find the equation of the sphere passes through the points (1,0,0),(0,1,0), (0,0,1) and centre

on the plane x+y+z=6

Solution: S: ------------------ (1)

As c (-u,-v,-w) on the plane x+y+z=6

Þ –u-v-w=6 ----------------------------------- (2)

(1, 0, 0): 1+2u+d=0 u=-1/2-d/2

(0, 1, 0): 1+2v+d=0 v=-1/2-d/2

(0, 0, 1): 1+2w+d=0 w=-1/2-d/2 ------------------ (3)

Put in (2)

=

d =

Put in (3): u=-2 v=-2 w=-2

Put in (1): S:

6. Find the equation of the sphere with centre (1, 2, 3) and touch the plane, x+2y+2z=1.

Solution: Centre = (-u,-v,-w) = (1, 2, 3)

U=-1, v=-2, w=-3

= (perpendicular distance from the centre of the sphere

to the plane)2

d= 14-

S:

S: .

7. Find the equation of the sphere through the circle , 2x+3y+4z=5 and the

point (1,2,3).

Solution: U: 2x+3y+4z-5=0.

S1: S+KU = 0

S1: ) + k(2x+3y+4z-5) =0 ------------------ (1)

As S1 passes through (1, 2, 3).



46

(1+4+9-9)+k (2+6+12-5) =0

K= =

Put in (1): ) (2x+3y+4z-5) =0

8. Find the equation of the sphere through the spheres

and having its centre on the plane 4x-5y-z=3.

Solution: Given spheres are:

S=

X2(1+K) +Y

2 (1+K) +Z

2 (1+K) +X (-2+4K) +Y (-3+5K) +Z (4-6K) + (8+2K) =0

2U=-2+4K, U=-1+2K

2V=-3+5K, V=-3+5K/2

2W=4-6K, W=2-3K

C= (-U,-V,-W) =

As centre lie on: 4x-5y-z=3

4-8k-5 - (-2+3k) =3

8-16k-15+25k+4-6k=6

3k=9 ÞK=3

(1) Þ 4(x+y+z) +10x+12y-14z+14=0

9. Find the centre, radius and area of the circle. , x+y+z=3

Solution: Given sphere is S:

U: x+y+z=3

U=0, v=0, w=0, d=-9

C= (-U,-V,-W) = (0, 0, 0)

r = =3

Radius of the circle R=NP=

CN= perpendicular from center ‘c’ to the plane

d.r. of CN=(1,1,1)

Equation of CN = k



47

X=k, y=k, z=k

Now N(x,y,z) satisfies the plane equation

K+k+k=3ÞK=1

.

10. Obtain the equation of the sphere having the circle , x+y+z=3

as the Great circle.


U: x+y+z-3=0

S1: S+KU = 0

------------------ (1)

2u=k, u=

2v=10+k, v=

2w=-4+k, w=

Centre = (-U,-V,-W) =

x+y+z=3 Þ =3

4

(1) Þ

11.Find the equation of the tangent plane to the sphere


U=1, v=2, w=-3, d=-6

Point of contact (x1,y1,z1) = ( )

Tangent plane is

Substitute the values: 2x+4y-10=0

12. Find the equation of the sphere through the circle , 2x+3y-

7z=10 and touch the plane x-2y+2z=1


U: 2x+3y-7z=10

S1: S+KU = 0



48

(1)

2u= , u=-2+k

2v= , v=

2w= , w=

C= (-U,-V,-W) =

As sphere touches the plane x-2y+2z-1=0

Radius =

K = ± 1

Substitute the value of k in (1)

13. Show that the plane 2x-2y+z+12=0 touches the sphere and

find the point of contact. (AU MAY/JUNE 2009)


U: 2x-2y+z+12=0

U=-1, v=-2, w=1, d=-3

C= (-U,-V,-W) = (1,2, -1)

r = =3

CN = perpendicular distance from the centre of the sphere to the

plane

CN =

r = perpendicular

The plane 2x-2y+z+12=0 touch the sphere d.r of CN = (2,-2, 1)

Equation of CN is k

X=2k+1, y=-2k+2, z=k-1 (1)

As N(x,y,z) lie on the plane 2x-2y+z+12=0

2(2k+1)-2(-2k+2) + (k-1) +12=0

K=-1

Substitute in (1), (-1, 4,-2)



49

14. Find the equation of the sphere that passes through the circle

, 3x-4y-6z+11=0 and cuts the sphere

orthogonally.


U: 3x-4y-6z+11=0

S1: S+KU = 0

(1)

U1= v1= w1= d1=

As sphere cuts the sphere Orthogonally.

U2=1, v2=2, w2=-3, d2=11

The condition of orthogonality is 2(U1U2 +, v1v2+, w1w2 = d1+ d2

-2+3k+6-8k+12+18k=17+11k

K=1/2

Put in (1):

2( )-x+2y-14z+23=0

15. Find the equation to the sphere passing through the circle

and cuts orthogonally the sphere

Solution:

The required sphere cuts orthogonally

Condition for orthogonal spheres is

Here

Hence equn (3) becomes

Sub k = -1 in (A) We get the required sphere



50

16. If any tangent to the sphere makes intercepts a,b,c on the co-ordinate

axes, prove that

Solution: The plane intercept form is:

Perpendicular distance from the centre of the plane = r =k

17. Find the equation of the sphere passes through the points (1,0,0),(0,1,0), (0,0,1) and its

radius as small as possible.

Solution: As sphere passes through (1, 0, 0), (0, 1, 0) and (0, 0, 1)

2u=2v=2w=-(1+d)

But r =

R=

As r is least Þ

d = -1/3

(Since it is least)

The sphere is 3( )-2(x+y+z)-1=0.

18. Show that the spheres =25, touch

externally and find their point of conduct.

Solution: Given the sphere S1=

u1=-9, v1=-12, w1=-20 d1=225 , c1= (9, 12, 20); r1= = 20

S2: =25

u2=0 , v2=0, w2=0 d1=-25 c2(0,0,0,) ;

r2= 5

C1C2=25 r1+r2=25

Spheres touch each other externally



51

The point of contact C divides C1C2 in the ratio 20: 5 = 4:1

The point of contact (9/5, 12/5,4)

19. Find the equation of the sphere which is tangential to the plane 2x + 2y -2z = 11 at (2,2,1)

and passes through the point (1,0,-1)

Solution: S: (1)

It passes through the point (1,0,-1)

2u+2w+ d= -2 (2)

Tangent pane at ( 2,2,1) is

x (2+u)+ y(2+v) +z(1+w) + 2u +2v +w + d = 0 (3)

Equation (2) & (3) represent the same line

= = k

2k-2, v= -2, w= -2k-1 d = -2k-2u-2v-w (4)

Using (4) in (2) we get k= -9/4

K=-9/4 in (4) gives u = -13/2, v=-2, w=7/2 , d=23/2

Equation (1) Þ

20. The circle on the sphere: has centre (2, 1, 2) find the

equation of the circle.

Solution: Given sphere:

whose center is C( -2,1,-4)

centre of circle N( 2,1,2)

dr of NC = 4,0,-6

Equation of the plane is ax+by+cz+d = 0 (1)

Equation (1) Becomes -4x -6z + d= 0 (2)

The point (2, 1, 2) is on the plane -8-12+d = 0 Þ d= 20

Equation (2) becomes -4x -6z+20 = 0 or -2x -3z+10 = 0

The equation of the circle is , -2x -3z+10 = 0

21. Find the equation of the sphere through the spheres

as the great circle

Solution: Given the spheres S1 : (1)

S2: (2)

The plane is S1-S2=0 U: 6x + 5y +16z-1=0 (3)



52

The required sphere is

S1+ kU= (4)

Centre C =

As the circle is the great CN = perpendicular distance from the centre of the sphere to the plane

CN =

Radius of the circle R=NP= = =

Area of circle πR2=6π.

To find centre of the circle N(x,y,z)

Circle centre of the sphere should lie on its own plane, gives

k =-197/61

Using k in (4)

22. Find the equation of the sphere having the points (2,-3,4) and (-1,5,7) as the ends of a

diameter. (AU MAY/JUNE 2009)

Solution: Given A (2,-3, 4) and B (-1, 5, 7)

S:

23. Find the equation of the sphere which passes through the circle and

x+2y+3z=3 and touch the plane 4x+3y=15 (AU MAY/JUNE 2009)


U: x+2y+3z=3

S1: S+KU = 0

(1)

2u= , u= ; 2v=2 , v= k ; 2w=3 , w= ;

C= (-u,-v,-w) =

As sphere touches the plane 4x+3y-15=0 ,We get k = -3

Substitute the value of k in (1)

24. Find the centre radius and area of the circle given by





54

UNIT – III DIFFERENTIAL CALCULUS

Curvature: The rate of bending of a curve in any interval is called the curvature of the curve in that

interval.

Curvature of a circle: The curvature of a circle at any point on it equals the reciprocal of its radius.

Radius of curvature: The radius of curvature of a curve at any point on it is defined as the reciprocal of the

curvature

Cartesian form of radius of curvature

2

3

2

2

2

1

÷÷ø

öççè

æ

úúû

ù

êêë

é÷ø

öçè

æ+

=

dx

yd

dx

dy

r

Parametric equation of radius of curvaturer =

Polar form of radius of curvature r =

Implicit form of radius of curvature r =

Centre of curvature: The circle which touches the curve at P and whose radius is equal to the radius of

curvature and its centre is known as centre of curvature.

