# differential-calculus-1-23.pdf

18. Apr 2023
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### differential-calculus-1-23.pdf

• 1. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 1 2011 Engineering Mathematics – I (10 MAT11) LECTURE NOTES (FOR I SEMESTER B E OF VTU) VTU-EDUSAT Programme-15 Dr. V. Lokesha Professor and Head DEPARTMENT OF MATHEMATICS ACHARYA INSTITUTE OF TECNOLOGY Soldevanahalli, Bangalore – 90
• 2. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 2 2011 ENGNEERING MATHEMATICS – I Content CHAPTER UNIT I DIFFERENTIAL CALCULUS – I UNIT II DIFFERENTIAL CALCULUS – II UNIT III DIFFERENTIAL CALCULUS – III
• 3. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 3 2011 UNIT - I DIFFERENTIAL CALCULUS – I Introduction: The mathematical study of change like motion, growth or decay is calculus. The Rate of change of given function is derivative or differential. The concept of derivative is essential in day to day life. Also applicable in Engineering, Science, Economics, Medicine etc. Successive Differentiation: Let y = f (x) --(1) be a real valued function. The first order derivative of y denoted by or y’ or y1 or ∆1 The Second order derivative of y denoted by or y’’ or y2 or ∆2 Similarly differentiating the function (1) n-times, successively, the n th order derivative of y exists denoted by or yn or yn or ∆n The process of finding 2nd and higher order derivatives is known as Successive Differentiation. nth derivative of some standard functions: 1. y = eax Sol : y1 = a eax y2 = a2 eax Differentiating Successively yn = an eax ie. Dn[eax] = an eax For, a =1 Dn[ex] = ex dx dy 2 2 dx y d n n dx y d
• 4. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 4 2011
• 5. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 5 2011
• 6. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 6 2011
• 7. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 7 2011
• 8. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 8 2011 Leibnitz’s Theorem : It provides a useful formula for computing the nth derivative of a product of two functions. Statement : If u and v are any two functions of x with un and vn as their nth derivative. Then the nth derivative of uv is (uv)n = u0vn + n C1 u1vn-1 + n C2u2vn-2 + …+n Cn-1un-1v1+unv0 Note : We can interchange u & v (uv)n = (vu)n, nC1 = n , nC2 = n(n-1) /2! , nC3= n(n-1)(n-2) /3! … 1. Find the nth derivations of eax cos(bx + c) Solution: y1 = eax – b sin (bx +c) + a eax cos (b x + c), by product rule. .i.e, y1 = eax ( ) ( ) [ ] c bx sin b c bx cos a + − + Let us put a = r cos θ , and b = r sin θ . ∴ 2 2 2 r b a = + and tan a / b = θ .ie., 2 2 b a r + = and θ = tan-1 (b/a) Now, [ ] ) c bx sin( sin r ) c bx cos( cos r e y ax 1 + θ − + θ = Ie., y1 = r eax cos ( ) c bx + + θ where we have used the formula cos A cos B – sin A sin B = cos (A + B) Differentiating again and simplifying as before, y2 = r2 eax cos ( ) c bx 2 + + θ . Similarly y3 = r3 e ax cos ( ) c bx 3 + + θ . ……………………………………… Thus ( ) c bx n cos e r y ax n n + + θ = Where 2 2 b a r + = and θ = tan-1 (b/a). Thus Dn [eax cos (b x + c)] = ( ) [ ] [ ] c bx a b n e b a ax n + + + − / tan cos ) ( 1 2 2
• 9. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 9 2011 2. Find the nth derivative of log 3 8 4 2 + + x x Solution : Let y = log 3 x 8 x 4 2 + + = log (4x2 + 8x +3) ½ ie., 2 1 y = log (4x2 + 8x +3) ∵ log xn = n log x 2 1 y = log { (2x + 3) (2x+1)}, by factorization. ∵ 2 1 y = {log (2x + 3) + log (2x + 1)} Now ( ) ( ) ( ) ( ) ( ) ( ) ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + − − + + − − = − − n n 1 n n n 1 n n 1 x 2 2 ! 1 n 1 3 x 2 2 ! 1 n 1 2 1 y Ie., yn = 2n-1 (-1) n-1 (n-1) ! ( ) ( ) ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + + + n n 1 x 2 1 3 x 2 1 3. Find the nth derivative of log 10 {(1-2x)3 (8x+1)5} Solution : Let y = log 10 {1-2x)3 (8x+1)5} It is important to note that we have to convert the logarithm to the base e by the property: 10 log x log x log e e 10 = Thus ( ) ( ) { } 5 3 e e 1 x 8 x 2 1 log 10 log 1 y + − = Ie., ( ) ( ) { } 1 x 8 log 5 x 2 1 log 3 10 log 1 y e + + − = ( ) ( )( ) ( ) ( ) ( ) ( ) ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + − − + − − − − = ∴ − − n n 1 n n n 1 n e n 1 x 8 8 ! 1 n 1 5 x 2 1 2 ! 1 n 1 . 3 10 log 1 y Ie., ( ) ( ) ( ) ( ) ( ) ( ) ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + + − − − − = − n n n n e n 1 n n 1 x 8 4 5 x 2 1 1 3 10 log 2 ! 1 n 1 y
• 10. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 10 2011 4. Find the nth derivative of e2x cos2 x sin x Solution : >> let y = e2x cos2 x sin x = e2x ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + 2 x 2 cos 1 sin x ie., 2 e y x 2 = (sin x + sin x cos 2x) = ( ) [ ] ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − + + x sin x 3 sin 2 1 x sin 2 e x 2 = ( )∵ x sin x 3 sin x sin 2 4 e x 2 − + sin (-x) = -sin x 4 e y x 2 = ∴ (sin x + sin 3x) Now ( ) ( ) { } x 3 sin e D x sin e D 4 1 y x 2 n x 2 n n + = Thus ( ) ( ) [ ] ( ) ( ) [ ] { } x 3 2 3 tan n sin e 13 x 2 1 tan n sin e 5 4 1 y 1 x 2 n 1 x 2 n n + + + = − − ( ) ( ) [ ] ( ) ( ) [ ] { } x 3 2 3 tan n sin 13 x 2 1 tan n sin 5 4 e y 1 n 1 n x 2 n + + + = ∴ − − 5. Find the nth derivative of e2x cos 3x Solution : Let y=e2x cos3 x = e 2x. 4 1 (3 cos x + cos 3x) Ie., y = 4 1 (3 e2x cos x + e2x cos 3x) 4 1 yn = ∴ {3Dn (e2x cos x) + Dn (e2x cos 3x)} ( ) ( ) [ ] ( ) ( ) [ ] { } x n e x n e y x n x n n 3 2 3 tan cos 13 2 1 tan cos 5 3 4 1 1 2 1 2 + + + = − − Thus ( ) ( ) [ ] ( ) ( ) [ ] { } x 3 2 3 tan n cos 13 x 2 1 tan n cos 5 3 4 e y 1 n 1 n x 2 n + + + = − −
• 11. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 11 2011 6. Find the nth derivative of ( )( ) 3 x 2 1 x 2 x2 + + Solution : ( )( ) 3 x 2 1 x 2 x y 2 + + = is an improper fraction because; the degree of the numerator being 2 is equal to the degree of the denominator. Hence we must divide and rewrite the fraction. 3 x 8 x 4 x 4 . 4 1 3 x 8 x 4 x y 2 2 2 2 + + = + + = for convenience. 4x2 +8x +3 1 2 2 3 8 3 8 4 4 − − + + x x x x ∴ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + − − + = 3 x 8 x 4 3 x 8 1 4 1 y 2 Ie., ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + + − = 3 x 8 x 4 3 x 8 4 1 4 1 y 2 The algebraic fraction involved is a proper fraction. Now ⋅ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + + − = 3 x 8 x 4 3 x 8 D 4 1 0 y 2 n n Let ( )( ) 3 x 2 B 1 x 2 A 3 x 2 1 x 2 3 x 8 + + + = + + + Multiplying by (2x + 1) (2x + 3) we have, 8x + 3 = A (2x + 3) + B (2x + 1) ................(1) By setting 2x + 1 = 0, 2x + 3 = 0 we get x = -1/2, x = -3/2. Put x = -1/2 in (1): -1 -1 + A (2) ⇒ A = -1/2 Put x = -3/2 in (1): -9 = B (-2) ⇒ B = 9/2 ∴ ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − − = 3 x 2 1 D 2 9 1 x 2 1 D 2 1 4 1 y n n n
• 12. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 12 2011 ( ) ( ) ( ) ( ) ( ) ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + − ⋅ + + − ⋅ − − = + + 1 n n n 1 n n n 3 x 2 2 ! n 1 9 1 x 2 2 ! n 1 1 8 1 ie., ( ) ( ) ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + + + − = + + + 1 n 1 n n 1 n n 3 x 2 9 ) 1 x 2 ( 1 8 2 ! n 1 y 7. Find the nth derivative of ) 2 ( ) 1 ( 4 + + x x x Solution : y = ) 2 ( ) 1 ( 4 + + x x x is an improper fraction. (deg of nr. = 4 > deg. of dr. = 2) On dividing x4 by x2 + 3 x + 2, We get y = ( x2 – 3x + 7 ) + ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + − − 2 3 14 15 2 x x x ∴ yn = Dn (x2-3x+7)-Dn ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + − 2 3 14 15 2 x x x But D = ( x2 – 3x + 7 ) = 2x – 3, D2 ( x2 – 3x + 7 ) = 2 D3( x2 – 3x + 7 ) = 0......... Dn ( x2 – 3x + 7 ) = 0 if n > 2 Hence yn = -Dn ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + + ) 2 ( ) 1 ( 14 15 x x x Now, let Dn ) 2 ( ) 1 ( 2 3 14 15 2 + + + = + + + x B x A x x x => 15x+ 14 = A(x+2) + B(x+ 1 ) Put x = - 1 ; - 1 = A ( 1 ) or A = - 1 Put x = - 2 ; - 16 = B ( - 1 ) or B = 16 Yn = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − 2 1 16 1 1 x D x D n n
• 13. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 13 2011 1 1 1 1 ( 1) ! 1 ( 1) ! 1 16 ( 1) ( 2) 1 16 ( 1) ! 2 ( 1) ( 2) n n n n n n n n n n n n x x y n n x x + + + + − − = − + + ⎧ ⎫ = − − > ⎨ ⎬ + + ⎩ ⎭ 8. Show that ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − − − − − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + n x x n x x dx d n n n n 1 3 1 2 1 1 log ! ) 1 ( log 1 Solution : Let y = x x x x 1 . log log = and let u = log x, v = x 1 We have Leibnitz theorem, (uv)n = uvn + v u v u n v u n n n C n C + + + − − .... 2 2 1 1 2 1 …… (1) Now, u = log x ∴un = n n x n )! 1 ( ) 1 ( 1 − − − x v 1 = ∴vn = 1 ! ) 1 ( + − n n x n Using these in (1) by taking appropriate values for n we get, Dn = n n n n x n x n x n x x x )! 1 ( ) 1 ( . 1 ! ) 1 ( .. log log 1 1 − − + − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + ( ) x x n x n x n n n n n n 1 . ! 1 ) 1 ( ...... )! 2 ( ) 1 ( 1 2 . 1 ) 1 ( 1 1 2 2 − − + + − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − + − − − Ie.. = log x 1 1 1 ! ) 1 ( ! ) 1 ( + − + − + − n n n n x n x n 1 1 1 2 )! 1 ( ) 1 ( .... 2 ! ) 1 ( + − + − − − + + − − n n n n x n x n ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − + + − − − − − − − − + − 1 1 2 1 1 2 )! 1 ( ) 1 ( .... 2 ) 1 ( ) 1 ( log ! ) 1 ( n n x x n n n Note : (-1)-1 = 1 ) 1 ( 1 ) 1 ( ; 1 1 1 2 2 = − = − − = − −
• 14. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 14 2011 Also n n n n n n 1 )! 1 ( )! 1 ( ! )! 1 ( = − − = − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − − − − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ∴ + n x x n x x dx d n n n n 1 ... 3 1 2 1 1 log ! ) 1 ( log 1 9. If yn= Dn (xn logx) Prove that yn = n yn-1+(n-1)! and hence deduce that yn = n ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + + + + n x 1 .... 3 1 2 1 1 log Solution : yn = Dn(xn log x) = Dn-1 {D (xn log x} = Dn-1 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + − x nx x x n n log 1 . 1 = Dn-1(xn-1) + nDn-1 (xn-1 log x} ∴ yn = (n-1)! +nyn-1. This proves the first part. Now Putting the values for n = 1, 2, 3...we get y1 = 0! + 1 y0 = 1 + log x = 1! (log x + 1 ) y2 = l! + 2y1 = l+2 (l + log x) ie., y2 = 21og x + 3 = 2(log x + 3/2) = 2! ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + 2 1 1 log x y3 = 2! + 3y2 = 2 + 3(2 log x + 3) ie., y3 = 61og x+ll = 6 (log x + ll/6) = 3! ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + + 3 1 2 1 1 log x ………………………………………………………………………….. ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + + + + = n x n yn 1 ... 3 1 2 1 1 log ! 10. If y = a cos (log x) + b sin ( log x), show that x2y2 + xy1 + y = 0. Then apply Leibnitz theorem to differentiate this result n times. or If y = a cos (log x) + b sin (log x ), show that x2yn + 2 + (2n+l)xyn + l+(n2+1)yn = 0. [July-03]
• 15. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 15 2011 Solution : y = a cos (log x) + b sin (log x) Differentiate w.r.t x ∴ y1 = -a sin (log x) x 1 + b cos (log x). x 1 (we avoid quotient rule to find y2) . => xy1 = - a sin (log x) + b cos (log x) Differentiating again w.r.t x we have, xy2 + 1 y1 = - a cos (log x) + b sin ( log x) x 1 or x2y2 + xy1 = - [ a cos (log x) + b sin (log x) ] = -y ∴ x2y2+xy1+y = 0 Now we have to differentiate this result n times. ie., Dn (x2y2) + Dn (xy1) + Dn (y) = 0 We have to employ Leibnitz theoreom for the first two terms. Hence we have, ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − + + − − ) ( . 2 . 2 . 1 ) 1 ( ) ( . 2 . ) ( . ) 2 2 2 1 2 2 y D n n y D x n y D x n n n { } 0 ) ( . 1 . ) ( . 1 1 1 = + + − n n n y y D n y D x ie., {x2yn + 2 + 2n x yn + 1 + n (n – 1)yn} + {xyn+1+nyn}+yn = 0 ie., x2yn + 2 + 2n x yn + 1 + n2yn - nyn + xyn+1+nyn+yn = 0 ie., x2yn + 2 + (2n+l)xy n+l + (n2+l)yn = 0 11. If cos-1 (y/b ) = log (x/n)n, then show that x2yn + 2 + (2n+l) xyn+l + 2n2yn = 0 Solution :By data, cos-1 (y/b) = n log (x/n) ∴log(am) = m log a => b y = cos [n log (x/n )] or y = b . cos [ n log (x/n)] Differentiating w.r.t x we get,
• 16. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 16 2011 y1 = -b sin [n log (x/n)] ( ) n n x n 1 / 1 ⋅ ⋅ ⋅ or xy1 = - n b sin [n log (x/n )] Differentiating w.r.t x again we get, xy2 + 1. 1 y = - n . b cos [ n log (x/n )] n n n x 1 . ) / ( 1 or x (xy2+y1) = n2b cos [n log (x/n) ] =-n2y, by using (1). or x2y2 +xyl + n2y = 0 Differentiating each term n times we have, D(x2y2) + Dn(xy1) + n2Dn (y) = 0 Applying Leibnitz theorem to the product terms we have, { } 0 . 1 . . 2 . 2 . 1 ) 1 ( . 2 . 2 1 1 2 2 = + + + ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − + + + + + n n n n n n y n y n xy y n n y x n y x ie x2yn+2 + 2 x yn+1 + n2yn + xy n+1+ nyn + n2yn=0 or x2 yn+2 + (2n + l) xyn+1 + 2n2yn = 0 12. If y = sin( log (x2 + 2 x + 1)), or [Feb-03] If sin-1 y = 2 log (x + 1), show that (x+l)2yn + 2 + (2n+1)(x+1)yn+l + (n2 + 4)yn = 0 Solution : By data y = sin log (x2 + 2 x + 1 ) ∴ y1 = cos log (x2 + 2 x + 1) 2 2 ) 1 ( 1 2 + + x x ie., y1 = cos log (x2 + 2 x + 1) 1 2 1 2 + + x x 2 (x + 1) ie., y1 = ) 1 ( ) 1 x 2 x log( cos 2 2 + + + x or (x + 1) y1 = 2 cos log (x2 + 2 x + 1 ) Differentiating w.r.t x again we get
