Mathematics First Year

Transcription

1 Ministry of Higher Education and Scientific Research University of Technology Chemical Engineering Department Mathematics First Year BY Dr. Asawer A.Alwasiti

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3 Mathematics Contents - Chapter one : Revision lines, equation of straight line, functions and graphs, even and odd functions, function in pieces, the absolute function, how to shift graph, sum difference product and quantitative of functions, composition of functions, inverse functions. - Chapter two : Quadratic Functions Circles, parabolas, ellipses, hyperbolas, eccentricity, rotating the coordinate aes, discriminate. - Chapter three: Transcendental Functions Power functions, eponential functions, logarithmic functions, natural logarithmic functions. - Chapter four : Trigonometric Functions Graphs of trigonometric functions, periodicity, even and odd trigonometric functions, Inverse trigonometric functions, hyperbolic trigonometric functions, inverse hyperbolic trigonometric functions. - Chapter Five: Limits and Continuity Properties, limits of trigonometric functions, limits involving infinity, limits of eponential functions, continuity. - Chapter Si: Derivatives Definition, differentiation by rules, second and higher order derivative, application, implicit functions, the chain rule, derivative of trigonometric functions, derivative of hyperbolic functions, derivative of inverse hyperbolic functions, derivative of eponential and logarithmic functions, hospitals rule, partial derivative. - Chapter seven: Integration Indefinite integration, integration of trigonometric functions, integration of inverse trigonometric functions, integration of logarithmic and eponential functions, integration of hyperbolic functions, integration of inverse hyperbolic functions, integration

4 methods; (substitution, by part, power trigonometric functions, trigonometric substitution, by part fraction), definite integration, applications; (area between two curves, length of curves, surface area, volumes). - Chapter eight: Determinants Definition, properties, solution of systems (Cramers rule). - Chapter nine: Polar Coordinates Definition, Cartesian versus polar coordinates, graphing in polar coordinate. - Chapter ten : Comple Number Definition, Graphing in comple number, Polar form of comple number, power of comple number, roots of comple number

5 University of Technology Mathematics Chemical Engineering Department Coordinates and Graphs in the Plain y II (-) negative quarter I (+) positive quarter III (-) negative quarter IV (+) positive quarter Chapter One Revision Dr. Asawer A. Alwasiti 1

6 University of Technology Mathematics Chemical Engineering Department Lines Increment When particle moves from one point ( 1, y 1 ) to (, y ), the increments are: = 1 and y =y y 1 Find the net changes in coordinates when particle moves from (4, -3) to (, 5). = -4 = - and y = 5- (-3) = 8 (, 5) y (4, -3) Chapter One Revision Dr. Asawer A. Alwasiti

7 m 1 m University of Technology Mathematics Chemical Engineering Department Slope of a line The slope of the straight line is the ratio of rise to sum, so when point p 1 ( 1, y 1 ) and p (, y ) are points on a nonverticle line L, the slope of L is m = y = y y1 1 P (, y ) P 1 ( 1, y 1 ) y Q (, y) Parallel and Perpendicular Lines L 1 L L L 1 m m 1 θ θ θ 1 θ 1 If L 1 //L then θ 1 =θ and m 1 =m If L 1 L then m 1 =-1/m Chapter One Revision Dr. Asawer A. Alwasiti 3

8 University of Technology Mathematics Chemical Engineering Department Equations of Straight Lines - Horizontal Lines The standard form of equation of horizontal lines is y = b - Vertical lines The standard form of equation of vertical lines is = a y L (, y) y (b, 0) L (, y) (a, 0) - Neither horizontal nor vertical Point slope equation The general form of point slope equation of the point ( 1, y 1 ) with slope m is : y = m ( 1 ) +y 1 Write an equation for the line thought (-, -1) to (3, 4). m = 1 4 = 3 5 = 1 5 Using (, y ) = (-, -1) y = 1( (-)) +(-1) y = + 1 Chapter One Revision Dr. Asawer A. Alwasiti 4

9 University of Technology Mathematics Chemical Engineering Department Slope intercept equation The general form of slope- intercept equation of line L with slope m and y- intercept b is y = m + b Find the slope and y-intercept of the line 8 + 5y = y = y = y = Slope = -8/5 and intercept is b=4 - General linear equation The general linear equation is A +By = C where both A and B are not zero Find the formula relating Fahrenheit and Celsius temperature, then find the Celsius equivalent of 90ºF and the Fahrenheit equivalent of -5 ºC. F = m ºC + b (linear relationship) The freezing point of water is T F =3 or T C = 0 and boiling point is T F =1 or T C =100. F = m ºC + b 3 = m(0) +b b = 3 and 1 = m (100) +3 m = 9/5 F = 9/5 C +3 or C = 5/9 (F-3) Chapter One Revision Dr. Asawer A. Alwasiti 5

10 University of Technology Mathematics Chemical Engineering Department When F = 90 ºF then C = 5/9 (90 3) 3. º and when C = -5 ºC then F = 9/5 (-5) +3 = 3 º H.W Q1) Find an equation for the line throught the point (p) perpendicular to L and then graph the equation a- p (1, ), L: +y =3 b- p (-, 4), L: = 5 Q) For what value of k are the two lines + ky =3 and + y = 1 if they are (a)parallel (b) perpendicular Q3) The pressure p eperienced by a driver under water is related to the drivers depth d by an equation of the form p = kd + 1 (k is a constant). When d=0 meters, the pressure is 1 atmospheres. The pressure at 100meter is about atmospheres. Find the pressure at 50 meters. Chapter One Revision Dr. Asawer A. Alwasiti 6

11 University of Technology Mathematics Chemical Engineering Department Functions and Graphs Functions A function is like a machine that assigns a unique output to every allowable input. input (domain) f y = f() output (range) y= f() in which : y = dependent variable = independent variable Does the equation represents a function y in terms of? Solve the above equation for y y = 1 - y + = 1 y = + SQRT(1 - ) or y = - SQRT(1 - ) For one value of we have two values of y and this is not a function. Chapter One Revision Dr. Asawer A. Alwasiti 7

12 University of Technology Mathematics Chemical Engineering Department Domain and Range Domain : is the set of real numbers over which may vary and makes the value of y true ( input to the function). Rang : is the set of all values of y that makes the value of true ( output of the function). : Find the domain and range of the following : a) y = b) y = SQRT( 1 - ) c) y = SQRT (4 - ) d) y = 1/ e) y = SQRT ( ) f) g() = SQRT ( - + 9) + 1 / ( - 1) a) y = domain : - <= <= + (-, + ) range : y >= 0 [0, + ) b) y = SQRT (1 ) domain : 1<= <= 1 [1, 1] range : 0<= y <= 1 [0, 1] c) y = SQRT (4 - ) domain : <= 4 ( -, 4] range : y>=0 [0, + ) d) y = 1/ domain : 0 range : y 0 e) y = SQRT ( - ) domain : >= [, + ) Chapter One Revision Dr. Asawer A. Alwasiti 8

13 University of Technology Mathematics Chemical Engineering Department range : y >= 0 [0, + ) g) f) g() = SQRT ( - + 9) + 1 / ( - 1) domain : [-3, 1) U (1, +3] range : Find the domain and range of the area of circle A= (π/4) D Domain D > 0 Range A > 0 H.W Find the domain and range of a) y = ( /4) +1 b) y = 1/SQRT( 4- ) c) f() = - 10 d) h() = e) g() = SQRT( - 1) and h() = + 1 Chapter One Revision Dr. Asawer A. Alwasiti 9

14 University of Technology Mathematics Chemical Engineering Department Even and Odd Functions - A function of y = f() is even function of if y = f(-) = f() - A function of y = f() is odd function of if y = f(-) =-f() a) f () = is even function f(-) = is symmetric about y-ais b) f() = SQRT( -1) is even function f(-) = SQRT ((-) -1) = SQRT ( -1) c) f() = 3 f(-) 3 = - 3 y = y = 3 H.W Are the function bellow even or odd? a) f() = 3 b) f() = +1 c) f() = 1/ Chapter One Revision Dr. Asawer A. Alwasiti 10

15 University of Technology Mathematics Chemical Engineering Department Function in Pieces While some functions are defined by single formula, others are defined by applying different formulas to different parts of their domain. Graph - if = 0 y = f()= if 0<<=1 1 if > 1 y = - y = 1 y = Write a formula for the function y = f() whose graph consists of the two lines (1, 1) (, 1) m 1 m Chapter One Revision Dr. Asawer A. Alwasiti 11

16 University of Technology Mathematics Chemical Engineering Department 1 0 m1 = = y- intercept b = 0 so y = 0<=<=1 1 0 m = = 1 at point (, 1) 1 y = -1 1<=<= f()= if 0<=<=1-1 if 1<=<= T he Absolute Function = SQRT( ) = if >=0 y = - if <0 For eample 3 = 3-3 = 3 0 = 0 P roperties Domain (-, ) Range [0, ) 1) -a = a ) a.b = a. b a a 3) b = b 4) a+b <= a + b Solve the equation f () = -3 =7-3 =7 3 = ± 7 for 3 =7 = 5 Chapter One Revision Dr. Asawer A. Alwasiti 1

17 University of Technology Mathematics Chemical Engineering Department and for 3 = -7 = - the equation -3 has two solutions; = 5 and = - How to Shift Graph 1- Vertical shift If y = f() then y = f() +c shift f() c units up and y = f() c shift f() c units down y = + y = + 1 y = - y = 1 - Horizontal shift If y = f() then y = f( +c ) shift f() c unit to left and y = f( c ) shift f() c unit to right y = ( + 1) y = y = ( 1) Chapter One Revision Dr. Asawer A. Alwasiti 13