Equation of circle of curvature: + =

Centre of curvaturee: = x – = y +

Evolute: The locus of the centre of curvature is called an evolute

Involute: If a curve C1 is the evolute of C2 , then C2 is said to be an involute of a curve C1.

Parametric equation of some standard curves

Curve Parametric form

Y2

= 4 ax (parabola) X = at2 , y =2at

(ellipse) X= a cosq , y =b sinq

(hyperbola) X= a secq , y = b tanq

= X= a cos3q , y = a sin

3q

Xy = c2 ( rectangular hyperbola) X = ct , y =

Envelope: A curve which touches each member of a family of curves is called envelope of that family

curves.



55

Envelope of a family of curves: The locus of the ultimate points of intersection of consecutive members of

a family of curve is called the envelope of the family of curves.

Properties of envelope and evolute

Property:1: The normal at any point of a curve is a tangent to its evolute touching at the corresponding

centre of curvature.

Property:2 The difference between the radii of curvature at two points of a curve

is equal to the length of the arc of the evolute between the two corresponding points.

Property:3: There is one evolute ,but an infinite number of involutes

Property:4 The envelope of a family of curves touches at each of its point. The corresponding member of

that family

Evolute as the envelope of normals: The normals to a curve form a family of straight lines.we know that

the envelope of the family of these normals is the locus of the ultimate points of intersection of consecutive

normals. But the centre of curvature of a curve is also the point of consecutive normals. Hence the envelope

of the normals and the locus of the centres of curvature are the same that is ,the evolute of a curve is the

envelope of the normals of the curve.

Part - A

1. Find the radius of curvature of y= at x=0

Solution:

y=

y1= at x= 0 y1=1

y2= at x= 0 y2=1

= =2

2. Find the radius of curvature of at x = on the curve y = 4 sin x – sin 2x

Solution:

y1=4 cosx – 2 cos 2x at x= y1=2

y2= at x = y2=-4

= =

3. Given the coordinates of the centre of curvature of the curve is given as 2a +3at2



56

-2at

3 Determine the evolute of the curve

Solution: 2a +3at2

t2=( )------------ 1

-2at3

t3= ------------ 2

( )3 = (

2

4( -2a)3=27a

2

The locus of the centre of curvature (evolute) is 4(x-2a)3=27a

2

4. Write the envelope of Am2+Bm+C=0, where m is the parameter and A, B and C are functions of x

and y. (NOV-08)

Solution: Given Am2+Bm+C=0……………………(1)

Differentiate (1) partially w.r.t. ‘m’

2Am+B=0 m=-B/2A………….(2)

Substitute (2) in (1) we get

A(-B/2A)2+B(-B/2A)+C=0

AB2/4A

2-B

2/2A+C=0

AB2-2AB

2+4A

2C=0

- AB2+4A

2C=0

Therefore B2-4AC=0 which is the required envelope.

5. Find the radius of curvature at any point of the curve y=x2. (NOV-07)

Solution: Radius of curvature

Given y=x2 y1= =2x and Y2 = =2

=

6. Find the envelope of the family of x sin a+ y cos a= p, abeing the parameter. (NOV-07)

Solution: Given x sin a + y cos a= p……………. (1)

Differentiate (1) partially w.r.t. ‘a’

X cos a- y sin a= o………………….(2)

Eliminate abetween (1) and (2)

X cos a = y sin a Þ Þ Tan a =



57

Sin cos

Substitute in (1)

x. + y. = p

= p

Squaring on both sides, x2 +y

2=p

2 which is the required envelope

7. What is the curvature of x2 +y

2 - 4x-6y+10=0 at any point on it . (JAN-06)

Solution: Given x2 +y

2 - 4x-6y+10=0

The given equation is of the form x2 +y

2 +2gx+2fy+c =0

Here 2g =-4 g=-2

2f =-6 f=-3

Centre C(2,3), radius r = = =

Curvature of the circle =

Therefore Curvature of x2 +y

2 - 4x-6y+10=0 is

8. Find the envelope of the family of straight lines y= mx , where m is the parameter (JAN-

06)

Solution: Given y= mx

(y-mx)2=m

2-1

Y2+m

2x

2 – 2mxy-m

2+1=0

m2 (x

2-1)-2mxy+y

2+1=0 which is quadratic in ‘m’

Here, A=x2-1 B=-2xy C=y

2+1

The condition is B2-4AC=0

4 x2y

2-4(x

2-1)(y

2+1)=0

4 x2y

2-4 x

2y

2-4x

2+4y

2+4=0

X2-y

2=4 which is the required envelope

9. Find the curvature of the curve 2x2 +2y

2 +5x-2y+1=0 (MAY-05,NOV-07)

Solution: Given 2x2 +2y

2 +5x-2y+1=0

÷2

x2 +y

2 +5/2x-y+1/2=0

Here 2g =5/2 g=5/4



58

2f=-1 f=-1/2 centre C (-5/4,1/2) radius r= = = =

Therefore Curvature of the circle 2x2 +2y

2 +5x-2y+1=0 is

10. State any two properties of evolute . (MAY-05)

Solution: (i) The normal at any point of a curve is a tangent to its evolute touching at the

corresponding contre of curvature. (ii) The difference between the radii of curvature at two points of a

curve is equal to the length of the arc of the evolute between the two corresponding points.

11. Define the curvature of a plane curve and what the curvature of a straight line. (JAN-05)

Solution: The rate at which the plane curve has turned at a point (rate of bending of a curve is called

the curvature of a curve. The curvature of a straight line is zero.

12. Define evolute and involute . (JAN-05)

Solution: The locus of centre of curvature of a curve (B1,B2,B3,…) is called evolute of the given

curve.

If a curve C2 is the evolute of a curve C1 ,then C1 is said to be an involute of a curve C2.

13. Find the radius of curvature of the curve x2 +y

2 -6x+4y+6=0 (NOV-08)

Solution: Given X2 +y

2 - 6x+4y+6=0

The given equation is of the form x2 +y

2 +2gx+2fy+c =0

Here 2g =-6 g=-3

2f =4 f=2

Centre C(3,-2), radius r = = =

Radius of Curvature of the circle = radius of the circle=

14. Find the envelope of the family of circles (x-a)2+y

2=4a,where a is the parameter.(MAY-07)

Solution: Given (x-a)2+y

2=4a

X2-2ax+a2

-4a+y2=0

a2-2a (x+2)+x

2+y

2=0 which is quadratic in


Here A=1 B=-2 (x+2) C= x2+y

2

4(x+2)2-4(x

2+y

2)=0

x2-4x+4- x

2-y

2=0



59

y2+4x=4 which is the required envelope.

15. Define evolute . (MAY-07)

Solution: The locus of centre of curvature ( , is called an evolute .

16. Find the envelope of the family of straight lines y=mx+ for different values of ‘m’.

Solution: Given y=mx+ (NOV-07, May 2009)

m2x-my+a=0 which is quadratic in ‘m’


Here A=x B=-y C=a

Y2-4ax = 0

There fore y2= 4ax which is the required envelope.

17. Find the envelope of the line +yt=2c, where‘t’ is the parameter. (NOV-02,05)

Solution: Given +yt=2c

Yt2-2ct+x=0 which is quadratic in ‘t’


Here A=y B=-2c C=x

C2-xy=0

Therefore xy=c2 which is the required envelope.

18. Find the radius of curvature of the curve y=c cosh(x/c)at the point where it crosses the y-axis.

Solution: Radius of curvature (NOV-05,May-09)

Given y=c cosh(x/c) and the curve crosses the y-axis. (i.e.)x=0 implies y=c.

Therefore the point of intersection is (0,c)

=c sin h(x/c)(1/c)=sin h (x/c)

(0,c)=sinh 0= 0

=cos h(x/c)(1/c)

(0,c)= cos h(0) (1/c) = 1/c



60

= c

19. Find the radius of curvature of the curve xy=c2at (c,c). (NOV-02)


Given xy=c2

x + y =0

= implies (c,c)=-1

=-

(c,c)= =-

20. Find the envelope of the family of straight lines y= mx , where m is the parameter

Solution: Given y= mx (Jan 09)

(y-mx)2=( )

Y2+m

2x

2 – 2mxy- m

2- =0

m2 (x

2- )-2mxy+y

2 =0 which is quadratic in ‘m’

Here, A=x2- B=-2xy C= y

2


4 x2y

2-4(x

2- )(y

2 - )=0

1 which is the required envelope

21. Write down the formula for radius of curvature in terms of parametric coordinate system. (May-

09)

Solution: Radius of curvature .

22. Define the circle of curvature at a point P(x1,y2) on the curve y = f(x). (Jan-09)

Solution: The circle of curvature is the circle whose centre is the centre of curvature and radius is the

radius of curvature. Therefore the equation of circle of curvature is



61

PART-B

1. Find the radius of curvature at the point on the curve .

Solution: Given x= ……………………(1) (NOV-07,MAY-08,MAY-09)

Y= …………………..(2)

Differentiate (1) and (2) w.r.t

=

=-sec2

=

Radius of curvature =

=3a sin

2. Find the radius of curvature of the curve at the point (-a, 0). (NOV-08)


Given

Differentiate w.r.t. ‘x’

2y

∞



62

\

\

3. Find the radius of curvature at the point (a,0)on the curve .(MAY-07)


Given

Differentiate w.r.t.’x’

2xy +y2.1=-3x

2 ………………(1)

∞

Therefore

Differentiate (2) w.r.t.’y’.

