P 0 (x 0, y 0 ), P 1 (x 1, y 1 ), P 2 (x 2, y 2 ), P 3 (x 3, y 3 ) The goal is to determine a third degree polynomial of the form,

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1 Bezier Curves While working for the Renault automobile compan in France, an engineer b the name of P. Bezier developed a sstem for designing car bodies based partl on some fairl straightforward mathematics. Although such a design would naturall take place in three dimensions, a two dimensional discussion will permit the reader to glimpse the power of Bezier s concept. A (cubic) Bezier curve is completel determined b four consecutive points: P 0 ( 0, 0 ), P (, ), P (, ), P 3 ( 3, 3 ) The goal is to determine a third degree polnomial of the form, which "fits" these points according to the following criteria, = a 3 + b + c + d () a) The polnomial is to pass through the endpoints, P 0 and P 3, and, b) The slope at the endpoints is to agree with the slope determined b the endpoint and the adjacent point. The polnomial does not pass through the two intermediar points, P and P, called control points but rather their role is to influence the shape of the curve passing through P 0 and P 3. Eample Suppose we wish to construct a Bezier curve through the points, P 0 ( 3, ), P (, 3)), P (, ) and P 3 (3, 3) The substitution of these coordinates into () would ield the following sstem of equations, 7a + 9b 3c + d = 7a + 9b + 3c + d = 3 7a 6b + c = 7a + 6b + c This sstem of equations can be solved with the help of IMAGE-Algebra & Geometr. The solutions for a, b, c and d ield the cubic, = Submitting this to IMAGE-Calculus, we obtain the following graph, c Chris D Arc = 5

2 IMAGE Function f(): f()=- 3/7- /+7/ c Chris D Arc

3 The Binomial Theorem Given n, r W, and the following definition from Combinatorial mathematics, ( ) n = r n! (n r)!r! where n! = n (n ) (n )... 3, we can state the Binomial Theorem as, n ( ) n (p + q) n = p n r q r r r=0 ( ) ( ) n n = p n 0 q 0 + p n q + 0 = n! n! n!0! pn + (n )!! pn q + = p n + np n q + ( ) n p n q n(n ) p n q q n ( ) n p n n q n n n! (n )!! pn q n! 0!n! qn Eamples a) (3 + ) 4 = = 4 ( ) 4 (3) 4 r r r r=0 ( ) 4 (3) ( ) 4 (3) 4 + = ( ) 4 (3) ( ) 4 (3) b) ( )3 = = 3 ( ) 3 3 r ( r )r r=0 ( ) ( 0 )0 + ( ) 3 3 ( ) + ( ) 3 3 ( ) + ( ) ( 3 )3 = Application: Probabilit A Binomial Eperiment is one in which we define two outcomes - one deemed a success which denote b the letter, p, and the other a failure, denoted b q. Tossing a coin is an obvious eample. Since each outcome, a head or a tail is equall likel, or will 50% of the time, we make the assignments, p = q =. If we perform the eperiment n times, the Binomial Theorem can suppl us with a complete probabilit distribution. Consider a coin tossed three times. Here is the probabilit distribution, (p + q) 3 = p 3 + 3p q + 3pq + q 3 ( + ( ) 3 ( ) ( )3 = + 3 = ) + 3 ( ) ( ) + ( ) 3 c Chris D Arc

