A text book of engineering mathematics volume i

Contents Chapters Basic Results and Concetps Unit I : Differential Calculus- I Chapter 1 : Successive Differentiation and Leibnitz's Theorem Chapter 2: Partial Differentiation Chapter 3:

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Basic Results and Concepts

3 Unit Prefixes Used

Multiples and Prefixes

P rho Ll caE- delta

cr sigma L cap sigma

1 rad = 57°17'45"

viii

loge 3 = 1.0986 logH)e = 0.4343

10 = 0.0174 rad

Quantily F.P.s System e.G.S System M.K.S System

Length foot (ft) centimetre (cm) metre(m)

Time second (sec) second (sec) second (sec)

Roots are equal if b2 - 4ac = 0

Roots are real and distinct if b2 - 4ac > 0

Roots are imaginary if b2 - 4ac < 0

Its nth term T n = arn- and sum S = n 1-r' '" 1-r S = - - (r < 1)

(iii) Numbers l/a, 1/(a + d), 1/(a + 2d), are said to be in Harmonic Progression (H.P.) (i.e., a sequence is said to be in H.P if its reciprocals are in A.P Its nth term

Tn =1/(a+n-1d).)

(iv) If a and b be two numbers then their

Arithmetic mean = ! (a + b), Geometric mean = jiili; Harmonic mean = 2ab/(a

(vi) Stirling's approximation When n is large n! - J21tn nn e-n

3 Permutations and Combinations

(i) Natural logarithm log x has base e and is inverse of ex

Common logarithm lOglOX = M log x where M = lOglOe = 0.4343

(ii) loga 1= 0; logaO = - oo(a > 1); loga a = 1

(iii) log (mn) = log m + logn; log (min) = log m -log n; log (mn) = n log m III GEOMETRY

1 Coordinates of a point: Cartesian (x ,y) and polar (r , 8)

Then x = r cos 8, Y = r sin 8

(XlI yd and (x2/Y2) = -+-(Y-2 -Y-l-)2-=-]

Points of division of the line joining (XlI Yl) and (X21 Y2) in the ration ml : m2 is

(ffilX2 + ffi 2Xl ffil + ffi2 I ffi lY2 + ffi2Yl ) ffi l + ffi2

In a triangle having vertices (XlI Yl), (X2, Y2) and (X31 Y3)

where a, b, c are the lengths of the sides of the triangle

(iv) Circumcentre is the point of intersection of the right bisectors of the sides of the triangle

(v) Orthocentre is the point of intersection of the perpendiculars drawn from the vertices to the opposite sides of the triangle

2 Straight Line

(i) Slope of the line joining the points (XlI Yl) and (X21 Y2) = Y2 - Yl

51 ope 0 f th e me ax + Y + c 1 b = 0 IS - -I.e a I - - - - -coeff , of X

b eoeff/of Y (ii) Equation of a line:

X2 - Xl

(a) having slope m and cutting an intercept c on y-axis is Y = mx + c

(b) cutting intercepts a and b from the axes is + .r = 1

a b

(c) passing through (XlI Yl) and having slope m is Y - Yl = m(x - Xl)

(d) Passing through (XlI Y2) and making an La with the X - axis is

Two lines are parallel if ml = m2

Two lines are perpendicular if mlm2 = -1

Any line parallel to the line ax + by + c = 0 is ax + by + k = 0

Any line perpendicular to ax + by + c = 0 is bx - ay + k = 0

(iv) Length of the perpendicular from (Xl, Yl)of the line ax + by + c = O is

(i) Standard equation of the parabola is y2 = 4ax

Its parametric equations are X = at2, y = 2at

Latus - rectum LL' = 4a, Focus is S (a,O)

Directrix ZM is X + a = O

xii

M b + ; :?I

o

II ctl

Its parametric equations are

x = a cos 8, Y = b sin 8

Eccentricity e = b2 / a2)

Latus - rectum LSL' = 2b2/ a

Foci S (- ae, 0) and S' (ae, 0)

Directrices ZM (x = - a/e) and Z'M' (x = a/e.)

