Mechanical Engineering Department Technology University For 1st class students, Exams of Engineering

Download Mechanical Engineering Department Technology University For 1st class students and more Exams Engineering in PDF only on Docsity! Lectures in MATHEMATICS For 1st class students Mechanical Engineering Department Technology University Prepared by Assistant Professor Dr. Nabel George Nacy and Lecture Dr. Laith Jaafer Habeeb References: 1) Thomas & Finney " Calculus and Analytic Geometry " (1988) , 7th edition , Addison Wesley. 2) Ford , S.R. and Ford , J.R. " Calculus " , (1963) McGraw-Hill. 3) J.K.Back house and S.P.T. Houldsworth " Pure Mathematics a First Course " (1979) , S1 Edition , Longman Group . ٢ m  y  y2  y1 where x  0 x x2  x1 - A line that goes uphill as x increases has a positive slope . A line that goes downhill as x increases has a negative slope . - A horizontal line has slope zero because ∆y = 0 . - The slope of a vertical line is undefined because ∆x = 0 . - Parallel lines have same slope . - If neither of two perpendicular lines L1 and L2 is vertical , their slopes m1 and m2 are related by the equation : m1 . m2 = -1 . Angles of Inclination: The angle of inclination of a line that crosses the x- axis is the smallest angle we get when we measure counter clock from the x-axis around the point of intersection . The slope of a line is the tangent of the line angle of inclination . m = tan Ф where Ф is the angle of inclination . - The angle of inclination of a horizontal line is taken to be 0 o . - Parallel lines have equal angle of inclination . x-axis L P2(x2 ,y2) y-axis ∆y P1(x1,y1) Ф ∆x Q(x ,y 2 1) ٢ EX-1- Find the slope of the line determined by two points A(2,1) and B(-1,3) and find the common slope of the line perpendicular to AB. Sol.- Slope of AB is: m  y2  y1    3  1   2 AB 2  x1  1  2 3 Slope of line perpendicular to AB is :  1  3 m AB 2 EX-2- Use slopes to determine in each case whether the points are collinear (lie on a common straight line ) : a) A(1,0) , B(0,1) , C(2,1) . b) A(-3,-2) , B(-2,0) , C(-1,2) , D(1,6) . x ٣ Sol. – a) m AB  1  0  1 0  1 an d mBC  1  1  0  m 2  0 AB The points A , B and C are not lie on a common straight line . b) m AB  0  ( 2 )  2  2  ( 3 ) , mBC  2  0  2  1  ( 2 ) , mCD  6  2  2 1  ( 1 ) Since mAB = mBC = mCD Hence the points A , B , C , and D are collinear . 1-3- Equations for lines : An equation for a line is an equation that is satisfied by the coordinates of the points that lies on the line and is not satisfied by the coordinates of the points that lie elsewhere . Vertical lines : Every vertical line L has to cross the x-axis at some point (a,0). The other points on L lie directly above or below (a,0) . This mean that : x  a ( x, y ) Nonvertical lines : That point – slope equation of the line through the point ( x1 , y1 ) with slope m is : y – y1 = m ( x – x1 ) Horizontal lines : The standard equation for the horizontal line through the point ( a , b ) is : y = b . The distance from a point to a line : To calculate the distance d between the point P(x1 , y1 ) and Q( x2 , y2 ) is : d  We use this formula when the coordinate axes are scaled in a common unit . To find the distance from the point P( x1 , y1 ) to the line L , we follow : 1. Find an equation for the line L' through P perpendicular to L : y – y1 = m' ( x – x1 ) where m' = -1 / m ( x  x )2  ( y  y )2 2 1 2 1 ٤ EX-7- Find the line through the point P(1, 4) with the angle of inclination Ф=60 o . Sol.