18 University of Technology Mathematics Chemical Engineering Department Sum, Difference, Product and Quotients of Function If f() and g() are two functions of with domain D f and D g respectively then: f() + g() = ( f + g )() f() g() = ( f g ) () g() f() = ( g f ) () f(). g() = ( f.g ) () f() / g() = ( f / g )() g() 0 g() / f() = ( g / f)() f() 0 If f() = and g() = 1- find ( f+g ), (f-g), (g-f), (f.g), (f/g) The domain of f() = is >=o The domain of g() = 1- is - <=<=+ (f+g)() = + 1- (f-g)() = - 1- Domain : 0<=<=1 Domain : 0<=<=1 (g-f)() = 1- - Domain : 0<=<=1 (f.g)() =. 1- Domain : 0<=<=1 f ( ) = g 1 Domain : 0<=<=1 g ( ) = 1 Domain : 0<=<=1 f ( the domain is the intersection) Graph f() = For < -1 Here + 1 < 0 and -3 >0 Chapter One Revision Dr. Asawer A. Alwasiti 14

19 University of Technology Mathematics Chemical Engineering Department so f() = = - (+1) (-3 ) y = - For -1<=<=3 Here +1 > =0 and -3 <=0 so f() = = +1 (-3) y = 4 For >3 Here +1 > 0 and -3 >0 so f() = = y = - - <-1 Hence f() = 4 1<=<=3 - >3 y = - y = - y = Chapter One Revision Dr. Asawer A. Alwasiti 15

20 University of Technology Mathematics Chemical Engineering Department Composition of Function f f() g g(f()) If f is a function of and its an input of another function g then we can link f and g together to form a new function denoted as (f o g)() or (g o f)(). f()= and g()=+1 find ( f o g )() ( f o g )() = ( + 1) ( f o g )() = 9 f() = and g() = 7. Find (g o f)() and (f o g)(). (g o f)() = 7 (g o f)() = -3 (f 7) o g)() = ( (f o g)() = 5 Chapter One Revision Dr. Asawer A. Alwasiti 16

21 University of Technology Mathematics Chemical Engineering Department Inverse Function One one function: function means each value of gives one value of y. Many one function: function means each value of gives many values of y. y = f() and its one one function then = f (y) Find the inverse function of f() = + Step 1: let y = f() y = + Step : solve for = y Step 3: interchange and y so the result function inverse is : f () = - for f() = + step 4: check f (f ()) = f (f()) = f() = +, f () = f (f ()) = ( -) +, f (f()) = ( + ) - =, = So the result is right Find the inverse function y =, >=0 epressed as a function of. Step 1: let y = f() Step : solve for y = SQRT(y) = SQRT( ) = = because >=0 Chapter One Revision Dr. Asawer A. Alwasiti 17

22 University of Technology Mathematics Chemical Engineering Department Step 3: interchange and y y = SQRT() a + b Find the inverse function of f ( ) = and check the result. A, b, c and d are c + d constants a + b y = cy + dy = a + b c + d dy b = a cy ( dy b) = a cy = a cy dy b a c a c y = f ( ) = d b d b Checking a + b a c c d f of + ( )( ) = = a + b d b c d + a c a + b d b f f ( ( )) = = a c c + d d b Note: a, b, c and d are not zero. H.W Find the inverse function of 3 1 f ( ) = Chapter One Revision Dr. Asawer A. Alwasiti 18

23 Chapter two Quadratic Functions Circles A circle is a set of points in a plain whose distance from a given fied point in the plain is a constant. The fied point is the center of the circle. The constant distance is the radius of the circle. The distance from (, y) to ( h, k) equals a ( h) + ( y k) = a The standard equation for the circle of radius a centered at the point ( h, k) is: ( h) + (y k) = a The standard equation for the circle of radius a and centered at the origin is : + y = a Find the center and radius of the circle ( -1 ) + ( y + 5) = 3 ( h) + (y k) = a so a = 3 a = SQRT(3) - h = -1 h = 1 - k = 5 k = 5 The center is the point (1, -5), the radius is SQRT(3). If the circle + y =5 is shifted two points or units to the left and three units up. Find the new equation of the circle. + y = a ( (-)) + ( y 3) =5 ( + ) + ( y 3) = 5 The radius is 5 and the center is ( -, 3) Chapter Two Quadratic functions Dr. Asawer A. Alwasiti 19

24 The points that lie inside the circle ( h) + ( y k) = a are the points less than ( a) units from ( h, k). they satisfy the inequality ( h) + ( y k) < a they make up the region we call interior of the circle. The circle eterior consists of the points that lie more than ( a) units from ( h, k). These points satisfy the inequality ( h) + ( y k) > a. The basic shape with aes in different position. ( h) + (y k) = 0 ( h, k) ( ) + (y) = a ( ) + (y k) = a ( h) + (y) = a ( k) ( h) The circle + y = 1 is not the graph of any function of. if we attempt to solve the equation for y we find that y = 1 y = ± SQRT ( 1- ) the graph of y = SQRT( 1- ) is the upper half of the circle + y = 1 and the graph y = - SQRT( 1 ) is the lower half. Chapter Two Quadratic functions Dr. Asawer A. Alwasiti 0

25 Parabola A parabola is the set of points in a plain that are equidistance from a given fied point and fied line. The fied point is the focus of the parabola. The fied line is the directri of the parabola. The point where a parabola crosses its ais, mid way between the focus and diretri is called the verte of the parabola. The standard form equations of the parabola lies at the origin is The standard form equations of the parabola at (h, k) are: ( h) ( y k) = focus lies on the y- ais to up 4 p ( y ( ( ( h) k) = 4 p focus lies on the y- ais to down ( y k) h) = 4 p focus lies on the - ais to the right ( y k) h) = 4 p focus lies on the - ais to the left y = 4 p Find the focus and the diretri of the parabola y =. 8 Chapter Two Quadratic functions Dr. Asawer A. Alwasiti 1

26 y = is 8 y = 4 p 4p = 8 p = Focus : (0, p) ( 0, ) Directri y = -p y = - Ellipses An ellipse is the set of points in a plane whose distance from two fied points in the plane have a constant sum. The two fied points are the foci of the ellipse. Standard equations form of the for ellipses centered at ( h, k) ( h) ( y k) Foci on the -ais + = 1 a b (a > b) centered to foci distance: c = a b foci : (h ± c, k) vertices (h ± a, k) ( h) ( y k) Foci on the y-ais + = 1 b a (a > b) centered to foci distance: c = a b foci : (h, k ± c) vertices (h, k ± a) Chapter Two Quadratic functions Dr. Asawer A. Alwasiti

27 in each case a is the semimajor ais and b is the semiminor ais. If the center is the origin point then h and k will be zero. (Major ais horizontal) The ellipse + y = semimajor ais : a = ± 4 semiminor ais : b = ± 3 center -to- focus distance c=sqrt(7) foci: (± SQRT(7), 0), vertices : (±4, 0), center (0, 0) (Major ais vertical) The ellipse + y = semimajor ais : a = ± 4 semiminor ais : b = ± 3 center -to- focus distance c=sqrt(7) foci: ( 0, (± SQRT(7)), vertices : (0, ±4), center (0, 0) Hyperbolas A hyperbola is the set of points in the plan whose distance from two fied points in the plane has a constant distance. The two fied points are the foci of the hyperbola. Chapter Two Quadratic functions Dr. Asawer A. Alwasiti 3

28 Standard equations form of hyperbola centered at ( h, k) Foci on the -ais ( h) ( y k) = 1 a b centered to foci distance: c = a + b foci : (h ± c, k) vertices (h ± a, k) b asymptotes : ( y k) = m ( h) a Foci on the y-ais ( y k) ( h) = 1 a b centered to foci distance: c = a + b foci : ( k, h ± c) vertices ( k, h ± a) a asymptotes : ( y k) = m ( h) b Eccentricity The eccentricity in both ellipse and hyperbola Eccentricity = distance between foci y The eccentricity of the ellipse + = 1 b a a>b is the number c a b e = = e < 1 a a y The eccentricity of the hyperbola = 1 b a a>b is the number c a b e = = e > 1 a a + The eccentricity in both ellipse and hyperbola is: Eccentricity = Distance between foci / distance between vertices The eccentricity of a parabola is e = 1 Chapter Two Quadratic functions Dr. Asawer A. Alwasiti 4

29 Graph the equation of quadratic + y + 4 6y = 1 Analysis the equation will gives ( + ) + ( y 3) = 5 The function is circle function with center = (-, 3) and radius =5 Graph 4 + 3y + 6y = 0 Analyzing this equation gives ( ) ( y + 1) The graph is = 1 y (, 3-1) (-1,-1) (,1) (5,-1) (-,- 3+1) Analyze the equation 4y = 0 The analyzing of this equation gives Chapter Two Quadratic functions Dr. Asawer A. Alwasiti 5

30 ( 1) ( y 1) 4 1 = 1 Graph the equation 4y + 16y = 5 Analyzing this equation gives ( y + ) ( 5) 9 4 The graph is = 1 y (5, 3-) (5,1) y=0.5 (5,-) (5,-5) y=-4.5 (5,- 3+) H.W 1) Graph the equations a y 18y = 116 b- (y-) =5(+4) ) Find the equation of the hyperbolic with foci at (0,0) and (0,4) if its passed through (1,9). Chapter Two Quadratic functions Dr. Asawer A. Alwasiti 6