(ii) Sum of the focal distances of any point on the ellipse is equal to the major axis i.e.,

7 Nature of the a Conic

The equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents

and a rectangular hyperbola if in addition, a + b = O

8 Volumes and Surface Areas

xv

Total Surface Area

6a2

2 (Ib + bh + hI)

41tr2 2nr (r + h)

1tr (r + 1)

2 Any t-ratio of (n 900 ± 8) = ± same ratio of 8, when n is even

= ± co - ratio of 8, when n is odd

270 360 -1 0

-00 0

The sign + or - is to be decided from the quadrant in which n 900 ± 8 lies

1

e.g., sin 5700 = sin (6 x 900 + 300) = -sin 300 = - 2'

tan 3150 = tan (3 x 900 + 450) = cot 450 = - 1

3 sin (A ± B) = sin A cos B ± cos A sin B

cos (A ± B) = cos A cos B ± sin A sin B

1 + tan A tan B 1 - tan A

5 sin A cos B = ! [sin (A + B) + sin (A - B)]

2

cos A sin B = ! [sin (A + B) - sin (A - B)]

2

1 coa A cos B = - [cos (A + B) + cos (A - B)]

7 a sin x + b cos x = r sin (x + 8)

a cos x + b sin x = r cos (x - 8)

xvi

1 + tan2 A

(iv) Area of = !bc sin A = Js(s - a) (s - b) (s - c) where s = !(a + b + c)

log (1 - x) = - x + "2 + "3 + 00 (iv) Gregory series

tan-1 x=x - -+ - - 00,tanh-1 x= -log =x+ -+ -+ 00

10 (i) Complex number: z = x + iy = r (cos a + i sin a) = rei6

(ii) Euler1s theorem: cos a + i sin a = ei9

(iii) Demoivre1s theorem: (cos a + isin a)n = cos na + i sin n a

11 (i) Hyperbolic functions: sin h x = eX - e-x ;cos h x = eX + e-x ;

tan h x = cot h x = sec h x = cosec h x ==

(ii) Relations between hyperbolic and trigonometric functions:

sin ix = i sin h x i cos h x = cos h x i tan ix = i tan h x

(iii) Inverse hyperbolic functions;

sinh-1x==log[x+j;;2;1];cosh-1x=log[x+Jx2 -1];tanh-1x==.:! log l+x

2 1 - x

xvii

cos{bx c) cos(bx+c+ntan-1b/a)

(vii) Leibnitz theorem: (UV)n

= Un + nC1Un-1Vl+ nC2Un_2V2 + + nCrUn_rVr + + nCnVn

3 Integration

(i) fxn dx = ; + 1 (n:!;- 1) f-; dx = logex

fex dx = eX fax dx = aX jlogea

(ii) fSin x dx = - cos x fcos x dx = sin x

ftan x dx = - log cos x fcot x dx = log sin x

fsec x dx = log(sec x + tan x) = log tan + %)

fcosec x dx = log(cosec x - cot x) = log tan (%)

(vi) fsin h x dx = cos h x

ftan h x dx = log cos h x

1 (i) IfR=xi+ y1 +zkthenr= IRI =J(x2 +y2 +z2)

(ii) PQ = position vector of Q-position vector of P

2 If A = a1 I + a2J + a3K, B = b1I + b2J + b¥, then

(i) Scalar product: A B = abcos8 = a1b1+ a2b2+ a3b3

(ii) Vector product: A x B = ab sin 8 N = Area of the parallelogram having A + B

(ii) If [A B C] = 0, then A, B, C are coplanar

(iii) Vector triple product A x (B x C) = (A C) B - (A B) C

(A x B) x C = (C A) B - (C B) A

xx

(iii) If curl F = 0 then F is called an irrotational vector

5 Velocity = dR/ dt; Acceleration = d2R/ dt2;

Tangent vector = dR/dt; Normal vector =

6 Green's theo<em , l' + ",dy) " I Ie (z - : 1 dxdy

Stoke's theorem: JF dR = JcurI F N ds

Gauss divergence theorem: JF N ds = JdiV F dv

7 C d' oor tna t e sys ems t

Polar coordinates Cylindrical (r,8) coordinates (p, <p , z) Coordinate x = r cos 8 x = p cos