4 Comple Numbers Attempts to find roots of the equation, + = 0 within the set of real numbers proves fruitless. A new set of numbers was created to satisf this equation and ones like it, the comple numbers, denoted as the set C. The set of comple numbers is defined as the set of binomial epressions of the form a + bi in which a, b R and i =, C = { a + bi a, b R} The Comple (Argand) Plane The Cartesian Plane, R R, or R, is suitable for plotting ordered pairs of the form (, ) where, R. The Comple Plane consists of the ordered pairs (, ) standing for + i. The Magnitude of a Comple Number Given the comple number z = a + bi, we define the magnitude of z, written as z, to be, z = a + b This can be easil seen as the distance from z to the origin of the Comple Plane. The Comple Conjugate Earl in our stud of algebra, we encountered the patterned factoring of a difference of squares, a b = (a b)(a + b) We defined the factors as a pair of conjugate binomials: same terms, opposite sign. Similarl, the conjugate of the comple number, z = a + bi is denoted and defined as z = a bi. The Arithmetic of Comple Numbers Given the two comple numbers, z = a + bi and z = c + di we define the four arithmetic operations as follows, Addition: Subtraction: Multiplication: z + z = (a + bi) + (c + di) = (a + c) + (b + d)i z z = (a + bi) (c + di) = (a c) + (b d)i z z = (a + bi) (c + di) = ac + adi + bci + bdi = (ac bd) + (ad + bc)i Division: For division, we rel on an approach similar to rationalizing the denominator and the comple conjugate, z = z z = a + bi z z z c + di c di c di (ac + bd) + (bc ad)i = c + d (ac + bd) (bc ad) = + c + d c + d i c Chris D Arc

5 The Square Root of a Comple Number For z we simpl perform z z, but what meaning can be given to z = a + b? For this we could determine the equivalence, a + bi = ( + i) () Epanding () unfolds as follows, Comparison ields the sstem of equations, a + bi = ( + i) ( + i) = ( ) + ()i = a = b Given an a and b the outcome of this sstem can be determined. Eamples Given z = 5 + i and z = + i, observe the following, z = 5 + i = ( 5) + () = = 69 = 3 z 5 + i = i z + i i ( 5 + 4) + (0 + )i = + 4 = i z i = ( + i)i = i + i = + i We proceed, z = 5 + i ( + i) = 5 + i ( ) + ()i = 5 + i Equating real and imaginar parts leads to = 5 and =. Solving simultaneousl unfolds as follows, = = 6 Substitution ields, ( ) 6 = 5 36 = = 0 ( + 9)( 4) = 0 From the roots, = ±, we determine = ±3. Thus 5 + i = + 3i or 5 + i = 3i. c Chris D Arc

6 ɛ δ Definition of a Limit Few statements in elementar mathematics appear as crptic as the one defining the limit of a function f () at the point = a, ɛ > 0 δ > 0 such that f () L < ɛ whenever a < δ (Translation: for ever epsilon greater than zero, there eists a delta greater than zero such that f () lies within epsilon units of the limit L whenever lies within delta units of a) With the aid of a graphic interpretation, the definition is less intimidating. f()+ L f() - f() a - a a+ The definition sas that the limit of f () at = a is L if and onl if an interval of the domain around a can be found for ever interval of the range around L. In other words, to defend a limit result, we must produce a value for δ in response to the challenge of an ɛ. Eample Should we have to defend the result, lim 4 5 = 3, we would proceed as follows, (4 5) 3 < ɛ 4 8 < ɛ 4 < ɛ < ɛ 4 Since we have determined that δ need onl be taken as 3 of an ɛ, the limit result of 3 is substantiated. c Chris D Arc

7 Eponent Rules (a) a b = a+b (b) a b = a b (c) () a = a a (d) ( ) a = a a (e) a = a (f) 0 = (g) n = n (h) m n = n m Logarithm Rules (a) log = log + log (b) log = log log (c) log = log (d) log = 0 (e) log e = ln (f) log = log a log a = ln ln (Change of Base Formula) (g) ln e = (h) = e ln c Chris D Arc