31 Rotting the coordinate aes to eliminate the cross-product term The quadratic equation A + By + Cy + D + E + F =0 The term By is called the cross product term Equations for rotating the coordinate aes = cosα ysinα y = sinα + y cosα A C cosα = B If we apply the rotation equations to the general quadratic equation, we obtain a new quadratic equation: A + B y + C y + D + E y + F = 0 The new coefficient are related to the old one by the equations: A = Acos α + Bcosα sinα + C sin α B = Bcosα + ( C A)sin α C = Asin α Bsinα cosα + C cos α D = Dcosα + E sinα E = Dsinα + E cosα F = F The and y aes are to rotated through the angle of (π/4) radius about the origin. Find an equation for the hyperbola y=9 in the new coordinates. y = 9 9 y 9 = 1 The graph is: Chapter Two Quadratic functions Dr. Asawer A. Alwasiti 7

32 The coordinate aes are to be rotated through an angle α to produce an equation for the curve + 3 y + y 10=0 that has no product term. Find α and the new equation. This equation has A=, B= 3, and C=1 cosα = 1/ 3 so α = π/6 the new equation is: + y = The curve is an ellipse with foci on the new y - ais H.W The coordinate aes are to be rotated through an angle α to produce an equation for the curve + y + y 6=0 that has no product term. Find α and the new equation. Chapter Two Quadratic functions Dr. Asawer A. Alwasiti 8

33 General Eponential Function Chapter Three Transcendental Functions Let a be a positive real number other than 1 the function F() = a Is the eponential function with base a The domain is all the values of while the range is y>0 As - y 0 As 0 y 1 As + y + (0,1) y = a Laws of eponents for a If a>0 and b>0, the following hold true for all real numbers a and y y + y a. a = a a y = a y a y y ( a ) = a = a a. b = ( a. b) a a = ( ) b b 3 - ( ) y The logarithmic Function: For any positive number a 1 log a is the inverse function of a. y=log a Chapter Three Transcendental Fnctions Dr. Asawer A. Alwasiti 9

34 the graph y = log a can be obtained by reflecting the graph of y = a across the 45. The domain is >0 while the range is all value of y. Rules for base a logarithms For any numbers y>0 and >0, the logarithm satisfies the following: 1- Product Rule: - Quotient Rule: log a (y) = log a + log a y log a y = log 1 3- Reciprocal Rule log a = log a y y 4- Power Rule: y log a a = y log a log The Natural Logarithms Function The natural logarithmic function is a logarithms function with base e not a. The number e is that number in the domain of the natural logarithm satisfying ln (e) =1 where : e =.718 so: log10 log y = ln = log e = = log10 e The domain is >0 while the range is all the values of y a y Properties of Logarithms For any numbers a>0 and >0, the natural logarithm satisfies the following: 1- product Rule: - Quotient Rule: 3- Reciprocal Rule ln a = ln a + ln ln a 1 ln = ln = ln a ln Rule () with a =1 Chapter Three Transcendental Fnctions Dr. Asawer A. Alwasiti 30

35 4- Power Rule: r ln = r ln r rational And a = e ln a = e ln a + 1 ln = ln( + 1) ln( 3) 3 The Natural Eponential Function -1 For every real number, e = ln = ep. The function ln, being an increasing function of with domain (0, ) and range (-, ), has an inverse ln -1 with domain (-, ) and range (0, ). The graph of ln - 1 is the graph of ln reflected across the line y =. so, e = ln -1 (1)=ep(1). Laws of eponents for e e. + e = e e 1 = e e 1 1 = e + 3- e 4- ( e ) 1 = e = ( e 1 ) 1 Impotent Note loga a = log a (a ) = ( > 0) ( all ) ln = e (all >0) ln (e )= (all ) Evaluating of log a the evaluating of log a is simplified by the observation that log a is a numerical multiple of ln. 1 log a =.ln = ln ln ln a Chapter Three Transcendental Fnctions Dr. Asawer A. Alwasiti 31

36 a- Graph the functions y=, y=3, y=10, for what values of is it true that >3 >10? b- Graph the functions y= -, y=3 -, y=10 -, for what values of is it true that - >3 - >10 -? : a- for <0 > 3 >10 for =0 = 3 = 10 =1 for >0 < 3 < 10 b- for <0 - < 3 - <10 - for =0 - = 3 - =10 - =1 for >0 - > 3 - >10 - Graph the function f()=log f ( ) = log = ln ln y = log (1,0) First order chemical reaction C A =C A0 e -kt,when t=t 1/ (half life) C A =0.5C Ao. Prove that half life depend only on k. C Ao e -kt = 0.5 C Ao So t 0.5 = ln k ln = k Chapter Three Transcendental Fnctions Dr. Asawer A. Alwasiti 3

37 Solve for a-ln =3t +5 b- e = 10 c- 3 loga 7 4 ( log5 5 = 5 ) log4 log solution a- = e 3t+5 b- = 1.15 c- = 1/5 H.W 1- Graph the functions below and state their domain and range. a- y = - +3 b- y = 3e - - Simplify a- 0.5ln( 4t 4 ) ln b- ln (e ln ) Chapter Three Transcendental Fnctions Dr. Asawer A. Alwasiti 33

38 Chapter Four Trigonometric Functions Trigonometric Functions When an angle of measure θ is placed in standard position at the center of a circle of radius r, the si trigonometric functions of θ are: Sine function: Cosine function: Tangent function: Cosecant function: Secant function: Cotangent function: sin θ = y r r cos θ = y y tan θ = 1 r csc θ = = sinθ y 1 r sec θ = = cosθ 1 cot θ = = tanθ y Conversion formula: 1 degree = π / 180 ( 0.0) radian 1 radian = 180 / π ( 57 ) degree Identities: sin θ + cos θ = 1 sec θ = 1 + tan θ sin θ = sinθ cosθ cos θ = cos θ sin θ 1+ cosθ cos θ = Chapter Four Trigonometric Fnctions Dr. Asawer A. Alwasiti 34

39 1 cosθ sin θ = sin( A ± B) = sin Acos B ± cos Asin B cos( A + B) = cos Acos B sin Asin B Periodicity A function f is periodic if there is a positive number p such that f (+p) = f (). the smallest such value of p is the period of f. cos (θ + π) = cos θ, sin (θ + π) = sin θ tan (θ + π) = tan θ, sec (θ + π) = sec θ sec (θ + π) = csc θ, cot (θ + π) = cot θ similarly cos (θ - π) = cos θ, sin (θ - π) = sin θ and so on. Graphs of trigonometric functions When we graph trigonometric functions in the coordinate plane, we denote the independent variable (radians) by instead of θ. Chapter Four Trigonometric Fnctions Dr. Asawer A. Alwasiti 35

40 Even and odd trigonometric functions The graph in the above figures suggest that cos θ and sin θ are even functions because their graphs are symmetric about the y-ais. The other four basic trigonometric functions odd. cos (-θ) = cos θ sin (-θ) = -sin θ sec(-θ)=1/cos (-θ)= 1/cos θ = sec θ Modeling temperature is π f ( ) = Asin ( c) + D B Where f is temperature and is the number of the day. Graph this model. is the period, C is the horizontal shift, and D is the B amplitude, is the A vertical shift. Chapter Four Trigonometric Fnctions Dr. Asawer A. Alwasiti 36

41 Inverse trigonometric functions and their graphs Chapter Four Trigonometric Fnctions Dr. Asawer A. Alwasiti 37

42 Important notes csc -1 = sin -1 (1/) sec -1 = cos -1 (1/) cot -1 = π/-tan -1 () Prove that cos(a-b)=cosa cosb + sina sinb Since cos( A + B) = cos Acos B sin Asin B So cos( A B) = cos Acos( B) sin Asin( B) = cosa cosb + sina sinb Chapter Four Trigonometric Fnctions Dr. Asawer A. Alwasiti 38

43 Hyperbolic Functions Identifies Chapter Four Trigonometric Fnctions Dr. Asawer A. Alwasiti 39

44 Inverse Hyperbolic Functions 1 y = sinh = ln( + + 1) - < < 1 y = cosh = ln( + 1) >= y = tanh 1 = ln < y = sec h = ln = cosh 0 < <= y = csc h = ln + = sinh y = coth 1 = ln 1 > 1 Graphing of Inverse Hyperbolic Functions Chapter Four Trigonometric Fnctions Dr. Asawer A. Alwasiti 40

45 Chapter Four Trigonometric Fnctions Dr. Asawer A. Alwasiti 41 Important notes Prove that cosh -1 sinh -1 = 1 1 = + e e e e Simplifying the above equation yield 1 Prove that sinh(+y) = sinh cushy + cosh sinhy = y y y y e e e e e e e e ) sinh( y e e y y + = = + Prove that y + = = 1 1 ln 1 tanh 1 y = tanh -1 y y y cosh sinh tanh = =

46 e e y = y Simplifying yield y = tanh 1 = 1 ln 1+ 1 H.W 1 1- Prove that sin cos y = {sin( + y) + sin( y)} t - If t = tanh(/), show that tanh = 1+ t 3 3- Prove that cosh 3 = 4cosh 3cosh Show that cosh = m (cosh + 1) Chapter Four Trigonometric Fnctions Dr. Asawer A. Alwasiti 4

47 Limits Chapter Five Limits and Continuity Definition: If the values of a function f of approach the value L as approaches c, we say f has limit L as approach c and we write: lim f ( ) = L c and read " the limit of f of approach c equals L" Properties of Limits Theorem 1 If L, M, c, and k are real numbers and lim f ( ) = L c and lim g( ) = M, then: c 1- Sum and Difference Rule: lim ( f ( ) + g( )) = L + M c - Product Rule lim ( f ( ). g( )) = L. M c 3- Quotient Rule f ( ) lim ( ) = c g( ) L M 4- Constant Multiple Rule lim ( k. f ( )) = k. L c 5- Power Rule r / s lim ( f ( )) = c L r / s where r and s are integers and s 0 Theorem Limits of Polynomial Function If P()= a n n + a n-1 n-1 + +a o, then lim ( ) ( )... n n 1 P = P c = anc + an 1c + + c Theorem 3 Limits of Rational functions If P() and Q() are two polynomials and Q(c) 0 them P( ) P( c) lim( ) = c Q( ) Q( c) Chapter Five Limits and Continuity Dr. Asawer A. Alwasiti 43 a o