transformations y = r sin e y = p sin

z=z Jacobian a(x , v) - = r 8(x, y, z) =p

D( r, 8) D(p, <p, z) (Arc - length)2 (dS)2 = (drF + r2 (ds)2= (d p)2 + p2

x = r sin8 cos

y = r sin 8 sin

Z = r cos 8 8(x, y, z) = r 2· 8 SIn 8(r, 8,<p)

(ds)2 = (dr)2 + r2 (d8)2 + (r sin 8)2 (dq, )2

dV = r2 sin 8 dr d8

dq,

Contents

Chapters

Basic Results and Concetps

Unit I : Differential Calculus- I Chapter 1 : Successive Differentiation and Leibnitz's Theorem

Chapter 2: Partial Differentiation

Chapter 3: Curve Tracing

Chapter 4 : Expansion of Function

Unit II: Differential Calculus - II Chapter 5: Jacobian

Chapter 6 : Approximation of Errors

Chapter 7 :Extrema of Functions of Several Variables

Chapter 8 : Lagranges Method of Undetermind Multipliers

Unit III: Matrices Chapter 9: Matrices

Unit IV : Multiple Integers Chapter 10 : Multiple Integres

Chapter 11 : Beta and Gamma Functions

Unit V: Vector Calculus Chapter 12: Vector Differential Calculus

Chapter 13 : Vector Integral Calculus

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UNIT-l

Differential Calculus-I

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Chapter 1 Succesive Differentiation and

Leibnitz's Theorm

Successive Differentiation

Definition and Notation :- If y be a function of x, its differential coefficient

dy / dx will be in general a function of x which can be differentiated The differential coefficient of dy / dx is called the second differential coefficient of y Similarly, the differential coefficient of the second differential coefficient is called the third differential coefficient, and so on The successive differential coefficients

of yare denoted by

dy d2y d3y

dx ' dx2 ' dx3

the nth differential coefficient of y being dny dxn

Alternative methods of writing the nth differential coefficient are

( dx y, Dny, Yn, dxn ,y(n) d )n dny

The Process to find the differential coefficient of a function again and again is called successive Differentiation

Thus, if y = f(x), the successive differential coefficients of f(x) are

dx' dx2 ' dx3 , dx"

These are also denoted by :

(i) yl, y2, y3 yn

Similarly, we can write

(a) If m = n, then Dn(ax+b)n = l!! .an

(-1)n1n (c) Dn (ax+br1

= I.!! an (ax + b)"+1

Y2 = m2 arnx (log a)2

Similarly, we can write yn = mn aRlX (log a)n

: Dn arnx = mn aRlX (log a)n

In particular,

Dn ax = ax (log a)n

4 nth Derivative of log (ax+b)

Lety = log (ax+b)

Succesive Differentiation and Leibnitz's Theorm

5 nth Derivative of sin

(ax+b):-Let y = sin (ax+b)

2 y3 = a3 sin (ax+b+ 3n )

2 Similarly, we can write

yn = an sin (ax+b+ nn)

2 : On sin (ax+b) = an sin(ax+b+ nn)

2

In particular,

On • sln X = sln x+-• ( nn)

2

6 nth Derivative of cos (ax+b) :

Lety = cos (ax+b)

: Yl = -a sin (ax+b)

n

= a cos (ax+b+-)

2 Y2 = -a2 sin (ax+b+ 2: ) 2

Yl = a eax sin (bx+c)+ b eax cos (bx+c)

put a = r cos<l> and b = r sin<l>

: r2 = a2+b2and <I> = tan-1 (bja)