• 17. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 17 2011 (x+1)y2 + 1 y1 = -2 sin log (x2 + 2x + 1) ) 1 ( 2 . ) 1 ( 1 2 + + x x or (x + 1)2y2 + (x+1) y1 = -4y or (x+l )2y2 + (x+l) y1 + 4y = 0 , Differentiating each term n times we have, Dn [(x + 1)2y2] +Dn [(x+ 1)y1] + Dn [y] = 0 Applying Leibnitz theorem to the product terms we have, ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − + + + + + + n n n y n n y x n y x . 2 . 2 . 1 ) 1 ( ). 1 ( 2 . ) 1 ( 1 2 2 + {(x+l) yn + 1+n. 1 .yn} + 4yn = 0 ie., (x+l)2yn + 2 + 2n (x+1)yn+1 + n2yn-nyn + (x+l)yn+l + nyn + 4yn = 0 ie., (x+l)2yn + 2 + (2n + l) (x + l) y n+ 1 + (n2 + 4)yn = 0 13. If = log ( ) 2 1 x x + + prove that (1 + x2) yn+2 + (2n + 1) xyn+1 + n2yn = 0 >> By data, y = log ( ) 2 1 x x + + ∴y1 = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ + + + + x x x x 2 . 1 2 1 1 ) 1 ( 1 2 2 Ie., 2 2 2 2 1 1 1 1 1 ) 1 ( 1 x x x x x x y + = + + + + + or 1 1 1 2 = + y x Differentiating w.r.t.x again we get 0 . 2 . ) 1 2 1 1 1 2 2 2 = + + + y x x y x or (1+x2)y2 + xy1 = 0 Now Dn [(l+x2)y2] + Dn[xy1] = 0 Applying Leibnitz theorem to each term we get,
• 18. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 18 2011 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − + + + + + n n n y n n y x n y x . 2 . 2 . 1 ) 1 ( . 2 . ) 1 ( 1 2 2 + [x . yn + 1+n .1 yn] = 0 Ie., (1 + x2) yn +2 + 2 n x yn + 1 + n2yn – nyn + xyn+l+ nyn = 0 or (l+x2)yn + 2 + (2n + l)xyn+1+n2yn = 0 14. If x = sin t and y = cos mt, prove that (l-x2)yn + 2-(2n+1)xyn+l + (m2-n2)yn = 0. [Feb-04] Solution : By data x = sin t and y = cos mt x = sin t => t = sin-1 x and y = cos mt becomes y = cos [ m sin-1x) Differentiating w.r.t.x we get y1 = - sin (m sin-1x) 2 1 x m − or 1 2 1 y x + = - m sin (m sin-1x) Differentiating again w.r.f .x we get, 2 1 1 2 2 2 1 ). sin ( cos ) 2 ( 1 2 1 1 x m x m m y x x y x − − = − − + − − or (1 -x2)y2-xyl = -m2y or (1 -x2)y2 –xy1 +m2y = 0 Thus (1-x2)yn+2-(2n+1)xyn+1+(m2-n2)yn=0 15. If x = tan ( log y), find the value of (l+x2)yn+1 + (2nx-l) yn+n(n-1)yn-1 [July-04] Solution : By data x = tan(log y) => tan-1 x = log y or y = etan-1 x Since the desired relation involves yn+1, yn and yn-1 we can find y1 and differentiate n times the result associated with y1 and y. Consider y = ⋅ − x e 1 tan ∴ y. = ⋅ − x e 1 tan 2 1 1 x + or (1 +x2)y1 = y Differentiating n times we have
• 19. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 19 2011 Dn[(l+x2)y1]=Dn[y] Anplying Leibnitz theorem onto L.H.S, we have, {(l+x2)Dn(y1) + n .2x .Dn-1 (y1) n n y y D n n = − + − )} ( . 2 . 2 . 1 ) 1 ( 1 2 Ie., (1+x2)yn+1+2n x yn + n (n-1) yn-1-yn=0 Or (l+x2)yn + 1 + (2nx-l)yn + n(n-l)yn-1 = 0
• 20. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 20 2011 Continuity & Differentiability Some Fundamental Definitions A function f (x) is defined in the interval I, then it is said to be continuous at a point x = a if A function f (x) is said to be differentiable at x = a if Ex : Consider a function f (x) is defined in the interval [-1,1] by f (x) = It is continuous at x = 0 But not differentiable at x = 0 Note : If a function f (x) is differentiable then it is continuous, but converse need not be true. Geometrically : (1) If f (x) is Continuous at x =a means, f (x) has no breaks or jumps at the point x = a Ex : Is discontinuous at x=0 (2) If f (x) is differentiable at x = a means, the graph of f (x) has a unique tangent at the point or graph is smooth at x = a 1. Give the definitions of Continuity & Differentiability: Solution: A function f (x) is said to be continuous at x = a, if corresponding to an arbitrary positive number ε, however small, their exists another positive number δ such that. ⏐f (x) – f (a)⏐ < ε, where ⏐x - a⏐ < δ It is clear from the above definition that a function f (x) is continuous at a point ‘a’. If (i) it exists at x = a (ii) f (x) = f (a) i.e, limiting value of the function at x = a is to the value of the function at x = a a x Lt → ⎩ ⎨ ⎧ ≤ ≤ ≤ ≤ − − = 1 0 0 1 x x x x x 0 ( ) ( ) lim '( ) I h f x h f a f a exists a h → + − = ∈ ) ( ) ( lim a f x f a x = → 1 1 0 ( ) 0 1 x f x x x − − ≤ ≤ ⎧ = ⎨ < ≤ ⎩
• 21. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 21 2011 Differentiability: A function f (x) is said to be differentiable in the interval (a, b), if it is differentiable at every point in the interval. In Case [a,b] the function should posses derivatives at every point and at the end points a & b i.e., Rf1 (a) and Lf1 (a) exists. 2. State Rolle’s Theorem with Geometric Interpretation. Statement: Let f (x) be a function is defined on [a,b] & it satisfies the following Conditions. (i) f (x) is continuous in [a,b] (ii) f (x) is differentiable in (a,b) (iii) f (a) = f (b) Then there exists at least a point C∈ (a,b), Here a < b such that f1 ( c ) = 0 Proof: Geometrical Interpretation of Rolle’s Theorem: Y y = f (x) P R A B A B f(a) Q S →f(a) → f(b) O x = a x = c x = b x =a c1 c2 c3 c4 x = b Let us consider the graph of the function y = f (x) in xy – plane. A (a,.f(a)) and B (b, f( b ) ) be the two points in the curve f (x) and a, b are the corresponding end points of A & B respectively. Now, explained the conditions of Rolle’s theorem as follows. (i) f (x) is continuous function in [a,b], Because from figure without breaks or jumps in between A & B on y = f (x). (ii) f (x) is a differentiable in (a,b), that means let us joining the points A & B, we get a line AB. ∴ Slope of the line AB = 0 then ∃ a point C at P and also the tangent at P (or Q or R or S) is Parallel to x –axis. ∴ Slope of the tangent at P (or Q or R or S) to be Zero even the curve y = f (x) decreases or increases, i.e., f (x) is Constant.
• 22. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 22 2011 f1 (x) = 0 ∴f1 (c) = 0 (iii) The Slope of the line AB is equal to Zero, i.e., the line AB is parallel to x – axis. ∴ f (a) = f (b) 3. Verify Rolle’s Theorem for the function f (x) = x2 – 4x + 8 in the internal [1,3] Solution: We know that every Poly nominal is continuous and differentiable for all points and hence f (x) is continuous and differentiable in the internal [1,3]. Also f (1) = 1 – 4 + 8 = 5, f (3) = 32 – 43 + 8 = 5 Hence f (1) = f (3) Thus f (x) satisfies all the conditions of the Rolle’s Theorem. Now f1 (x) = 2x – 4 and f1 (x) = 0 ⇒ 2x – 4 = 0 or x = 2. Clearly 1 < 2 < 3. Hence there exists 2t (1,3) such that f1 (2) = 0. This shows that Rolle’s Theorem holds good for the given function f (x) in the given interval. 4. Verify Rolle’s Theorem for the function f (x) = x (x + 3) in the interval [-3, 0] Solution: Differentiating the given function W.r.t ‘x’ we get = ∴ f1(x) exists (i.e finite) for all x and hence continuous for all x. Also f (-3) = 0, f (0) = 0 so that f (-3) = f (0) so that f (-3) = f (0). Thus f (x) satisfies all the conditions of the Rolle’s Theorem. Now, f1 (x) = 0 ⇒ = 0 Solving this equation we get x = 3 or x = -2 Clearly –3 < -2 < 0. Hence there exists –2∈ (-3,0) such that f1 (-2) = 0 This proves that Rolle’s Theorem is true for the given function. 2 x e − 2 2 2 1 ) 3 2 ( 2 1 ) 3 ( ) ( x x e x e x x x f − − + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + = 2 2 ) 6 ( 2 1 x e x x − − − − 2 2 ) 6 ( 2 1 x e x x − − − −
• 23. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 23 2011 5. Verify the Rolle’s Theorem for the function Sin x in [-π, π] Solution: Let f (x) = Sin x Clearly Sinx is continuous for all x. Also f1 (x) = Cos x exists for all x in (-π, π) and f (-π) = Sin (-π) = 0; f (π) = Sin (π) = 0 so that f (-π) = f (π) Thus f (x) satisfies all the conditions of the Rolle’s Theorem . Now f1 (x) = 0 ⇒ Cos x = 0 so that X = ± Both these values lie in (-π,π). These exists C = ± Such that f1 ( c ) = 0 Hence Rolle’s theorem is vertified. 6. Discuss the applicability of Rolle’s Theorem for the function f (x) = ІxІ in [-1,1]. Solution: Now f (x) = ⏐x⏐= x for 0 ≤ x ≤ 1 -x for –1 ≤ x ≤ 0 f (x) being a linear function is continuous for all x in [-1, 1]. f(x) is differentiable for all x in (- 1,1) except at x = 0. Therefore Rolle’s Theorem does not hold good for the function f (x) in [-1,1]. Graph of this function is shown in figure. From which we observe that we cannot draw a tangent to the curve at any point in (-1,1) parallel to the x – axis. Y y = ⏐x⏐ x -1 0 1 2 π 2 π
• 24. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 24 2011 Exercise: 7. Verify Rolle’s Theorem for the following functions in the given intervals. a) x2 – 6x + 8 in [2,4] b) (x – a)3 (x – b)3 in [a,b] c) log in [a,b] 8. Find whether Rolle’s Theorem is applicable to the following functions. Justify your answer. a) f (x) = ⏐x – 1 ⏐ in [0,2] b) f (x) = tan x in [0, π] . 9. State & prove Lagrange’s (1st ) Mean Value Theorem with Geometric meaning. Statement: Let f (x) be a function of x such that (i) If is continuous in [a,b] (ii) If is differentiable in (a,b) Then there exists atleast a point (or value) C∈ (a,b) such that. i.e., f (b) = f (a) + (b – a) f1 (c) Proof: ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + + x b a ab x ) ( 2 a b a f b f c f − − = ) ( ) ( ) ( 1 [b,f(b)] [a,f(a)] x y a c b
• 25. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 25 2011 Define a function g (x) so that g (x) = f (x) – Ax ---------- (1) Where A is a Constant to be determined. So that g (a) = g (b) Now, g (a) = f (a) – Aa G (b) = f (b) – Ab ∴ g (a) = g (b) ⇒ f (a) – Aa = f (b) – Ab. i.e., ---------------- (2) Now, g (x) is continuous in [a,b] as rhs of (1) is continuous in [a,b] G(x) is differentiable in (a,b) as r.h.s of (1) is differentiable in (a,b). Further g (a) = g (b), because of the choice oif A. Thus g (x) satisfies the conditions of the Rolle’s Theorem. ∴ These exists a value x = c sothat a < c < b at which g1 ( c ) = 0 ∴Differentiate (1) W.r.t ‘x’ we get g1 (x) = f 1 (x) – A ∴ g1 ( c ) = f1 ( c )- A (∵ x =c) ⇒ f 1 ( c ) - A = 0 (∵g1 ( c ) = 0) ∴f1 ( c ) = A -------------- (3) From (2) and (3) we get (or) f (b) = f (a) + (b – a) f1 (c) For a < c < b Corollary: Put b – a = h i.e., b = a + h and c = a + θ h Where 0 < θ < 1 Substituting in f (b) = f (a) + (b – a) f1 ( c ) ∴ f (a + h) = f (a) + h f (a + θ h), where 0 < θ < 1. a b a f b f A − − = ) ( ) ( a b a f b f c f − − = ) ( ) ( ) ( 1
• 26. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 26 2011 Geometrical Interpretation:- Since y = f (x) is continuous in [a,b], it has a graph as shown in the figure below, At x = a, y = f (a) At x = b, y = f (b) Y B Y B P α A C A Q x X 0 a c b 0 Figure (ii) Figure (i) Slope of the line joining the points A (a,f(a)) and B ( (b,f (b)) Is (∴ Slope = m = tan θ) = tan α Where α is the angle mode by the line AB with x – axis = Slope of the tangent at x = c = f1 ( c ), where a < c < b Geometrically, it means that there exists at least are value of x = c, where a < c < b at which the tangent will be parallel to the line joining the end points at x = a & x = b. Note: These can be more than are value at which the tangents are parallel to the line joining points A & B (from Fig (ii)). a b b f b f − − ) ( ) (
• 27. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 27 2011 10. Verify Lagrange’s Mean value theorem for f(x) = (x – 1) (x – 2) (x – 3) in [0,4]. Solution: Clearly given function is continuous in [0,4] and differentiable in (0,4), because f (x) is in polynomial. f (x) = (x – 1) (x – 2) (x – 3) f (x) = x3 – 6x2 + 11x – 6 and f (0) = 03 – 6(0)2 + 11 (0) – 6 = -6 f(4) = 43 – 6 (4)2 + 11 (4) – 6 = 6 Differentiate f (x) W.r.t x, we get F1 (x) = 3x2 – 6x + 11 Let x = c, f1 ( c) = 3c2 – 6c + 11 By Lagrange’s Mean value theorem, we have ∴3c2 – 6c + 11 = 3 ⇒ 3c2 – 6c + 8 = 0 Solving this equation, we get C = 2 ± ∈ (0,4) Hence the function is verified. 11. Verify the Lagrange’s Mean value theorem for f (x) = logx in [1,e]. Solution: Now Logx is continuous for all x > 0 and hence [1,e]. Also which exists for all x in (1,e) Hence f (x) is differentiable in (1,e) ∴by Lagrange’s Mean Value theorem, we get ) 0 4 ( ) 0 ( ) 4 ( ) ( ) ( ) ( 1 − − = − − = f f a b a f b f c f 3 4 ) 6 ( 6 = − − = 3 2 x x f 1 ) ( 1 =