48 3 t 8 Find lim t 3 t 4 3 t 8 lim = 3 t 3 t Find lim lim * = 3 4 H.W Find the limits of + h 1- lim h 0 h 3 - lim lim 0 + Limits of Trigonometric Functions 1- limsinθ = 0 θ 0 - limcosθ = 1 0 θ limsinθ θ lim tanθ = = = 0 θ 0 limcosθ 1 θ 0 sinθ 4- lim = 1 θ 0 θ sinθ 5- lim = 0 θ ± θ Chapter Five Limits and Continuity Dr. Asawer A. Alwasiti 44

49 sin( θ ) Find lim θ 0 θ sin( θ ) = lim = 1 θ 0 θ tanθ Find lim θ 0 θ = 1 sinθ lim. = 1 1 = 1 θ 0 θ cos θ cosh Show that lim 1 = 0 0 h h 1 cosθ Since sin θ = So we can rewrite the limits as sinθ = lim.sinθ = 0 h 0 θ sin lim h 0 ( h / ) h sin Show that lim = sin /5 sin lim. = lim 0 5 /5 5 0 = (/5) H.W Find sin 3 1- lim 0 - lim 0 sin Chapter Five Limits and Continuity Dr. Asawer A. Alwasiti 45

50 Limits Involving Infinity If 1 f ( ) = It could be seen from the graph: 1- as approaches 0 from right then 1/ tends to + - as approached 0 from the left then 1/ tends to - 3- as approached from the right (+ ), then 1/ tends to 0 4- as approached from the right (- ), then 1/ tends to 0 so 1 1 lim = 0, lim = And lim = +, lim = lim (5 + ) = lim5+ lim = 5 π 3 lim = π = 0 1 Find limsin( ) Let t = 1/ So t 0 + as So 1 lim sin( ) = 0 0 Chapter Five Limits and Continuity Dr. Asawer A. Alwasiti 46

51 Vertical and Horizontal asymptotes - A line y = b is a horizontal asymptote of the graph function y= f() if either; lim f ( ) = b or lim f ( ) = b - A line = a is a vertical asymptote of the graph of the function y = f() if either; lim f ( ) = ± a + or lim f ( ) = ± a + 3 lim / lim = 1 1+ / asymptote) + 3 lim = + asymptote) i.e y = 1 ( horizontal i.e. = - ( vertical Limits of Rational Functions 1- Degree of Denominator and Numerator of same degree lim 3 + y = lim = Degree of Denominator greater than Numerator 11 + lim 3 1 Chapter Five Limits and Continuity Dr. Asawer A. Alwasiti 47

52 lim 1 3 = 0 3- Degree of Denominator less than Numerator 3 lim lim = H.W 1- find a- lim b- lim ± Find the vertical and horizontal asymptotes of: + 3 a- lim b- lim Limits of Eponential Functions The curve y = e The line y = 0 is a horizontal asymptote of the graph so: 1- lim e = 0 e - lim = 0 3- lim = 0 e Chapter Five Limits and Continuity Dr. Asawer A. Alwasiti 48

53 Find lim e 0 1 Let t = 1/ so t - So lim e 0 1 = 0 as 0 - lim( + 1) lim( + + 1)* = lim + 3 lim = 0 1 = lim 1+ 1/ 5 1/ 3 1/ 5 1/ 3 = 1 H.W Find the limits and vertical and horizontal asymptote if eists: lim sin - lim + 5 s s 3t lim( )( ) s 4- lim s s (try to graph this) t 1 Chapter Five Limits and Continuity Dr. Asawer A. Alwasiti 49

54 Continuity A function f() is continuous at =c if and only if it meets the following three conditions: 1- f(c) eists ( c lies in the domain of f) - lim f ( ) eists ( f has a limit as approaches c) c i.e lim f ( ) lim f ( ) = lim f ( ) c + = c c 3- lim f ( ) =f(c) ( the limit equals the function value) c In other meaning: a function y = f() that can be graphed over each interval of its domain with one continuous motion of the pen is a continuous function. The graph y = 1/ is continuous at every value of ecept =0. It has a point of discontinuity at =0 The following types of functions are continuous at every point in their domain: 1- polynomials - rational functions: they have points of discontinuity at the zero of their denominator. 3- Root functions ( 4- Trigonometric functions 5- Inverse trigonometric functions 6- Eponential functions 7- Logarithmic functions y = n, n positive integer great than 1) * the inverse function of any continuous function is continuous. Chapter Five Limits and Continuity Dr. Asawer A. Alwasiti 50

55 Definition: The function f - f ( + ) f ( ) f ( ) = lim 0 Chapter Si Derivatives defined by the formula is called the derivative with respect to of the function f. The domain of f - consists of all for which the limit eists. Geometric meaning of the derivative: f - is the function value at and it s the slope of the line to y=f() at. Common notification for derivatives: y- ( y prime) dy ( dy by d) d d [] d (the derivative with respect to (w.r.t)) Differentiation by Rules Let u =f() and v = g() d [ c] = 0 d 1- c = constant d - [ a] = a a= constant d d n n 1 3- [ ] = n d d d d 4- [ u ± v] = [ u] ± [ v] d d d 5- d d[ u] d[ v [ u. v] = v + u ] d d d d d v [ u] u [ v] d u 6- [ ] = d d d v v Chapter si Derivative Dr. Asawer A. Alwasiti 51

56 7-8- d u d d u d = nu n n 1 = nu du d n du d n 1 Find the derivative of s = ds dt = (t + 3) * = 8t + 1 s=(t+3) ( 1)( Find the derivative of r = 4 ) r = dr d = Does the curve y= have any horizontal tangent? If so, where? y - = solving the above equation when =0 for we get =0, =1, =-1 at =0 (0,) at =1 (1,1) at =-1 (-1,1) Find the derivative for y= ( + 1) 3 ( 3-1) Let u=( +1) 3 and v= ( 3-1) so y - = uv - +vu - Chapter si Derivative Dr. Asawer A. Alwasiti 5

57 P so University of Technology Mathematics Chemical Engineering Department y - = 6 ( +1) ( 3-1)( 3 +-1) H.W Find the derivative of a- y = ( 1 1) 5 b- 1 y = ( ( + 7) 6 ) Second and Higher Order Derivative dy y = d (first order) = d dy d y y = ( ) = d d d (second order) 3 d d y d y y = ( ) = 3 d d d (third order) n d n 1 y = ( y) d (nth order) Application Velocity is the derivative of the distance. Speed is the absolute value of velocity. velocity = Speed ds/dt = ( v(t = Speed Acceleration is the derivative of velocity with respect to time. dv d s a ( t) = = dt dt The distance of a ball falls freely from rest is proportional with time as s=4.9tp a- How long did it take the ball bearing to fall the first 14.7m? b- What is the velocity, speed and acceleration after second? a- s = 4.9tP 14.7=4.9tP t = ± 3 second t= 3 ( time increase from t=0 so we ignore the negative root) b- velocity at any time Chapter si Derivative Dr. Asawer A. Alwasiti 53

58 P P after University of Technology Mathematics Chemical Engineering Department ds v( t) = = 9. 8t dt Velocity after second = v() = 19.6 m/s Speed = = m/s Acceleration at any time d s a ( t) = = 9.8m / s dt Acceleration after second = 9.8 m/sp A dynamite blast blows heavy rock straight up with a lunch velocity of 160 ft/sec. It reaches a height of s=160t -16tP time (sec) a- How height does the rock ago? b- what are thw velocity and speed of the rock when its 56 ft above the ground on the way up? on the way down? c- what is the acceleration of the rock at any time t during its flight (after the blast)? d- when does the rock hit the ground again? ds dt a- v = = 160 3t ft/sec at v=0 t=5 sec SRmaR = s(5) = 400ft b- s(t) = tp at s = 56 ft then the time will be and 8 second v() = () = 96 ft/s v(8) = (8) = - 96 ft /s At both instants, the rocks speed is 96 ft/s Chapter si Derivative Dr. Asawer A. Alwasiti 54

59 dv dt c- a = = 3 ft / sec ( the acceleration is always downword) d- at s=0 then time will be 10 sec The Chain Rule If dy d And y = f(t) and = g(t), then = dy dt / d dt or dy d = dy. dt dt d d y d dy d = / or d dt d y d dy =. d dt d Find dy/d if y = SQRT(1-t ) and = t y = dy d dy dt = d / dt / = t(1 t Find d y / d if = t t and y = t t 3 ) y = dy d = dy / dt d / dt 1 3t = 1 t dy dt = 6t + 6t (1 t) d y d = dy / dt d / dt = 6t + 6t 3 (1 t) H.W Find d y / d for a- y = t, = 3-3t b- y = SQRT(t), = SQRT(t+1) Chapter si Derivative Dr. Asawer A. Alwasiti 55

60 Implicite Function Différentiation We can find the derivative of implicit functions in two steps Step 1: Differentiate both sides of the equation with respect to, treating y as a differentiable function of Step : Solve for dy/d Find dy/d for y +5y 6 = 0 dy dy y y 1 = 0 d d dy 1 5y = d y + 5 Find the second derivative of the function 3 3y = 0 dy 6 6y = 0 d dy = d y 4 d y = 3 d y y where y 0 Show that the point (, 4) lies on the curve 3 + y 3-9y = 0. Then find the tangent and normal to the curve there (see figure) Point (,4) lies on the curve because **4=0 To find the slope of the curve at (,4) Chapter si Derivative Dr. Asawer A. Alwasiti 56

