Higher Mathematics Notes

St Andrew s Academ Mathematics Department Higher Mathematics Notes 018-19 Page 1 of 107

Circle: The equation g f c Higher Maths Formulae List 0 represents acircle centre (-g, -f) and radius g f c The equation a b r represents a circle centre (a, b) and radius r Scalar Product: a. b a b cos, where θ is the angle between a and b or a. b a1b 1 ab a3b3 where a1 a a a 3 and b1 b b. b 3 Trigonometric Formulae: sin A B cos A B = sin Acos B cos AsinB = cos Acos A sin AsinB sina = sin Acos A cos A = cos A sin A = cos A 1 = 1 sin A Table of Standard Derivatives: f f' sina acos a cos a a sina Table of Standard Integrals: f f d sina 1 cosa C a cosa 1 sina C a Page of 107

The Straight Line Revision from National 5 The graph of = m + c is a straight line, where m is the gradient and c is the -intercept. Gradient is a measure of the steepness of a line. The gradient of the line joining points A ( 1, 1) and B (, ) is given b: m AB 1 1 A (1, 1) 1 B (, ) 1 Eample 1: Find: a) the gradient and -intercept of the line = + 5 b) the equation of the line with gradient 4 and -intercept (0, -) c) the gradient of the line joining P (-, 4) and Q (3, -1) d) the gradient of the line 3 + 4 11 = 0 Equation of a Straight Line: b = m( a) Points A (a, b) and P (, ) both lie on a straight line. The gradient of the line b m. Rearranging this gives: a b = m( a) P (, ) NTE: when ou are asked to find the equation of a straight line, it is fine to leave it in this form (unless ou are specificall asked to remove the brackets). A (a, b) Eample : Find the equations of the lines: a) through (4, 5) with m = b) joining (-1, -) and (3, 10) c) parallel to the line + 4 = 0 and passing through the point (, -3) Page 3 of 107

Eample 3: Find the equation of the line through (-5, -1) with m =, giving our answer in the 3 form A + B + C = 0. Equation of a Straight Line = m + c AND A + B + C = 0 Eample 4: Sketch the line 5 4 = 0 b finding the points where it crosses the and aes. The Angle with the -ais The gradient of a line can also be described as the angle it makes with the positive direction of the ais. As the -difference is PPSITE the angle and the -difference is ADJACENT to it, we get: m AB tanθ (where θ is measured CLCKWISE from the -ais) A (1, 1) θ θ 1 B (, ) 1 Eample 5: Find the angle made with the positive direction of the -ais and the lines: a) = 1 b) = 5 3 c) joining the points (3, -) and (7, 4) Gradients of straight lines can be summarised as follows: a) lines sloping up from left to right have positive gradients and make acute angles with the positive direction of the -ais b) lines sloping down from left to right have negative gradients and make obtuse angles with the positive direction of the -ais c) lines with equal gradients are parallel d) horizontal lines (parallel to the -ais) have gradient zero and equation = a e) vertical lines (parallel to the -ais) have gradient undefined and equation = b Page 4 of 107

Collinearit If three (or more) points lie on the same line, the are said to be collinear. Eample 6: Prove that the points D (-1, 5), E (0, ) and F (4, -10) are collinear. Perpendicular Lines If two lines are perpendicular to each other (i.e. the meet at 90 ), then: m 1m = -1 Eample 7: Show whether these pairs of lines are perpendicular: a) + + 5 = 0 7 = 0 b) 3 = 5 3 = + 9 c) = - 5 6 = 10 3 When asked to find the gradient of a line perpendicular to another, follow these steps: 1. Find the gradient of the given line. Flip it upside down 3. Change the sign (e.g. negative to positive) Eample 8: Find the gradients of the lines perpendicular to: a) the line = 3-1 b) a line with gradient = -1.5 c) the line + 5 = 0 Eample 9: Line L has equation + 4 + = 0. Find the equation of the line perpendicular to L which passes through the point (-, 5). Page 5 of 107

Midpoints and Perpendicular Bisectors The midpoint of a line lies eactl halfwa along it. To find the coordinates of a midpoint, find halfwa between the and coordinates of the points at each end of the line (see diagram). The coordinate of M is halfwa between - and 8, and its coordinate is halfwa between 5 and -3. In general, if M is the midpoint of A ( 1, 1) and B (, ): M 1, 1 The perpendicular bisector of a line passes through its midpoint at 90. Eample 10: Find the perpendicular bisector of the line joining F (-4, ) and G (6, 8). A (-, 5) To find the equation of a perpendicular bisector: M B (8, -3) Find the gradient of the line joining the given points Find the perpendicular gradient (flip and make negative) Find the coordinates of the midpoint Substitute into b = m( a) Lines Inside Triangles: Medians, Altitudes & Perpendicular Bisectors In a triangle, a line joining a corner to the midpoint of the opposite side is called a median. A A line through a corner which is perpendicular to the opposite side is called an altitude. A A line at 90 to the midpoint is called a perpendicular bisector. A C The medians are concurrent (i.e. meet at the same point) at the centroid, which divides each median in the ratio :1 The median alwas divides the area of a triangle in half. A solid triangle of uniform densit will balance on the centroid. B The altitudes are concurrent at the orthocentre. The orthocentre isn t alwas located inside the triangle e.g. if the triangle is obtuse. For all triangles, the centroid, orthocentre and circumcentre are collinear. B The perpendicular bisectors are concurrent at the circumcentre. The circumcentre is the centre of the circle touched b the vertices of the triangle. B Page 6 of 107