• 28. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 28 2011 ⇒ C = e – 1 ⇒ 1 < e - < 2 < e Since e∈ (2,3) ∴ So that c = e – 1 lies between 1 & e Hence the Theorem. 12. Find θ for f (x) = Lx2 + mx + n by Lagrange’s Mean Value theorem. Solution: f (x) = Lx2 + mx + n ∴f1 (x) = 2 Lx + m We have f (a + h) = f (a) + hf1 (a + θ h) Or f (a + h) – f (a) = hf1 (a + θ h) i.e., { (a + h)2 + m (a + h) + n} – { a2 + ma + n} = h {2 (a + θ h) + m} Comparing the Co-efficient of h2, we get 1 = 2θ ∴θ = ∈ (0,1) Exercise: 13. Verify the Lagrange’s Mean Value theorem for f (x) = Sin2x in 14. Prove that, < tan-1 b – tan-1 a < if 0 < a < b and reduce that 15. Show that in 16. Prove that Where a < b. Hence reduce . c e c e Log Loge 1 1 1 1 1 1 = − ⇒ = − − 2 1 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 , 0 π 2 1 b a b + − 2 1 a a b + − 6 1 4 3 4 tan 25 3 4 1 + < < + − π π 1 2 < < x Sinx π ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 2 , 0 π 2 1 1 2 1 1 b a b a Sin b Sin a a b − − < − < − − − − 15 1 6 4 1 3 2 1 6 1 − < < − − π π Sin
• 29. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 29 2011 17. State & prove Cauchy’s Mean Value Theorem with Geometric meaning. Proof: The ratio of the increments of two functions called Cauchy’s Theorem. Statement: Let g (x) and f (x) be two functions of x such that, (i) Both f (x) and g (x) are continuous in [a,b] (ii) Both f (x) and g (x) are differentiable in (a,b) (iii) g1 (x) ≠ 0 for any x ε (a,b) These three exists at least are value x = c ε (a,b) at which Proof: Define a function, φ (x) = f (x) – A. g (x) ------------------ (1) So that φ (a) = φ (b) and A is a Constant to be determined. Now, φ (a) = f (a) – A g (a) φ (b) = f (b) – A. g (b) ∴ f (a) – A g (a) = f (b) – A. g (b) -------------------- (2) Now, φ is continuous in [a,b] as r.h.s of (1) is continuous in [a,b] and φ (x) is differentiable in (a,b) as r.h.s of (1) is differentiable in [a,b]. Also φ (a) = φ (b) Hence all the conditions of Rolle’s Theorem are satisfied then there exists a value x = c ∈ (a,b) such that φ1 ( c ) = 0. Now, Differentiating (1) W.r.t x, we get φ1 (x) = f1 (x) – A.g1 (x) at x = c ∈ (a,b) ) ( ) ( ) ( ) ( ) ( ) ( 1 1 a g b g a f b f c g c f − − = ) ( ) ( ) ( ) ( a g b g b f b f A − − = ⇒
• 30. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 30 2011 ∴φ1 ( c ) = f1 ( c ) – A g1 (c) 0 = f1 ( c) – A g1 ( c ) (∵ g1 (x) ≠ 0) --------------- (3) Substituting (3) in (2), we get , where a < c < b Hence the proof. 18. Verify Cauchy’s Mean Value Theorem for the function f (x) = x2 + 3, g (x) = x3 + 1 in [1,3] Solution: Here f (x) = x2 + 3, g (x) = x3 + 1 Both f (x) and g (x) are Polynomial in x. Hence they are continuous and differentiable for all x and in particular in [1,3] Now, f1 (x) = 2x, g1 (x) = 3x2 Also g 1 (x) ≠ 0 for all x ∈ (1,3) Hence f (x) and g (x) satisfy all the conditions of the cauchy’s mean value theorem. Therefore , for some c : 1 < c < 3 i.e., i.e., Clearly C = lies between 1 and 3. Hence Cauchy’s theorem holds good for the given function. ) ( ) ( 1 1 c g c f A = ⇒ ) ( ) ( ) ( ) ( ) ( ) ( 1 1 a g b g a f b f c g c f − − = ) ( ) ( ) 1 ( ) 3 ( ) 1 ( ) 3 ( 1 1 c g c f g g f f = − − 2 3 26 2 28 4 12 c = − − 6 1 2 6 13 3 1 13 2 = = ⇒ = C c 6 1 2
• 31. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 31 2011 19. Verify Cauchy’s Mean Value Theorem for the functions f (x) Sin x, g (x) = Cos x in Solution: Here f (x) = Sin x, g (x) = Cos x so that f1 (x) = Cos x ,g1 (x) = - Sinx Clearly both f (x) and g (x) are continuous in , and differentiable in Also g1 (x) = -Sin x ≠ 0 for all x ∈ ∴ From cauchy’s mean value theorem we obtain for some C : 0 < C < i.e., i.e., -1 = - Cot c (or) Cot c = 1 ∴ C = , clearly C = lies between 0 and Thus Cauchy’s Theorem is verified. Exercises: 20. Find C by Cauchy’s Mean Value Theorem for a) f (x) = ex, g (x) = e-x in [0,1] b) f (x) = x2, g (x) = x in [2,3] 21. Verify Cauchy’s Mean Value theorem for a) f (x) = tan-1 x, g (x) = x in b) f(x) = log x, g (x) = in [1,e] ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 , 0 π ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 2 , 0 π ( ) 2 , 0 π ( ) 2 , 0 π ) ( ) ( ) 0 ( 2 ) 0 ( 2 1 1 c g c f g g f f = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π π 2 π Sinc Cosc − = − − 1 0 0 1 4 π 4 π 2 π ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 1 , 3 1 x 1
• 32. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 32 2011 Generalized Mean Value Theorem: 22. State Taylor’s Theorem and hence obtain Maclaurin’s expansion (series) Statement: If f (x) and its first (n – 1) derivatives are continuous in [a,b] and its nth derivative exists in (a,b) then f(b) = f(a) + (b – a) f1 (a) + f11 (a) + ---------+ fn-1(a) + Where a < c < b Remainder in Taylor’s Theorem: We have f (x) = f (a) + (x – a) f 1 (a) + f 11 (a) + --------- + f n-1 (a) + f n [a + (x – a) θ} f (x) = S n (x) + R n (x) Where R n (x) = f n [a + (x – a) θ] is called the Largrange’s form of the Remainder. Where a = 0, R n (x) = f n (θx), 0 < θ < 1 Taylor’s and Maclaurin’s Series: We have f (x) = Sn (x) + Rn (x) ∴ If Thus converges and its sum is f (x). This implies that f (x) can be expressed as an infinite series. i.e., f (x) = f (a) + (x – a) f 1 (a) + f 11 (a) + ---------- to ∞ This is called Taylor’s Series. ! 2 ) ( 2 a b − )! 1 ( ) ( 1 − − − n a b n ) ( ! ) ( c f n a b n n − ! 2 ) ( 2 a x − )! 1 ( ) ( 1 − − − n a x n n a x n ∠ − ) ( ! ) ( n a x n − ! n xn ) ( )] ( ) ( [ x R Lim x S x f Lim n n n n ∞ → ∞ → = − ) ( ) ( 0 ) ( x S Lim x thenf x R Lim n n n n ∞ → ∞ → = ) (x S Lim n n ∞ → ! 2 ) ( 2 a x −
• 33. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 33 2011 Putting a = 0, in the above series, we get F (x) = f (0) + x f 1 (0) + f 11 (0) + --------- to ∞ This is called Maclaurin’s Series. This can also denoted as Y = y (0) + x y 1 (0) + y 2 (0) + -------- + y n (0) ---------- to ∞ Where y = f (x), y1 = f 1 (x), -------------- y n = f n (x) 23. By using Taylor’s Theorem expand the function e x in ascending powers of (x – 1) Solution: The Taylor’s Theorem for the function f (x) is ascending powers of (x – a) is f (x) = f (a) + (x – a) f 1 (a) + f 11 (a) + ------------ (1) Here f (x) = e x and a = 1 f 1 (x) = e x ⇒ f 1 (a) = e f 11 (x) = e x ⇒ f 11 (a) = e ∴ (1) becomes e x = e + (x – 1) e + e + ------------- = e { 1 + (x – 1) + + --------} 24. By using Taylor’s Theorem expand log sinx in ascending powers of (x – 3) Solution: f (x) = Log Sin x, a = 3 and f (3) = log sin3 Now f 1 (x) = = Cotx, f 1 (3) = Cot3 f 11 (x) = - Cosec 2x, f 11 (3) = - Cosec 23 f 111 (x) = - 2Cosecx (-Cosecx Cotx) = 2Cosec 3x Cotx ∴ f 111 (3) = 2Cosec 33 Cot3 ∴ f (x) = f (a) + (x – a) f 1 (a) + f 11 (a) + f 111 (a) + ------------ ∴Log Sinx = f (3) + (x – 3) f 1(3) + f 11 (3) + f 111 (3) + ----------- = logsin3 + (x – 3) Cot3 + (-Cosec23) + 2 Cosec 33Cot3 + --- ! 2 ) ( 2 x ! 2 ) ( 2 x ! ) ( n x n 2 ) ( 2 ∠ − a x 2 ) 1 ( 2 − x 2 ) 1 ( 2 − x Sinx Cosx ! 2 ) ( 2 a x − ! 3 ) ( 3 a x − ! 2 ) 3 ( 2 − x ! 3 ) 3 ( 3 − x ! 2 ) 3 ( 2 − x ! 3 ) 3 ( 3 − x
• 34. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 34 2011 Exercise: 25. Expand Sinx is ascending powers of 26. Express tan –1 x in powers of (x – 1) up to the term containing (x – 1) 3 27. Apply Taylor’s Theorem to prove e x + h = e x Problems on Maclaurin’s Expansion: 28. Expand the log (1 + x) as a power series by using Maclaurin’s theorem. Solution: Here f (x) = log (1 + x), Hence f (0) = log 1 = 0 We know that = , n = 1,2,--------- Hence f n (0) = (-1) n-1 (n – 1) ! f 1 (0) = 1, f 11 (0) = -1, f 111 (0) = 3!, f 1v (0) = - 3! Substituting these values in f (x) = f (0) + x f 1 (0) + f 11 (0) + -------- + f n (0) + ------------ ∴ log (1+x) = 0 + x . 1 + (-1) + 2! + - 3! + --------------- = x - + - + --------- This series is called Logarithmic Series. ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 2 π x ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − − − − − − − − + + + + ! 3 ! 2 1 3 2 h h h { } ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + = + = − − x dx d x dx d x f n n n n n 1 1 ) 1 log( ) ( 1 1 n n n n ) 1 ( )! 1 ( ) 1 ( 1 + − − − ! 2 2 x ! n xn ! 2 2 x ! 3 3 x ! 4 4 x 2 2 x 3 3 x 4 4 x
• 35. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 35 2011 29. Expand tan –1 x by using Macluarin’s Theorem up to the term containing x 5 Solution: let y = tan –1 x, Hence y (0) = 0 We find that y 1 = which gives y 1 (0) = 1 Further y 1 (1 + x 2) = 1, Differentiating we get Y 1 . 2x + (1 + x 2) y 2 = 0 (or) (1 + x2) y 2 + 2xy 1 = 0 Hence y 2 (0) = 0 Taking n th derivative an both sides by using Leibniz’s Theorem, we get (1 + x 2) y n + 2 + n . 2xy n +1 + . 2. y n + 2xy n – 1 + n.2.y n = 0 i.e., (1 + x 2) y n + 2 + 2 (n +1) x y n + 1 + n (n + 1) y n = 0 Substituting x = 0, we get, y n + 2 (0) = -n (n + 1) y n (0) For n = 1, we get y 3 (0) = - 2y 1 (0) = - 2 For n = 2, we get y 4 (0) = - 2 .3.y 2 (0) = 0 For n = 3, we get y 5 (0) = - 3.4.y 3 (0) = 24 Using the formula Y = y (0) + x y 1 (0) + y 2 (0) + y 3 (0) + --------- We get tan –1 x = x - + - ------------- Exercise: 30. Using Maclaurin’s Theorem prove the following: a) Secx = 1 + + + -------- b) Sin –1 x = x + + + -------------- c) e x Cos x = 1 + x - + --------------- d) Expand e ax Cos bx by Maclaurin’s Theorem as far as the term containing x 3 2 1 1 x + 2 . 1 ) 1 ( − n n ! 2 2 x ! 3 3 x 3 3 x 5 5 x ! 2 2 x ! 4 5 4 x 6 3 x 40 3 5 x 3 3 x
• 36. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 36 2011 Exercise : Verify Rolle’s Theorem for (i) in , (ii) in (iii) in . Exercise : Verify the Lagrange’s Mean Value Theorem for (i) in (ii) in Exercise : Verify the Cauchy’s Mean Value Theorem for (i) and in (ii) and in (iii) and in ) cos (sin ) ( x x e x f x − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 4 5 , 4 π π 2 / ) 2 ( ) ( x e x x x f − = [ ] 2 , 0 2 sin 2 ( ) x x f x e = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ 4 5 , 4 π π ( ) ( 1)( 2) f x x x x = − − 1 0, 2 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ 1 ( ) f x Tan x − = [ ] 0,1 ( ) f x x = 1 ( ) g x x = 1 ,1 4 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ 2 1 ( ) f x x = 1 ( ) g x x = [ ] , a b ( ) f x Sin x = ( ) g x Cos x = [ ] , a b
• 37. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 37 2011 UNIT – II DIFFERENTIAL CALCULUS-II Give different types of Indeterminate Forms. If f (x) and g (x) be two functions such that and both exists, then If = 0 and = 0 then Which do not have any definite value, such an expression is called indeterminate form. The other indeterminate forms are , 00, ∞0 and 1 ∞ 1. State & prove L’ Hospital’s Theorem (rule) for Indeterminate Forms. L’Hospital rule is applicable when the given expression is of the form or Statement: Let f (x) and g (x) be two functions such that (1) = 0 and = 0 (2) f1 (a) and g1 (a) exist and g1 (a) ≠ 0 Then Proof: Now , which takes the indeterminate form . Hence applying the L’ Hospitals theorem, we get ) (x f Lim a x→ ) (x g Lim a x→ ) ( ) ( ) ( ) ( x g Lim x f Lim x g x f Lim a x a x a x → → → = ) (x f Lim a x→ ) (x g Lim a x→ 0 0 ) ( ) ( = → x g x f Lim a x ∞ − ∞ ∞ × ∞ ∞ , 0 , 0 0 ∞ ∞ ) (x f Lim a x→ ) (x g Lim a x→ ) ( ) ( ) ( ) ( 1 1 x g Lim x f Lim x g x f Lim a x a x a x → → → = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = → → ) ( 1 1 ) ( ) ( x f x g Lim x g x f Lim a x a x 0 0
• 38. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 38 2011 If then 1 i.e If = 0 or ∞ the above theorem still holds good. 2. Evaluate form Solution: Apply L’Hospital rule, we get = 1 3. Evaluate Solution: = form Apply L’ Hospital rule = 2 1 1 2 1 2 1 ) ( ) ( ) ( ) ( )] ( [ ) ( )] ( [ ) ( ) ( ) ( ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = − − = → → → x g x f x f x g Lim x f x f x g x g Lim x g x f Lim a x a x a x 2 1 1 ) ( ) ( ) ( ) ( ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = → → x g x f Lim x f x g Lim a x a x ∞ ≠ ≠ → and x g x f Lim a x 0 ) ( ) ( ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = → → ) ( ) ( ) ( ) ( 1 1 x g x f Lim x f x g Lim a x a x ) ( ) ( ) ( ) ( 1 1 x g x f Lim x g x f Lim a x a x → → = ) ( ) ( x g x f Lim a x→ 0 0 = → x Sinx Lim a x 1 1 1 1 = = → θ Cos Cosx Lim a x 1 = ∴ → x Sinx Lim a x Cotx Sinx Lim a x log → Cotx Sinx Lim a x log → ∞ ∞ − = ∞ = 0 log 0 0 log Cot Sin xCotx xCo Co x Cosce Lim a x sec sec 2 2 − − →
• 39. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 39 2011 Exercise: 1 Evaluate a) b) c) d ) 4. Explain ∞ - ∞ and 0 × ∞ Forms: Solution: Suppose = 0 and in this case - g(x) = 0 × ∞, reduce this to or form Let form Or form L’ Hospitals rule can be applied in either case to get the limit. Suppose = ∞ and in this case form, reduce this or form and then apply L’Hospitals rule to get the limit 0 2 1 = = → Cotx Lim a x 0 log = ∴ → Cotx Sinx Lim a x x x Lim x tan 0 → x x x Lim 1 0 ) 1 ( + → x a Lim x x 1 − ∞ → a x a x Lim n n x − − →0 ) (x f Lim a x→ ∞ = → ) (x g Lim a x ) (x f Lim a x→ 0 0 ∞ ∞ [ ] 0 0 ) ( 1 ) ( ) ( ). ( = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ = → → x g x f Lim x g x f Lim a x a x [ ] ∞ ∞ = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ = → → ) ( 1 ) ( ) ( ). ( x f x g Lim x g x f Lim a x a x ) (x f Lim a x→ ∞ = → ) (x g Lim a x [ ] ∞ − ∞ = → ) ( ). ( x g x f Lim a x 0 0 ∞ ∞