61 3 dy d dy d + 3y 3y = y (,4) = dy d 3 4 / 5 9( dy d + y) = 0 the tangent at (,4) is the line thought (,4) with slope 4/5 4 1 The equation of the tangent will be y = The normal to the curve at (,4) is the line perpendicular to the tangent, so the slope will be (-5/4) And the equation is 5 13 y = + 4 Assume that the radius r and the height of a cone are differentiable functions of t and let V be the volume of the cone. Find an equation that relates dv/dt, dr/dt and dh/dt. π V = r h 3 dv π = ( r dt 3 dv π = ( r dt 3 dh dr + r h dt dt dh dr + rh ) dt dt Chapter si Derivative Dr. Asawer A. Alwasiti 57

62 Derivative of Trigonometric Functions If u = f() then y = sin u y = cos u y = tan u y = cot u y = sec u y - = cos u.du/d y - = -sin u.du/d y - = sec u.du/d y - = -csc u.du/d y - = sec u. tan u.du/d y = csc u y - = -csc u.cot u. du/d y + sin y = 0 dy dy ( + y) + cos y. d d dy y = d + cos y = 0 Find the slope of the line tangent to the curve y = sin 5 at the point where =π/3. y - = 5sin 4. cos the tangent line has slope dy d = π 4 = 45 3 A body hanging from a spring is starched 5 units beyond its rest position and released at time t=0 to bob up and down. Its position at any later time is s = 5 cost What are its velocity and acceleration at time t? Position s = 5 cost Velocity v(t) = -5sint Chapter si Derivative Dr. Asawer A. Alwasiti 58

63 Acceleration a(t) = -5cost H.W Find the derivative for sin y = 1 y = sin y = + sin y - Find the second derivative of y = sec 5 Derivative of Inverse Trigonometric Functions 1 y = cos dy 1 1 =. = d 1 / y = csc + 1 dy 1 = ( ) + csc d = csc 1 1 Chapter si Derivative Dr. Asawer A. Alwasiti 59

64 1 1 y = tan ( ) + 1 dy 1 ( + 1)(1) ( 1)( 1) = 1 d ( 1) 1 ( ) dy = d ( + 1) + ( 1) H.W Find the derivative of 1 a- y sin + y = tan b- y = sec Derivative of Hyperbolic Function y = tanh(1 + ) dy = (sec h (1 + d y = sec h dy = sec h d ) tanh + sec h Chapter si Derivative Dr. Asawer A. Alwasiti 60

65 Dérivative of Inverse Hyperbolic Function Find the derivative of y = cosh -1 (sec) dy d dy d = = sec 1.sec.tan sec 1 Proof that d d sinh 1 = 1 1+ Let y = sinh -1 so = sinh y d dy = cosh y dy d = 1 cosh y cosh sinh = 1 cosh y = dy 1 = d 1+ sinh y dy 1 = d 1+ d 1 1 sinh = d sinh y Chapter si Derivative Dr. Asawer A. Alwasiti 61

66 If y = tanh -1 u y = tanh -1 u dddd dddd = ssssssh yy Proof that dy d u = tanhy dddd dddd = dddd dddd. dddd dddd dddd dddd = 1 ssssssh yy. dddd dddd dddd dddd = 1 1 uu. dddd dddd 1 = 1 u du. d Derivative of Eponential and Logarithmic Function If u = f() 1- - u y = b u y = e 3- y = log u b 4- y = ln u dy d u du = b.ln b. d dy du = e u d. d dy d dy d sin y = ln y = sin.ln 1 dy 1. = cos.ln + sin. y d dy sin sin = ( + cos.ln ) d 1. = u 1. = u 1 ln du d du. b d Chapter si Derivative Dr. Asawer A. Alwasiti 6

67 1+ y = ln 1 dy 1 = d 1 tan y = e + ln dy 1 sec y. = e + d dy 1 e + 1 = d sec y y = dy d 3 = 3 sin y = π dy = π d H.W sin.ln 1 ln y = (ln(1 + ) ln(1 )).lnπ.cos Find the derivative for a) y = b) y = ln( sin c) y = e + 3) How rabidly will the fluid level inside a vertical cylindrical tank drop if we pump the fluid out at the rate of 300 L/min? dv/dt = L/min = -3 m 3 /min the rate is negative because the volume is decreasing Chapter si Derivative Dr. Asawer A. Alwasiti 63

68 V = πr h dv dh = πr dt dt dh 3 = dt πr The fluid level will drop at a rate of 3/πr m/min A hot air balloon rising straight up from a level field is tracked by a range finder 500ft from the lift off point. At the moment the range finders elevation angle isπ/4, the angle is increasing at the rate of 0.14 rad/min. How fast is the balloon rising at that moment? θ = f(t) = the angle in radius that range finder makes with the ground y = f(t) = the height in feet of the balloon y tanθ = y = 500.tanθ 500 dy dθ = 500sec θ. dt dt dy = 140 ft / min dt The balloon is rising at the rate of 140 ft/min A police cruiser, approaching a right- angled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.6m north of the intersection and the car is 0.8m to the east. The police determined with radar that the distance between them and the car is Chapter si Derivative Dr. Asawer A. Alwasiti 64

69 increasing at 0 mph. If the cruiser is moving at 60 mph at the instance of the measurement, What is the speed of the car? = position of the car at time t y = position of the cruiser at time t s = distance between car and the cruiser at time t at = 0.8m and y = 0.6m, ds/dt = 0 mph s = + y s = + y ds d dy s = + y dt dt dt ds 1 d dy = ( + y ) dt s dt dt ds 1 d dy = ( + y ) dt + y dt dt d = 70mph dt So the car speed is 70 mph and dy/dt = -60 mph ( because y is decreasing) Water runs into a conical tank at the rate of 9 ft 3 /min. The tank stands point down and has a height of 10 ft and a base radius of 5ft. How fast is the water level rising when the water is 6ft deep? V = volume of the water in the tank at time t = radius of the surface of the water at time t y = depth of the water in the tank at time t Chapter si Derivative Dr. Asawer A. Alwasiti 65

70 y = 5 10 = 1 π V = y 3 π y V = y 3 dv π dy = y dt 4 dt dy = 0.3 ft / min dt y Hence, the water level is rising at about 0.3 ft/min. Application of Derivatives on Limits: Hopitals Rule Intermediate forms of L ' Hopitals, Rule: - Intermediate form 0/0 f ( ) lim a g( ) f ( a) = where f(a) = g(a) = 0 g ( a) 3 sin Find lim 0 3 cos = lim 0 1 sin Find lim 3 0 = 1 cos = lim 0 3 sin = lim 0 6 cos 1 = lim = Chapter si Derivative Dr. Asawer A. Alwasiti 66

71 H.W Find lim 0 - sin lim π 1+ cos - Other intermediate form /, 0/, /- f ( ) f lim = a g( ) g ( a) ( a) where f(a) = g(a) = tan lim π 1+ tan sec = lim π sec = lim( ) 0 sin sin = lim 0 sin 1 cos = lim 0 cos + sin = 0 = 0 + ln e lim( ) 0 + e = lim 0 + e. + e = lim = = 1 1 Theorem If lim ln f ( ) = a L Then lim a ln f ( ) f ( ) = lime = a e L Chapter si Derivative Dr. Asawer A. Alwasiti 67

72 lim(1 o 0 ) 1 = f ( ) = (1 1 1 ln f ( ) = ln(1 ) ln(1 ) limln f ( ) = lim 0 0 = lim limln f ( ) = + lim = lim f ( ) = e ) H.W Find 1- cos 0.5 lim π / 3 π / 3 - lim + Chapter si Derivative Dr. Asawer A. Alwasiti 68

73 Partial Derivative Let f(,y) be a function of two variables, then we define the partial derivative as: ff = = lim ff(+h,yy) ff(,yy) h oo h ff yy = = lim ff(,yy+h) ff(,yy) h oo h Algebraically, we can think of partial derivative of a function with respect to as the derivative of the function with y held constant. ff(, yy) = 3yy yy ff = = 3yy 4 ff yy = = 6 Find f and f y if ff(, yy) = ff = = ff yy = = H.W Find f and f y for yyyyyyyyyy (yy + cccccccc) cccccccc (yy + cccccccc) yy yy=cccccccc 1) ff(, yy) = ) ff(, yy) = + yy yy Higher Order Partial Derivative: Just as with function of one variable, we can define second derivatives for functions of two variables. For function of two variables we have four types: Chapter si Derivative Dr. Asawer A. Alwasiti 69

74 ff oooo ff, ff oooo ff ff yyyy oooo ff, ff yyyy oooo ff Theorem Let f(,y) be a function with continues second order derivative, then f y = f y let f(,y) = ye then f = ye, f y = e f = ye, f y = e, f yy = 0, f y = e so f y = f y ff(, yy) = + yyee ff = = cccccccc + yyee ff = ff = yyee ff = = ff = ssssssss + ee ff yy = = + ee ff yyyy = ff yy = ff yyyy = = ff = ssssssss + ee Chapter si Derivative Dr. Asawer A. Alwasiti 70

75 Functions of More Than Two Variables Suppose that f(,y,z) = y -yz is a function of three variables, then we can define the partial derivatives in such the same way as we defined the partial derivatives, so we have: ff = = yy ff yy = = zz ff zz = = yy H.W Find f, f y, f z for f(,y,z) = sin(y+3z) The Chain Rule for Functions of Two and Three Variable: If w = f(,y) and = f(t), y = f(t) =. dddd dddd +. dddd dddd In similar way if w=f(, y, z) And = f(t), y = f(t), z=f(t) =. dddd dddd +. dddd dddd +. dddd dddd Use the Chain Rule to find the derivative of w = y with respect to t along the path = cost, y = sint. What is the derivatives value when t=π/? = yy = ssssssss, = = cccccccc dddd dddd = ssiiiiii, dddd dddd = cccccccc Chapter si Derivative Dr. Asawer A. Alwasiti 71