Yl = r eax [cos<l> sin (bx+c) +sin<l> cos (bx+c)]

Similarly, we can obtain

Y2 = r2 eax sin (bx+c+ 2<1»

y3 = r3 eax sin (bx+c+ 3<1»

Continuingthis process n times, we get

yn = rn eax sin (bx+c+n<l»

: On {eax sin (bx+c)} = rn eax sin(bx+c+n<l»

where r = (a2+b2)1/2 and <I> = tan-1

8 nth Derivative of eax cos (bx+c) :

Proceeding exactly as above, we get

On (eax cos(bx+c)}= rn eax cos (bx+c+n<l»

where r and <I> have the same meaning as above

Example 1: if y = 1 2 ' find yn

1-5x+6x Solution:

1

y = 1- 5x+6x2

(U.P.T.U.2005)

Succesive Differentiation and Leibnitz's Theorm

Succesive Di(ferentiation and Leibnitz's Theorm

Leibnitz's Theorem

:-Statement :-If u and v are functions of x then

Dn (uv)=(Dn u).v + nCl Dn-l u Dv + nC2 Dn-2 u D2v + + nCr Dn-r u Drv + +u.Dnu

Proof:- We shall prove this theorem by mathematical induction method By differentiation, we have

D(uv) = (Du) v + u (Dv)

=(Du) v + (ICl U ) (Dv)

Thus the theorem is true for n = 1

Let the theorem be true for n = m i.e

Dm (uv) = (Dmu) v + mel Dm-l u Dv + mC2 Dm-2 u D2v + +mCr Dm-ru Drv + + uDmv

Differentiating the above once again, we get

Dm+l(uv) = (Dm+lu) v+Dmu DV+mCl{Dmu Dv+Dm-lu D2v}+me2{Dm-lu.D2v+ Dm-2u D3v}+ +mer{Dm-r+lu Drv+Dm-ru Dr+lv}+ + +{Du Dmv+uDm+lv}

Rearranging the terms, we get

Dm+l(uv) = (Dm+lu)v+(1+mCl) Dmu.Dv+(mCl+mC2) Dm-lu D2v+ + (mCr +mer+l) Dm-ru Dr+lv+ +u Dm+lv

Now using mCr + mCr+l = m+1Cr+l, we have

Dm+l(uv)= (Dm+lu).v +m+1CIDmu.Dv+ +m+1Cr+1Dm-ru.Dr+l v+ +

u.Dm+lv

Thus we have seen that if the theorem is true for n = m, it is also true for n = m+l Therefore by principle of induction, the theorem is true for every positive integral value ofn

Example 5: If y = cos(m sin-I x) prove that

(1_X2) yn+2 - (2n+1) Xyn+l + (mL n2)Yn =0

Solution: y = cos (m sin-I x)

Differentiating again, we have (1-x2) 2Y1Y2+Y12 (-2x)+m2 2YYI =0

or (1 - X2) Y2- Xyl + m2y = 0

Differentiating n times by Leibnitz's theorem, we get

yn+2 (1-x2)+ nCl yn+l (-2x) + nC2 yn(-2) - {Yn+l x+ nC1 yn.1} +m2 yn =0

or (1-x2)Yn+2 - 2nx yn+l - n(n-1)Yn - xy n+l - nyn + m2Yn =0

or (1-x2)Yn+2 - (2n+1) xy n+l + (m2-n2) = 0 Hence proved

I If asin-lx h th

Differentiating both sides with respect to x, we get

Y12(-2x) + 2YlY2(1-X2) =a2 2YYl

(ii)

Differentiating each term of this equation n times with respect to x, we get Dn[(1-x2)Y2] - Dn(x Yl) - a2 Dn(y) =0