- m  tan   tan60  3 y  4  3( x  1 )  y  3 x  4   EX-8- The pressure P experienced by a diver under water is related to the diver's depth d by an equation of the form P = k d + 1 where k a constant . When d = 0 meters , the pressure is 1 atmosphere . The pressure at 100 meters is about 10.94 atmosphere . Find the pressure at 50 meters. Sol.- At P = 10.94 and d = 100 → 10.94 = k(100)+1 → k = 0.0994 P = 0.0994 d + 1 , at d = 50 → P = 0.0994 * 50 + 1 = 5.97 atmo. 3 ٥ 1-4- Functions : Function is any rule that assigns to each element in one set some element from another set : y = f( x ) The inputs make up the domain of the function , and the outputs make up the function's range. The variable x is called independent variable of the function , and the variable y whose value depends on x is called the dependent variable of the function . We must keep two restrictions in mind when we define functions : 1. We never divide by zero . 2. We will deal with real – valued functions only. Intervals : - The open interval is the set of all real numbers that be strictly between two fixed numbers a and b : ( a,b)  a  x  b - The closed interval is the set of all real numbers that contain both endpoints : [ a,b ]  a  x  b - Half open interval is the set of all real numbers that contain one endpoint but not both : [ a,b )  a  x  b ( a,b]  a  x  b Composition of functions : suppose that the outputs of a function f can be used as inputs of a function g . We can then hook f and g together to form a new function whose inputs are the inputs of f and whose outputs are the numbers : ( go f )( x )  g( f ( x )) EX-9- Find the domain and range of each function : a ) y  c ) y  , b ) , d ) y  1 x  2 y  2  x  4 9  x 2 x ٥ Sol. - a ) x  4  0  x  4  Dx : x  4 , Ry : y  0 b ) x  2  0  x  2  Dx : x  2 y  1  x  2 x  1  2  y Ry : y  0 c ) 9  x 2  0   3  x  3  Dx : 3  x  3 y  9  x 2  x  ∓ sin ce 9  y 2  0   3  y  3 sin ce y  0  Ry : 0  y  3 9  y 2 ٧  1 1 Note that – A function F( t ) has a limit at point C if and only if the right hand and the left hand limits at C exist and equal . In symbols : lim F( t )  L  t C lim F( t )  L t C an d lim F( t )  L t C   The limit combinations theorems : 1 ) limF1 ( t ) ∓ F2 ( t )  lim F1 ( t ) ∓ lim F2 ( t ) 2 ) limF ( t )* F ( t )  lim F ( t )* lim F ( t ) 2 2 3 ) lim F1 ( t )   lim F1 ( t ) wher e lim F ( t )  0 F2 ( t )  lim F   2 ( t ) 2 k 4 ) lim k * F1 ( t ) k * lim F1 ( t ) 5 ) lim Sin  1  0  provided that   is measure d in radius The limits ( in 1 – 4 ) are all to be taken as t→C and F1( t ) and F2( t ) are to be real functions . Thm. -1 : The sandwich theorem : Suppose that f ( t )  g( t )  h( t ) for all t  C in some interval about C and that f( t ) and h( t ) approaches the same limit L as t→C , then : Infinity as a limit : lim g( t )  L t C 1. The limit of the function f( x ) as x approaches infinity is the number L: lim x f ( x )  L . If , given any ε > 0 there exists a number M such that ٧ for all x : M  x  f ( x )  L   . 