8 Factorial Notation Locate a button on our calculator with either n! (or!) etched on it (it could possibl be a second function). This is read as n factorial, and it s mathematical definition is, Given n W, n! = { ; n=0 n (n ) (n )... 3 ; n>0 This is to sa that if ou enter a 5 and then press n!, our calculator will displa a result of 0, (5 4 3 ). Eamples a) 7! = = 5040 b) 7! 5! = 7 6 5! 5! = 4 8! c) 5!3! = ! 5! 3 = = 56 Application: Permutations There is a branch of discrete mathematics dealing with arrangements and combinations of items. One formula from this area eists which will tell us the number of arrangements that can be made of sa, r objects taken from within a larger pool of sa, n items. This formula is, n! P (n, r) = () (n r)! The following eploits this formula. Assume that there are 7 students ving for the four student council positions, President, Vice-President, Secretar and Treasurer. How man different was could the election turn out? We are faced are with the problem of picking 4 people from a pool of 7. Using () above we find, P (7, 4) = 7! (7 4)! = ! = = 840 3! Application: Combinations In the previous eample, we sa order is important. B this we mean that if after the election the elected Secretar and Treasurer echanged roles, we would have had a different arrangement! With combinations however, order is not important, and hence an echange of duties from the four elected would not be taken as a new combination. For combinations, we modif () to obtain, C(n, r) = ( ) n = r n! (n r)!r! Using our election eample one last time, asking how man combinations of 4 people can be made from a pool 7 of is like asking how man different was could a team of 4 students be made, without regard to which role the pla. For this we use () and find, C(7, 4) = ( ) 7 = 4 7! (7 4)!4! = ! = 7 5 = 35 3!4! () c Chris D Arc

9 More Eamples a) 4! 4 3! = = 4 3! 3! b) 5! 5 5! = = 5 5! 5! c) 0! ! = 99! 99! d) 0 9! = 0! e) 90 8! = 0 9 8! = 0! f) 30 4! = 6 5 4! = 6! = 0 00 = 0 00 c Chris D Arc

10 Graphing Fundamentals The abilit to generate the graph of a function is a useful skill to master. Indeed, the burden of having to find a solution to a problem can be lessened significantl if one can epress the contet visuall. This stud introduces ver little in the wa of new material, instead it will squeeze greater understanding from the concepts ou currentl possess. The ke to rapid, successful graphing is the appreciation of the inseparable links between algebra and geometr. The sincere student should find these pages well worth the effort. Basic Shapes Arithmetic Transformations Sketching functions requires little more than a firm grasp of the three areas depicted above. We will start b reviewing Basic Shapes. Basic Shapes This section will identif a number of fundamental families of functions. Your responses to the questions and the corresponding graphic should be committed to memor. Constant Functions, f () = k, k R The graph of a constant function is a horizontal line. With a straight edge, sketch and label the following constant functions, (a) f () = 0 (b) g() = (c) h() = (d) State the domain of each function. (e) State the range of each function. (f) Identif an smmetr observed for this famil c Chris D Arc

11 Power Functions, f () = n, n N (a) Identif the common smmetrical nature of the functions depicted in the adjacent graphic. (b) Can ou suggest and label the equation of each? (c) State the common domain and range of each member of this famil, D = { R R = { R (a) Identif the common smmetrical nature of the functions depicted in the adjacent graphic. (b) Can ou suggest and label the equation of each? (c) State the common domain and range of each member of this famil, D = { R R = { R Trigonometric Functions, f () = sin, g() = cos (a) Identif the common smmetrical nature of the functions depicted in the adjacent graphic. (b) Can ou suggest and label the equation of each? (c) State the common domain and range of each member of this famil, D = { R R = { R (a) Identif the common smmetrical nature of the functions depicted in the adjacent graphic. (b) Can ou suggest and label the equation of each? (c) State the common domain and range of each member of this famil, D = { R R = { R c Chris D Arc

12 QuickSketch A competent mathematician must be a skilled geometer. Whereas techniques in algebra provide ou with the micro precision required to defend the accurac of our results to an arbitrar number of decimal places, the purpose of geometr is to provide ou with the macro or global insight into a concept. The foundation of the geometric method rests with a sound understanding of fundamental algebraic operations. This worksheet has been prepared in the hope of pointing ou in the right direction. Square: f () = One of the first operations ou learn in mathematics is that of squaring. The number line presented below crudel summarizes this algebraic transformation. Note an value(s) that do not change under the squaring transformation (invariant point). Appl this summar across the domain of f () = to produce the graph of f () =, Invariant = 0 - c Chris D Arc