• 40. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 40 2011 5. Evaluate Solution: Given = ∞ - ∞ form ∴ Required limit = form Apply L’Hospital rule. = 6. Evaluate Solution: Given limit is ∞ - ∞ form at x = 0. Hence we have Required limit = = form Apply L’ Hospital’s rule = = form Apply L’ Hospitals rule ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + − → 2 0 ) 1 log( 1 x x x Lim x ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + − → 2 0 ) 1 log( 1 x x x Lim x 0 0 1 log( 2 0 = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + − → x x x Lim x x x Lim x 2 1 1 1 0 + − → 2 1 ) 1 ( 2 1 2 1 0 0 = + = + = → → x Lim x x x Lim x x ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − → Cotx x Lim x 1 0 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − → Sinx Cosx x Lim x 1 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − → 0 0 0 xSinx xCosx Sinx Lim x Sinx xCosx xSinx Cosx Cosx Lim x + + − →0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = + → 0 0 0 Sinx xCosx xSinx Lim x
• 41. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 41 2011 = = 7. Evaluate tan x log x Solution: Given limit is (0 × - ∞) form at x = 0 ∴ Required limit = form Apply L’ Hospitals rule = = form Apply L Hospitals rule = 8. Evaluate Solution: Given limit is (∞ × 0) form at x = 1 ∴ Required limit = form Apply L’ Hospitals rule Cosx xSinx Cosx Sinx xCosx Lim x + − + →0 0 2 0 0 2 0 = = − 0 → x Lim ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ∞ ∞ − = → Cotx x Lim x log 0 x Co x Lim x 2 0 sec 1 − → ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − → 0 0 2 0 x x Sin Lim x 0 1 2 0 = − → SinxCosx Lim x x x Sec Lim x log . 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ → π 0 0 2 log 1 x Cos x Lim x π → π π π 2 2 . 2 1 1 − = − = → x Sin x Lim x
• 42. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 42 2011 Exercise: 2 Evaluate a) b) c) d) 9. Explain Indeterminate Forms , , , Solution: At x = a, takes the indeterminate form (i) if f(x) = 0 and g (x) = 0 (ii) if f(x) = 1 and g (x) = ∞ (iii) if f(x) = ∞ and g(x) = 0 and (iv) if f(x) = 0 and f (x) = ∞ In all these cases the following method is adopted to evaluate Let L = so that Log L = [g(x) log f (x)] = 0 × ∞ Reducing this to and applying L’ Hospitals rule, we get Log L = a Or L = ea ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − → x x x Lim x log 1 1 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − → a x Cot x a Lim x 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − → Sinx Secx Lim x 1 1 2 π x a Lim x x ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ∞ → 1 1 0 0 ∞ 1 0 ∞ ∞ 0 [ ] ) ( ) ( x g x f 0 0 a x Lim → a x Lim → ∞ 1 a x Lim → a x Lim → 0 ∞ a x Lim → a x Lim → ∞ 0 a x Lim → a x Lim → a x Lim → [ ] ) ( ) ( x g x f a x Lim → [ ] ) ( ) ( x g x f a x Lim → ∞ ∞ or 0 0
• 43. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 43 2011 10. Evaluate x Sinx Solution: let L = x Sinx ⇒ 00 form at x = 0 Hence Log L = Sinx log x ⇒ 0 × ∞ form ∴ form Apply L’ Hospital rule, = form Apply L’ Hospitals rule we get = ∴ L = 0 11. Evaluate Solution: let L = is 1∞ form form Apply L’ Hospitals rule = 0 → x Lim 0 → x Lim 0 → x Lim ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∞ ∞ = = → → Coscex x Lim Sinx x Lim LogL x x log 1 log 0 0 ( ) 0 0 tan . 1 0 0 x x Sinx Lim CoscexCotx x Lim x x − = − → → −∞ = − → 1 tan sin 2 0 x Cosx x xSec Lim x 0 1 1 = ∞ = = = ⇒ −∞ = ∴ ∞ ∞ − e e L LogL x x x Lim − → 1 1 1 ) ( x x x Lim − → 1 1 1 ) ( ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ∴ → 0 0 log 1 1 1 x x Lim LogL x 1 1 1 1 1 1 − = − = − → → x Lim x Lim x x
• 44. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 44 2011 ∴ Log L = -1 ⇒ L = 12. Evaluate Solution: let L = form form form Apply L’ Hospitals rule = form Apply L’ Hospital rule, we get Log L = e e 1 1 = − 2 1 0 tan x x x x Lim ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ → ∞ → ≡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 1 tan 2 1 0 x x x x Lim ) 0 ( tan log 1 2 0 × ∞ ≡ ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∴ → x x x Lim LogL x ( ) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≡ ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ = ∴ → 0 0 tan log 2 0 x x x Lim LogL x ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ − → x x x x xSec x x Lim x 2 tan tan 1 2 2 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = → 0 0 tan 2 1 3 2 0 x x x xSec Lim LogL x 2 2 2 2 0 3 tan 2 2 1 x x Sec x x xSec x Sec Lim x − + = → ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = → x x x Sec Lim x tan ) ( 3 1 2 0 3 1 3 1 e L = ∴
• 45. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 45 2011 Exercise: 3 Evaluate the following limits. a) b) c) d) Evaluate the following limits. Cotx x Secx Lim ) ( 0 → ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ → ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − a x a x a x Lim 2 tan 2 π x e x Lim x x − + → 1 0 ) 1 ( 2 ) ( 0 x x Cosax Lim ∞ → 3 3 0 0 2log(1 ) log(1 ) ( ) lim ( ) lim sin sin x x x x e e x x i ii x x x − → → − − + + 2 0 2 2 1 sin cos log(1 ) logsin ( ) lim ( ) lim tan ( ) 2 x x x x x x iii iv x x x π π → → + − + − − 1 2 2 0 2 cosh log(1 ) 1 sin sin ( ) lim ( ) lim x x x x x x x v vi x x π − → → + − − + 2 2 0 (1 ) ( ) lim log(1 ) x x e x vii x x → − + + ( ) 0 ( ) lim 2 cot( ) ( ) lim cos cot x a x x i x a ii ecx x a → → ⎛ ⎞ − − − ⎜ ⎟ ⎝ ⎠ 2 2 0 2 1 ( ) lim tan sec ( ) lim cot 2 x x iii x x x iv x x π π → → ⎡ ⎤ ⎛ ⎞ − − ⎜ ⎟ ⎢ ⎥ ⎣ ⎦ ⎝ ⎠ [ ] 2 0 2 1 1 ( ) lim ( ) lim 2 tan sec tan x x v vi x x x x x x π π → → ⎡ ⎤ − − ⎢ ⎥ ⎣ ⎦ 2 2 1 0 0 1 cos ( ) lim(cos ) ( ) lim 2 b x x x x x i ax ii → → ⎛ ⎞ + ⎜ ⎟ ⎝ ⎠
• 46. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 46 2011 Evaluate the following limits. 2 1 1 2 log(1 ) 1 0 sin ( ) lim(1 ) ( ) lim x x x x x iii x iv x − → → ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ ( ) cot tan 0 0 ( ) lim(sin ) ( ) lim 1 sin x x x x v x iv x → → + ( ) 2 tan 2 cos 0 4 ( ) lim(cos ) ( ) lim tan x ec x x x vii x viii x π → → 1 0 1 ( ) lim( ) ( ) lim 1 3 x x x x x x x ax a b c ix x ax →∞ → ⎛ ⎞ + + + ⎜ ⎟ − ⎝ ⎠ 2 2 1 0 0 1 cos ( ) lim(cos ) ( ) lim 2 b x x x x x i ax ii → → ⎛ ⎞ + ⎜ ⎟ ⎝ ⎠ 2 1 1 2 log(1 ) 1 0 sin ( ) lim(1 ) ( ) lim x x x x x iii x iv x − → → ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ ( ) cot tan 0 0 ( ) lim(sin ) ( ) lim 1 sin x x x x v x iv x → → + ( ) 2 tan 2 cos 0 4 ( ) lim(cos ) ( ) lim tan x ec x x x vii x viii x π → → 1 0 1 ( ) lim( ) ( ) lim 1 3 x x x x x x x ax a b c ix x ax →∞ → ⎛ ⎞ + + + ⎜ ⎟ − ⎝ ⎠
• 47. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 47 2011 Polar Curves If we traverse in a hill section where the road is not straight, we often see caution boards hairpin bend ahead, sharp bend ahead etc. This gives an indication of the difference in the amount of bending of a road at various points which is the curvature at various points. In this chapter we discuss about the curvature, radius of curvature etc. Consider a point P in the xy-Plane. r = length of OP= radial distance θ = Polar angle ( r, θ)→ Polar co-ordinates Let r = f (θ) be the polar curve x = r Cos θ y = r Sin θ Relation (1) enables us to find the polar co-ordinates ( r, θ) when the Cartesian co-ordinates ( x, y) are known. Expression for arc length in Cartesian form. Proof: Let P (x,y) and Q (x + δx, Y + δy) be two neighboring points on the graph of the function y = f (x). So that they are at length S and S + δs measured from a fixed point A on the curve. Y = f (x) Q T (Tangent) δs δy A δx N O δx X From figure, , ∠TPR = ψ and PR =δx, RQ = δy S PQ δ = ∩ S AP = ∩ ( ) 2 2 1 , tan (1) y r x y x θ − −−−−−− = + =
• 48. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 48 2011 ∴ Arc PQ = δS From ∆le PQR, we have [Chord PQ]2 = PR2 + QR2 [Chord PQ]2 = (δx)2 + (δy)2 (∵ from figure) When Q is very close to point P, the length of arc PQ is equal to the length of Chord PQ. i.e arc PQ = Chord PQ = δs ∴ (δs)2 = (δx)2 + (δy)2 -------- (1) ÷ (δx)2, we get When Q → P along the curve, δx → 0, δs → 0 i.e., ∴ --------------(2) Similarly, dividing (1) by δy and taking the limit as δy → 0, we get ------------(3) 2 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ x y x s δ δ δ δ 2 0 2 0 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∴ → → x y Lim x s Lim x x δ δ δ δ δ δ 2 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ dx dy dx ds 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = dx dy dx ds 2 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = dy dx dy ds
• 49. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 49 2011 Expressions for & Trace a tangent to the curve at the point P, it makes an angle ψ with the x – axis. From ∆le PRT, we have tan ψ = Q Equation (2) becomes, Y Tangent T T A P ψ R ψ P R ψ O M N X = ∴ = Sec ψ ( Or) = Cos ψ and equation (3) becomes = = Cosec ψ ∴ = Cosecψ or = Sin ψ Derive an expression for arc length in parametric form. Solution: Let the equation of the curve in Parametric from be x = f (t) and y = g (t). We have, dx ds dy ds dx dy dx ds ψ 2 tan 1+ ψ ψ sec 2 = = Sec dx ds ds dx dy ds ψ 2 1 Cot + ψ 2 sec Co = dy ds ds dy
• 50. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 50 2011 (δs)2 = (δx)2 + (δy)2 ÷ by (δt)2, we get Taking the limit as δt → 0 on both sides, we get (Or) ------------ (4) Derive an expression for arc length in Cartesian form. Solution: Let P (r,θ) and Q (r + δr, θ + δθ) be two neighboring points on the graph of the function r = f (θ). So that they are at lengths S and s + δs from a fixed point A on the curve. ∴ PQ = (S + δs) – s = δs Draw PN ⊥ OQ From ∆le OPN, Q Tangent N r + δr P φ r δθ θ X O 2 2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ t y t x t s δ δ δ δ δ δ 2 0 2 0 2 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ → → → t y Lim t x Lim t s Lim t t t δ δ δ δ δ δ δ δ δ 2 2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∴ dt dy dt dx dt ds 2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = dt dy dt dx dt ds
• 51. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 51 2011 i.e., = Sin δθ (or) PN = r sinδθ and = Cos δθ i.e = Cos δθ (or) ON = r Cos δθ When Q is very close to P, the length of arc PQ as equal to δs, where δs as the length of chord PQ. In ∆le PQN, (PQ)2 = (PN)2 + (QN)2 but PN = r sin δθ (∵Sin δθ ≈ δθ) And QN = OQ – ON = (r + δr) – r Cos δθ = r + δr – r ( ∵Cos δθ ≈ 1) ∴ QN = δr And (PQ)2 = (PN)2 + (QN)201 (δS)2 + (rδθ)2 + (δr)2 ---------------- (5) ÷ by (δθ)2 we get When Q → P along the curve δθ → 0 as δS → 0 and δr → 0 δθ Sin OP PN = r PN OP ON r ON 2 2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ δθ δ δθ δ r r S
• 52. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 52 2011 i.e i.e., --------------- (6) Similarly equation (5) ÷ by (δr)2 and taking the limits as δr → 0 we get + 1 i.e + 1 ∴ --------------- (7) Note: Angle between Tangent and Radius Vector:- We have, Tan φ = r i.e., r = r . ∴Sinφ = r and Cos φ = 2 0 2 2 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ → → δθ δ δθ δ δθ δθ r Lim r S Lim 2 2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ θ θ d dr r d ds 2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = θ θ d dr r d ds 2 0 2 2 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ → → r r Lim r S Lim r r δ δθ δ δ δ δ 2 2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ dr d r dr ds θ 1 2 2 + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = dr d r dr ds θ dr dθ = φ φ Cos Sin dr dθ ds dθ dr ds = φ φ Cos Sin ds dr ds d r θ ds dθ ds dr
• 53. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 53 2011 Find and for the following curves:- 1) y = C Cos h Solution: y = C Cos h Differentiating y w.r.t x. we get = Sin h = Cosh Again Differentiating y w.r.t y we get 1 = C Sin h i.e = Cosech 2) x3 = ay2 Solution : x3 = ay2 Differentiating w.r.t y and x separately we get dx ds dy ds ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ c x ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ c x dx dy ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ c x 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ∴ dx dy dx ds ( ) ( ) c x Cosh c x 2 2 sinh 1 = + ≡ dx ds ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ c x ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ c x C 1 dy dx dy dx ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ c x ( ) c x h Co dy ds 2 sec 1 + = ∴