2. The limit of f( x ) as x approaches negative infinity is the number L : lim x f ( x )  L . If , given any ε > 0 there exists a number N such that for all x : x  N  f ( x )  L   . The following facts are some times abbreviated by saying : a) As x approaches 0 from the right , 1/x tends to ∞ . b) As x approaches 0 from the left , 1/x tends to -∞ . c) As x tends to ∞ , 1/x approaches 0 . d) As x tends to -∞ , 1/x approaches 0 . Continuity : Continuity at an interior point : A function y = f( x ) is continuous at an interior point C of its domain if : lim xC f ( x )  f ( C ) . Continuity at an endpoint : A function y = f( x ) is continuous at a left endpoint a of its domain if : lim xa  f ( x )  f ( a ) . A function y = f( x ) is continuous at a right endpoint b of its domain if: lim t b f ( x )  f ( b ) . ٨  x Continuous function : A function is continuous if it is continuous at each point of its domain . Discontinuity at a point : If a function f is not continuous at a point C , we say that f is discontinuous at C , and call C a point of discontinuity of f . The continuity test : The function y = f ( x ) is continuous at x = C if and only if all three of the following statements are true : 1) f ( C ) exist ( C is in the domain of f ) . 2) lim xC f ( x ) exists ( f has a limit as x→C ) . 3) lim xC f ( x )  f ( C ) ( the limit equals the function value ) . Thm.-2 : The limit combination theorem for continuous function : If the function f and g are continuous at x = C , then all of the following combinations are continuous at x = C : 1 ) f ∓ g 2 ) f .g 3 ) k .g k 4 ) go f , fo g 5 ) f / g provided g( C )  0 Thm.-3 : A function is continuous at every point at which it has a derivative . That is , if y = f ( x ) has a derivative f ' ( C ) at x = C , then f is continuous at x = C . EX-12 – Find : 5 x 3  8 x 2 x 3  a 3 1 ) lim x0 3 x 4  16 x 2 , 2 ) lim xa x 4  a 4 3 ) lim Sin5 x , 4 ) lim tan 2 y x0 Sin3 x y0 3 y 5 ) lim Sin2 x , 6 )  1  lim 1 Cos x0 2 x 2  x   x  ٩      lim   Cos0  1 12 ) x0  x  lim Si x0  x  0 x   1 EX-13- Test continuity for the following function :  x 2  1  1  x  0  2 x 0  x  1   f ( x )   1 x  1    2 x  4 1  x  2   0 2  x  3  Sol.- We test the continuity at midpoints x = 0 , 1 , 2 and endpoints x = -1 , 3 . At x  0  f ( 0 )  2* 0  0 lim x0 f ( x )  lim( x 2  1 )  1 x0 lim x0 f ( x )  lim 2 x  0  x0 lim x0 f ( x ) Since lim f ( x ) x0 doesn' t exist Hence th e function discontinuous at x  0             n Cos(tan x )  Sin Cos(tan0 )  Sin Cos0   Sin  2   2   2  2 ٠١ At x  1  f ( 1 )  1 lim x1 f ( x )  lim 2 x  2 x1 lim x1 f ( x )  lim( 2 x  4 )  2  x 1 lim x1 f ( x )  lim x  1 f ( x ) Since lim x 1 f ( x )  f ( 1 ) Hence th e function is discontinuous at x  1 At x  2  f ( 2 )  2* 2  4  0 lim x2 f ( x )  lim( 2 x  4 )  0 x2 lim x2 f ( x )  lim 0  0  x2 lim x2 f ( x )  lim x  2 f ( x ) Since lim f ( x )  x2 f ( 2 )  0 Hence th e function is continuous at x  2 At x  1  f ( 1 )  ( 1 )2  1  0 lim x1 f ( x )  lim ( x 2  1 )  0  x1 f ( 1 ) Hence th e function is continuous at x  1 At x  3  f ( 3 )  0 lim x3 f ( x )  lim 0  0  x3 f ( 3 ) Hence th e function is continuous at x  3         ٠١  EX-14- What value should be assigned to a to make the function : f ( x )    x 2  1 x  3  continuous at x = 3 ? Sol. –  2ax x  3 lim f ( x )  lim f ( x )  lim( x 2  1 )  lim 2ax  8  6a  a  4 x  3 x  3 x  3 x 3 3 ٢١ 11. The coordinates of the vertices A, B and C of the triangle ABC are (-3,7) , (2,19) and (10,7) respectively : a) Prove that the triangle is isosceles. b) Calculate the length of the perpendicular from B to AC , and use it to find the area of the triangle . (ans.:12,78) 12. Find the equations of the lines which pass through the point of intersection of the lines x - 3y = 4 and 3x + y = 2 and are