13 Cube: f () = 3 Use the number line presented at the left below to summarize the general results for cubing real numbers (note an invariant points). Appl this summar across the domain of f () = to produce the graph of f () = 3, Invariant = 0 - Square Root: f () = The inverse operation to squaring is taking the square root. Summarize the algebraic effects of taking the square root of a real number on the number line and appl these transformations across the domain of f () = to produce the graph of f () =, Invariant = 0 - c Chris D Arc

14 Cube Root: f () = 3 Use the number line presented at the left below to summarize the general results for taking the cube root of real numbers. Appl this summar across the domain of f () = to produce the graph of f () = 3, Invariant = 0 - Absolute Value: f () = One of the simplest algebraic transformations involves the absolute value of a number. The result is simpl the distance the number is from 0 (distance can t be negative). Summarize the algebraic effects of taking the absolute value of a real number and appl these transformations across the domain of f () = to produce the graph of f () =, Invariant = 0 - c Chris D Arc

15 Reciprocal: f () = Use the number line presented at the left below to summarize the general results for taking the reciprocal of a real number. Appl this summar across the domain of f () = to produce the graph of f () =, Invariant = 0 - Eample : f () = The graph of this function should be seen as appling the reciprocal transformation to the function, g() = Recall the details of the reciprocal transformation on the number line left below to produce the graph of f () =, g()= 0 - Invariant c Chris D Arc

16 Eample : f () = 3 + The graph of this function should be seen as appling the absolute value transformation to the function, g() = 3 +. Recall the details of the absolute value transformation on the number line left below to produce the graph of f () = 3 +, g()= Invariant Eample 3: f () = (sin ) The graph of this function should be seen as appling the squaring transformation to the function, g() = sin Recall the details of the squaring transformation on the number line left below to produce the graph of f () = (sin ), 0 - g()= sin( ) Invariant c Chris D Arc

17 Eercises. Develop the graphs of the given functions using the techniques presented. a) f () = b) f () = 3 = 3 c) f () = d) f () = 3 e) f () = 4 f) f () = sin. State the domain and range of each of the functions in. c Chris D Arc

18 Indefinite Integration: An Interesting Evaluation Using the definition of absolute value, we recall the derivative of =, 0 as, = d d = d d = d d = d d d d = = = The integral d, can be epressed as d and we can proceed with Integration b Parts b letting u =, dv = d d = = d d = d = d Adding d to both sides ields, d = = + C Demonstration Determine the area under the function, f () =, on the partition, [, 3]. 3 [ ] 3 3 A = d =.5 = = = 5u 0 3 c Chris D Arc

19 Applications of the Definite Integral: Arc Length A formula introduced in intermediate mathematics, d = ( ) + ( ) () ields the straight line distance, d, between two points. In addition to straight segments, we can also determine the length of an arc of a circle, l, with the aid of, θ π = a πr = l πr However, we must turn to the definite integral for a general method of quantifing non-linear measures. In our earlier stud of limits, ou were asked to conceptualize the circle as the limit of a n-sided regular polgon where n. In particular, the circumference of the circle arises from the sum of the infinite number of infinitesimall short sides of the regular polgon. We can adapt this concept to the problem of finding the length of an arc of a function. From our stud of differentials, we are aware that can be approimated b d where d = f () d () Using a Riemann Sum approach, we can incorporate concepts () and () into the development of an integral formula for the length, s, of an arc of the function, f (). ds s d = d We must first develop ds as an epression for s. The Pthagorean relationship, can be approimated through the use of differentials, substituting (), we simplif as follows, ( s) ( ) + ( ) (ds) = (d) + (d) (ds) = (d) + (f () d) (ds) = ( + [f ()] ) (d) ds = + [f ()] d The formula for the length of an arc of the function, f (), over the continuous closed interval [a, b] is completed as follows, s = lim n s n = lim n = lim n = b a n s i= n + [f ()] d i= + [f ()] d c Chris D Arc