76 PR R+siny PR R+siny and are University of Technology Mathematics Chemical Engineering Department =. dddd dddd +. dddd dddd = cccccctt ll tt=ππ/ = 1 Find the partial derivative of w if w =y+z, = cost, y = sint, z =t What the derivatives value at t=0? = yy( ssssssss) + + (1)(1) = UImplicit Differentiation Suppose f(,y) and its partial derivatives frr fryr continues and the equation f(,y)=0 defines y as a differentiable function of. Then at any point where fryr 0: ll tt=0 = dddd dddd = ff ff yy Find dy/d if P f(,y) = P -y =y dddd dddd = ff ff yy Chapter si Derivative Dr. Asawer A. Alwasiti 7

77 Chapter Seven Integration It s a mathematics we use to find lengths, area and volumes of irregular shapes; to calculate the average values of functions; and to predict future population sizes and future costs of living. Two kinds of integral: 1- Identified integral - Definite Integral 1-Indefinite Integral: If f - () is a derivative then the ant derivative of f - () is called the indefinite integral of f - () with respect to and is denoted by: ff()dddd = ff() + cc Integration formula: 1- dddd dddd = uu() + cc dddd - aaaa() dddd dddd = aa uu() dddd dddd dddd 3- uu() ± vv() dddd = uu() dddd ± vv()dddd uu nn () dddd 4- dddd = uu()nn +1 + cc dddd nn+1 nn nn +1 dddd = + cc nn+1 Find y if dddd = dddd yy y" > 0 dddd yy = dddd yy = 3 + cc dddd = ( ) 1 3( + 1)dddd 3 3 Chapter seven Integration Dr. Asawer A. Alwasiti 73

78 = 1 ( ) 3 + c At any point (,y) of curve dd yy = dd 1 6 when =-1, y= -11 and when =1, y= -3. Find the equation of the curve and the equation of the tangent and normal at (1,-3). dd yy dd = (1 6)dddd dddd dddd = ( cc)dddd y= 4-3 +c+d at (-1.-11) and (1,-3) c = 4 and d=-5 Hence, the equation of the curve is y= dddd = dddd at (1,-3) The slope of the tangent is and the slope of the normal is -1/ The equation of the tangent is y=+5 The slope of the normal is y = --5 As a function of elapsed time t, the velocity v(t) of a body falling from rest in a vacuum near the surface of the earth satisfies the differential equation dddd = 9.8, dddd with initial condition v=0 at t=0 find v as a function of t. dv =9.8dt v = 9.8t +c (0,0) v=9.8 t m/s Chapter seven Integration Dr. Asawer A. Alwasiti 74

79 H.W 1- Speed of the body movement straight line is given by v=6(t-t ), find the distance of the body from a fied point after two second. - Find the equation of the curve whose slope is ( 1/ ) and passing through (,5) 3- Find y if dddd dddd = +1 yy 1 4-Find r if dddd dddd y 1 = (zz + 1)3 Integration of Trigonometric Functions 1- ssssssss dddd dddd - cccccccc dddd dddd 3- ssssss uu dddd dddd 4- cccccc uu dddd dddd 5- ssssssss. tttttttt dddd dddd 6- cccccccc. cccccccc dddd dddd = cccccccc + cc = ssssssss + cc = tttttttt + cc = cccccccc + cc = ssssssss + cc = cccccccc + cc cccccc. dddd cccccc dddd = ssssss 3 ssssss 3. cccccc. dddd Let u= sin so du=cos.d (power function) = uu 3. dddd 1 cccccc (7 = ssssss (7 Chapter seven Integration Dr. Asawer A. Alwasiti 75

80 = ssssss = (1 cccc = tttttt 4 = tttttt 4 = tttttt7 14 University of Technology Mathematics Chemical Engineering Department = 1 7 tan(7 cccccc. dddd = 1 + cccc = ss ssssss 3. dd = ssssss. = 1 6 cccccc3 tttttt 4. ss UIntegration of Inverse Trigonometric Functions 1- dddd aa +uu = 1 aa ssssss 1 uu aa + cc uu < aa - dddd aa +uu = 1 aa tttttt 1 uu aa + cc dddd 3- = 1 uu uu aa aa ssssss 1 uu + cc aa uu > aa Chapter seven Integration Dr. Asawer A. Alwasiti 76

81 P so University of Technology Mathematics Chemical Engineering Department u=p 3 du=3p P.d du/3 = P P.d dddd dddd 3 1 uu 3 ( + ) = ( + = ( + = 3ssssss 1 Let u = du=d and a=5 dddd = 1 () = 1 10 tttttt 1 H.W Find: 1- cccccccc.dddd 1 ssssss - ( 1).dddd 1 ( 1) 4 3-4cccccc3yy. dddd 4- csc + ππ. cot + ππ. dddd dddd 9 Chapter seven Integration Dr. Asawer A. Alwasiti 77

82 UIntegration of Basic Logarithmic and Eponential Functions 1- dddd uu = llll uu + cc - ee uu dddd = ee uu + cc 3- aa uu dddd = 1 llllll. aauu + cc aa 1, aa > 0 dddd.dddd = 3 +4 llll cc Since > 0 = ln (3P P+4)+c ee dddd = 4ee3 3 ee +1 = 4 ee3 ee +1 = 4 ee = ee 1 ssssssss cccccccc dddd + (1 + llllll uu = 1 + = dddd uu = Chapter seven Integration Dr. Asawer A. Alwasiti 78

83 ee ee ee + ee = llll ee + ee ee +ee +ll = ee. ee ee llllll uu = ee = ee uu. dddd UIntegration of Hyperbolic Functions cccccch5. dd = cccccch5 ssssssh5 Chapter seven Integration Dr. Asawer A. Alwasiti 79

84 = ee (ee ) University of Technology Mathematics Chemical Engineering Department ssssssh llllll uu = dd = cccccch ssssssh. dddd = dddd ee + ee UIntegration of Inverse Hyperbolic Functions cccccccc 1 + ssssss Chapter seven Integration Dr. Asawer A. Alwasiti 80

85 dddd + 6 = + dd = ( + ( + 1 ) = ( + 1 = ( + 1 UIntegration Methods U1- Integration by Substitution Let u= 4-P P so du = -d d=-du/ 4 dd = dddd/ uu Let u=sin(p P+1) (du=cos(p P+1).4.d)*4 16d.cos(P P+1) =4du ssssss 3 ( Chapter seven Integration Dr. Asawer A. Alwasiti 81

86 uu 3. 4dddd = tttttt dd University of Technology Mathematics Chemical Engineering Department U- Integration by Parts sin(llllll) llllll uu = llllll = cos(llllll u =f(), v=g() dd (uu. vv) = dddd uuuuuu = uu. llllll. dddd Let u =ln du=d/ dv=d v= llllllllll llllllllll = llllllll = Chapter seven Integration Dr. Asawer A. Alwasiti 8

87 dddd = 1 tttttt 1. dd University of Technology Mathematics Chemical Engineering Department tttttt 1. dd llllll uu = tttttt ee dddd llllll uu = dddd = dd ee dddd = Take ssssssss. ee. dddd ee cccccccc. dd llllll uu = ee dddd = ee ee cccccccc. dd llllll uu = ee dddd = ee ee cccccccc ee cccccccc. dddd = ee ssssssss + ee. cccccccc ee cccccccc. dddd = 1 (ee ssssssss + ee. cccccccc) + cc ssssssss. dddd llllll uu = Chapter seven Integration Dr. Asawer A. Alwasiti 83

88 ssssssss. dddd = cccccccc +. dddd Take. dddd ssssssss. dddd = cccccccc + + cccccccc + cc dddd = llllll uu = dddd = dd ln(cccccccc). ssssssss. dddd ln(cccccccc). ssssssss. dddd = cccccccc. ln(cccccccc) ccccssss ssssssss dddd cccccccc ln(cccccccc). ssssssss. dddd = (1 ln(cccccccc))cccccccc + cc llllll uu = ln dddd = H.W. 1- ln( + 1). dddd - ee. cccccc3. dddd 3- Power Trigonometric Functions tan 4. dddd = tan. tan. dddd = tan. (sec 1). dddd = tan. sec. dddd (sec 1). dddd = 1 3 tttttt3 tttttttt + + cc Chapter seven Integration Dr. Asawer A. Alwasiti 84

89 tan. sec 3. dddd = sec 3. (sec 1). dddd = sec 5. d sec 3. dddd For sec 3. dddd = ssssssss. ssssss. dddd sec 3. dddd = ssssssss. tttttttt tttttt. ssssssss. dddd llllll uu = sec dddd = ss = ssssssss. tttttttt ssssss 3. dddd + ssssssss. dddd sec 3. dddd = 1 ssssssss. tttttttt 1 llll ssssssss + tttttttt + cc For sec 5. dddd = ssssss 3. ssssss. dddd sec 5. dddd = sec 3. tttttttt tttttttt. 3ssssss. ssssssss. tttttttt. dddd llllll uu = sec dddd = 3 = sec 3. tttttttt 3 tttttt. ssssss 3. dddd = sec 3. tttttttt 3 (ssssss 1). ssssss 3. dddd = sec 3. tttttttt 3 ssssss 5. dddd + 3 ssssss 3. dddd 4 sec 5. dddd = sec 3. tttttttt 3[ 1 ssssssss. tttttttt + 1 llll ssssssss + tttttttt ] tan. sec 3. dddd = 1 4 sec3. tttttttt 1 8 ssssssss. tttttttt 1 llll ssssssss + tttttttt + cc 8 cccccc 3. cccccc 5. dddd = cccccc. cccccc 4. cccccccc. cccccccc. dddd cccccc 3. cccccc 5. dddd = (cccccc 1). cccccc 4. cccccccc. cccccccc. dddd Chapter seven Integration Dr. Asawer A. Alwasiti 85