respectively parallel and perpendicular to the line 3x + 4y = 0 . (ans.:4y+3x+1=0;3y-4x+7=0) 13. Through the point A(1,5) is drawn a line parallel to the x-axis to meet at B the line PQ whose equation is 3y = 2x - 5 . Find the length of AB and the sine of the angle between PQ and AB ; hence show that the length of the perpendicular from A to PQ is 18/√13 . Calculate the area of the triangle formed by PQ and the axes . (ans.:9,2/√13,25/12) x 2  2 14. Let y  x 2  1 , express x in terms of y and find the values of y for which x is real . (ans.: x  ∓ 15. Find the domain and range of each function : ; y  2 or y  1 ) a ) y  1 1  x 2 , b ) y  1 1  x , c ) y  1 3  x ( ans.: a )x,0  y  1; b )x  0 , y  0 ; c )x  3, y  0 ) 16. Find the points of intersection of x2 = 4y and y = 4x . (ans.:(0,0),(16,64)) y  2 y  1 ٢١ 17. Find the coordinates of the points at which the curves cut the axes : a ) y  x 3  9 x 2 , b ) y  ( x 2  1 )( x 2  9 ) , c ) y  ( x  1 )( x  5 )2 (ans.:a)(0,0);(0,0),(9,0);b)(0,9);(1,0),(-1,0),(3,0),(-3,0);c)(0,25);(-1,0),(5,0)) 18. Let f(x) = ax + b and g(x) = cx + d . What condition must be satisfied by the constants a , b , c and d to make f(g(x)) and g(f(x)) identical ? (ans.:ad+b=bc+d) 19. A particle moves in the plane from (-2,5) to the y-axis in such away that ∆y = 3*∆x . Find its new coordinates . (ans.:(0,11),(0,-1)) 20. If f(x) = 1/x and g(x)=1/√x , what are the domain of f , g , f+g , f-g , f.g , f/g , g/f , fog and gof ? What is the domain of h(x) = g(x+4) ? ( ans.: x  0 ,x  0 ,x  0 ,x  0 ,x  0 ,x  0 ,x  0 ,x  0 ,x  0 ;x  4 ) ٣١    0   21. Discuss the continuity of :  x  1 x for x  0     x 3 f ( x )    for 0  x  1   1  for 1  x  2   1 for  for x  2  x  2   (ans.: discontinuous at x=0,2 ; continuous at x=1) 22. Evaluate the following limits : a ) lim x  Sinx b ) lim 1  Sinx x  c ) lim 2 x  5 x x   d ) lim x x.Sinx x 0 tan 3 x x  ( x  Sinx )2 e ) lim 1  x f ) lim  2 x x 1 1  x x1 x 2  x g ) lim( n  n ) (ans.:a)1/2, b)0, c)1/3, d)0, e)1/2, f)-1/2√2, g)0) 23. Suppose that : f(x) = x3 – 3x2 -4x +12 and   f ( x ) h( x )   x  3 for x  3    . Find : a) all zeros of f .  k for x  3 x  1 n 2  1 ١ Logarithm function : If a is any positive number other than 1 , then the logarithm of x to the base a denoted by : y = logax where x > 0 At a = e = 2.7182828… , we get the natural logarithm and denoted by : y = ln x Let x , y > 0 then the properties of logarithm functions are : 1. y = ax ↔ x = logay and y = ex ↔ x = ln y . 2. logex = ln x . 3. logax = ln x / ln a . ٢ 2 4. ln (x.y) = ln x + ln y . 5. ln ( x / y ) = ln x – ln y . 6. ln xn = n. ln x . 7. ln e = logaa = 1 and ln 1 = loga1 = 0 . 8. ax = ex. ln a . 9. eln x = x . The graph of the function y = ln x is : y 1.5 1 0.5 0 X -0.5 -1 -1.5 -2 0 1 2 e 3 4 5 Application of exponential and logarithm functions : We take Newton's law of cooling : T – TS = ( T0 – TS ) et k where T is the temperature of the object at time t . TS is the surrounding temperature . T0 is the initial temperature of the object . k is a constant . EX-1- The temperature of an ingot of metal is 80 oC and the room temperature is 20 oC . After twenty minutes, it was 70 oC . a) What is the temperature will the metal be after 30 minutes? b) What is the temperature will the metal be after two hours? c) When will the metal be 30 oC? ٢ Sol. : T  TS  ( T0  TS )e tk  50  60 e 20 k  k  ln 5  ln 6 20  0.0091 a ) T  20  60e 30( 0.0091 )  60 * 0.761  