• 54. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 54 2011 3x2 = 2 ay and 3 x2 = 2 ay i.e., = and = We know that = = = = and = (∴x3 = ay2) 3. y = log cos x Solution . y= log cos x Differentiating w.r.t x and y separately, we get = ( – Sin x) = - tan x i.e., = - tan x and 1 = (- Sin x) dy dx dx dy dy dx 2 3 2 x ay dx dy ay x 2 3 2 dx ds 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + dx dy 2 2 2 3 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ay x 2 2 2 2 2 3 4 9 1 4 9 1 y a x ay y a x x + = + dx ds a x 4 9 1+ 2 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = dy dx dy ds 2 2 3 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + x ay 2 1 2 1 4 2 2 9 4 1 9 4 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = x a x y a dy ds dx dy Cosx 1 dx dy Cosx 1 dy dx
• 55. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 55 2011 i.e., 1 = - tan x (Or) = - Cot x We have = and & = = ∴ = Sec x and Find for the following Curves:- 1. x = a (Cos t + t Sin t), y = a (Sin t = t Cos t) 2. x = a Sec t , y = b tan t 3. x = a , y = a Sin t Solution of 1 Given x = a (Cos t + t Sin t), y = a (Sin t – t Cos t ) Differentiating x & y W.r.t ‘t’, we get = a (-Sin t + Sin t + t Cos t) = a t cos t and = a (Cos t – Cos t + t Sin t) = at Sin t dy dx dy dx dx ds 2 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + dx dy 2 1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = dy dx dy ds x dx ds 2 tan 1+ = ∴ dy ds x Cot2 1+ x Sec2 dx ds 2 1 2 ) 1 ( x Cot dy ds + = dt ds ( ) 2 tan log t Cost + dt dx dt dx dt dy dt dy 2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∴ dt dy dt dx dt ds
• 56. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 56 2011 = = at Solution of 2 x = a Sec t, x = b tan t ∴ = a Sec t tan t, = b Sec2t We have = (a2Sec2 t tan2t + b2 Sec4t) = [a2 Sec2 t (Sec2t – 1) + b2 Sec4 t] = [a2 Sec4 t – a2 Sec2 t + b2 Sec4 t] = [(a2 + b2) Sec4 t – a2 Sec2t] Solution of 3 x = a (Cos t + log tan ), y = a Sin t Differentiating x and y w.r.t ‘t’ we get = a , = a Cos t = , = = a Cot t t Sin t a t Cos t a 2 2 2 2 2 2 + dt ds dt dx dt dy 2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∴ dt dy dt dx dt ds 2 1 2 1 2 1 dt ds 2 1 2 t dt dx ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ + − 2 1 . 2 2 tan 1 int 2 t Sec t S dt dy 2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∴ dt dy dt dx dt ds 2 1 2 2 2 2 2 2 tan 2 2 int ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + − t Cos a t t Sec S a ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + − 2 tan 4 2 2 tan 2 int 2 4 2 2 2 2 t t Sec a t t Sec S a a
• 57. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 57 2011 Find and for the following curves 1. r = a (1 – Cos θ) 2 .r2 = a2 Cos 2θ 3. r = a eθCot α Solution of 1 r = a (1 – Cos θ) Differentiating r w.r.t θ we get = a Sin θ Hence = = {a2 (1 – Cos θ)2 + a2 Sin2 θ} = {a2 (1 – 2 Cos θ + Cos2θ) + a2 Sin2 θ} = {a2 – 2a2 Cos θ + a2} = {2a2 – 2a2 Cos θ} = a{ 2(1 – Cos θ)} = a {2 (2 Sin2 )} = 2 a Sin And = = θ d ds θ d dr θ d dr θ d ds 2 1 2 2 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + θ d dr r 2 1 2 1 2 1 2 1 2 1 2 θ 2 1 θ d ds 2 θ dr ds 2 1 2 2 1 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + dr d r θ 2 1 2 2 2 2 ) 1 ( 1 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − + θ θ Sin a Cos a
• 58. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 58 2011 = = = Solution of 2 r2 = a2 Cos 2θ Differentiating W.r.t ‘θ’ we get 2r = -a2 Sin 2θ . 2 r = -a2 Sin 2θ = Hence = = = = = 2 1 2 2 2 ) 1 ( ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − + θ θ θ Sin Cos Sin { } 2 2 2 2 2 2 2 ) 1 ( 2 2 1 θ θ θ θ θ θ θ Cos Sin Sin Sin Sin Sin Cos = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − dr ds 2 1 θ Cos θ d dr θ d dr θ d dr r Sin a βθ 2 2 θ d ds 2 1 2 2 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + θ d dr r 2 1 2 2 2 2 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + r Sin a r θ 2 1 2 2 4 2 2 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + r Sin a r θ 2 1 2 2 4 2 2 2 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + θ θ θ Cos a Sin a Cos a 2 1 2 2 4 2 4 2 2 2 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + θ θ θ Cos a Sin a Cos a
• 59. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 59 2011 = Or = (∵r2 = a2 Cos 2θ) And = = = = = = = Cosec 2θ Solution 3 r = aeθCot α , here α is constant Differentiating w.r.t ‘θ’ we get = a eθCot α . Cot α Hence = = {a2 e2θCotα + a2 e2θCot α cot2 α} = a eθCot α {1 + Cot2α} = a eθCot α {Cosce2α} θ θ 2 2 2 1 2 4 Cos a Cos a a = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ θ d ds r a r a 2 2 1 2 4 = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ dr ds 2 1 2 2 1 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + dr d r θ 2 1 2 2 2 2 1 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + θ Sin a r r 2 1 2 4 4 2 1 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + θ Sin a r 2 1 2 4 2 4 2 4 2 sin 2 2 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + θ θ θ a Cos a Sin a 2 1 2 4 4 2 sin ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ θ a a θ 2 1 Sin θ d dr θ d ds 2 1 2 2 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + θ d dr r 2 1 2 1 2 1
• 60. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 60 2011 = a eθCot α Cosce α and = = = {1 + tan2 α} = {Sec2 α} = Sec α Exercises: Find and to the following curves. 1. rn = an Cos nθ 2. r (1 + Cos θ) = a 3. rθ = a Note: We have Sin φ = r and Cos φ = ∴ = Cosφ = (1 – Sin2φ) = Since P = r Sin φ. = θ d ds dr ds 2 1 2 2 1 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + dr d r θ 2 1 2 2 2 2 2 1 1 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ + α α θ α θ Cot Cot Cot e a e a 2 1 2 1 dr ds θ d ds ds dθ ds dr ds dr 2 1 2 1 2 2 1 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − r p ds dr r p r 2 2 −
• 61. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 61 2011 = Q r+δr P(x,y) φ OR = P r O X P R Sin φ = ∴ P = r Sin φ Prove that with usual notations ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = dr d r θ φ tan Let P (r, θ) be any point on the curve r = f (θ) r OP and P O X = θ = ∴ ∧ Let PL be the tangent to the curve at P subtending an angle ψ with the positive direction of the initial line (x – axis) and φ be the angle between the radius vector OP and the tangent PL. That is φ = ∧ L P O From the figure we have dr ds ∴ 2 2 p r r − r p OP OR =
• 62. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 62 2011 θ + φ = ψ (Recall from geometry that an exterior angle is equal to the sum of the interior opposite angles) ( ) θ + φ = ψ ⇒ tan tan θ φ φ + φ ψ tan tan - 1 tan tan tan or …(1) Let (x, y) be the Cartesian coordinates of P so that we have, X = r cos θ , y = r sin θ Since r is a function of θ , we can as well regard these as parametric equations in terms of θ . We also know from the geometrical meaning of the derivative that PL tangent the of slope dx dy tan = = ψ ie., d dx d dy tan θ θ = ψ since x and y are function of θ ie., ( ) ( ) θ = ′ θ ′ + θ θ ′ + θ = θ θ θ θ = ψ d dr r where cos r sin r - sin r cos r cos r d d sin r d d tan Dividing both the numerator and denominator by θ ′ cos r we have, θ ′ θ ′ + θ ′ θ − θ ′ θ ′ + θ ′ θ = ψ cos r cos r cos r sin r cos r sin r cos r cos r tan Or θ ′ θ + ′ = ψ tan . r r - 1 an t r r tan …(2) Comparing equations (1) and (2) we get ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ θ = φ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ θ = ′ = φ dr d r or tan d dr r r r tan
• 63. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 63 2011 Prove with usual notations 2 4 2 2 d dr r 1 r 1 p 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ θ + = or r 1 u where d du u 1 2 2 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = θ p Proof : Let O be the pole and OL be the initial line. Let P (r, θ ) be any point on the curve and hence we have OP = r and θ = ∧ P O L Draw ON = p (say) perpendicular from the pole on the tangent at P and let φ be the angle made by the radius vector with the tangent. From the figure θ = = ∧ ∧ P O L 90 P N O Now from the right angled triangle ONP OP ON sin = φ ie., φ = = φ sin r p or r P sin we have p = r sin φ …(1) and θ = φ d dr r 1 cot …(2) Squaring equation (1) and taking the reciprocal we get,
• 64. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 64 2011 φ = φ = cosec r 1 p 1 ie., sin 1 . r 1 p 1 2 2 2 2 2 2 Or ( ) φ + = cot 1 r 1 p 1 2 2 2 Now using (2) we get, ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ θ + = 2 2 2 2 d dr r 1 1 r 1 p 1 Or 2 4 2 2 d dr r 1 r 1 p 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ θ + = …(3) Further, let u r 1 = Differentiating w.r.t. θ we get, 2 2 4 2 d du d dr r 1 d du d dr r 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ θ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ θ ⇒ θ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ θ − , by squaring Thus (3) now becomes 2 2 2 d du u p 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ θ + = …(4) 1. Find the angle of intersection of the curves: ( ) ( ) θ = θ + = cos - 1 b r & cos 1 a r Solution : r = a (1 + cos θ) : r = b(1 – cos θ) ) cos (1 log a log r log θ + + = ⇒ : log r = log b + log (1 – cos θ) Differentiating these w.r.t. θ we get cos 1 sin - 0 d dr r 1 θ + θ + = θ : cos 1 sin 0 d dr r 1 θ − θ + = θ ( ) ( ) ( ) /2 cos 2 /2 cos /2 sin 2 cot 2 1 θ θ θ − = φ : ( ) ( ) ( ) /2 sin 2 /2 cos /2 sin 2 cot 2 2 θ θ θ = φ ie., ( ) ( ) /2 /2 cot /2 tan - cot 1 θ + π = θ = φ : ( ) /2 cot cot 2 θ = φ /2 /2 1 θ + π = φ ⇒ : /2 2 θ = φ
• 65. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 65 2011 ∴ angle of intersection = /2 /2 - /2 /2 - 2 1 π θ θ π φ φ = + = Hence the curves intersect orthogonally. 2. S.T. the curves ( ) ( ) θ = θ + = sin - 1 a r & sin 1 a r cut each other orthogonally Solution : log r = log a + log (1 + sin θ) : log r = log a + log (1 – sin θ) Differentiating these w.r.t θ we get θ + θ = θ sin 1 cos d dr r 1 : θ − θ = θ sin 1 cos - d dr r 1 ie., sin 1 cos cot 1 θ + θ = φ : sin 1 cos - cot 2 θ − θ = φ We have θ θ = φ θ θ + = φ cos - sin - 1 tan and cos sin 1 tan 2 1 1 - cos - cos cos sin - 1 tan . tan 2 2 2 2 2 1 = θ θ = θ − θ = φ φ ∴ Hence the curves intersect orthogonally. 3. Find the angle of intersection of the curves: r = sin θ + cos θ, r = 2 sin θ Solution : log r = log (sin θ + cos θ) : r = 2 sin θ ( ) θ + θ = ⇒ cos sin log r log : log r = log 2 + log (sin θ) Differentiating these w.r.t θ we get θ + θ θ θ = θ cos sin sin - cos d dr r 1 : sin cos d dr r 1 θ θ = θ ie., ( ) ( ) θ + θ θ θ = φ tan 1 cos tan - 1 cos cot 1 : θ = φ ⇒ θ = φ cot cot 2 2 ie., ( ) θ + π = φ ⇒ θ + π = φ /4 /4 cot cot 1 1 /4 - /4 - 2 1 π = θ θ + π = φ φ ∴ The angle of intersection is 4 / π
• 66. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 66 2011 4. Find the angle of the curves: r = a log θ and r = a/ log θ Solution : r = a log θ : r = a/ log θ ( ) θ + = ⇒ log log a log r log : ( ) θ = log log - a log r log Differentiating these w.r.t q, we get, θ θ = θ . log 1 d dr r 1 : θ θ − = θ . log 1 d dr r 1 ie., log 1 cot 1 θ θ = φ : log 1 - cot 2 θ θ = φ Note : We can not find 2 1 and φ φ explicitly. θ θ = φ ∴ log tan 1 : θ θ = φ log tan 2 Now consider, θ θ = φ log tan 1 : θ θ = φ log - tan 2 Now consider, ( ) 2 1 2 1 tan tan 1 log 2 - tan φ φ θ θ φ φ + = ( )2 log 1 log 2 θ θ − θ θ = ......…(1) We have to find θ by solving the given pair of equations : θ = θ = a/log r and log a r Equating the R.H.S we have a θ = θ log a log ie., ( ) e log or 1 1 log 2 = θ ⇒ θ = = θ Substituting ( ) get we 1 in e = θ ( ) 2 2 1 e - 1 2e tan = φ − φ ( ) 1 e log = ∵ ∴ angle of intersection e tan 2 e - 1 2e tan - 1 - 2 1 - 2 1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = φ φ 5. Find the angle of intersection of the curves: r = a (1 – cos θ ) and r = 2a cos θ Solution : r = a (1 – cos θ) : r = 2a cos θ Taking logarithms we have, Log r = log a + log (1 – cos θ) : log r = log 2a + log (cos θ) Differentiating these w.r.t θ, we get,
• 67. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 67 2011 cos - 1 sin d dr r 1 θ θ = θ : cos sin - d dr r 1 θ θ = θ ie., ( ) ( ) ( ) θ = φ θ θ θ = φ tan - cot : /2 sin 2 /2 cos /2 sin 2 cot 2 2 1 ie., ( ) /2 cot cot 1 θ = φ : ( ) θ + π = φ /2 cot cot 2 /2 1 θ = φ ⇒ : θ + π = φ /2 2 2 2 / - /2 - /2 2 1 θ π θ π θ φ φ + = = − ∴ …(1) Now consider ( ) θ = θ = cos 2a cos - 1 a r Or ( ) 1/3 cos or 1 cos 3 -1 = θ = θ Substituting this value in (1) we get, The angle of intersection ( ) 1/3 cos . 1/2 2 / -1 + π 6. Find the angle of intersection of the curves : θ = a r and θ = / a r Solution : θ = a r : θ = / a r θ + = ⇒ log a log r log : θ = log - a log r log Differentiating these w.r.t θ, we get, θ = θ 1 d dr r 1 : θ = θ 1 - d dr r 1 ie., θ = φ 1 cot 1 : θ = φ 1 - cot 2 or θ = φ tan 1 : θ = φ - tan 2 Also by equating the R.H.S of the given equations we have 1 1 or a/ a 2 ± = θ ⇒ = θ θ = θ When and 1 - tan 1, tan 1, 2 1 = φ = φ = θ When 1 tan 1, - tan 1, - 2 1 = φ = φ = θ . /2 - 1 - tan . tan 2 1 2 1 π = φ φ ⇒ = φ φ ∴ The curves intersect at right angles.