Eample 11: A triangle has vertices P (0, ), Q (4, 4) and R (8, -6). a) Find the equation of the median through P. b) Find the equation of the altitude through R. To find the equation of a median: Find the midpoint of the side opposite the given point Find the gradient of the line joining the given point and the midpoint Substitute into b = m( a) Distance between Two Points To find the equation of an altitude: Find the gradient of the side opposite the given point Find the perpendicular gradient (flip and make negative) Substitute into b = m( a) The distance between an two points A ( 1, 1) and B (, ) can be found easil b Pthagoras Theorem. If d is the distance between A and B, then: A (1, 1) B (, ) 1 d 1 1 1 Eample 1: Calculate the distance between: a) A (-4, 4) and B (, -4) Eample 13: A is the point (, -1), B is (5, -) and C is (7, 4). Show that BC = AB. b) X (11, ) and Y (-, -5) Page 7 of 107

Past Paper Eample 1: The vertices of triangle ABC are A(7, 9), B(-3, -1) and C(5, -5) as shown: The broken line represents the perpendicular bisector of BC a) Show that the equation of the perpendicular bisector of BC is 5 A B C b) Find the equation of the median from C c) Find the co-ordinates of the point of intersection of the perpendicular bisector of BC and the median from C. Past Paper Eample : The line GH makes an angle of 30 with the -ais as shown in the diagram opposite. H What is the gradient of GH? 30 G 0 Page 8 of 107

Sets and Functions A set is a group of numbers which share common properties. Some common sets are: Natural Numbers N = {1,, 3, 4, 5, } Whole Numbers W = {0, 1,, 3, 4, 5,..} Integers Z = {..,-3, -, -1, 0, 1,, 3,..} Rational Numbers Q = all integers and fractions of them (e.g. ¾, -⅝, etc) Real Numbers R = all rational and irrational numbers (e.g.,, etc. ) Sets are written inside curl brackets. The set with no members { } is called the empt set. means is a member of, e.g. 5 {3, 4, 5, 6, 7} means is not a member of, e.g. 5 {6, 7, 8} A function is a rule which links an element in Set A to one and onl one element in Set B. Set A Set B Set A Set B Domain Range This shows a function Domain Range This does not show a function The set that the function works on is called the domain; the values produced are called the range. For graphs of functions, we can think of the domain as the values, and the range as the values. This means that an operation which produces more than one answer is not considered a function. For eample, since 4 = and -, f() = is not considered a function. Eample 1: Each function below is defined on the set of real numbers. State the range of each. a) f( ) = sin b) g( ) = c) h( ) = 1 When choosing the domain, two cases MUST be avoided: a) Denominators can t be zero b) Can t find the square root of a negative value 1 e.g. For f() = 5, -5, i.e. { R : -5} e.g. For g() = 3, 3, i.e. { R : 3 } Eample : For each function, state a suitable domain. a) g( ) = 3 b) p() = 5 c) f( ) = 1 Page 9 of 107

Composite Functions In the linear function = 3 5, we get b doing two acts: (i) multipl b 3; (ii) then subtract 5. This is called a composite function, where we do a function to the range of another function. e.g. If h() is the composite function obtained b performing f() on g(), then we sa Eample 3: f() = 5 + 1 and g() = 3 +. h() = f(g()) ( f of g of ) a) Find f(g(-1)) b) Find f(g()) c) Find f(f()) d) Find g(f()) NTE: Usuall, f(g()) and g(f()) are NT the same! Eample 4: f() = + 1, g() = + 6 a) Find formulae for: b) Solve the equation f(g()) = g(f()) (i) f(g()) (ii) g(f()) Eample 5: f() = 3, -1. Find an epression for f(f( )), as a fraction in its simplest form. 1 Page 10 of 107

Past Paper Eample: Functions f and g are defined on a set of real numbers b a) Find epressions for: (i) f(g()) 3 g 4 f (ii) g(f()) b) Show that f(g()) + g(f()) = 0 has no real roots Page 11 of 107