Therefore radius of curvature



63

(since the radius of curvature is non-negative)

4. Find the curvature of the parabola y2=4x at the vertex. (NOV-07)


Given; y2=4x


2y = 4

=2/y

∞

Therefore

Differentiate (1) w.r.t.’y’.

Therefore =2

Curvature K=1/ =1/2

5. Find the radius of curvature of the curve 27ay2= 4x

3 at the point where the tangent of the curve

makes an angle 450 with the X- axis.

Solution; Let (x1,y1) be the point on the curve at which the tangent makes an angle 450

with the X- axis.

(x1,y1) =Tan 45o=1-------------------------------- (1)

Given 27ay2= 4x

3


54ay =12x2 =



64

(x1,y1) = = -----------------------------------(2)

(x1,y1) =Tan 45o=1=

Gives -----------------------------(3)

As ( x1,y1) lies on the curve 27ay2

1= 4x3

1 ---------------------------------(4)

Using gives x1= 3a

And using (3) gives y1= 2a

Y1 at (3a,2a)= 1

Y2=

Y2= =1/6a

Therefore radius of curvature

6. Find the evolute of the rectangular hyperbola xy=c2. (JAN-06,NOV-08)

Solution: The equation of the given curve is xy=c2……….(1)

The parametric form of (1) is

X=ct; y=

=c; =c =-

Y1=

Y2=

=

The co-ordinates of the center of curvature Is

Where

………………………..(2)



65

=

………………………………(3)

Eliminating’ between (2) and (3),

(2)+(3)gives

……………………………(4)

(2)-(3)gives

…………………..(5)

(4)2/3

-(5)2/3

gives

=

Therefore

The locus of centre of curvature is

which is the required evolute of the rectangular hyperbola xy=c2.

7. Find the radius of curvature for the curve r=a(1+cos )at is a constant.

Solution: Given r=a(1+cos ) (NOV-07,08)

r’= -a sin and r’’ = cos

The radius of curvature in polar form is =′

′ ′′

=

= =

at is



66

Also,

Therefore, = .

8. Considering the evolute as the envelope of normals, find the evolute of the parabola x2=4ay.

Solution: Given x2=4ay (NOV-08)

The parametric equations are x=2at, y=at2

=2a and =2at

m =

We know that the equation of normal to the curve is y-y1= (x-x1)

y-at2= (x-2at) yt-at

3=-x+2at

x+yt=at3+2at……………….(1)

Differentiate (1) partially w.r.t.’t’we get

Y=3at2+2a t

2

Substitute the value of ‘t’ in (1)

y +x=a +2a

x = =

Squaring on both sides, we get

which is the required evolute.

9. Obtain the evolute of the parabola y2=4ax. (NOV-07)

Solution: Given y2=4ax………………………(1)

The parametric equations are x= at2, y=2at

=2at =2a

=y1

Y2=



67

= =

The co-ordinates of the center of curvature Is

Where

……………………..(2)

………………………………(3)

Eliminating ‘t’ between (2)and(3),

(2) gives

(3) gives

The locus of centre of curvature is

which is the required evolute.

10. Find the equation of the envelope of . (NOV-02,07)

Solution: Given that …………………(1)

And ……………..(2)

Differentiate (1)and(2) w.r.t ‘b’

………………..(3)

2a +2b=0…………………………(4)

(3)gives ……………..(5)

(4)gives ……………………(6)

From (5)and (6)



68

Substitute in (2) we get,

Therefore which is the required envelope.

11. Find the equation of circle of curvature of the parabola y2=12x at the point (3,6).

Solution: The equation of circle of curvature is (NOV-07,08,JAN-09)

Where,

Given y2=12x

Differentiate w.r.t.’x’ we get

2y =12 implies

Y1= (3,6)=1

Y2= (3,6)=-1/6

( can not be negative)

=3

Therefore, the equation of circle of curvature is

12. Find the radius of curvature at ‘t’ on x=etcost,y=e

tsint. (JAN-06)

Solution: Radius of curvature ′ ′

′ ′′ ′ ′′

Given



69

X’=

Y’=

X’’=

Y’’=

The radius of curvature is ′ ′

′ ′′ ′ ′′

=

13. Find the evolute of the ellipse . (MAY-05,07)

Solution: The given curve is

The parametric equations are x=acos ,y=bsin

Y1=

Y2=

Y2

The Co-ordinate of centre of curvature is

Where

= acos -

= acos

=

………………….(1)



70

= +

………………………(2)

Eliminating ‘ ’ between (1) and (2),we get

we know that,

The locus of is which is the evolute of the ellipse

14. Find the envelope of where l and m are connected by and a,b are constants. (MAY –

05, NOV-05)

Solution: Given that ………………..(1) …………………(2)

Differentiating (1) w.r.t.’m’. we get

…….(3)

Differentiating (2) w.r.t.’m’

…………..(4)



71

From (3) and (4)

Substitute in equation (2) ,

which is the required envelope.

15. Find the points on the parabola at which the radius of curvature is 4 . (MAY – 05)

Solution: Given …………….(1)

Let, P (a,b) be the point on the curve at where

Differentiate (1) w.r.t. ‘x’

Y1=2y

Y2=

But, hence 8

a+1=2 a=1, The points are (1,2),(1,-2).



72

16. Considering the evolute of a curve as the envelope of its normals find the evolute of .

(NOV-02,05,MAY-05)

Solution: The given curve is

The parametric equations are x=acos ,y=bsin

m=

We know that the equation of the normal is y-y1= (x-x1)

,we get

…………..(1)

Differentiate (1) partially w.r.t.’ ’, we get

Substitute in equation (1),we get



73

which is the required evolute of the ellipse.

17. Find the circle of curvature at (3,4)on xy=12. (JAN-05)

Solution: The equation of circle of curvature is

Where,

Given xy=12

Differentiate w.r.t.’x’ we get

x implies

Y1= (3,4)=-4/3

Y2= (3,6)

=3

Therefore, the equation of circle of curvature is .

18. Find the curvature for . (JAN-05)

Solution: Given

r’= =

r’’ =





75

20. Find the radius of curvature at the origin of the cycloid x = a (q + sin q) and y = a( 1- cos q).

(MAY’07, Nov ’08)

Given: x = a (q + sin q) , y = a( 1- cos q).

x’ =a( 1 + cos q) y’ = a( sin q)

x” = -a sin q y” = a cos q

The radius of curvature is

= = 4a cos

At q= 0

21. Find the envelope of the straight lines represented by the equation x cos α + y sin α = a sec α , α is

the parameter (Nov’ 07)

Solution: Given x cos α + y sin α = a sec α

Divided by cos α

x + y tan α = a sec2α

x + y tan α = a ( 1 + tan2α)

a tan2α – y tan α + a –x = 0

which is quadratic in tan α

A = a, B = -y C = a-x

The envelope is given by B2 – 4AC = 0

y2

= 4a(a-x) which is the required envelope

22. Prove that the evolute of the curve x = a (cos q + log tan ), y=a sin q is the catenary y = a cosh

( Nov ’05)

Solution : x = a (cos q + log tan )

x’ = a(-sin q + ) = a cot q cos q

y=a sin q

y’ = a cos q

y1 = = tan q

y2 = =¼ (sec 4 q sin q)



76

= a (cos q + log tan )- ( 1 + tan 2

=a log tan ………………..(1)

= a sin q+ ( 1 + tan 2

……………….(2)

Eliminate ‘q’from (1) and (2)

tan = ………………….(3)

=

y = a cosh which is the required evolute

23. Obtain the equation of the evolute of the curve x= .(May-09)

Solution: Given x=

= tan

tan = tan = =

The co-ordinates of centre of curvature is

=

…………………….(1)

=



77

= …………………………….(2)

Eliminating from equations (1) & (2) we get,

The locus of centre of curvature is which Is the required evolute.

24. Find the envelope of the straight line ,where a and b are parameters that are connected

by the relation a+b = c. (May-09)

Solution: Given ……………….(1)

And a+b = c……………………..(2)

Differentiate Eqns (1) and (2) partially w.r.t ‘b’

- =0

…………………………(3)

………………………………(4)

Equate (3) & (4) we get

and

a= b=

Substitute in eqn (2)

, which is the required evolute.

25. Find the envelope of the family of lines subject to the condition a+b =1.(Jan-09)

Solution: Given ……………….(1)

And a+b = 1……………………..(2)

Differentiate Eqns (1) and (2) partially w.r.t ‘b’

- =0





79

= =

Substitute in eqn (1) we get,

ax -by =

=

, which is the required evolute of the given curve.