45.6 oC  T  65.6 oC b ) T  TS  60e 120( 0.0091 )  60 * 0.335  20.1 oC  T  40.1 oC c ) 10  60e 0.0091 t  0.0091 t   ln 6  t  3.3 hrs . ٤ 11 ) Sin  Sin Sin  Sin  2 Sin     2  2Cos     2 .Cos     2 .Sin     2 12 ) Cos Cos  Cos  Cos  2Cos     2  2 Sin     2 .Cos     2 .Sin     2 θ 0 Π / 6 Π / 4 Π / 3 Π / 2 Π Sinθ 0 1/2 1/√2 √3/2 1 0 Cosθ 1 √3/2 1/√2 1/2 0 -1 tanθ 0 1/√2 1 √3 ∞ 0 Graphs of the trigonometric functions are : 1.5 1 -2Л -Л 0.5 0 -0.5 Л 2Л ٤ -1 -1.5 y  Sinx Dx : x Ry : 1     y  1 ٥ 1.5 1 -2Л -Л 0.5 0 -0.5 Л 2Л -1 -1.5 y  Cosx Dx : x Ry : 1     y  1 ٦ y  Secx D : x  2n  1  x 2 Ry : y  1 or y  1 1 -2π -π 0 π 2π -1 ٧ y  csc x Dx : x  n Ry : y  1 or y  1 Where n  0 ,∓1,∓2 ,∓3,...... EX-2 - Solve the following equations , for values of θ from 0o to 360o inclusive . a) tan θ = 2 Sin θ b) 1 + Cos θ = 2 Sin2 θ Sol.- a ) tan  2 Sin  Sin  Cos   2 Sin  either or  Sin ( 1  2Cos )  0 Sin  0    0 o ,180 o ,360 o Cos  1    60o ,300o 2 1 -2π -π 0 π 2π -1 ٧ Therefore the required values of θ are 0o,60o,180o,300o,360o . b ) 1  Cos  2.Sin 2  1  Cos  2( 1  Cos 2 )  ( 2Cos 1 )( Cos  1 )  0 either Cos  1   2  60 o ,300 o or Cos  1    180 o There the roots of the equation between 0o and 360o are 60o,180o and 300o . ٩ i ) x  a.Sin  Sin  x  Csc  a a x y  b tan  tan  y  Cot  b b y Since Csc 2  Cot 2  1  a x 2  b 2  y 2 ii ) x  2 Sec   Cos   2 x 2  4 y  Cos 2  x y  Cos 2   Sin 2  2 y  4  x 2  4   x 2 y  8  x 2 x 2 x 2 EX-7- If tan2θ – 2 tan2β = 1 , show that 2 Cos2θ – Cos2β = 0 . Sol. – tan 2  2 tan 2   1  Sec 2  1  2( Sec 2   1 )  1  Sec 2  2 Sec 2   0  1 Cos 2  2  0 Cos 2   2Cos 2  Cos 2   0 Q.E.D. EX-8- If a Sinθ = p – b Cosθ and b Sinθ = q + a Cosθ .Show that : a2 +b2 = p2 +q2 Sol.- p  a.Sin  b.Cos an d q  b.Sin  a .Cos p 2  q 2  ( aSin  bCos )2  ( bSin  aCos )2  a 2 ( Sin 2 Cos 2 )  b 2 ( Cos 2  Sin 2 )  a 2  b 2 x 2 1 ٩ 5 4 A -3 13 5 B 12 EX-9- If Sin A = 4 / 5 and Cos B = 12 / 13 ,where A is obtuse and B is acute . Find , without tables , the values of : a) Sin ( A – B ) , b) tan ( A – B ) , c) tan ( A + B ) . Sol. - ٠١ a ) Sin( A  B )  SinA .CosB  CosA.SinB  4 . 12 3 . 5  63 5 13 5 13 65 b ) tan( A  B )  tan A  tan B 1  tan A.tan B  4  5  3 12 1  4 . 5   63 16 3 12 c ) tan( A  B )  tan A  tan B 1  tan A.tan B  4  5  3 12  33 1  4 . 5 56 3 12 EX-10 – Prove the following identities: a ) Sin( A  B )  Sin( A  B )  2.SinA .CosB b ) tan A  tan B  Sin( A  B ) CosA .CosB c ) Sec( A  B )  SecA.SecB .CscA .CscB CscA .CscB  SecA.SecB d ) Sin 2  Cos 2  1  Cot  ١١  3 4 24 ∓ Sin2 2.Sin .Cos 2. .( ∓ )  ∓ 5 5 25 Cos2 Cos 2  Sin 2  ( 4 )2 5  ( 3 )2  7 5 25 ٢١ ∓ ∓ ٣١ θ 5 θ ٢١ ٣١ -5 -١ 1 θ θ -√3 2 2 -√3 b) Sin2 2.Sin .Cos  2( ∓ 5 ).( 12 )  ∓ 120 13 13 169 Cos2 Cos 2  Sin 2  ( 12 )2 13  ( 5 )2 13  119 169 c) Sin2 2 Sin .Cos  2(  3 ).( ∓ 1 )   3 Cos2   Cos 2   Sin 2 2  ( 1 )2 2 2  (  2 3 )2 2   1 2 EX-12- Solve the following equations for values of θ from 0o to 360o inclusive: ٢١ a) Cos 2θ + Cos θ + 1 = 0 , b) 4 tan θ . tan 2θ = 1 Sol.- ٤١ π 0 -π y  Cos 1 x Dx : 1  x  1 Ry : 0  y  180 y  tan 1 x Dx : x Ry : 90   y  90 -1  1 - ٥١ y  Cot 1 x Dx : x Ry : 0   y       y  Sec 1 x  -1 1 - 2π π 0 -π ٥١ Dx :  x  1 R : 0  y   , y     y 2 ٧١ 4  3 3 3 EX-13- Given that   Sin 1 3 , find : 2 Csc ,Cos , Sec , tan , and ,Cot  Sol.- ١   Sin 1 3  Sin  2 3  x  r   1 2 y Csc  2 ,Cos  1 2 , Sec  2 , tan   ,Cot  1 EX-14 – Evaluate the following expressions : a ) Sec( Cos 1 1 ) 2 b ) Sin 1 1  Sin 1 ( 1 ) c ) Cos 1 (  Sin  ) 6 Sol.