• 68. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 68 2011 7. Find the pedal equation of the curve: ( ) 2a cos - 1 r = θ Solution : ( ) 2a cos - 1 r = θ ( ) 2a log cos - 1 log r log = θ + ⇒ Differentiating w.r.t θ, we get θ θ = θ = θ θ + θ cos - 1 sin - d dr r 1 or 0 cos - 1 sin d dr r 1 ( ) ( ) ( ) ( ) /2 cot - /2 sin 2 /2 cos /2 sin 2 - cot 2 θ = θ θ θ = φ ∴ ie., ( ) ( ) /2 - /2 - cot cot θ = φ ⇒ θ = φ ∴ Consider φ = sin r p ( ) ( ) /2 sin r - p or /2 - sin r p θ = θ = ∴ Now we have, ( ) 2a cos - 1 r = θ …(1) ( ) /2 sin r - p θ = …(2) We have to eliminate θ from (1) and (2) (1) can be put in form ( ) 2a 2 / sin 2 . 2 = θ r ie., ( ) a /2 sin r 2 = θ But p/-r = sin (θ/2), from (2) ar p or a r p r 2 2 2 = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∴ Thus p2 = ar is the required pedal equation. 8. Find the pedal equation of the curve: r2 = a2 sec 2θ Solution : r2 = a2 sec 2θ ) 2 (sec log a log 2 r log 2 θ + = ⇒ Differentiating w.r.t θ , we get, θ = θ θ θ θ = θ 2 tan d dr r 1 ie., 2 sec 2 tan 2 sec 2 d dr r 2 ie., ( ) θ π = φ ⇒ θ π = φ 2 - /2 2 - /2 cot cot
• 69. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 69 2011 Consider ( ) θ = θ π = ∴ φ = 2 cos r p ie., 2 - /2 sin r p sin r p Now we have, θ = 2 sec a r 2 2 …(1) θ = 2 cos r p …(2) From (2) p/r = cos 2θ or r/p = sec 2 θ Substituting in (1) we get, ( ) r/p a r 2 2 = or pr = a2 Thus pr = a2 is the required pedal equation. 9. Find the pedal equation of the curve: rn = an cos nθ Solution : rn = an cos nθ ( ) θ + = ⇒ n cos log a log n r log n Differentiating w.r.t q we get θ = θ θ θ = θ n tan - d dr r 1 ie., n cos n sin n - d dr r n ( ) θ + π = φ ⇒ θ + π = φ ∴ n /2 n /2 cot cot Consider φ = sin r p ( ) θ = θ = π = ∴ n cos r p ie., n /2 sin r p Now we have, θ = n cos a r n n …(1) θ = n cos r p …(2) ( ) p/r a r is (2) of e consequenc a as (1) n n = ∴ Thus rn + 1 = pan is the required pedal equation. 10. Find the pedal equation of the curve: rm = am (cos mθ + sin mθ) Solution : rm = am (cos mθ + sin mθ) Differentiating w.r.t θ, we get, θ + θ θ + θ = θ m sin m cos m cos m m sin m - d dr r m ie., ( ) ( ) θ + θ θ θ = θ + θ θ θ = θ m tan 1 m cos m tan - 1 m cos m sin m cos m sin - m cos d dr r 1 ( ) θ + π = φ ⇒ θ + π = φ ∴ m /4 m /4 cot cot
• 70. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 70 2011 Consider φ = sin r p ( ) θ + π = ∴ m /4 sin r p ie., ( ) ( ) [ ] θ π + θ π = m sin /4 cos m cos /4 sin r p ie., ( ) θ + θ = m sin m cos 2 r p (we have used the formula of sin (A + B) and also the values ( ) ( ) ) 2 1/ /4 cos /4 sin = π = π Now we have, ( ) θ + θ = m sin m cos a r m m …(1) ( ) θ + θ = m sin m cos 2 r p …(2) Using (2) in (1) we get, p a 2 r or r 2 p . a r m 1 m m m = = + Thus p a 2 r m 1 m = + is the required pedal equation. 11. Establish the pedal equation of the curve: θ + θ = n cos b n sin a r n n n in the form ( ) 2 n 2 n 2 2n 2 r b a p + = + Solution : We have θ + θ = n cos b n sin a r n n n ( ) θ + θ = ⇒ n cos b n sin a log r log n n n Differentiating w.r.t θ we get θ + θ θ θ = θ n cos b n sin a n sin nb - n cos na d dr r n n n n n Dividing by n, θ + θ θ θ = φ n cos b n sin a n sin b - n cos a cot n n n n Consider φ = sin r p Since φ cannot be found, squaring and taking the reciprocal we get, ( ) φ + = φ = cot 1 r 1 p 1 or cosec r 1 p 1 2 2 2 2 2 2 ( ) ( ) ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ θ + θ θ θ + = ∴ 2 n n 2 n n 2 2 n cos b n sin a n sin b - n cos a 1 r 1 p 1
• 71. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 71 2011 ( ) ( ) ( ) ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ θ + θ θ θ + θ + θ = 2 n n 2 n n 2 n n 2 2 n cos b n sin a n sin b - n cos a n cos b n sin a r 1 p 1 e., i ( ) ( ) ( ) ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ θ + θ θ + θ + θ + θ = 2 n n 2 2 n 2 2 2 n 2 2 2 n cos b n sin a n sin n cos b n cos n sin a r 1 p 1 e., i (product terms cancels out in the numerator) ( )2 n n n 2 n 2 2 2 n cos b n sin a b a . r 1 p 1 ., ie θ + θ + = ( ) , r b a . r 1 p 1 or 2 n n 2 n 2 2 2 + = by using the given equation. ( ) 2 n 2 n 2 n 2 2 r b a p + = + ∴ is the required pedal equation. 12. Define Curvature and Radius of curvature Solution: A Curve Cuts at every point on it. Which is determined by the tangent drawn. Y y = f (x) Tangent s P (x,y) A O ψ X Let P be a point on the curve y= f (x) at the length ‘s’ from a fixed point A on it. Let the tangent at ‘P’ makes are angle ψ with positive direction of x – axis. As the point ‘P’ moves along curve, both s and ψ vary. The rate of change ψ w.r.t s, i.e., as called the Curvature of the curve at ‘P’. The reciprocal of the Curvature at P is called the radius of curvature at P and is denoted by ρ. ds dψ ψ ψ ρ d ds ds d = = ∴ 1
• 72. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 72 2011 (or) Also denoted ∴K = K is read it as Kappa. 13. Derive an expression for radius of curvature in Cartesian form. Solution :(1) Cartesian Form: Y = f (x) P T A S ψ dy ψ ψ P dx R X O Let y = f (x) be the curve in Cartesian form. We know that, tan ψ = (From Figure) ----------- (1) Where ψ is the angle made by the tangent at P with x – axis. Differentiating (1) W.r.t x, we get Sec2ψ . = i.e., = Sec2 ψ = (1 + tan2 ψ) = (1 + tan2 ψ) . = (1 + tan2 ψ) = . . from eqn (1) ψ ρ d ds = ∴ ds dψ ρ = 1 K 1 = ρ ds dψ dx dy dx dψ 2 2 dx y d 2 2 dx y d dx dψ dx dψ dx dψ dx ds ρ 1 2 1 2 1 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + dx dy ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + 2 1 dx dy ρ 1 2 1 2 1 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + dx dy
• 73. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 73 2011 = ∴ ρ = ∴ ρ = ( 1+ y1 2 )3/2 ……………(1) y2 Where y1 = , y2 = This is the formula for Radius of Curvature in Cartesian Form. 14. Show that the Curvature of a Circle at any point on it, is a Constant Tangent PP P A ψ X T Solution: Consider a Circle of radius r. Let A be a fixed point and ‘P’ be a given point on the circle such that arc AP = S. Let the angle between the tangent to the Circle at A and P be ψ. Then clearly AOP = ψ. ∴ AP = rψ i.e., S = rψ This is the intrinsic equation of the circle. Differentiating w.r.t S we get 1 = r Or K = = ∴ K = which is Constant Hence the Curvature of the Circle at any point on it is constant. ρ 1 2 3 2 1 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + dx dy 2 3 2 2 2 1 dx y d dx dy ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + dx dy 2 2 dx y d ds dψ ds dψ r 1 r 1 O r
• 74. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 74 2011 15. Derive an expression for radius of curvature in parametric form. Solution: We have ρ = (1 + y 1 2 )3/2 y2 Let x =f (t), y = g (t) be the curve in Parametric Form. Then y1 = Y2 = = = = . . = ∴ Y2 = Substituting Y1 and Y2 in equation 1. ρ = , we get ρ = -------------------- (2) Equation (2) is called the Radius of Curvature in Parametric Form. • • = = X Y dt dx dt dy dx dy 2 2 dx y d dx d ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ dx dy dx d ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ • • X Y dt d ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ • • X Y dx dt dt d ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ • • X Y ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ dt dx 1 dt d ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎝ ⎛ • • X Y • X 1 ( )2 x x y y x − • ) ( 1 X 3 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − • • x x y y X ( ) 2 2 3 2 1 1 y Y + ( ) ( ) ( )3 2 3 2 1 x x y y x x Y − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + ( ) ( ) [ ] x y y x y x − + = 2 3 2 2 ρ
• 75. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 75 2011 16. Derive an expression for radius of curvature in polar form. Solution: Let r = f (θ) be the curve in the Polar Form. We know that, Angle between the tangent and radius vector, Tan φ = r = r. i.e., tan φ = Differentiate w.r.t ‘θ’ we get Sec2 φ . = = = = From figure ψ = θ + φ Differentiating w.r.t θ, we get dr dθ θ d dr 1 ( ) θ d dr r θ φ d d 2 2 2 . ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − θ θ θ θ d dr d r d r d dr d dr ( ) ( ) ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ = ∴ 2 2 2 2 2 1 θ θ θ φ θ φ d dr d r d r d dr Sec d d ( ) ( ) ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ − + 2 2 2 2 2 tan 1 1 θ θ θ φ d dr d r d r d dr ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + 2 2 2 2 2 2 1 1 θ θ θ θ d dr d r d r d dr d dr r θ φ d d 2 2 2 2 2 r d dr d r d r d dr + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ θ θ θ
• 76. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 76 2011 = 1 + ∴ = 1 + = Also we know that = Now, ρ= = . = . ρ ----------- (3) Where r1 = , r2 = Equation (3) as called the radius of curvature in Polar form. 17. Derive an expression for radius of curvature in pedal form. Solution: Let p = r Sin φ be the curve in Polar Form. We have p = r Sin φ Differentiating p W.r.t r, we get = Sin φ + r Cos φ θ ψ d d θ φ d d θ ψ d d 2 2 2 2 2 r d dr d r d r d dr + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ θ θ θ θ ψ d d 2 2 2 2 2 2 2 r d dr d r d r d dr r + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + θ θ θ θ d ds 2 1 2 2 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + θ d dr r ψ d ds θ d ds ψ θ d d 2 1 2 2 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + θ d dr r 2 2 2 2 2 2 2 θ θ θ d r d r d dr r r d dr − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ( ) 2 2 1 2 2 3 2 1 2 2 rr r r r r − + + = θ d dr 2 2 θ d r d dr dp dr dφ