20 Questions. Confirm the formula for the circumference of a circle through the use of the definite integral. 3. Determine the length of the arc of the parabola = over the interval [0,]. (Ans units) Determine the total length of the hpoccloid of four cusps whose equation can be given b =. (Ans. 6 units) - - c Chris D Arc

21 π The set of real numbers, R, is a union of the rational numbers, Q, and the irrational numbers, Q, R = Q Q Your first investigation of the qualitative properties of the circle produced a formula for it s circumference, C, C = πr which can be reepressed as, π = C r = C d From this point on, we recognize π as the constant ratio of the circumference of a circle to it s diameter. Since π Q, we can not know an eact value for this constant. A crude approimation of this constant is taken as 3.4. A slightl better epression is fraction, An even better rational approimation is 3. All sorts of interesting computational techniques have been devised, including this one handed to me b Dr. Skalinski, π Using series, it is possible to obtain as close an approimation to the value of π as ou wish. Man series eist, two of which are presented below. These can be evaluated b hand, but if ou know a computer language, it is fairl painless and quite rewarding to write a program to sum the terms of each series. Series A Series B π = 4 n=0 ( ) n n + π = n=0 ( ) 4 6 n 8n + 8n + 4 8n + 5 8n + 6 Each successive term of these infinite series is smaller than it s predecessor. We can interpret this to mean that the difference between our approimate value and the theoretical values of π is shrinking. One could deem one series superior to another based on how fast this difference shrinks. Which of the two series presented is superior according to this criteria? c Chris D Arc

22 Factoring: Differences and Sums The following results are well known, and, a b = (a b)() () a b = (a b)(a + b) () The question is now asked, "Can we etend these results to factor a n b n for n >, n R?". Let us eamine the case for n = 3. For n = and n =, we observe the first factor to be a b. We could assume this pattern continues, and test our assumption through division: (a 3 b 3 ) (a b). Our assumption is correct, and we discover, Putting (), () and (3) together reveals a pattern that can easil be verified, a 3 b 3 = (a b)(a + ab + b ) (3) a b = (a b)() a b = (a b)(a + b) a 3 b 3 = (a b)(a + ab + b ) a 4 b 4 = (a b)(a 3 + a b + ab + b 3 ) To confirm whether the previous factoring is correct, we would simpl multipl. Generalizing this result produces, n a n b n = (a b)(a n + a n b + a n 3 b ab n + b n ) = (a b) a n r b r (4) r=0 We shall now eamine the factoring of sums. In searching for potential factors of the epression, +, there are man avenues one could take. M personal favourite is to think geometricall. In the stud, Factors and Roots, we were reminded of the inseparable link between algebra and geometr. Bringing this to bear on our immediate problem, we can easil visualize the graph of the function f () = +. Since the graph of the function fails to make contact with the -ais, we conclude there are no roots, hence, no linear factors. To this point, we have, + = ( + )() (5) + no factors (6) For n = 3, we could visualize the function, f () = 3 +. Our mind s ee spots one (and onl one) -intercept, leading to a single linear factor. It would not be too great a leap to speculate that the root might be -, so we could hazard a guess that one factor of 3 + might be +. Division ields the following, 3 + = ( + )( + ) (7) A few seconds of mental imager rules out linear factors for the epression, 4 +. The same logical approach we pursued to arrive at (7) can be used to discover factors in the case of n = 5, 5 + = ( + )( ) (8) We can now put (5), (6), (7) and (8) together to build a general result for factoring a sum for odd n, that can be verified, n a n + b n = (a + b)(a n a n b + a n 3 b... ab n + b n ) = (a + b) ( ) r a n r b r (9) r=0 c Chris D Arc