27. Find the radius of curvature of the curve + = at . (Jan-09)

Solution: Given + =




80

Radius of curvature = =



81

Unit 4 Functions of several Variables

· Problems based on Partial Derivatives

· Problems based on Euler`s Theorem

· Problems based on Total Derivatives-Differentiation of Implicit Function

· Problems based on Jacobian

· Problems based on Taylor`s and Laurent Series

· Problems based on Maxima and Minima for Functions of Two Variables

· Problems based on Lagrangian Multiplier

Partial Derivatives

Partial Derivatives: Let be a function of two Variables x and y, If we keep y as a constant and

Vary x alone , then z is a function of x only ,

The derivative of z w.r.to x, treating y as a constant is called the partial derivatives w.r.to x and it is denoted by

the symbols

Notation:

Successive Partial Differentiation: let , then being the function of x and y can further be

differentiation partially with respect to x and y.

Problems:

1. If u = find

Solution:

2. Find if where and

Solution:



82

3. If show that

Solution: Given

Adding (1), (2),and (3) we get

4. show that

Solution: Let ,

---------------- (1)

---------------- (2)

---------------- (3)

(1)+ (2)+ (3)

5. If , Show that

Solution: Given

z is a homogeneous function of degree n=2

As z is homogeneous function of order n=2, it satisfies the Eulers equation



83

6. If u = , find

Sol: given u =

Euler`s Theorem for Homogeneous Function

Euler`s Theorem: If u be a homogeneous function of degree n an x and y then

Problems:

1. Verify Eulers theorem for the function

Solution: Given

This is a homogeneous function of degree 2.

Adding (1) and (2) we get

Hence Eulers theorem is verified.

2. If , Show that

Solution: Given

As z is a homogeneous function of order n = 1, it satisfies the Eulers theorem

---------------- (1)

(1) ---------------- (2)

By Eulers theorem of second order

′ ---------------- (3)

, ′ put in (3)



84

3. If u = , prove that

Sol: given =

is a homogeneous function of degree 0.

Hence by Euler’s theorem, we have x u = 0

Total Derivatives-Differentiation of Implicit Function

Total Derivative: If , then we can express u as a function of t

alone by substituting the value of x and y in Thus we can find the ordinary derivatives which is

called the total derivative of u distinguish it from partial derivatives

Now to find the without actually substituting the values of x and y in we establish the following

formula

Problems:

1. Find if where and

Solution:

2. If where and find and

Solution:

3. Find if

Solution: Let

4. Find

Ans :



85

5. If z be a function of x and y and u and v are other two variables, such that

show that

Solution: z may be represented as the function of u,v

(1)

Similarly

(2)

(1)+(2)

6.

Ans: Here Z is a composite function of u and v

……………………………………(1)

………………………………………..(2)

Now

Sub these values in (1) & (2) We get

………………………………..(3)

Now

Which implies ……………..(4)

(3)x(4) We get

……………………………..(A)

Similarly we get ………………………(B)

(A)+(B) Gives



86

2 2 2 22

2 2 2 2

22 2 2

2

7. ( ) cos , sin

( ) cos , sin

1

r r

r

If u is a function x and y and x and y are functions of r and givenby a x e y e

u u u ushownthat e

x y r

b x r y r provethat

z z z zan

x y r r r

q q q

q

q q

-

= =

ì ü¶ ¶ ¶ ¶+ = +í ý

¶ ¶ ¶ ¶î þ= =

ì ü¶ ¶ ¶ ¶ì ü ì ü ì ü+ = +í ý í ý í ý í ý¶ ¶ ¶ ¶î þ î þ î þî þ

2 2 2 2

2 2 2 2 2

1 1.

[

( ) , .

cos ; sin ;

sin ; cos

;

r r

r r

z z z z zd

x y r r r

Solution

a Hereu is a functionof x y whichis given as a functionof r and

x xe x e y

r

y ye y e x

r

x xx

r

q q

q

q qq

q qq

ì ü ì ü ì ü ì ü¶ ¶ ¶ ¶ ¶ì ü+ = + +í ý í ý í ý í ý í ý¶ ¶ ¶ ¶ ¶î þî þ î þ î þ î þ

¶ ¶= = = - = -

¶ ¶¶ ¶

= = = =¶ ¶

¶ ¶\ =¶

;

. . ...(1)

y yy x and y

r

u u x u y u uNow x y

r x r y r x y

q q¶ ¶

= - = =¶ ¶ ¶

¶ ¶ ¶ ¶ ¶ ¶ ¶= + = +

¶ ¶ ¶ ¶ ¶ ¶ ¶

. . ...(2)

(1), ,

...(3)

u u x u y u uy x

x y x y

From we get

x yr x y

q q q¶ ¶ ¶ ¶ ¶ ¶ ¶

= + = - +¶ ¶ ¶ ¶ ¶ ¶ ¶

¶ ¶ ¶= +

¶ ¶ ¶

2

2

2 2 2 22 2

2 2 2

(2) , ,

...(4)

,

2 ...(5)

From we get

y xx y

u u u uNow x y x y

r r r x y x y

u u u ux xy y

r x x y y

q¶ ¶ ¶= - +

¶ ¶ ¶

æ öæ ö¶ ¶ ¶ ¶ ¶ ¶ ¶æ ö= = + +ç ÷ç ÷ç ÷¶ ¶ ¶ ¶ ¶ ¶ ¶è ø è øè ø

¶ ¶ ¶ ¶= + +

¶ ¶ ¶ ¶ ¶

2

2

2 2 2 22 2

2 2 22 ...(6)

u u u uy x y x

x y x y

u u u uy xy x

x x y y

q q q

q

æ öæ ö¶ ¶ ¶ ¶ ¶ ¶ ¶æ ö= = - + - +ç ÷ç ÷ç ÷¶ ¶ ¶ ¶ ¶ ¶ ¶è ø è øè ø

¶ ¶ ¶ ¶= - +

¶ ¶ ¶ ¶ ¶



87

2 2 2 22 2 2 2

2 2 2 2

2 2 2 22 2 2

2 2 2 2

2 2 2 22

2 2 2 2

(5) (6) ,

( ) ( )

( )

. .,

( ) cos ; sin ; sin ; cos

r

r

Adding and we get

u u u ux y x y

r x y

u u u ux y e

x y x y

u u u ui e e

x y r

x x y yb r r

r r

q

q

q q q qq q

-

¶ ¶ ¶ ¶+ = + + +

¶ ¶ ¶ ¶

é ù é ù¶ ¶ ¶ ¶= + + = +ê ú ê ú¶ ¶ ¶ ¶ë û ë û

é ù¶ ¶ ¶ ¶+ = +ê ú¶ ¶ ¶ ¶ë û

¶ ¶ ¶ ¶= = - = =

¶ ¶ ¶ ¶¶

2 22

2

22

. cos . sin .

sin cos

1cos sin sin cos

z z x z y z z

r x r y r x y

z z x z y z zr r

x y x y

z z z z z z

r r x y x y

z z

x y

q q

q qq q q

q q q qq

¶ ¶ ¶ ¶ ¶ ¶= + = +

¶ ¶ ¶ ¶ ¶ ¶ ¶

¶ ¶ ¶ ¶ ¶ ¶ ¶= + = - +

¶ ¶ ¶ ¶ ¶ ¶ ¶

ì ü ì ü¶ ¶ ¶ ¶ ¶ ¶ì ü ì ü+ = + + - +í ý í ý í ý í ý¶ ¶ ¶ ¶ ¶ ¶î þ î þ î þ î þ

ì ü¶ ¶ì ü= +í ý í ý¶ ¶î þ î þ

2

2

2 2 2

2 2

2 2 22 2

2 2

2

2

cos sin

cos sin

cos sin 2sin cos

sin cos

z z

r r r

z z

r x y

z x z y z x z y

x r y x r x y r y r

z z z

x y x y

z z

z zr r

x y

q q

q q

q q q q

q q q

q qq

¶ ¶ ¶æ ö= ç ÷¶ ¶ ¶è ø

é ù¶ ¶ ¶= +ê ú¶ ¶ ¶ë û

é ù é ù¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶= + + +ê ú ê ú¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶ë û ë û

¶ ¶ ¶= + +

¶ ¶ ¶ ¶

¶ ¶ ¶æ ö= ç ÷¶ ¶ ¶è ø

é¶ ¶ ¶= - +ê¶ ¶ ¶ë

2 2

2cos sin

z z x z yr r

x x y xq q

q q

ùúû

ì ü¶ ¶ ¶ ¶ ¶= - - +í ý

¶ ¶ ¶ ¶ ¶ ¶î þ

2 2

2sin cos

z z z yr r

y yq q

q qé ù¶ ¶ ¶ ¶

- + +ê ú¶ ¶ ¶ ¶ë û

2 2 22 2 2 2 2

2 2sin cos 2 sin cos cos sin

z z z z zr r r r

x y x y x yq q q q q q

æ ö¶ ¶ ¶ ¶ ¶= + - - +ç ÷¶ ¶ ¶ ¶ ¶ ¶è ø





89

Solution:

( )( )

( ) ( ) ( )

( )( )

( ) ( )( )( )( )

1 2 3

1 1 2 1 2 3 1 2 3

1 2 3

1 1 1

1 2 3

1 2 3 2 2 2

1 2 3 1 2 3

3 3 3

1 2 3

2 1

2 3 1 3 1 2

1 1 2

2

1 2

, ,4. , 1 , 1 , 1 .