- a ) Sec( Cos 1 1 )  Sec   2 2 3 b ) Sin 1 1  Sin 1 ( 1 )    2  (    2 )   c ) Cos 1 (  Sin  )  Cos 1 (  1 )  2  6 2 3 3 ٢ 3 ٧١ EX-15- Prove that : Sol. a ) Sec 1 x  Cos 1 1 x b ) Sin 1 (  x )   Sin 1 x a ) Let y  Sec 1 x  x  Secy  x  1 Cosy  y  Cos 1 1  Sec 1 x  Cos 1 1 x x b ) Let  y   Sin 1 x  x  Sin(  y )  x   Siny y  Sin 1 (  x )  Sin 1 (  x )   Sin 1 x ٨١ 2-4- Hyperbolic functions : Hyperbolic functions are used to describe the motions of waves in elastic solids ; the shapes of electric power lines ; temperature distributions in metal fins that cool pipes …etc. The hyperbolic sine (Sinh) and hyperbolic cosine (Cosh) are defined by the following equations : 1. Sinhu  1 ( eu  eu ) 2 an d Coshu  1 ( eu  eu ) 2 Sinhu eu  e u Coshu eu  eu 2. tanhu  Coshu  eu  e u an d Cothu  Sinhu  eu  eu 3. Sechu  1   Coshu 2 eu  e u an d Cschu  1   Sinhu 2 eu  e u 4. Cosh2u  Sinh2u  1 5. tanh2 u  Sech2u  1 an d Coth2u  Csch2u  1 6. Coshu  Sinhu  e u 7. Cosh( u )  Coshu and and Coshu  Sinhu  e u Sinh( u )   Sinhu 8. Cosh0  1 an d Sinh0  0 9. 10. Sinh( x  Cosh( x  y )  Sinhx .Coshy  Coshx.Sinhy y )  Coshx .Coshy  Sinhx.Sinhy 11. Sinh2 x  2.Sinhx .Coshx 12. Cosh2 x  Cosh 2 x  Sinh 2 x 13. Cosh 2 x  Cosh2 x  1 2 an d Sinh 2 x  Cosh2 x  1 2 y=Sinhx ٩١ EX-16- Let tanh u = - 7 / 25 , determine the values of the remaining five hyperbolic functions . Sol.- ٠٢ Cothu  1 tanh u   25 7 tanh 2 u  Sech2 u  1  49 62 5  Sech2 u  1  Sechu  24  25 Coshu   tanh u  1  25 Sechu 24 Sinhu Coshu   7  25 Sinhu  Sinhu   7 25 24 24 Cschu  1 Sinhu   24 7 EX-17- Rewrite the following expressions in terms of exponentials . Write the final result as simply as you can : Sol.- a ) 2Cosh(ln x ) c ) Cosh5 x  Sinh5 x b ) tanh(ln x ) d ) ( Sinhx  Coshx )4 a ) 2Cosh(ln x )  2. eln x  e  ln x 2  x  1 x e ln x  e ln x x  1 x x 2  1 b ) tanh(ln x )  e ln x  e ln x  1   x  x x 2  1 c ) Cosh5 x  Sinh5 x  e5 x  e 5 x 2  e x  e  x  e5 x  e 5 x 2 e x  e x  4  e5 x d ) ( Sinhx  Coshx )4     2 ٠٢   e 4 x 2    EX-18- Solve the equation for x : Cosh x = Sinh x + 1 / 2 . Sol. - Coshx  Sinhx  1  e  x  1   x  ln 1  ln 2  x  ln 2 2 2 EX-19 – Verify the following identity : a) Sinh(u+v)=Sinh u. Cosh v + Cosh u.Sinh v b) then verify Sinh(u-v)=Sinh u. Cosh v - Cosh u.Sinh v Sol.- ٢٢ ١ y  Sinh1 x Dx : x y  Cosh1 x Dx : x  1 Ry : y Ry : y  0 y  tanh1 x Dx : 1  x  1 y  Coth1x Dx : x  1 or x  1 Ry : y Ry : y  0 -1 ١ -١ ١ 1 0 ٢٢ y  Sech 1 x Dx : 0  x  1 y  Csch1 x Dx : x  0 Ry : y  0 Ry : y  0 ٣٢ x2  1 x 2  1 1  x x x Some useful identities : 1. Sinh1 x  ln( x  2. Cosh1 x  ln( x  x 2  1 ) x 2  1 ) 3. tanh1 x  1  1  x   2 .ln 1  x    4. Coth1 x  1  x  1   tanh1 1 2 .ln x  1  x   1  1  1  x 2  1 1 5. Sech x  ln   Cosh     6. Csch1  x  ln  x 2  1   Sinh 1 1      EX-21 - Derive the formula : Sinh1 x  ln( x  ) Sol.- Let y  Sinh  1 x  x  Sinhy   e y  e  y 2  x   e 2 y  1 2e y  e 2 y  2 x .e y  1  0 2 x ∓ e y  2  e y  x ∓ either or y  ln( x  y  ln( x  x 2  1 ) x 2  1 ) 4 x 2  4 x 2  1 x x ٤٢ ( ans. : a ) ∓ 3 ,∓ 7 ;b ) ∓ 4 ,∓ 3 ;c ) ∓ 5 ,∓ 12 ) 4 4 5 5 13 13 7. If Sin A = 3/5 and Sin B = 5/13 , where A and B are acute angles , find without using tables , the values of : a) Sin(A+B) , b) Cos(A+B) , c) Cot(A+B) (ans.: 56/65; 33/65; 33/56) 8. If tan A = -1/7 and tan B = 3/4 , where A is obtuse and B is acute , find without using tables the value of A – B . (ans.: 135 ) 9. Prove the following identities : ٥٢ i ) Sec 2  Csc 2  Sec 2 .Csc 2 ii ) Sin2 ( 1  Sec 2 )  Sec 2  Cos 2 iii ) 1  Sin 1  Sin  ( Sec  tan )2 iv ) Sec  Sin  tan2   Cos 2  Sec  Sin v ) vi ) Cos( A  B )  Cos( A  B )  tan B Sin( A  B )  Sin( A  B ) CosB  CosA.Cos( A  B )  SinA.Sin( A  B ) vii ) tan( A  B  C )  tan A  tan B  tanC  tan A.tan B.tanC 1  tan B.tanC  tanC .tan A  tan A.tan B If A,B,C ar e angles of a triangle,show that : tan A  tan B  tanC  tan A.tan B.tanC 1          tan x.Sin2h viii ) tan( x 2 h ) tan( x h ) tan x Cos 2 x  Sin2h ix ) tan x  1  Cos2 x 1  Cos2 x x ) Sin4 A  Sin2 A Cos4 A  Cos2 A  1  tan 2 A xi ) Sin4  Cos 4  1 ( Cos4  3 ) 4 xii ) 4 Sin3 A.Cos3 A  4Cos 3 A.Sin3 A  3Sin4 A 3 tan A  tan3 A xiii ) tan 3 A    1  3 tan2 A xiv ) Cos 1 (  x )    Cos 1 x xv ) ٥٢ Cot 1 x    2  tan1 x xvi ) Cosh( u  v )  Coshu.Coshv  Sinhu.Sinhv and the n verify xvii ) xviii ) xix ) xx ) Cosh( u  v )  Coshu.Coshv  Sinhu.Sinhv Coshu.Sinhv  1 Sinh( u  v )  Sinh( u  v ) 2 Sinhu.Sinhv  1 Cosh( u  v )  Cosh( u  v ) 2 Cosh3u  Coshu  4 Sinh2u.Coshu  4Cosh3u  3Coshu ( Coshx  Sinhx )n  Coshnx  Sinhnx 10. If u  1  Sin , prove that 1  1  Sin   and deduce formula for Sinθ , Cos u Cos Cosθ , tanθ in terms of u. (ans.:(u2-1)/(u2+1); 2u/(u2+1);(u2-1)/(u2+1)) ٧٢ a ) Sin( Cos 1 1 ) 2 b ) Csc( Sec 1 2 ) c ) Cot( Cos 1 0 ) d ) Sin1 1  Sin1 ( 1 ) e ) Cos( Sin1 0.8 ) f ) Cos1 (  Sin  ) 6 ( ans.: 1 / 2 ;2 / 3 ;0 ; ;0.6 ;2 / 3 ) 20. Find the angle α in the below graph ( Hint : α+β = 65o ) : (ans.: 42.2) 21. Let Sech u = 3/5 , determine the values of the remaining five hyperbolic functions . ( ans. : Coshu  5 / 3; Sinhu  ∓4 / 3; tanhu  ∓4 / 5 ;Cothu  ∓5 / 4 ;Cschu  ∓3 / 4 ) 22. Rewrite the following expressions in terms of exponentials , write the final result as simply as you can : a ) Sinh( 2.ln x ) b ) 1 65o β ١٢ ٠٥ α ٧٢ Cosh2u Coshx  Sinhx c ) Cosh3 x  Sinh3 x d ) ln( Coshx  Sinhx )  ln( Coshx  Sinhx ) (ans.:(x4-1)/(2x2); ex ; e--3x ; 0 ) 23. Solve the equation for x ; tanh x = 3/5 . (ans.: ln 2 ) 24. Show that the distance r from the origin O to the point P(Coshu,Sinhu) on the hyperbola x2 – y2 = 1 is r  . 25. If θ lies in the interval        2 2 and Sinh x = tan θ . Show that : Cosh x = Sec θ , tanh x =Sin θ , Coth x = Csc θ , Csch x = Cot θ , and Sech x = Cos θ . 26. Derive the formula : tanh1 x  1 ln 1  x ; x  1 2 1  x 27. Find : limCosh1 x  ln x . (ans.: ln 2 ) x 1 2 x  3 ( 2 x  3 )3    Chapter three Derivatives Let y = f ( x ) be a function of x . If the limit : dy    f ' ( x )  lim f ( x  x )  f ( x )   lim y dx x0 x xo x exists and is finite , we call this limit the derivative of f at x and say that f is differentiable at x . EX-1 – Find the derivative of the function : f ( x )  Sol.: 1   f' ( x )  lim f ( x  x )  f ( x )  lim 2( x  x )  3 x0 x x0 x  lim 2 x  3  2( x  x )  3 . 2 x  3  2( x  x )  3 x0 x. 2( x  x )  3 2 x  3 2 x  3  2( x  x )  3  lim ( 2 x  3 )  ( 2( x  x )  3 ) x0 x.  2( x  x )  3  2 2 x  3(    2( x  x )  3 ) 1 ( 2 x  3 )( 2 x  3  2 x  3 ) 1 2 x  3 1 2 x  3 4  dy dx   12 x 2  12 x 4  12 x 5 e ) y ( x  1 )( x 2  x  1 )   x 3 dy  x ( x  x  1 )  ( x  1 )( 2 x  1 ) 3 x ( x  1 )( x  x  1 )   3 3 2 2 2 dx x6 x 4 f ) dy  dx 2 x( x 2  x  2 )  ( x 2  1 )( 2 x  1 ) ( x 2  x  2 )2  x 2  2 x  1 ( x 2  x  2 )2 5 The Chain Rule: 1. Suppose that h = go f is the composite of the differentiable functions y = g( t ) and x = f( t ) , then h is a differentiable functions of x whose derivative at each value of x is : dy  dy  dx dx dt dt 2. If y is a differentiable function of t and t is differentiable function of x , then y is a differentiable function of x : y  g( t ) an d t  f ( x )  dy  dy * dt dx dt dx EX-3 – Use the chain rule to express dy / dx in terms of x and y : a ) y   b ) y  t 2 t 2  1 1 t 2  1  t  1 2 and t   and x  1 c ) y    t  1 and x  t 2  1 at t  2 Sol.