Example 7: If y = (x2 -1)", Prove that

(x2.-1)Yn+2 +2XYn+l - n(n+l)Yn =0

Differentiating again both sides with respect to x, we get

Yl(2x) + yz(xL1) = 2n(xYl+Y)

or [(xL1)Yn+2 +nyn+l (2x)+ yn(2)]+ 2(1 - n)[Xyn+l+nYn] - 2nyn=0

or (X2 -1)Yn+2 + 2XYn+l - n(n+1)Yn =0 Hence proved

Succesive Di[ferentiation and Leibnitz's Theorm

Example 8: If cos-1 (t) Prove that

X2Yn+2 + (2n+1)xYn+1 + 2n2Yn =0

Solution: Given cos-1 (y jb) = log(xj n)n

or Xyl = -b n sin{n log (x/ n)}

Again differentiating both sides with respect to x, we get

XY2 + y1 = -bn cos {n } n _ ( 1 ) .!

or X2Y2 + Xy1 = - n2b cos { n log (x/n)}

= -n2y, from (i)

or x2 yn+2 + (2n+1) XYn+l + 2n2 yn =0 Hence proved

Example 9: If x= cos h y] Prove that

(X2 -1)Y2 + Xyl - m2y =0

and (X2 -1) yn+2 + (2n+1) xyn+l + (n2 - m2) yn =0

Solution: Here x = cos h )log y]

(x2 1) 2Yl y2 + (2X)Yl2 = m 2 2YYl

Differentiating (iii) n times by Leibnitz's theorem, we get

n(n -1)

[(X2 1)Yn+2 + n (2X)Yn+2+ (2) yn] + [xy n+1 + n(1)YnJ - m2 yn =0

or (x2-1) yn+2 + (2n+1) Xyn+l + (n2 - m2) yn =0 Hence Proved

Example 10: If yl/m +y -l/m = 2x Prove that (X2 -1) yn+2 + (2n +1)XYn+l+(n2 - m2) =0

which is the same result as (ii)

Hence for both the values of y given by (i) we get Y12 (X2 -1) = m2y2

Differentiating both sides of this with respect to x, we get

Y12 (2x) + 2Y1Y2 (X2 -1) = m2.2YY1

or Y2 (x2 -1) + Xyl - m2y =0

(i)

Succesive Ditferentiation and Leibnitz's Theorm

Differentiating each term of this equation n times (by Leibnitz's theorem) with respect to x, we get

Dn{Y2(x2_1)}+Dn(xYl) - m2 Dn(y) =0

n(n -1)

or {(X2 -1) yn+2 +nyn+l (2x) + yn (2)} + {XYn+l + nYn.1} - m2 yn =0

1.2

or (x2 -1) yn+2 + (2n +1) Xyn+l + (n2 - m2) yn =0 Hence Proved

Example 11 : If y = (X2 -1)n Prove that (X2 -1)Yn+2 + 2XYn+l - n(n+1)Yn =0

Differentiating again both sides with respect to x, we get

Yl(2x) +Y2 (x2 -1) = 2n [XYl +y]

which Proves the first part

or -d { (1_x2)_n + n(n+1) Pn=O Henceproved dP }

Problem 12 : If y = sin-1 x , find (yn)o

Solution: Given y = sin-1 x

Differentiating both sides of (iii) n times with respect to x by Leibnitz's theorem,

we get

n(n-1) [yn+2(1-x2)+n.yn+l(-2x)+ yn (-2)] - [XYn+l +n.1 yn] =0

or (1-X2)Yn+2 - (2n+1)XYn+l - n2Yn =0 (iv)