, ,

:

, ,

, ,

1 0 0

1 0

1 1

1

y y yFind the valueof if y x y x x y x x x

x x x

Solution

y y y

x x x

y y y y y y

x x x x x x

y y y

x x x

x x

x x x x x x

x x x

x x

Hen

¶= - = - = -

¶

¶ ¶ ¶¶ ¶ ¶

¶ ¶ ¶ ¶=

¶ ¶ ¶ ¶

¶ ¶ ¶¶ ¶ ¶

-

= - -

- - -

= - - -

= -

.cethe solution

( )( )

( )( )

2 2 2, ,

5. , , , ., ,

:

, , exp , , , ,

, ,

, ,

2 2 2

1 1 1

x y zIf u xyz v x y z w x y z find J

u v w

Solution

Sinceu v ware liciltly given in terms of x y z we first evaluate

x y zJ

u v w

u u u

x y zyz zx xy

v v vWe knowthat J x y z

x y z

w w w

x y z

¶= = + + = + + =

¶

¶¢ =

¶

¶ ¶ ¶¶ ¶ ¶

¶ ¶ ¶¢ = =

¶ ¶ ¶

¶ ¶ ¶¶ ¶ ¶

( ) ( ) ( )( ) ( ) ( )

2 2 2 2 2 2

2 2 2 2 2 2

2

2

yz y z zx x z xy x y

yz y z zx x z xy x y

x y x z xy xz y z yz

= - - - + -

= - - - + -é ùë û

é ù= - - + + -ë û



90

( ) ( ) ( )

( ) ( )

( ) ( ) ( )( )( )( )( )( )( )

( )( )

( )( )( )

2 2 2

2

2

2

2

2

2

sin , 1, ,

, ,

, ,

1

2

x y z x y z yz y z

y z x x y z yz

y z y z x x z x

y z z x y x

x y y z z x

By u g JJ we get

x y zJ

u v w

x y y z z x

é ù= - - - + -ë û

é ù= - - + +ë û

= - - - -é ùë û= - - -

= - - - -

¢ =

¶=¶

-=

- - -

( )( )

2 3 3 1 1 21 2 3

1 2 3

1 2 3 1 2 3

2 3 3 1 1 21 2 3

1 2 3

1 1 1

1 2 3

1 2 3 2 2 2

1 2 3 1 2 3

3

6. , , .

, , , , 4.

Pr :

: , , .

, ,

, ,

x x x x x xIf y y y

x x x

Showthat the Jacobianof y y y with respect to x x x is

oof

x x x x x xGiven y y y

x x x

y y y

x x x

y y y y y yWe knowthat

x x x x x x

y

x

= = =

= = =

¶ ¶ ¶¶ ¶ ¶

¶ ¶ ¶ ¶=

¶ ¶ ¶ ¶

¶¶

3 3

1 2 3

y y

x x

¶ ¶¶ ¶

2 3 3 2

2

1 1 1

3 3 1 1

2

2 2 2

2 1 1 2

2

3 3 3

2 3 3 1 1 2

2 3 3 1 1 22 2 2

1 2 3

2 3 3 1 1 2

1

x x x x

x x x

x x x x

x x x

x x x x

x x x

x x x x x x

x x x x x xx x x

x x x x x x

-

= -

-

-

= -

-

2 2 2

1 2 3

2 2 2

1 2 3

1 1 1

1 1 1

1 1 1

x x x

x x x

-

= -

-



91

( ) ( ) ( )1 1 1 1 1 1 1 1

0 2 2 4

.Hencethe proof

=- - - - - + +

= + + =

7. If and . Evaluate with out actual substitution.

Solution: and

8. If . Find the Jacobian .

Solution:



92

( )

2 2

2 2

( , )23. cos , sin ,

( , )

cos sin

cos sin

sin cos

( , )

( , )

cos sin

sin cos

cos sin

cos sin

x yIf x r y r find

r

Solution

Given x r y r

x y

r r

x yr r

x x

x y rWe knowthat

y yr

r

r

r

r r

r r

q qq

q q

q q

q qq q

qq

qq qq q

q q

q q

¶= =

¶

= =

¶ ¶= =

¶ ¶¶ ¶

= - =¶ ¶

¶ ¶¶ ¶ ¶=

¶ ¶¶¶ ¶

-=

= +

= + =

9. If ,

show that they are not independent. And also find the relation between them.

Sol: will not be independent if

Hence, = = 0.

Hence are not independent.

To find the relation between them:

Given

and

Now,

Þ

Þ , which is the required relation.

Taylor`s Series and Laurent Series

The Taylor’s series expansion of in the power of and is

Problems:



93

1. Find the Taylor’s series expansion of in the power of x and y upto third degree terms.

Sol: The Taylor’s series expansion of in the power of and is

+….…..(1)

,

, = = 1

= =

= , = 1

,

= 0

= 2

+

= ……..

2. Using Taylors series expansion express in powers of x and y upto second degree

terms at

Solution:



94

Taylors series is

3. Expand as Taylors series up to second degree terms

Solution:

Taylors series is

2

2

4. ' var exp 3 2 1 2.

exp ( , )

( , ) ( , ) ( ) ( , ) ( ) ( , )

1( ) ( , ) 2( )(

2!

x y

xx

Use Taylor s series oftwo iables to and x y y in powers of x and y

Solution

We knowthat the ansionof f x y in powers x a and y bis givenby

f x y f a b x a f a b y b f a b

x a f a b x a y

+ - + -

- -

= + - + -

+ - + - - 2

3 2 2

3

) ( , ) ( ) ( , )

1[( ) 3( ) ( ) 3( )( )

3!

( ) ] ....(1)

xy yy

xxx xyy xyy

yyy

b f a b y b f a b

x a f x a y b f x a y b f

y b f

é ù+ -ë û

+ - + - - + - -

+ -

2

2

1, 2

( , ) 3 2 ( 1, 2) 6

2 ( 1,2) 4

3 ( 1,2) 4

2 ( 1,2) 4

2 ( 1,2) 2

0 ( 1,2) 0

0 ( 1,2) 0

x x

y y

xx xx

xy xy

yy yy

xxx xxx

Here a b

Now f x y x y f

f xy f

f x f

f y f

f x f

f f

f f

= - =

= + - \ - =

= - = -

= + - =

= - =

= - = -

= - =

= - =

2 ( 1,2) 2xxy xxyf f= - =



95

0 ( 1,2) 0

0 ( 1,2) 0

s in (1) ,

xyy xyy

yyy yyy

f f

f f

Substituting thesevalue we get

= - =

= - =

2 2 2

2

2 2

13 2 6 ( 1)( 4) ( 2)(4) [( 1) (4) 2( 1)( 2)( 2) ( 2) (0)]

2!

1[0 3( 1) ( 2)(2) 0 0]

3!

6 4( 1) 4( 2) 2( 1) 2( 1)( 2) 2( 1) ( 2)

.

x y y x y x x y y

x y

x y x x y x y

Hencethe solution

+ - = + + - + - + + + + - - + -

+ + + - + +

= - + + - + + - + - + + -

1

1

2 2

2

2 2

5. tan (1,1).

( , ) tan

1( , ) .

1

1, (1,1)

2

x

x

yExpand in the neighbourhood of

x

solution

yLet f x y

x

yf x y

y x

x

yf

x y

-

-=

æ ö= -ç ÷è ø+

= - = -+

2

2

2 2

2 2 2

1 1( , ) .

1

1, (1,1)

2

( , ) ( 1)( ) .2

y

y

xx

f x yy x

x

xf

x y

f x y y x y x-

=+

= =+

= - - +

( )( )

( )

22 2

2 2

22 2

2 1, (1,1)

2

1 2( , )

xx

xy

xyf

x y

x y x xf x y

x y

= =+

+ -=

+

( )2 2

22 2

2 2 2

(1,1) 0

( , ) ( 1)( ) .2

xy

yy

y xf

x y

f x y x x y y-

-= =

+

= - +



96

( )

( )

22 2

1

2 2

2

2 1(1,1)

2

tan ( , )

1(1,1) ( 1) (1,1) ( 1) (1,1) ( 1) (1,1) 2( 1)( 1) (1,1) ( 1) (1,1) ...

2!

sin : 2

1 1 1 1( 1) ( 1) ( 1) . 2( 1)(

4 2 2 2! 2

yy

x y xx xy yy

xyf

x y

yf x y

x

f x f y f x f x y f y f

u g cor

x y x x yp

-

= - = -+

\ =

é ùé ù= + - + - + - + - - + - +ë û ë û

é ùæ ö= - - + - + - + - -ç ÷ê úè øë û2

1 2 2

11)0 ( 1) .....

2

1 1 1 1tan ( 1) ( 1) ( 1) ( 1) ...

4 2 2 4 4

y

yx y x y

x

Hence the solution

p-

é ùæ ö+ - - +ç ÷ê úè øë û

\ = - - + - + - - - +

Maxima and Minima and Lagrangian Multiplier

Defn: Maximum Value

Defn: Minimum Value

Defn: Extremum Value

Defn: Lagrangian Multiplier

Suppose we require to find the maximum and minimum values of where x,y,z are subject to a

constraint equation

We define a function where

Which is independent of x,y,z

Problems:

1. Examine for its extreme values

Solution: Given

At maximum point and minimum point

The points may be maximum points or minimum points.