90 cccccc 3. cccccc 5. dddd = cccccc 6. cccccccc. cccccccc. dddd cccccc 4. cccccccc. cccccccc. dddd = 1 7 cccccc cccccc5 + cc H.W tttttt 3. ssssss. dddd 5- Trigonometric Substitution This method enables us to replace the binomials: 1- aa + bb uu bbbb uu = aa bb tttttttt - aa bb uu bbbb uu = aa bb ssssssss 3- bb uu aa bbbb uu = aa bb ssssssss 1- If the binomials in the form of aa + bb uu dddd (4 + ) Let = tanθ so d = sec θ.dθ dddd (4 + ) = sec θ. dθ (4 + 4tttttt θθ) = sec θ. dθ (4 + 4tttttt θθ) = sec θ. dθ 16(1 + tttttt θθ) Since = tanθ tanθ=/ ssssssθθ = ssssssss. cccccccc = 1 dθ 8 ssssss θθ = cccccc dddd = ccccccθθ dddd 8 = 1 16 θθ + 1 ssssssθθ + cc 3 θ=tan -1 (/) Chapter seven Integration Dr. Asawer A. Alwasiti 86

91 = dddd (4 + ) = 1 16 tttttt cc - If the binomial in the form of aa bb uu uu 16 uu. dddd Let u = 4sinθ du = 4cosθ.dθ uu 16 uu. dddd = 16ssssss θθ. 4cccccccc 16 16ssssss θθ. dddd = ssssss θθ. dddd = 1 ccccccθθ. dddd = 8 θθ 1 ssssssθθ + cc Since u = 4sinθ θ=sin -1 (u/4) ssssssθθ = ssssssss. cccccccc = 1 8 uu 16 uu uu 16 uu. dddd = 8 uu ssssss uu 16 uu + cc 3- If the binomial in the form of bb uu aa dddd 5 Let = 5secθ d = 5secθ.tanθ.dθ dddd 5 = 5ssssssss. tttttttt. dddd 5ssssss θθ 5 Chapter seven Integration Dr. Asawer A. Alwasiti 87

92 ssssssss. tttttttt. dddd = tttttttt = ssssssss. dddd = llll sssscccc + tttttttt + cc Since secθ =/5 θ = sec -1 (/5) tttttttt = 5 5 dddd 5 = llll cc 5 4- Integration by Part Fraction a- If the denominator has two linear factor ( + 1)( 3) dddd 5 3 ( + 1)( 3) = AA = AAAA 3AA + BBBB + BB Hence, A= and B= ( + 1)( 3) BB 3 3 dddd = dddd dddd dddd = llll llll 3 + cc ( + 1)( 3) b- If the denominator has a repeated linear factor ( + ) dddd ( + ) = AA + + BB ( + ) = AAAA + (AA + BB) Hence, A=6 and B=-5 Chapter seven Integration Dr. Asawer A. Alwasiti 88

93 dddd = ( + ) + dddd + 5BB ( + ) dddd dddd = 6llll + + ( + ) + + cc c- If the denominator has a quadratic factor + 4 ( + 1)( 1) dddd + 4 AAAA + BB ( = + 1)( 1) ( + 1) + CC 1 + DD ( 1) + 4 = (AAAA + BB)( 1) + cc( + 1)( 1) + DD( + 1) HHHHHHHHHH, AA =, BB = 1, CC =, D= ( dddd = + 1)( 1) ( + 1) dddd + 1 dddd + 1 ( 1) dddd + 4 ( + 1)( 1) dddd = llll tttttt 1 llll cc d- If the numerator is higher than the degree of denominator dddd First we have to divide this function and this results: dddd + 4 = dddd = ( + 1)( 1) dddd Take : + 4 AAAA + BB ( = + 1)( 1) ( + 1) + CC 1 + DD ( 1) Chapter seven Integration Dr. Asawer A. Alwasiti 89

94 From the previous eample we get: A=, B=1, C=-1, D= dddd H.W Find: = + + llll tttttt 1 llll cc dddd ( 1)( )( 3) dddd ( 1)( ++1) dddd - Definite Integral Properties of Definite Integral: bb bb 1- kk. ff()dddd = kk ff()dddd kk = cccccccccccccccc aa aa bb aa - ff() gg() dddd = ff()dddd gg() dddd bb 3- ff()dddd = aa bb aa aa bb ff()dddd bb 4- IIII aa = bb ttheeee ff(). dddd=0 aa bb 5- ff(). dddd aa bbbb + ff(). dddd bb bb aa cc = ff(). dddd aa Find the area bounded by the -ais and the parabola y = = 6 So = -3 and = Area = A = Area = 05/6 3 ff(). dddd = (6 )dddd 3 Chapter seven Integration Dr. Asawer A. Alwasiti 90

95 Find the area bounded by the -ais and the parabola y = -4, =3, = Area = A = ff(). dddd = ( 4)dddd A = -/7 AAAAAAAA = = 3 3 Find the total area and net area bounded by the curve y = 3-4 and the -ais. 0 = 3-4 So =0 and = ± 0 AA 1 = ( 3 4). dddd = AA = ( 3 4). dddd = 4 0 Total area = AA 1 + AA = = 8 Net area = AA 1 + AA = 4 4 = 0 H.W 4 Find the area bounded by y = tan from =0 and =π/4e Application of Definite Integral. 1- Area between two curves If y1 = f 1 () and y = f () are continuous for a b and f 1 () f () for same interval then the area between y 1 and y bb aa bb aa = ff 1 (). dddd ff (). dddd = [ff 1 () ff ()]dddd bb aa Chapter seven Integration Dr. Asawer A. Alwasiti 91

96 Find the area bounded by the parabola y = 6 and y = - 6 = - So = 0 y = 0 And = 4 y = 8 4 AA = [ff 1 () ff ()]dddd 0 4 = [(6 ) ( )]dddd 0 AA = 64 3 Find the area of the region enclosed by the parabola y = and the curve y = -. Let [ff 1 () = ff ()] = - Then = -1 and = AA = [ff 1 () ff ()]dddd 1 = [ + ]dddd 1 AA = 9 Find the area bounded on the right by the line y = -, on the left by the parabola = y and from below by the -ais. Chapter seven Integration Dr. Asawer A. Alwasiti 9

97 1 st method Integration with respect to y y = y+ then y = and y=-1 the point y= is the only point lie in region AA = [ff 1 (yy) ff (yy)]dddd 0 4 = [yy + yy ]dddd 0 AA = 10 3 nd method Integration with respect to =y y = ± = f 1 () y = - y = - = f = - Then = 1 and =4 () The point =1 does not satisfy the equation = - For 0, [ff 1 (yy) ff (yy)] = 0 = For 4, [ff 1 (yy) ff (yy)] = + AAAAAAAA = dddd + dddd AA = Lengths of Curves in the Plain If the function f has a continuous first derivative through the interval a b, the length of the curve y = f() from a to b is: Chapter seven Integration Dr. Asawer A. Alwasiti 93

98 bb LL = 1 + dddd dddd. dddd aa Find the length of the curve yy = 1 aa(ee + ee aa ) from =0 to = a. bb LL = 1 + dddd dddd. dddd aa yy = 1 aa ee aa + ee aa yy = 1 ee aa + ee aa bb LL = ee aa + ee aa. dddd aa LL = 1 aa ee + 1 uuuuuuuu ee Find the length of the curve y = /3 between =1 and = 8 bb LL = 1 + dddd dddd. dddd aa yy = /3 yy = 3 1/3 This give no answer at =0, so we will use as f(y) = yy 3 dddd dddd = 3 yy Chapter seven Integration Dr. Asawer A. Alwasiti 94

99 At =-1 then y = 1 At = 8 then y =4 We have two intervals: 0 y 1 and 4 y 0 LL = LL 1 + LL 1 = 1 + dddd 4 dddd. dddd dddd dddd. dddd 0 1 = =105 unit 4 yy. dddd yy 0 0. dddd 3- Area of Surface of Revolution If the function f has a continuous first derivative through the interval a b, the area of surface generated by revolving the curve will be: bb SS = ππππ 1 + ( dddd dddd ). dddd aa bb SS = ππππ 1 + ( dddd dddd ). dddd aa (revolution about -ais) (revolution about y-ais) Find the surface area of revolution generated by revolving about -ais the area of the parabola y = 1 from =0 to =3. bb SS = ππππ 1 + dddd dddd. dddd aa y = 1 dddd dddd = 6 yy Chapter seven Integration Dr. Asawer A. Alwasiti 95

100 3 SS = ππππ yy. dddd 0 S = 43.9π square unit 3 = (1 + 36) 1. dddd 0 Find the area of the surface of revolution generated by revolving about y-ais of = y 3 from y=0 to y =1. bb SS = ππππ 1 + dddd dddd. dddd 3 aa =y dddd dddd = 3yy bb SS = ππyy (3yy ). dddd aa S = 1.13 π square unit 4- Volumes of Solids a- Disk Method The volume of the solid generated by revolving the region between the graph of a continuous function and aes are: bb vvvvvvvvvvvv = ππ(ff()) dddd aa bb vvvvvvvvvvvv = ππ(ff(yy)) dddd aa if the revolution about -ais if the revolution about y-ais Chapter seven Integration Dr. Asawer A. Alwasiti 96