• 77. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 77 2011 But Sin φ = r , Cos φ= ∴ = r + r = r + . = r + r = r (θ + φ) = r ∴ = r = r. ∵ = ∴ ρ= r --------------- (4) Equation (4) is called Radius of Curvature of the Curve in Polar Form. 18. Find the radius of curvature at (x, y) for the curve ay2 = x3. Solution: Given ay2 = x3 -------- (1) is in Cartesian form. We have, Radius of curvature in Cartesian form. ------------ (2) Differentiating (1) w.r.t x we get, ds dθ ds dr dr dp ds dθ ds dr dr dφ ds dθ dr dφ ds dr ds dθ ds dφ ds d ds dψ dr dp ds dψ dr dp ρ 1 ds dψ ρ 1 dp dr ( ) 2 2 3 2 1 1 y y + = ρ
• 78. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 78 2011 a 2y Differentiating y1 w.r.t ‘x’ we get Substitute Y1 and Y2 in (2), we get. 19. Find the radius of curvature at (x,y) for the curve y = c log Sec Solution: Given y = c log Sec -------- (1) Differentiating (1) w.r.t x, we get = c tan . = tan Differentiating y1 W.r.t x we get = Substitute y1 and y2 in ∴ ρ = a x a x a x ay x y dx dy x dx dy 2 3 2 3 2 3 3 2 1 3 2 2 1 2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = = = ⇒ = a x y 2 3 1 = ∴ x a x a dx y d y 4 3 2 1 2 3 2 2 2 = = = ( ) a x a x 6 9 4 2 3 + = ρ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ c x ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ c x ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = c x c x Sec c x Sec c y tan . 1 . 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ c x c 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ c x ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = c x Sec y 2 2 c 1 c 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ c x Sec2 ( ) 2 2 3 2 1 1 y y + = δ ( ) [ ] ( ) ( ) [ ] ( ) c x Sec c c x Sec c x Sec c c x 2 2 3 2 2 2 3 2 1 1 tan 1 = + = δ ( ) c x cSec
• 79. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 79 2011 20. Find the radius of curvature at the point ‘t’ on the curve x = a (t + Sint), y = a (1 – Cost). Solution: Given Curves are in Parametric Form ∴ Radius of Curvature, ------ (1) Differentiating the given Curves W.r.t t, we get = a (1 + Cost) = = a Sint Differentiating W.r.t ‘t’ we get = - a Sint, = a Cost Substitute , , and in (1), we get = = 21. Find the Radius of Curvature to x = , y = a Sint at t. Solution: Here x = , y = a Sin t =a { -sint + } =a { -sint + } = a { (1-sin 2t) / sint } = a Cos2t / sint ( ) ( ) ( ) x y y x y x − + = 2 3 2 2 ρ dt dx x = y dt dy x y x y y x ( ) int) int( ) 1 ( int ) 1 ( 2 3 2 2 2 aS aS aCost Cost a S a Cost a − − + + + = ρ { } { } t Sin t Cos Cost a t Sin t Cos Cost a 2 2 2 2 3 2 2 3 2 1 + + + + + { } { } 2 2 2 . 8 2 2 2 2 . 2 ) 1 ( ) 1 ( 2 2 3 2 2 3 2 2 3 t Cos t Cos a t Cos t Cos a Cost Cost a = = + + 2 4 t aCos = ρ ( ) { } 2 tan log t Cost a + ( ) { } 2 tan log t Cost a + ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ + − = 2 1 . 2 . 2 tan 1 int 2 t Sec t S a dt dx 2 / cos 2 / sin 2 1 t t 2 / sin 2 1 t dt dx
• 80. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 80 2011 = = tant Differentiating W.r.t ‘x’ we get Substitute & in , we get i.e., ∴ρ = a Cot t. 22. Find the Radius of Curvature to at Solution: Given -------------- (1) Differentiating (1) w.r.t to ‘x’ we get i.e., (From (1)) ----------- (2) aCost dx dy and = dt dx dt dy dx dy / / = ∴ t t t a aCost sin / cos cos t dx dy tan = ∴ int 1 . 1 2 2 2 2 2 2 S t aCos t Sec dt dx t Sec dx dt t Sec dx y d = = = int 1 4 2 2 tS Sec a dx y d = = dx dy 2 2 dx y d ( ) 2 2 3 2 1 1 y y + = ρ ( ) aCott S Cost a tS Sec t aSec tS Sec a t = = = + = int int int 1 tan 1 4 3 4 2 3 2 ρ a y x = + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 4 , 4 a a a y x = + 0 2 1 2 1 = + dx dy y x ( ) x x a x y dx dy y − − = − = = 1 x a y − =1 1
• 81. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 81 2011 Also -------------- (3) At the given point Then = - 1 & ∴ Substitute y1 and y2 in ρ = ρ 23. Show that for the Cardioids r = a (1 + Cosθ), ρ2 / r 2 is a constant Solution: r = a (1 + Cosθ) = - a Sinθ We have, , is Pedal Equation. = a2Sin2θ = = 2 3 2 3 2 2 2 2 2 1 x a x a dx y d y = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − = = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 4 , 4 a a 2 1 1 a a y − = ( ) a a a y 4 4 2 1 2 3 2 = − = ( ) 2 2 3 2 1 1 y y + = ρ ( ) 2 4 2 2 4 2 4 ) 1 ( 1 2 3 2 3 2 a a a a = = = − + 2 a = θ d dr 2 4 2 2 1 1 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = θ d dr r r P 4 2 1 1 r r + 4 2 2 2 2 4 2 2 2 ) 1 ( r Sin a Cos a r Sin a r θ θ θ + + = + 4 2 ) 1 ( 2 r Cos a θ +
• 82. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 82 2011 (∵ r = a (1 + Cosθ) Differentiating w.r.t ‘P’ we get 2P = Now, And 24. Find the Radius of Curvature of the Curve Solution: Given Differentiating w.r.t to P, we get ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∴ a r r a P 4 2 2 2 1 θ Cos a r + = ∴ 1 3 2 2 1 r a P = a r P 2 3 2 = ∴ dp dr r a 2 3 2 1 2 3 4 r ap dp dr = r ap dp dr r 3 4 = = ρ 9 8 2 . 9 16 9 16 1 3 3 2 2 2 2 2 2 a a r r a r p a r r = = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = ρ 2 2 2 2 2 2 1 1 1 b a r b a P − + = 2 2 2 2 2 2 1 1 1 b a r b a P − + = dp dr r b a P 2 1 2 2 2 3 − = − r p b a dp dr 3 2 2 = ∴ 3 2 2 3 2 2 . . p b a r p b a r dp dr r = = = ∴ ρ
• 83. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 83 2011 25. Find the Radius of Curvature at (r,θ) on r = Solution: Given r = Differentiating w.r.t to ‘θ’ we get We have Differentiating above result w.r.t to‘P’ we get θ a θ a θ θ θ θ θ r a a d dr − = − = − = 1 . 2 θ θ r d dr − = 2 4 2 2 1 1 1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = θ d dr r r P 2 2 2 2 2 4 2 1 1 1 1 θ θ r r r r r + = + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = 2 2 2 2 2 2 1 1 1 1 1 1 θ θ θ r r P 1 2 + = ∴ θ θ r P 2 2 2 2 2 . 1 . 1 r a r a r a r a r r P + = + = + = θ θ ( ) 2 2 2 2 2 2 2 2 1 . . 1 r a dp dr r r a a r dp dr a r a + / + / − + = dp dr r a a r a r a r a ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − + = + 2 2 2 2 2 2 2 . dp dr r a a r a r a r a 2 2 2 2 2 2 2 ). ( + − + = +
• 84. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 84 2011 Thus, Exercises: (1) Find the Radius of the Curvature at the point (s, ψ) on S = a log tan (2) Find the Radius of the Curvature of xy = C2 at (x,y) (3) Find the Radius of the Curvature of xy3 = a4 at (a,a) (4) Find the Radius of Curvature at the point θ on x = C Sin 2θ (1 + Cos 2θ), y = C Cos 2θ (1 – Cos 2θ) (5) If ρ1 and ρ2 are the radii of curvature at the extremities of any chord of the cardiode r = a (1 + Cosθ) which posses through the Pole prove that ( ) dp dr a r a r a 3 2 2 2 2 = + + ( ) dp dr a r a = + 3 2 3 2 2 ( ) 3 2 3 2 2 . . a r a r dp dr r + = = ρ ( ) 2 3 2 2 3 r a a r + = ∴ ρ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + 2 4 ψ π 9 16 2 2 2 2 1 a = + ρ ρ
• 85. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 85 2011 DIFFERENTIAL CALCULUS – III PARTIAL DIFFERENTIATION Introduction :- Partial differential equations abound in all branches of science and engineering and many areas of business. The number of applications is endless. Partial derivatives have many important uses in math and science. We shall see that a partial derivative is not much more or less than a particular sort of directional derivative. The only trick is to have a reliable way of specifying directions ... so most of this note is concerned with formalizing the idea of direction So far, we had been dealing with functions of a single independent variable. We will now consider functions which depend on more than one independent variable; Such functions are called functions of several variables. Geometrical Meaning Suppose the graph of z = f (x,y) is the surface shown. Consider the partial derivative of f with respect to x at a point (x0, y0). Holding y constant and varying x, we trace out a curve that is the intersection of the surface with the vertical plane y = y0. The partial derivative fx(x0,y0). measures the change in z per unit increase in x along this curve. That is, fx(x0, y0) is just the slope of the curve at (x0, y0). The geometrical interpretation of fy(x0, y0). is analogous.
• 86. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 86 2011 Real-World Applications: Rates of Change: In the Java applet we saw how the concept of partial derivative could be applied geometrically to find the slope of the surface in the x and y directions. In the following two examples we present partial derivatives as rates of change. Specifically we explore an application to a temperature function ( this example does have a geometric aspect in terms of the physical model itself) and a second application to electrical circuits, where no geometry is involved. I. Temperature on a Metal Plate The screen capture below shows a current website illustrating thermal flow for chemical engineering. Our first application will deal with a similar flat plate where temperature varies with position. * The example following the picture below is taken from the current text in SM221,223: Multivariable Calculus by James Stewart.