97

At and r = 12 > 0

is a minimum point

Minimum value =

At

The points are saddle points.

At and r = -12 < 0

is a maximum point

maximum value = .

2. In a plane triangle ABC ,find the maximum value of .

Solution:

At the maximum point and minimum point

Solving these equations

A+B+C = π

At ,

and r < 0

The point is a maximum point.

Maximum value = .

3. Find the volume of the largest rectangular parallelepiped that can be inscribed in the ellipsoid .

Solution: The given ellipsoid is ---------------- (1)

The volume of the parallelepiped is ---------------- (2)

At the max point or min point



98

---------------- (3)

---------------- (4)

---------------- (5)

Solve the equation

(3)x+ (4)y+(5)z Þ

Put in (3)

Similarly,

Put in (2) Max volume = .

4. Find the dimensions of the rectangular box without a top of maximum capacity, whose surface is 108 sq. cm

Solution: Given Surface area

---------------- (1)

The volume is ---------------- (2)


---------------- (3)

---------------- (4)

---------------- (5)

Solve the equation

(3)x - (4)y

(3)x - (5)z

Put in (1)

y =6, z =3

The dimension of the box, having max capacity is Length=6cm, Breadth = 6cm, Height = 3cm.

5. The temperature T at any point (x, y, z) in space is T = 400xy . Find the highest temperature on the surface of

the unit sphere

Solution: ---------------- (1)

---------------- (2)




99

---------------- (3)

---------------- (4)

---------------- (5)

Solve the equation

(3)x+ (4)y+(5)z

1600

Put in (3) and (4) we get

The highest temperature is = 50.

2 2 2 2

2 2 2 2

2 2

2 2

6.Pr , cos sin , sin cos

( )

var cos sin ,

sin cos , int

z z z zovethat where x u v y u v

x y u v

OR

By changing independent iablesu and v to x and y by means of the relations x u v

z zy u v showthat transforms

u v

a a a a

a a

a a

¶ ¶ ¶ ¶+ = + = - = +

¶ ¶ ¶ ¶

= -

¶ ¶= + +

¶ ¶

2 2

2 2.

z zo

x y

¶ ¶+

¶ ¶

( )

:

.

cos sin

cos sin

cos sin (1)

Solution

Here z is a composite functionof u and v

z z x z y

u x u y u

z z

x y

or z zx x y

u x y

a a

a a

a a

¶ ¶ ¶ ¶ ¶= × + ×

¶ ¶ ¶ ¶ ¶

¶ ¶= +

¶ ¶

æ ö¶ ¶ ¶= +ç ÷¶ ¶ ¶è ø

¶ ¶ ¶Þ º + ®

¶ ¶ ¶



100

( )

2

2

,

sin cos

sin cos

sin cos (2)

(1) (2).

cos sin

z z x z yAlso

v x v y v

z z

x y

or z zv x y

v x y

Now we shall makeuseof theequivalanceof operators as givenby and

z z

u u u

u

a a

a a

a a

a a

¶ ¶ ¶ ¶ ¶= × + ×

¶ ¶ ¶ ¶ ¶

¶ ¶= - +

¶ ¶

æ ö¶ ¶ ¶= - +ç ÷¶ ¶ ¶è ø

¶ ¶ ¶Þ º - + ®

¶ ¶ ¶

¶ ¶ ¶æ ö= ç ÷¶ ¶ ¶è ø

¶ ¶= +

¶2 2 2 2

2 2

2 2

2 2 2 22 2

2 2 2

2

2

cos sin ( sin (1) (2) )

cos cos sin sin cos sin

., cos 2cos sin sin (3)

sin cos

z zu g and

y x y

z z z z

x x y y x y

z z z zie

u x x y y

z zSimilarly

v v v

x

a a

a a a a a a

a a a a

a

æ öæ ö¶ ¶+ç ÷ç ÷¶ ¶ ¶è øè ø

¶ ¶ ¶ ¶= + + +

¶ ¶ ¶ ¶ ¶ ¶

¶ ¶ ¶ ¶= + + ®

¶ ¶ ¶ ¶ ¶

¶ ¶ ¶æ ö= ç ÷¶ ¶ ¶è ø

¶= - +

¶sin cos ( sin (3) (4) )

z zu g and

y x ya a a

æ öæ ö¶ ¶ ¶- +ç ÷ç ÷¶ ¶ ¶è øè ø

2 2 2 22 2

2 2

2 2 2 22 2

2 2 2

2 2 2 2

2 2 2 2

sin sin sin cos

sin 2 sin cos (4)

(3) (4),

.

z z z zcos cos

x x y y x y

z z z zcos

v x x y y

Adding and

z z z z

u v x y

Hence the proof

a a a a a a

a a a a

¶ ¶ ¶ ¶= - - +

¶ ¶ ¶ ¶ ¶ ¶

¶ ¶ ¶ ¶= - + ®

¶ ¶ ¶ ¶ ¶

¶ ¶ ¶ ¶+ = +

¶ ¶ ¶ ¶

( )

( )

2 2

2 2

2

1 17. max , .

:

1 1,

12

Investigatethe ima of the functions f x y x xy yx y

Solution

Given f x y x xy yx y

fx y

x x

= + + + +

= + + + +

¶= + -

¶



101

2

12

fx y

y y

¶= + -

¶

2

2 3

2

2 3

2

3 2

2

3 2

2

3 2 2

3 2 2

22

22

1

1: max min ,

10 2 0 . ., 2 1 0 ...(1)

10 2 0 . ., 2 1 0 ...(2)

(1) 2 0 ...(3)

(2) 2 0 ...(4)

f

x x

f

y y

f

x y

Step For a imumor imum we must have

fx y i e x x y

x x

fx y i e y xy

y y

y x y x y y

x xy x y x

Fr

¶= +

¶¶

= +¶

¶=

¶ ¶

¶= Þ + - = + - =

¶¶

= Þ + - = + - =¶

´ Þ + - =

´ Þ + - =2 2 3(3) 2 ...(5)

(5) (4)

om we get x y y x y

Substituting in we get

= -

{ }{ }

3 3

3 3

2 2

2 2 0

. ., 2 2 0

. ., 2 ( ) ( ) 0

. ., 2 ( )( ) ( ) 0

( ) 2 ( ) 1 0

( )2 ( ) 1 0

xy y x y x

i e x y xy y x

i e xy x y x y

i e xy x y x y x y

x y xy x y

x y or xy x y

+ - - =

- - + =

- + - =

- + + - =

- + + =

\ = + + =

3 3

3 3

3

3

3 3

3 3

, 2 1 0

2 1 0

. .,3 1

1

3

1 1

3 3

1 1, int

3 3

when x y theequation x x y

gives x x

i e x

x

x y

Hence is a critical po

= + - =

+ - =

=

=

= \ =

æ öç ÷ç ÷è ø

3 31 1

2 : , ,3 3

Step Atæ öç ÷ç ÷è ø



102

2

2

2 2

2

22 2 2

2 2

2

2

22 8

13

8 ; 1

. 8 8 1 0

8 0

f

x

f f

y x y

f f f

x y x y

fand

x

¶= + =

¶

¶ ¶= =

¶ ¶ ¶

æ ö¶ ¶ ¶\ - = ´ - >ç ÷¶ ¶ ¶ ¶è ø

¶= >

¶

3 3

4

3

1 1( , ) min int ,

3 3

min 3 .

f x y has a imum at the po and

the imumvalueis

æ ö\ ç ÷ç ÷

è ø

2 2

2 2

2

2

2

2

2

8. min ( , ) .

( , ) ...(1)

2 ...(2)

2 ...(3)

2 ...(4)

2 ...(5)

1 ...(6)

Find the imumvalueof the function f x y x y xy ax by

Solution

Given f x y x y xy ax by

fx y a

x

fy x b

y

f

x

f

y

f

x y

= + + + +

= + + + +

¶= + +

¶¶

= + +¶

¶=

¶¶

=¶

¶=

¶ ¶

1:

0 2 ...(7)

Step For miimumvalueof the function

fx y a

x

¶= Þ + +

¶

0 2 ...(8)f

y x by

¶= Þ + +

¶

(7) (8)

2

2

Solving and we get

x y a

x y b

+ = -

+ = -



103

. ., 4 2 2

2

3 2

2

3

2(8)

3

2

3

2

(4 2 )

2

2

i e x y a

x y b

x b a

b ax

b aSubstituting x in we get

b ab

y

b a

a b

+ = -

+ = -

= -

-=

-=

-ì ü- +í ýî þ=

- -=

= -

2 2 2

2 2

22 2 2

2 2

2int , 2

3

22 : , 2

3

2 , 2 1

. 2 2 1

3 0

2min , 2

3

b aThecritical po is a b

b aStep At a b

f f fand

x y x y

f f f

x y x y

b aWe have imumvalueof the function at a b

-æ ö\ -ç ÷è ø

-æ ö-ç ÷è ø

¶ ¶ ¶= = =

¶ ¶ ¶ ¶

æ ö¶ ¶ ¶\ - = ´ -ç ÷¶ ¶ ¶ ¶è ø

= >

-æ ö\ -ç ÷è ø

( ) ( ) ( )

( )[ ]

22

3: min ( , )

22 ( , ).