putting x = 0 in (i), (ii), (iii) and (iv) we get

(y)o = sin-1 0 =0; (Yl)O =1; (Y2)0 =0

If n is even, putting n = 2, 4, 6 , we get

(Y4)0 =22 (Y2)0 = 0 : (Y2)0 =0

(Y6)0 = 42(Y4)0 = 0, : (Y4)0 =0

In this way we can show that for all even values of n, (Yn)O =0 Answer

If n is odd, putting n=1,3,5,7 (n-2) in (v) we get

(Y3)0 = (Yl)O = 1; (ys)o = 32 (Y3)0 = 32.1, (Y7)0 = 52.(ys)o = 52.32.1 etc

(yn)o = (n -2)2 (yn -2)0

= (n -2)2 (n -4)2 (Yn -4)0

= (n - 2)(n -4)2 52,32, 12 Answer

Example 13 : If y = [log{x+ J(1 + x2) }P, show that

(Yn+2)0 = - n2 (yn)O, hence find (Yn)O

Solution: Let y = [log x+ J(1 + x2) F (i)

:.y, =2[IOg{X+J(1+X')}]{ 1 )[1+

2 1+x

or y, = [J(l !x,JIOg{X+J(l +x')}]

Succesive Differentiation and Leibnitz's Theorm

or (1+x2)Yn+2 + (2n+1) x)'n+t +n2Yn = 0 (iv)

putting x =0 in (iv) we have

Now from (i) putting x =0, we get

(y)o = [log (0+1)]2 = (log 1)2 =0

: From (ii) putting x =0, we get

= 4(y)0 =0 or (yt)o =0

Hence when n is odd (Yn)O =0 Answer

And if n is even, as before

(yn+2)O = (m2 - n2) (yn)O (iv)

If n is odd, putting n = 1,3, 5, 7 (n-2), we get

When n = 1, (Y3)O = (m2 -12) (Yl)O

And from (i) and (ii), putting x =0, we get

(y)o = 1; = m2

= m2 or (Yl)O = m : (Y3) 0 = (m2-12) m

when n = 3, from (iv) we get (ys)o = (m2 -32)(Y3)O

If n is even, putting n = 2,4,6,8, (n-2) is (iv), we get

when n =2, (Y4)O = (m2 - 22) (Y2)O

And from (iii) putting x =0, we get

(Y2)O +0+m2 (y)o = 0 or (Y2)O = m2 (y)o = m2

:.(Y4)O = (m2 - 22)m2

When n =4, from (iv) we get (Y6)O = (m2 - 42) (Y4)O

or (Y6)O = (m2 - 42) (m2 - 22) m2

Succesive Differentiation and Leibnitz's Theorm

when n =6, from (iv), we get (ys)o = (m2 - 62) (Y6)0

2 If Y = sin (a sino! x) prove that

(1-x2) yn+2 -(2n +1) xyn+! - (n2-a2) yn = 0

3 If Y = a cos (log x) + b sin (log x) show that

X2Yn+2 + (2n +1) xyn+l (n2+1) Yn = 0

4 If Y = sin (m sino! x) prove that

(1-x2) yn+2 - (2n+1) XYn+! + (m2 n2)y n = 0

5 If Y = emcos 'x, show that

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Chapter 2 Partial Differentiation

Partial differential coefficients: The Partial differential coefficient of f(x,y) with respect to x is the ordinary differential coefficient of f(x,y) when y is regarded as a constant It is written as

- and - are also denoted by fx and fy respectively ox oy

The partial differential coefficients of fx and fy are fxx, fxy, fyx, fyy

Solution: The given relation is

u = log(x3 + y3+ x3 - 3xyz)

(V.P.T.V 2004, B.P.S.C 2007)

(V.P.P.C.S.2oo3)

Differentiate it w.r.t x partially, we get

=eXYZ (2xyz+1) + yz exYZ (X2yZ+X)

= eXYZ [2xyz +1 + x2y2z2 + xyz]

= eXYZ (1 +3xyz + X2y2Z2) Hence Proved

Example 3 : If -2 + -2 + -2 = 1, prove that I

a +u b +u c +u + + u; = 2 (xu x +yUy + zUz)

Solution: Given +-a +u b +u c +u + -{ + -:: = 1 (i)

where u is a function of x/y and z,

Differentiating (i) partially with respect to x, we get