101 Find the volume of the solid generated by revolving the first quadrant area bounded by the parabola y = and its latus rectum =4 about the -ais. bb vvvvvvvvvvvv = ππ(ff()) dddd aa vvvvvvvvvvvv = ππ dddd 0 = 8π cubic unit The region enclosed by the semicircle yy = aa + and the -ais is revolved about the -ais to generate a sphere. Find the volume of the sphere. bb vvvvvvvvvvvv = ππ(ff()) dddd aa aa vvvvvvvvvvvv = ππ( aa + ) dddd aa = 4 3 ππ aa cccccccccc uuuuuuuu Find the volume of the solid generated by revolving the first quadrant area bounded by the parabola yy = Chapter seven Integration Dr. Asawer A. Alwasiti 97

102 and the lines y=1 and =4 about the line y=1. 4 vvvvvvvvvvvv = ππ( 1) dddd 1 = 7 ππ cccccccccc uuuuuuuu 6 The region between the curve = 1 generated a solid. Find the volume of the solid. bb vvvvvvvvvvvv = ππ(ff(yy)) dddd aa 4 vvvvvvvvvvee = ππ( 1 yy ) dddd 1 = ππππππ cccccccccc uuuuuuuu yy 1 y 4, is revolved about the y-ais to Find the volume of the solid generated by revolving the first quadrant area bounded by the parabola yy = 8 and its latus rectum = about the latus rectum. bb vvvvvvvvvvvv = ππ(ff(yy)) dddd aa 4 vvvvvvvvvvvv = ππ( ) dddd 4 vvvvvvvvvvvv = ππ( yy 8 ) dddd aa bb = 56 ππ cccccccccc uuuuuuuu 16 Chapter seven Integration Dr. Asawer A. Alwasiti 98

103 b- Washer Method Washer method for calculating volumes bb vvvvvvvvvvvv = ππ(rr () rr ())dddd aa R() = outer radius, r() = inner radius The region bounded by the curve y = +1 and the line y = -+3 is revolved about the -ais to generated a solid. Find the volume of the solid. + 1 = -+3 so =- and = 1 outer radius : R() = - +3 inner radius : r() = +1 bb vvvvvvvvvvvv = ππ(rr () rr ())dddd aa 1 vvvvvvvvvvvv = ππ(( + 3) ( + 1) )dddd = 117 ππ cccccccccc uuuuuuuu 5 The region bounded by the curve y = and the line y = in the first quadrant is revolved about the y-ais to generated a solid. Find the volume of the solid. = Chapter seven Integration Dr. Asawer A. Alwasiti 99

104 so =0 y=o and = y = 4 outer radius : R(y) = y inner radius : r(y) = y/ bb vvvvvvvvvvvv = ππ(rr (yy) rr (yy))dddd aa 4 vvvvvvvvvvvv = ππ(( yy) ( yy ) )dddd 0 = 8 ππ cccccccccc uuuuuuuu 3 The region bounded by the curve y = and the line y = in the first quadrant is revolved about the line = parallel to the y-ais. Find the volume swept out. = so =0 y=o and = 1 y = 4 outer radius : R(y) = y/ inner radius : r(y) = - y bb vvvvvvvvvvvv = ππ(rr (yy) rr (yy))dddd aa 4 vvvvvvvvvvvv = ππ(( yy ) ( yy) )dddd 0 = 8 ππ cccccccccc uuuuuuuu 3 Chapter seven Integration Dr. Asawer A. Alwasiti 100

105 c- Cylindrical Shells Alternative to Washer The Shell Method (Ais the y-ais) Suppose y = f() is continuous throughout an interval a b that does not cross the y-ais. Then, the volume of the solid generated by revolving the region between the graph of f and the interval a b about the y-ais is found by integrating πf() with respect to from a to b. bb vvvvvvvvvvvv = ππ(ssheeeeee rrrrrrrrrrrr)(ssheeeeee heeeeeehtt). dddd = ππππππ(). dddd aa The region bounded by the curve y =, the -ais and the line = 4 is revolved about the y-ais to generated a solid. Find the volume of the solid. bb aa vvvvvvvvvvvv = ππ(ssheeeeee rrrrrrrrrrrr)(ssheeeeee heeeeeehtt). dddd = ππππππ(). dddd aa bb aa bb 4 vvvvvvvvvvvv = ππππ. dddd 0 = 18 ππ cccccccccc uuuuuuuu 5 Chapter seven Integration Dr. Asawer A. Alwasiti 101

106 The region bounded by the curve y =, the -ais and the line = 4 is revolved about the -ais to generated a solid. Find the volume of the solid. y = 0 to y = cylinder radius = y cylinder height = 4 y bb vvvvvvvvvvvv = ππ(ssheeeeee rraaaaaaaaaa)(ssheeeeee heeeeeehtt). dddd = aa vvvvvvvvvvvv = ππππ(4 yy ). dddd 0 = 8ππ cccccccccc uuuuuuuu The region bounded by the curve y =, the y-ais and the line y = 1 is revolved about the line = to generated a solid. Find the volume of the solid. bb vvvvvvvvvvvv = ππ(ssheeeeee rrrrrrrrrrrr)(ssheeeeee heeeeeehtt). dddd = aa 1 vvvvvvvvvvvv = ππ( )(1 ). dddd 0 = 13 ππ cccccccccc uuuuuuuu 6 Chapter seven Integration Dr. Asawer A. Alwasiti 10

107 Chapter Eight Determinates A rectangular array of numbers like: AA = Is called a matri of order by 3 because it has rows and 3 columns. For matri A: a 11 =, a 1 = 1, a 13 = 3 a 1 =1, a = 0, a 3 = - Note: The first subscript denotes the numbers of the rows and the second subscript the number of columns. A matri with the same number of rows and columns is a square matri. A determinate: is a scalar-valued function whose domain is a set of square matri. For matri of order For matri of order AA = aa 11 aa 1 aa 1 aa AA = aa 11 aa 1 aa 1 aa det(a)= A = aa 11 aa 1 aa 1 aa = a 11 a - a 1 a 1 For matri of 3 order aa 11 aa 1 aa 13 AA = aa 1 aa aa aa 31 aa 3 aa 33 aa 11 aa 1 aa 13 det(a)=dddddd aa 1 aa 31 aa aa 3 aa = aa 33 A =aa 11 aa aa 3 aa 3 aa aa 1 aa 1 aa 3 33 aa 31 aa + aa 13 aa 1 aa 33 aa 3 Chapter eight Determinates Dr. Asawer A. Alwasiti 103 aa 31

108 = a 11 A 11 a 1 A 1 + a 13 A 13 in which A 11, A 1, A 13 are the minors of a 11, a 1, a 13 respectively. A =a 11 (a a 33 a 3 a 3 ) a 1 (a 1 a 33 - a 3 a 31 ) + a 13 (a 1 a 3 a a 33 ) Evaluate the determinate of the matri 3 4 AA = det(a) = A = H.W =6 ssssssss cccccccc tttttttt Evaluate ans. = -1 cccccccc ssssssss cccccccc Properties of Determinates 1- The value of the determinate is unchanged if all the corresponding rows and columns are interchanged. i.e. aa 11 aa 1 aa 1 aa = aa 11 aa 1 aa 1 1 AA = Transport matri aa AA = dddddd 4 3 = = Chapter eight Determinates Dr. Asawer A. Alwasiti 104

109 1 3 dddddd 4 1 = = If all the elements of a row or a column of a determinate are multiplied by the same number k, then the value of the determinate is multiplied by k = = If any two rows or any two columns of a determinate are interchanged then the sign of the value of the determinate is changed. 1 3 AA = 3 1 iiiiii dddddddddddddddddddddd iiii Prove that det(b) = -36 if BB = det(b)= = The determinate is zero if a- Two rows or columns of a matri is identical Find the determinate of the matri 1 AA = det(a)= A = = 0 Chapter eight Determinates Dr. Asawer A. Alwasiti 105

110 b- Every element in any row or any column is zero Cramers Rule a 11 + a 1 y =b 1 a 1 + a y =b If D 0 So D = = bb 1 aa 1 bb aa DD D y =y = aa 11 bb 1 aa 1 bb DD 1 3 AA = det(a)= A = = AA = det(a)= A = =0 Solve the system 3 y =9 solution D = = 7 = DD y = DD yy = DD 7 = DD 7 + y = -4 = = 3 Chapter eight Determinates Dr. Asawer A. Alwasiti 106

111 Systems of three equations in three unknowns work, it can be solved in the same way: a 11 + a 1 y + a 13 z = b 1 a 1 + a y + a 3 z = b a 31 + a 3 y + a 33 z = b 3 Solve the equations by Cramers rule = = = 0 4 DD = 1 = = = = 1 6 = = = DD DD = = 1 bb 1 aa 1 aa 13 DD bb aa aa 3 bb 3 aa 3 aa 33 DD yy DD = yy = 1 aa 11 bb 1 aa 13 DD aa 1 bb aa 3 aa 31 bb 3 aa 33 DD DD = = 1 aa 11 aa 1 bb 1 DD aa 1 aa bb aa 31 aa 3 bb 3 Chapter eight Determinates Dr. Asawer A. Alwasiti 107

112 H.W 1 Solve = = = = 0 ans. 1 =1, =-1, 3 =-1, 4 =1 - Solve 3y + = z z = 8 5y 3z - 1 = y ans. =3, y=-1, z= Chapter eight Determinates Dr. Asawer A. Alwasiti 108

113 Definition: p(r, θ) Chapter Nine Polar Coordinate r: the direct distance from 0 to p. θ: the direct angle from the initial ray to the segment op. Chapter nine Polar Coordinate Dr. Asawer A. Alwasiti 109