3

2 2 22 2 2

3 3 3

2 22 2 2

3 3

Step The imumvalueof f x y is obtained by putting

b ax and y a bin f x y

b a b a b aMinimumvalue a b a b a b a b

b a b aa b a a b a b b

-= = -

- - -æ ö æ ö æ ö= + - + - + + -ç ÷ ç ÷ ç ÷è ø è ø è ø

é ù- -æ ö æ ö= + - + + - - +ç ÷ ç ÷ê úè ø è øë û



104

2 22 2

2 2 2 2

2 2

2 4 5( 2 )( 0

3 3

4 5 8 102 2

9

4 5 8 9 27 18

9

13 23

9

b a a ba b a b

ab b a aba ab ab b

ab b a a ab b

b a ab

- -æ ö é ù= + - -ç ÷ ê úè ø ë û

- - += + - - +

- - + - +=

+ -=

2 6 5 5 3 49. tan

3 2 2 2 1 6

2 6 5...(1)

3 2 2

5 3 4...(2)

2 1 6

int (3 2, 2 6, 2 5)

int sec

x y z x y zFind the shortest dis cebetweenthelines and

Solution

x y zLet

x y z

Any po onthe first lineis P and

any po onthe ond lin

l

m

l l l

- - - - - += = = =

- - -

- - -= = =

- -- - +

= = =-

+ - + - +

2 2 2

2 2 2

2 2

2 2 2

2 2

(2 5, 3, 6 4)

(2 3 5 2 ) (6 2 3 ) (5 2 4 6 )

. ., 17 41 32 66 114 99

( , ) 17 41 32 66 114 99

34 32 66

32 82 114

34, 82, 32

eisQ

PQ

i e PQ

Let f

f

f

f f f

m m m

l m l m l m

l m lm l m

l m l m lm l m

l ml

l mm

l m l m

+ + - -

\ = + - - + - - - + - + +

= + - - + +

= + - - + +

¶= - -

¶¶

= - + +¶

¶ ¶ ¶= = = -

¶ ¶ ¶ ¶

max min ' '

0, 0

. ., 34 32 66 0

32 82 114 0

For a imumor a imumof f we should have

f f

i e

l ml ml m

¶ ¶= =

¶ ¶

- - =

- + + =

2

1, 1

(1, 1) 34 0

Solving thesetwoequations we get

fAt

l m

l m

= = -

¶- = >

¶ ¶



105

22 2 2

2 2. 0

f f f

l m l mæ ö¶ ¶ ¶

- >ç ÷¶ ¶ ¶ ¶è ø

2

2

(1, 1) ( , ) min .

. ., (1, 1) , min .

(1, 1), 17 41 32 66 114 99

9

9 3

At the function f has imum

i e At PQ has imum which gives the shortest length

At PQ

Shortest length PQ

l m\ -

-

- = + + - - +

=

\ = = =

2 2 2

2 2 2

2 2 2

1 1 110. min 1

1 1 11

' '

1 1 1( , , ) ( ) 1 ...(1)

, ,

Find the imumvalueof x y z subject to theconditionx y z

Solution

Let f x y z

gx y z

Let the auxillary function F be

F x y z x y zx y z

By Lagranges method thevalues of x y z

l

+ + + + =

= + +

= + + -

æ ö= + + + + + -ç ÷

è ø

3

2

3

2

3

2

3 3 3

' ' min

0 2 0 ...(2)2

0 2 0 ...(3)2

0 2 0 ...(4)2

1 1 10 1 0 ...(5)

(2) , (3) (4)

2

. .,

for which f is imum areobtained

by the following eqations

Fx x

x x

Fy y

y y

Fz z

z z

F

x y z

From and we get

x y z

i e x y

l l

l l

l l

l

l

¶= Þ - = Þ =

¶¶

= Þ - = Þ =¶

¶= Þ - = Þ =

¶¶

= Þ + + - =¶

= = =

=

1

3

...(6)2

zlæ ö= = ç ÷è ø

(6) (5)

31 3

Substituting in we get

or xx

= =



106

2 2 2 2 2 2

31 3

31 3

(3,3,3) int min . min

3 3 3 27

or yy

or zz

is the po where imumvalueoccur The imumvalueof

x y z is

= =

= =

\

+ + + + =

3

2

11. tan , tan

72 . .

tan 2 , , .

2 . . 72

. ., 2 72

Athinclosed rec gular box is to haveoneedgeequal to twicetheother and a cons t

valume m Find theleast surface area of thebox

Solution

Let the sides of the rec gular boxbe x x y

Thenvolumeis x x y

i e x y

i

=

=2

2

2

2

. ., 36 ...(1)

2(2 . ) 2(2 . ) 2( . )

4 6 ...(2)

min

4 6

36

e x y

The surface area is givenby

S x x x y x y

x xy

Now we haveto find the imum surface area

S x xy under thecondition

x y

=

= + +

= +

= +

=

( ) ( )

2

2

2 2

4 6 ,

36

' '

( , ) 4 6 36 ...(3)

, , ' ' min

0 8 6 2 0 ...(4)

Let f x xy

g x y

Let the auxillary function F be F f g

F x y x xy x y

By Lagranges method thevalues of x y z for which F is mumis obtained

fromthe following equations

Fx y xy

x

l

l

l

= +

= -

= +

= + + -

¶= Þ + + =

¶

20 6 0 ...(5)

0 0 0

Fx x

y

F

z

l¶

= Þ + =¶

¶= Þ =

¶



107

2 2

2

2

0 36 0 36 ...(6)

6(5) ...(7)

6(6)

36...(8)

Fx y x y

From we get x

Substituting x in we get

yx

l

l

l

l

¶= Þ - = Þ =

¶

= -

= -

= =

( )1

1

12. max

. ., ( , , ) ...(1)

0 0 ...(2)

0 0 ...(3)

0

m n p

m n p

m n p

m n p

n m p

m n

Find the imumvalueof x y z when x y z a

Solution

Let f x y z and g x y z a

Let the auxillary function F f g

i e F x y z x y z x y z a

Fmx y z

x

Fny x z

y

Fp x y

z

l

l

l

l

-

-

+ + =

= = + + -

= +

= + + + -

¶= Þ + =

¶¶

= Þ + =¶

¶= Þ

¶1 0 ...(4)

0 0 ...(5)

pz

Fx y z a

l

l

- + =

¶= Þ + + - =

¶

2

2 2 3

2

6, (4)

486 12 0 8

. ., 2

2 (7) (8)

3, 4.

min (3,4)

min 4(3) 6(3)(4)

108

Substituting x y in we get

i e

Substituting in and we get

x y

S has a imumvalue at

The imumvalueof S

ll

l l ll

ll

= - =

- + - = Þ =-

= -

= -

= =

\

= +

=

1

1

(2) , (3) (4) ,

m n p

n m p

From and we get

mx y z

ny x z

l

l

-

-

- =

- =

1m n pp x y zl -- =



108

1 1 1. .,

. .,

m n p n m p m n pi e mx y z ny x z p x y z

m n pi e

x y z

m n p m n p

x y z a

- - -= =

= =

+ + + += =

+ +

( )

max

max

n n pm n p

m n p

Hence imumvalueof f occurs when

amx

m n p

any

m n p

apz

m n p

The imumvalueof

m n pf a

m n p

+ ++ +

\

=+ +

=+ +

=+ +

=+ +







111

I =

2. Evaluate: over the region in the first quadrant of the circle x2+y

2=1.

Solution: In the given region, y varies from 0 to and x varies from 0 to 1.

I =

dx

Put x = sinθ. Then dx = cosθdθ. q varies from 0 to π/2.

3. Evaluate:

ò òúúû

ù

êêë

é+

1

0

22 )(

x

x

dydxxyyx

Solution : Let I= ò òúúû

ù

êêë

é+

1

0

22 )(

x

x

dydxxyyx = ò òúúû

ù

êêë

é+

1

0

22 )(

x

x

dydxxyyx = ò ÷÷ø

öççè

æ+

1

0

322

32dx

xyyxx

x

= ò úû

ùêë

é÷÷ø

öççè

æ+-÷÷

ø

öççè

æ+

1

0

442/33

323

.

2dx

xxxyxx =

1

0

2/74

5

5

6

5

)3)(2

7(8 úú

û

ù

êê

ë

é÷ø

öçè

æ-+xxx

= )0(6

1

21

2

8

1-÷ø

öçè

æ ++ = ÷ø

öçè

æ -+168

281621=

56

3

168

9=



112

DOUBLE INTEGRATION IN POLAR COORDINATES:

To evaluate , we first integrate w.r.to r between the limits and . Keeping is

fixed and the resulting expression is integrated w.r.to from to .

In this integral and and ,

PROBLEMS BASED ON POLAR FORMS USING DOUBLE INTEGRATION:

PART A

1. Evaluate: ∞

Solution: ∞

∞

2. Evaluate:

Solution: Let I =

PART B

1. Evaluate: over the cardioids r = a (1+cosθ).

Solution: The limits of r: 0 to a (1+cosθ) and The limits of θ: 0 to π.