• 87. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 87 2011 Suppose we have a flat metal plate where the temperature at a point (x,y) varies according to position. In particular, let the temperature at a point (x,y) be given by, where T is measured in oC and x and y in meters. Question: what is the rate of change of temperature with respect to distance at the point (2,1) in (a) the x-direction? and (b) in the y-direction ? Let's take (a) first. What is the rate of change of temperature with respect to distance at the point (2,1) in (a) the x-direction? What observations and translations can we make here? Rate of change of temperature indicates that we will be computing a type of derivative. Since the temperature function is defined on two variables we will be computing a partial derivative. Since the question asks for the rate of change in the x-direction, we will be holding y constant. Thus, our question now becomes: What is at the point (2,1)? Conclusion : The rate of change of temperature in the x-direction at (2,1) is degrees per meter; note this means that the temperature is decreasing ! 2 2 ( , ) 60 /1 T x y x y = + + dT dx 2 2 2 2 1 2 2 2 2 ( , ) 60 /1 60(1 ) 60(2 )(1 ) 20 (2,1) 60(4)(1 4 1) 3 T x y x y x y T x x y x T x − − − = + + = + + ∂ = − + + ∂ ∂ − = − + + = ∂ 20 3 −
• 88. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 88 2011 Part (b): The rate of change of temperature in the y-direction at (2,1) is computed in a similar manner. Conclusion : The rate of change of temperature in the y-direction at (2,1) is degrees per meter; note this means that the temperature is decreasing ! II. Electrical Circuits: Changes in Current The following is adapted from an example in a former text for SM221,223 Multivariable Calculus by Bradley and Smith. * In an electrical circuit with electromotive force (EMF) of E volts and resistance R ohms, the current, I, is I=E/R amperes. Question: (a) At the instant when E=120 and R=15 , what is the rate of change of current with respect to voltage. (b) What is the rate of change of current with respect to resistance? (a) Even though no geometry is involved in this example, the rate of change questions can be answered with partial derivatives. we first note that I is a function of E and R, namely, I(E,R) = ER-1 2 2 2 2 1 2 2 2 2 ( , ) 60 /1 60(1 ) 60(2 )(1 ) 10 (2,1) 60(2)(1 4 1) 3 T x y x y x y T y x y x T x − − − = + + = + + ∂ = − + + ∂ ∂ − = − + + = ∂ 10 3 −
• 89. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 89 2011 The rate of change of current with respect to voltage = the partial derivative of I with respect to voltage, holding resistance constant is when E=120 and R=15 , we have verbal conclusion : If the resistance is fixed at 15 ohms, the current is increasing with respect to voltage at the rate of 0.0667 amperes per volt when the EMF is 120 volts. Part (b): What is the rate of change of current with respect to resistance? Using similar observations to part (a) we conclude: The partial derivative of I with respect to resistance, holding voltage constant = when E=120 and R=15 , we have Conclusion : If the EMF is fixed at 120 volts, the current is decreasing with respect to resistance at the rate of 0.5333 amperes per ohm when the resistance is 15 ohms. Key Words :- Then the partial derivative of z w.r.t x is given by The partial derivative of z w.r.t y is given by 1 I R E − ∂ = ∂ 1 15 0.0667 I E − ∂ = ≈ ∂ 1 I ER E − ∂ = ∂ 1 (120,15) 120(15) 0.5333 I E − ∂ = − ≈ − ∂ 0 ( , ) ( , ) lim x x z f x x y f x y z x x δ δ δ → ∂ + − = = ∂ 0 ( , ) ( , ) lim y y z f x y y f x y z y y δ δ δ → ∂ + − = = ∂
• 90. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 90 2011 1. If ( ) by ax sin e u by - ax + = show that u ab 2 a - = ∂ ∂ ∂ ∂ y u x u b Solution : ( ) by ax sin e u by - ax + = ( ) ( ) by ax sin a.e a . by ax cos e x u by - ax by - ax + + + = ∂ ∂ ∴ ( ) u a by ax cos e a x u e., i by - ax + + = ∂ ∂ …(1) Also ( ) ( ) ( ) by ax sin e b - b . by ax cos e by - ax by - ax + + + = ∂ ∂ x u ( ) bu by ax cos e b y u ., ie by - ax − + = ∂ ∂ …(2) Now y u a x u b ∂ ∂ = ∂ ∂ by using (1) and (2) becomes ( ) ( ) abu by ax cos abe - abu by ax cos abe by - ax by ax + + + + = − = 2 abu Thus abu 2 y u a - x u b = ∂ ∂ ∂ ∂ 2. If ( ) that prove , by - ax f e u by ax+ = abu 2 y u a x u b = ∂ ∂ = ∂ ∂ Solution : ( ) by - ax f e u by ax+ = , by data ( ) ( ) by - ax f ae a by - ax f . e x u by ax by ax + + + ′ = ∂ ∂ Or ( ) u a by - ax f . e a by ax + ′ = ∂ ∂ + x u …(1) Next, ( ) ( ) ( ) by - ax f e b b - . by - ax f e y u by ax by ax + + + ′ = ∂ ∂ Or ( ) ba by - ax f e b y u by ax + ′ − = ∂ ∂ + …(2) Now consider L.H.S y u a x u b ∂ ∂ + ∂ ∂ =
• 91. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 91 2011 ( ) { } ( ) { } bu by - ax f be - a au by - ax f ae b by ax by ax + ′ + + ′ = + + ( ) ( ) abu by - ax f e ab - abu by - ax f e ab by ax by ax + ′ + ′ = + + = 2abu = R.H.S Thus u b a 2 y u a x u b = ∂ ∂ + ∂ ∂ 3. If 2 2 2 z y x log u + + = , show that ( ) 1 z u y u x u z y x 2 2 2 2 2 2 2 2 2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ + + Solution : By data ( ) 2 2 2 2 2 2 z y x log 2 1 z y x log u + + = + + = The given u is a symmetric function of x, y, z, (It is enough if we compute only one of the required partial derivative) 2 2 2 2 2 2 y 2 . 1 . 2 1 z x x x z y x x u + + = + + = ∂ ∂ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + ∂ ∂ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ ∂ ∂ = ∂ ∂ 2 2 2 2 z y u u x x x x x x2 ie., ( ) ( ) ( )2 2 2 2 2 2 2 2 2 2 2 2 2 2 z y z y z y 2 . - 1 z y + + − + = + + + + x x x x x x ( )2 2 2 2 2 2 2 2 2 z y z y u + + − + = ∂ ∂ ∴ x x x …(1) Similarly ( )2 2 2 2 2 2 2 2 2 z y x z u + + − + = ∂ ∂ ∴ x y y …(2) ( )2 2 2 2 2 2 2 2 2 z y y x u + + − + = ∂ ∂ x z z …(3) Adding (1), (2) and (3) we get, ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 z y 1 z y z y u u u + + = + + + + = ∂ ∂ + ∂ ∂ + ∂ ∂ x x x z y x Thus ( ) 1 z u y u u z y 2 2 2 2 2 2 2 2 2 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ + + x x
• 92. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 92 2011 4. If u = log (tan x + tan y + tan z), show that, sin 2x ux + sin 2y uy + sin 2z uz = 2 Solution : u = log (tan x + tan y + tan z) is a symmetric function. z tan y tan tan sec u 2 + + = x x x ( ) z tan y tan tan . sec cos sin 2 u 2 sin 2 + + = x x x x x x Or z tan y tan tan tan 2 u 2 sin + + = x x x x …(1) Similarly z tan y tan tan tan 2 u 2 sin + + = x y y y …(2) z tan y tan tan tan 2 u 2 sin + + = x z z y …(3) Adding (1), (2) and (3) we get, ( ) ( ) 2 z tan y tan x tan z tan y tan tan x 2 u 2z sin u 2y sin u 2 sin z y = + + + + = + + x x Thus 2 u 2z sin u 2y sin u 2x sin z y x = + + 5. If u = log (x3 + y3 + z3 – 3xyz) then prove that z y x 3 z u y u x u + + = ∂ ∂ + ∂ ∂ + ∂ ∂ and hence show that ( )2 2 z y 9 - u + + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ x z y x Solution : u = log (x3 + y3 + z3 – 3xyz) is a symmetric function yz 3 - z y 3yz - 3 u 3 3 3 2 x x x x + + = ∂ ∂ …(1) yz 3 - z y 3zx - 3 u 3 3 3 2 x x y y + + = ∂ ∂ …(2) yz 3 - z y 3xy - 3 u 3 3 3 2 x x z z + + = ∂ ∂ …(3) Adding (1), (2) and (3) we get,
• 93. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 93 2011 ( ) ( ) yz 3 z y z yz y z y 3 z u u u 3 3 3 2 2 2 x x x x x y x − + + − − − + + = ∂ ∂ + ∂ ∂ + ∂ ∂ Recalling a standard elementary result, ( )( ) ca bc ab c b a c b a abc 3 c b a 2 2 2 3 3 3 − − − + + + + = − + + We have, ( ) ( )( ) x x x x x x x x z yz y z y z y z yz y z y 3 z u y u u 2 2 2 2 2 2 − − − + + + + − − − + + = ∂ ∂ + ∂ ∂ + ∂ ∂ Thus z y 3 z u y u u + + = ∂ ∂ + ∂ ∂ + ∂ ∂ x x Further u z y 2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ x u z y z y ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ = x x z u y u u z y ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ = x x z y 3 z y ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ = x x , by using the earlier result. ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + ∂ ∂ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + ∂ ∂ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + ∂ ∂ = z y 3 z z y 3 y z y x 3 x x x ( ) ( ) ( ) ( )2 2 2 2 z y 9 z y 3 z y 3 z y 3 + + − = + + − + + + − + + + − = x x x x Thus ( )2 2 z y 9 - u z y x + + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ x
• 94. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 94 2011 6. If ( ) θ = θ = = sin r y , cos r and r f u x , prove that ( ) ( ) r r 1 r 2 2 2 2 f f y u x u ′ + ′ ′ = ∂ ∂ + ∂ ∂ Solution :Observing the required partial derivative we conclude that u must be a function of x, y. But u = f( r) by data and hence we need to have r as a function of x, y. Since x = r cos θ, y = r sin θ we have x2 + y2 = r2. ( ) 2 2 y x r where r f u have we + = = ∴ ( ) ( ) ( ) and . r r - r r r u 2 2 2 2 3 2 2 x f x f x ′ ′ + ′ = ∂ ∂ ( ) ( ) ( ) . r r - r r r u 2 2 2 2 3 2 2 y f y f y ′ ′ + ′ = ∂ ∂ Adding these results we get, ( ) ( ) { } ( ) ( ) 2 2 2 2 2 2 3 2 2 2 2 y r r f y - 2 r r f u u + ′ ′ + + ′ = ∂ ∂ + ∂ ∂ x x x y x ( ) ( ) ( ) ( ) r r r 1 r . r r r . r r 2 2 2 3 f f f f ′ ′ + ′ = ′ ′ + ′ = Thus ( ) ( ) r r 1 r 2 2 2 2 f f y u x u ′ + ′ ′ = ∂ ∂ + ∂ ∂ 7. Prove that nu y u y x u x = ∂ ∂ + ∂ ∂ Proof : Since u = f (x, y) is a homogeneous function of degree n we have by the definition, ( ) y/x g x u n = …(1) Let us differentiate this w.r.t x and also w.r.t.y ( ) ( ) x x x x x y/ g n y - . y/ g . x u 1 - n 2 n + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ′ = ∂ ∂ ∴ ie., ( ) ( ) x x x x x y / g n y/ g y u 1 - n 2 - n + ′ = ∂ ∂ …(2) Also ( ) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ′ = ∂ ∂ x x x y 1 . y/ g . u n
• 95. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 95 2011 ie., ( ) y/ g . u 1 - n x x y ′ = ∂ ∂ …(3) Now consider y x x ∂ ∂ + ∂ ∂ u y u as a consequence of (2) and (3) ( ) ( ) [ ] ( ) [ ] x x x x x x x y / g y y/ g n y/ g y - 1 - n 1 - n 2 - n ′ + + ′ = ( ) ( ) ( ) x x x x x x y/ g y y/ g n y/ g y 1 - n n 1 n ′ + + ′ − = − ( ) y/ g x . n n x = = n u, by using (1) Thus we have proved Euler’s theorem u n yu u ; u n y u y u y = + = ∂ ∂ + ∂ ∂ x x x x 8. Prove that ( )u 1 - n n y 2 2 2 = ∂ ∂ ∂ + ∂ ∂ y x u x x u x 2 2 Proof : Since u = f (x, y) is a homogeneous function of degree n, we have Euler’s theorem u n y u y u = ∂ ∂ + ∂ ∂ x x …(1) Differentiating (1) partially w.r.t. x and also w.r.t y we get, x x x x x ∂ ∂ = ∂ ∂ ∂ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ u n y u y u 1. u 2 2 2 …(2) Also, y u n y u . 1 y u y y u 2 2 2 ∂ ∂ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ ∂ x x …(3) We shall now multiply (2) by x and (3) by y. and u n y u y u x u 2 2 x x x x x x x 2 2 ∂ ∂ = ∂ ∂ ∂ + ∂ ∂ + ∂ ∂ y u ny y u y y u y y u y 2 2 2 2 ∂ ∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ ∂ x x Adding these using the fact that x ∂ ∂ ∂ y u 2 = x ∂ ∂ ∂ y u 2 we get,
• 96. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 96 2011 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ ∂ + ∂ ∂ y u y u n y u y u y u y y u y 2 u 2 2 2 2 2 2 2 x x x x x x x x ie., ( ) ( ) 1 using by , u n n u n y u y y u y 2 u 2 2 2 2 2 2 2 = + ∂ ∂ + ∂ ∂ ∂ + ∂ ∂ x x x x or ( ) ( )u 1 - n n nu - nu n y u y y u y 2 u 2 2 2 2 2 2 2 = + ∂ ∂ + ∂ ∂ ∂ + ∂ ∂ x x x x Thus ( )u 1 - n n y u y y u y 2 u 2 2 2 2 2 2 2 + ∂ ∂ + ∂ ∂ ∂ + ∂ ∂ x x x x ie., ( )u 1 - n n u y u y 2 u yy 2 y 2 = + + x xx x x 9. If y z z y z y + + + + + = x x x u show that 0 z u z y u y x u x = ∂ ∂ + ∂ ∂ + ∂ ∂ Solution : (Observe that the degree is 0 in every term) y z z y z y u + + + + + = x x x We shall divide both numerator and denominator of every term by x. ( ) { } x x x x x x x x z/ , y/ g / y 1 z 1 / z / y z/ y/ 1 u = + + + + + = ⇒ u is homogeneous of degree 0. ∴ n = 0 We have Euler’s theorem, u n z u z y u y u = ∂ ∂ + ∂ ∂ + ∂ ∂ x x Putting n = 0 we get, 0 z u z y u y u = ∂ ∂ + ∂ ∂ + ∂ ∂ x x 10. If 3 y that show y log 4 4 = ∂ ∂ + ∂ ∂ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + = y u x u x y x x u Solution : we cannot put the given u in the form xn g (y/x) ( ) ( ) ( ) ( )⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ + + = + + = + + = ∴ x x x x x x x x x / y 1 y/ 1 y/ 1 / y 1 y y e 4 3 4 4 4 4 4 u ie., eu = x3 g (y/x) ⇒ eu is homogeneous of degree 3 ∴ n = 3 Now applying Euler’s theorem for the homogeneous function eu
• 97. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 97 2011 We have ( ) ( ) u u u e n y e y e = ∂ ∂ + ∂ ∂ x x ie., u u u 3e y u e y u e = ∂ ∂ + ∂ ∂ x x Dividing by eu we get 3 y u y u = ∂ ∂ + ∂ ∂ x x 11. If ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + = y x y x u 3 3 1 - tan show that (i) x ux + y uy = sin 2 u (ii) x2uxx + 2 x y uxy + y2uyy = sin 4 u – sin 2 u Solution : (i) data by y - y tan u 3 3 1 - ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = x x ( ) ( ) ( ) ( )⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ − + = + = + = ⇒ x / y 1 y/ 1 y/ - 1 / y 1 y - y u tan 3 2 3 3 3 3 3 x x x x x x x x ie., tan u = x2/g (y/x) ⇒ tan u is homogeneous of degree 2. Applying Euler’s theorem for the function tan u we have, ( ) ( ) 2 n ; u tan . n y u tan y u tan = = ∂ ∂ + ∂ ∂ x x ie., u tan 2 y u u sec y u u sec 2 2 = ∂ ∂ + ∂ ∂ x x or u 2 sin u sin u cos 2 u cos u sin u cos 2 u sec u tan 2 y u y u 2 2 = = = = ∂ ∂ + ∂ ∂ x x 2u sin yu u y = + ∴ x x (ii) We have xux + y uy = sin 2 u …(1) Differentiating (1) w.r.t x and also w.r.t y partially we get x x x xx x u . 2u cos 2 yu u . 1 u y = + + …(2) And x uyx + yuyy + 1 . uy = 2 cos 2u . uy …(3) Multiplying (2) by x and (3) by y we get, x x x xx x x x x u u. 2 cos 2 u y u u y 2 = + +
• 98. Engineering Mathematics – I Dr. V. Lokesha 10 MAT11 98 2011 y y yy yx y y x u u. 2 cos 2 u y u u y 2 = + + Adding these by using the fact that uyx = uxy, we get ( ) ( ) y y yy 2 y 2 u y u 2u cos 2 yu y u y u y 2 u + = + + + + x x x xx x x x x By using (1) we have, 2u sin - u 2 sin 2u cos 2 u y u y 2 u yy 2 y 2 = + + x xx x x (since sin 2θ = 2 cos θ sin θ, first term in the R.H.S becomes sin 4u) Thus x2uxx + 2 x y uxy + y2 uyy = sin 4 u – sin 2u 12. If ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = x z , z y , y x f u Prove that 0 z u z y u y x u x = ∂ ∂ + ∂ ∂ + ∂ ∂ >> here we need to convert the given function u into a composite function. Let ( ) x x z r , z y q , y p where r q, p, f u = = = = ie., ( ) ( ) { } z y, x, u z y, x, r q, p, u → ⇒ → → x x x x ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ = ∂ ∂ ∴ r r u q q u p p u u ie., ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ + ∂ ∂ = ∂ ∂ 2 z - . r u 0 . q u y 1 . p u u x x r u z - p u y u ∂ ∂ ∂ ∂ = ∂ ∂ ∴ x x x x …(1) Similarly by symmetry we can write, p u x - q u z u y ∂ ∂ ∂ ∂ = ∂ ∂ y y y …(2) q u y - r u x u z ∂ ∂ ∂ ∂ = ∂ ∂ z z z …(3) Adding (1), (2) and (3) we get 0 z u z y u y u x = ∂ ∂ + ∂ ∂ + ∂ ∂ x