23 Surface Area of a Solid of Revolution Consider a region bounded b the function, f (), the -ais and the lines = a and = b. Sweeping this area around the -ais generates an object figure known appropriatel as a solid of revolution. For such objects we have previousl eplored the formula for the volume through the use of the definite integral, b V = π [f ()] d a We now turn to the development of a formula for the surface area of the solid of revolution. From a Riemann Sum perspective, the formula for volume was developed b considering the sum of an infinite number of clindrical volumes whose base radii were dictated b f (). For surface area, we return to the notion of the varing clindrical forms but consider instead their lateral areas. The lateral area of a clinder can be treated as a rectangle with dimensions πr which can adjusted to πf () ds where ds is our epression for a slice of arc length from a previous investigation, ds = + [f ()] d Our formula for surface area is built as follows, SA = lim n sa n = lim n = lim n = π n sa i= n π f () + [f ()] d i= b a f () + [f ()] d Questions. Confirm the formula for the surface area of a sphere through the use of the definite integral.. Find the surface area of a zone of altitude h of a sphere of radius r. (A zone is the portion of a sphere included between two parallel planes; the altitude of a zone is the distance between the two planes) 3. Confirm the formula for the surface area of a right circular cone. c Chris D Arc

24 The Area of a Triangle Numerous techniques eist for determining the area of a triangle. Four such approaches are reviewed in this document.. Arithmetic. Trigonometr 3. Heron s Formula 4. Analtic Geometr The best method depends on the information provided.. Arithmetic To each side of a triangle, an altitude can be determined. Letting the side be the base, b, and its altitude be h, the area, A, of the triangle can be epressed as A = bh () The rationale for () is straightforward. An triangle can define a rectangle containing it whose area is bh. The triangle consumes one-half of the area of the rectangle. h b c Chris D Arc

25 . Trigonometr Given the lengths of two sides of a triangle and the measure of the contained angle, a height can be determined using trigonometr. a h C b In the triangle above we have, sin C = h a h = a sin C () Substituting () into () ields a convenient formula for the area, A given two sides and the contained angle A = ab sin C (3) 3. Heron s Formula Should the lengths of the three sides of a triangle be readil known, Heron s formula suppling a value for the area, A, is hand, A = s(s a)(s b)(s c) (4) where a, b, and c are the lengths of the sides of the triangle and s is the semi-perimeter. s = (a + b + c) (5) a c b c Chris D Arc

26 4. Analtic Geometr The Arithmetic approach eamined earlier ielded an area of bh for a side taken as the base, b, and the corresponding height, h. This formula can be adapted to to develop an approach using coordinate geometr. P (,) P (,) h S b P 3( 3,3) Let P (, ), P (, ), and P 3 ( 3, 3 ) be the vertices of an triangle. Let P 3 P = b be the base of the triangle. Let P S = h be the height of the triangle. We have where b = ( ) + ( ) (6) = 3 and = 3 We can now develop and epression for h using the formula for the distance, d, from a point, (, ) to a line, A + B + C = 0, d = A + B + C A + B (7) The equation of the line through P and P 3 can be developed as follows, Substituting (6) and (8) into (7) ields an epression for h. = 3 ( ) ( ) 3 = ( ) ( ) 3 ( ) ( ) + ( ) 3 ( ) 3 = 0 (8) h = ( ) ( ) + ( ) 3 ( ) 3 ( ) + ( ) The area of P P P 3 can epressed as P P P 3 = bh = ( ) + ( ) ( ) ( ) + ( ) 3 ( ) 3 ( ) + ( ) = ( ) ( ) + ( ) 3 ( ) 3 (9) c Chris D Arc

27 Substituting 3 for and 3 for in (9) ields a final formula for the area of our triangle. P P P 3 = ( 3 ) ( 3 ) + ( 3 ) 3 ( 3 ) 3 = = ( (0) The following pattern can simplif the remembrance of the above formula. Points Down Products Up Products (, ) (, ) (, ) 3 (, ) Sum of the Down Products = Sum of the UpProducts = Eample Area = (Sum of the Down Products) (Sum of the Up Products) Find the area of ABC with coordinates A(3, ), B(, ) and C(, 3). Area ABC = ( ) ( 9 ) = = 8 The area of the triangle is 8 square units. A 3 B - - C -3 A 3 B(-,-) C(,-3) A(3,) c Chris D Arc