I = θ

= θ

Put 1+cosθ = t then –sinθ dθ = dt





114

CHANGE THE ORDER OF INTEGRATION:

The double integral will take the form when the order of

integration is changed. This process of converting a given double integral into its equivalent double integral by

changing the order of integration is often called change of order of integration. To effect the change of order of

integration, the region of integration is identified first, a rough sketch of the region is drawn and then the new

limits are fixed.

PART A

1. Find the limits of integration in the double integral

Solution: The limits are: y varies from 0 to 1 and x varies from 0 to 1-y.

2. Change the order of integration

Solution: The given region of integration is bounded by y=0, y=a, x=y & x=a.

After changing the order, we have, I =

3. Change the order of integration for the double integral

Solution:

PART B

1. Change the order of integration in I = and hence evaluate it.

Solution: Let I =

The given region of integration is bounded by x=0, x=1, y=x2 and x+y=2.

In the given integration x is fixed and y is varying.

So, after changing the order we have to keep y fixed and x should vary.

After changing the order we’ve two regions R1 & R2

I = I1+I2

I =



115

=

2. Evaluate by changing the order of integration.

Solution: The given region is bounded by x=0, x=1, y=x and x2+y

2=2.

I =

After changing the order we’ve,

The region R is splinted into two regions R1& R2.

In R1: limits of x: 0 to y & limits of y: 0 to 1

In R2: limits of x: 0 to & limits of y: 1 to

I = I1+I2

I1 =

I2 =

I =

3. Evaluate by changing the order of integration in

Solution: Let I =

The given region of integration is bounded by x=0, x=4, y = , y2 = 4x



116

After changing the order we’ve

Limits of x: y2/4 to 2√y

Limits of y: 0 to 4

I = = 16/3.

4. Change the order of integration in ò ò-

=1

0

2

2

),(

x

x

dydxyxfI

Solution: Given ò ò-

=1

0

2

2

),(

x

x

dydxyxfI

The given region of integration is bounded by x=0, x=1, y=x2 and x+y=2

In the given integration x is fixed and y is varying

So, after changing the order we have to keep y fixed and x should vary.

After changing the order we have two regions R1 & R2

I = I1 + I2

ò ò ò ò-

+=1

0 0

2

1

2

1

),(),(

y x

dxdyyxfdxdyyxfI

PROBLEMS BASED ON AREA AS A DOUBLE INTEGRAL:

v Area of the region R in Cartesian form is given by

R R

dxdy or dydxòò òò

v Area of the region R in polar form is given by

R

rdrdqòò

PART A

1. Find the smaller area bounded by y = 2-x and x2+y

2=4.

Solution: Required area =

PART B

1. Find the area of the region outside the inner circle r=2cos and inside the outer circle r=4 cos by

double integration.

Solution: Required Area =

=2



117

2. Find the area of the circle of radius ‘a’ by double integration.

Solution: Transforming Cartesian in Polar coordinates

(i.e.) x=rcosθ & y=rsinθ. Then dxdy = rdrdθ

limits of θ: 0 to and limits of r: o to θ

Required Area = 2xupper area

=2 θ

= θ

3. Find over the area bounded between the circles r = 2sinθ & r= 4sinθ.

Solution: In the region of integration, r varies from r=2sinθ& r= 4sinθ and θ varies from0to π.

I = θ

θ

θθ

4. Find the area enclosed by the ellipse

Solution: Area of the ellipse = 4 x area of the first quadrant =4

5. Find the area inside the circle r=asinq but lying outside the cardiod r=a(1-cosq)

Solution: Given curves are r=asinq and r =a(1-cosq)

The curves intersect where a sin q = a (1-cosq)

Þ a sin q = a –a cosq Þ a sin q + a cosq = a Þ sin q + cosq =1

Þ 2

1cos

2

1sin

2

1=+ qq Þ

2

1cos

4cos

4cossin =+ q

pq

pq



118

Þ 4

sin2

1)

4sin(

ppq ==+

4)(

44

pp

ppq -=+Þ or

Þ 44

)(0p

pp

qq -=+= or Þ 224

2 ppp

ppq =-=-=

2)(0

pqq ==Þ or

\The required area = ò ò-

4/

0

sin

)cos1(

p q

q

qa

a

rdrd = ò-

÷÷ø

öççè

æ2/

0

sin

)cos1(

2

2

p q

q

qdr

a

a

= ( )ò -+-2/

0

222

cos2cos1(sin2

p

qqqq da

= ( )ò +--2/

0

222

cos21cossin2

p

qqqq da

= ( ) úû

ùêë

é+---ò

2/

0

222

cos21coscos12

p

qqqq da

= ( )ò -2/

0

22

cos2cos22

p

qqq da

= ( )ò -2/

0

22

coscos2.2

p

qqq da

= ( ) úû

ùêë

é- ò

2/

0

22/

0

2 cossin

pp qqq da = ú

û

ùêë

é÷ø

öçè

æ +- ò

2/

0

2

2

2cos11

p

qq

da

= úû

ùêë

éúû

ùêë

é +-2/

0

2

2

2sin

2

11

pq

qa = úû

ùêë

é-÷ø

öçè

æ +- 0022

112 p

a

= 4

)4(

41

22 pp -

=úû

ùêë

é -a

a

6.Find by double integration, the area enclosed by the curves

Ans:

Sub (1) in (2) we get

Therefore the point of intersection of (1)&(2) is (0,0) and (4a,4a)

x Varies from 0 to 4a and y varies from

The required Area =





120

=

=

Hence

Therefore the integral will become,

PROBLEMS BASED ON TRIPLE INTEGRATION

PART – A

1. Evaluate:

Solution: Let I =



121

2. Evaluate:

Solution: Let I=

3. Evaluate:

Solution: Let I=

4. Evaluate:

Solution: Let I =

PART – B

1. Evaluate ò ò ò+

++a x yx

zyx dzdydxe

log

0 0 0

Solution : ò ò ò+

++a x yx

zyx dzdydxe

log

0 0 0

= ò ò +++a x

yxzyx dydxe

log

0 0

0][ = ò ò ++ -a x

yxyx dydxee

log

0 0

)(2 )(



122

= ò ÷÷ø

öççè

æ- +

+a x

yxyx

dxee

log

0 0

)(2

2= ò ú

û

ùêë

é÷÷ø

öççè

æ--÷

ø

öçè

æ -a

xx

xx dxee

ee

log

0

224

22

1

= ò ÷÷ø

öççè

æ+-

a

xxx

dxeee

log

0

24

2

3

2=

a

xxx

eee

log

0

24

2

3

2úû

ùêë

é+-

= ÷ø

öçè

æ +--÷ø

öçè

æ +- 14

3

8

1

4

3

8

1 loglog2log4 aaa eee

= 8

3

4

3

8

1 24 -+- aaa

2. Evaluate ( )ò ò ò ++a b c

dxdydzzyx0 0 0

222

Solution : ò ò úû

ùêë

é++=

a b c

dydzxzxyx

I0 0 0

223

3 = ò ò ú

û

ùêë

é++

a b

dydzczcyc

0 0

223

3

= ò úû

ùêë

é++

a b

dzcyzcyyc

0 0

233

33 = ò ú

û

ùêë

é++

a

dzcbzcbbc

0

233

33

=

a

cbzzcbbzc

0

333

333úû

ùêë

é++ =

333

333 cbaacbbac++ = ][

3

222 abcabc

++

3. Find the volume bounded by the cylinder x2+y

2=4 and the planes y+z=4 and z=0.

Solution: The limits are:

Z varies from: 0 to 4-y

X varies from: - to

Y varies from: -2 to 2.

Required volume = 2 = 2

= 2 dy= 2

=2

= 8 since y is an odd function.

= 16 =16

= 16x2x = 16π

4. Find the volume of the tetrahedron bounded by the plane and the coordinate plane.

Solution: The limits are:



123

X varies from 0 to a

Y varies from 0 to b

Z varies from o to c

Required Volume =

=

= c

= c

= c

= =

= =

4. Find the volume of the sphere x2+y

2+z

2=a

2 using triple integral.

Solution: Required Volume = 8 x volume in the positive octant = 8

Limits of integration are:

Z varies from 0 to

Y varies from 0 to

X varies from 0 to a

Volume = 8 = 8

= 8

= 8

= 8 = 2π = 2π

= 2π = 4πa3

3cu.units

5. Find the Volume of the ellipsoid

Solution: Required Volume = 8 x Volume in the first octant

Limits of Integration are:



124

Z varies from 0 to c 1x2

a2

y2

b2

Y varies from 0 to b 1x2

a2

X varies from 0 to a.

Volume = 8 dzdydxc 1

x2

a2y2

b2

0

b 1x2

a2

0

a

0

= 8 z 0

c 1x2

a2y2

b2b 1x2

a2

0

a

0dydx

= 8c 1x2

a2

y2

b2

b 1x2

a2

0dydx

a

0

=8c

bα2 y2α

0dydx

a

0 where α b 1

x2

a2

=8c

b

y

2α2 y2 α2

2sin 1 y

α 0

aa

0dx

=8c

b

π

4b2 1

x2

a2

a

0dx

=2πbc 1x2

a2

a

0dx

=2πbc xx3

3a20

a

=2πbc aa

3

= 4πabc

3cu units



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UNIT I: MATRIX CHARACTERISTIC EQUATION