-   2 x  1 4t  1 6 1 2x  1 d ) y  1  1 t and t  1 1  x at x  2 a) y  t 2  dy   t 2  1 dt 2t( t 2  1) 2t.t 2 ( t 2  1)2  2t ( t 2  1)2 1 dt 1  1 1 t  ( 2x  1)2   dx .( 2x  1) 2 .2  2 2x  1 dy  dy . dt   2t .  2 2x  1 .  1 dx dt dx ( t 2  1)2 ((2x  1)  1)2 2( x  1)2 1 2x  1 9    f lim about its derivate , if it has one . The procedure is formally identical with that used before , that is : d 2 y     d  dy  d     f ( x  x )  f ( x ) dx 2 dx  dx  dx x 0 x if the limit exists . This derivative is called the second derivative of y with respect to x . It is written in a number of ways , for example, d 2 f ( x ) y'' , f ''(x) , or dx 2 . In the same manner we may define third and higher derivatives , using similar notations . The nth derivative may be written : y( n ) , f ( n ) d n y ( x ) , dxn . EX-4- Find all derivatives of the following function : y = 3x3 - 4x2 + 7x + 10 Sol.- dy dx  9 x 2  8 x  7 d 2 y , dx 2  18 x  8 d 3 y dx 3  18 d 4 y , dx 4  0  d 5 y dx 5  .... ( x )  10 2 Ex-5 – Find the third derivative of the following function : y  1  x Sol.- dy   1  3 x 1 dx x 2 2 d 2 y  2  3  1 x 2 dx 2 x 3 4 d 3 y   6  3  3 x 2  d 3 y   6  3 dx 3 x 4 8 dx 3 x 4 8 x3 x 3 11 Implicit Differentiation: If the formula for f is an algebraic combination of powers of x and y . To calculate the derivatives of these implicitly defined functions , we simply differentiate both sides of the defining equation with respect to x . dy EX-6- Find dx for the following functions: a ) x 2 .y 2  x 2  y 2 b ) (x  y)3  ( x  y )3  x 4  y 4 x  y c )  2 at P(3,1) d) xy  2x - 5y  2 at P(3,2) x  2 y Sol. 2 dy 2 dy d y x  xy2 a ) x ( 2 y dx )  y ( 2 x )  2 x  2 y dx  dx  x 2 y  y b ) 3(x  y)2 ( 1  dy dx )  3( x  y )2 ( 1  dy dx )  4 x 3  4 y 3 dy dx  dy  4 x 3  3( x  y )2  3( x  y )2  dy  2 x 3  3 x 2  3 y 2 dx 3( x y )2  3( x  y )2  4 y 3 dx 6 xy  2 y 3 (x - 2y)(1 - dy )  ( x  y )( 1  2 dy ) c ) dx dx  0  dy    14 d lnu  1 . du dx u.lna dx dx u dx EX-8 – Find dy for the following functions : dx a ) y  log10e b ) y  log5 ( x  1 ) 2 c ) y  log2 ( 3 x  1 ) 2 3 d ) y  ln(x 2  2 )2 3 2 5 e ) y  ln(xy)  1 f) y    (2x 3  4 )3 .( 2 x 2  3 )2 (7 x 3  4 x  3 )2 Sol. – x 15   d y a ) y  log10 e x  y  x log10 e  dx  log10 e  lne ln 10  1 ln 10 b ) y  log5 ( x  1 )2  2 log ( x  1 )  dy dx  2 ( x  1 )ln 5 c ) y  3 log2 ( 3 x 2  1 )  dy  dx 3 3 x 2  1 . 6 x ln 2  18 x ( 3 x 2  1 )ln 2 dy  2 2 2 48 xln( x 2  2 )2 d) dx  3 2 ln( x  2 ) .2 x  x 2  2 x 2  2 e ) y  lnx  lny  1  dy  1  1 . dy  0  dy   y dx x y dx dx x( y  1 ) f ) lny  2 ln( 2 x3  4 )  5 ln( 2 x2  3 )  2 ln(7 x3  4 x  3 ) 3  1 . dy  2 . 2 6 x2 3  5 . 4 x  2. 21x2  4 3 y dx 3 2 x  4 2 2 x  3 7 x  4 x  3  dy   2 x2      5 x     21x2  4   dx 2 y 2 x3  4 2 x2  3 7 x3  4 x  3     Trigonometric functions : If u is any differentiable function of x , then : 2 5 16 9 ) 10 ) d sinu dx d cosu dx  cosu. du dx   sin u. du dx 11 ) d tanu dx  sec 2 u. du dx 12 ) d cotu dx  csc 2 u. du dx 13 ) d secu dx  secu.tanu. du dx 14 ) d cscu dx  cscu.cotu. du dx d y EX-9- Find dx for the following functions :