Recurrence Relations A recurrence relation is a rule which produces a sequence of numbers where each term is obtained from the previous one. Recurrence relations can be used to solve problems involving sstems which grow or shrink b the same amount at regular intervals (e.g. the amount of mone in a savings account which grows b 3.5% p/a, the volume of water left in a pool if 10% evaporates each da, etc). Recurrence relations are generall written in one of two forms: Un 1 aun b In both cases, a term is found b multipling the previous term b a constant a, then adding (or subtracting) another constant b. R U n means the n th term in the sequence (i.e. U 7 aun- b would be the 7 th term, etc). U 0 ( U zero ) is the starting point of the sequence, e.g. the amount of mone put into an account before interest is added. Un 1 Eample 1: A sequence is defined b the recurrence relation U n+1 = 3U n +, U 0 = 4. Eample : A sequence is defined b the recurrence relation U n = 4U n 1 3, where U 0 = a. Find the value of U 4. Find an epression for U in terms of a. Finding a Formula Recurrence relations can be used to describe situations seen in real life where a quantit changes b the same percentage at regular intervals. The first thing to do in most cases is find a formula to describe the situation. Eample: Jennifer puts 5000 into a high-interest savings account which pas 7.5% p/a. Find a recurrence relation for the amount of mone in the savings account. Solution: Starting amount = 5000 After 1 ear: amount = starting amount + 7.5% (i.e. 107.5% of starting amount) = 1.075 starting amount Eample 3: Find a recurrence relation to describe: a) The amount left to pa on a loan of 10000, with interest charged at 1.5% per month and fied monthl paments of 50. Recurrence relation is: U n+1 = 1.075U n (U 0 = 5000) b) The amount of water in a swimming pool of volume 750,000 litres if 0.05% per da is lost to evaporation, but 350 litres etra is added dail. Eample 4: Bill puts lotter winnings of 10000 in a bank account which pas 5% interest p/a. After a ear, he decides to spend 0000 per ear from the mone in the account. a) Find a recurrence relation to describe the amount of mone left each ear. Page 1 of 107

b) How much mone will there be in the account after five ears? c) After how man ears will Bill s mone run out? Limits of Recurrence Relations Some recurrence relations produces sequences which tend towards a limit, i.e. as the number of terms increases, the sequence gets closer and closer to a certain value without actuall meeting it. The graph opposite shows two sequences generated b the recurrence relation U n+1 = 0.75U n + 100, one where U 0 = 100, the other where U 0 = 600. In both cases, irrespective of the value of U 0, as U n approaches infinit, the sequences tend to a limit of 400. How can we find the value of these limits without plotting a graph? 100 1000 800 600 400 00 0 0 5 10 15 0 5 30 For U n+1 = au n + b, a limit eists when -1 < a < 1 If a limit eists, its value is independent of the value of U 0 Eample 5: For the recurrence relation U n+1 = 0.6U n 0, a) State whether a limit eists, and if so b) Find the limit. Eample 6: A man plants some trees as a boundar between his house and the house net door. Each ear, the trees are epected to grow b 0.5m. To counter this, he decides to trim them b 0% per ear. a) To what height will the trees eventuall grow? Page 13 of 107

b) His neighbour is unhapp that the trees are too tall, and insists the grow no taller than m high. What is the minimum percentage the must be trimmed each ear to meet this condition? Solving Recurrence Relations to Find a and b If we have three consecutive terms in a sequence, we can find the values of a and b in the recurrence relation which generated the sequence using simultaneous equations. Eample 7: A sequence is generated b a recurrence relation of the form U n+1 = au n + b. In this sequence, U 1 = 8, U = 3 and U 3 = 38. Find the values of a and b. Past Paper Eample: Marine biologists calculate that when the concentration of a particular chemical in a loch reaches 5 milligrams per litre (mg/l) the level of pollution endangers the lives of the fish. A factor wishes to release waste containing this chemical into the loch, and supplies the Scottish Environmental Protection Agenc with the following information: 1. The loch contains none of the chemical at present.. The compan will discharge waste once per week which will result in an increase in concentration of.5 mg/l of the chemical in the loch. 3. The natural tidal action in the loch will remove 40% of the loch ever week. a) After how man weeks at this level of discharge will the lives of the fish become endangered? b) The compan offers to install a cleaning process which would result in an increase in concentration of onl 1.75 mg/l of the chemical in the loch, and claim this will not endanger the lives of the fish in the long term. Should permission be given to allow the compan to discharge waste into the loch using this revised process? Justif our answer. Page 14 of 107

Graphs of Functions Sketching a Quadratic Graph (Revision) To sketch a quadratic graph: Find the roots (set = 0) Find the intercepts (set = 0) Find the turning point ( value is halfwa between roots; sub. into formula to find ) Eample 1: Sketch and annotate the graph of = 15 Eample : Sketch and annotate the graph of 4 4 Note: when quickl sketching a quadratic graph, the roots and shape ( happ or sad face) are enough. Page 15 of 107

Graphs of Functions Sketching Graphs (Revision) In the eam, diagrams are provided whenever the question involves a graph. However, this is not the case when working from the tetbook: it is therefore important that we are able to sketch basic graphs where necessar, as often the question becomes simpler when ou can see it. Eample 3: in the spaces provided, make a basic sketch of the graph(s) of the function(s) stated. a) = + 1 b) 3 + 4 1 = 0 c) = -1 and = 5 d) = and = 4 e) = - 4 f) = ( ) and = Eample 4: Sketch and annotate the graph of = - 8 Eample 5: Sketch and annotate the graph of = ( + 3) + 1 Page 16 of 107

Eample 6: Sketch the graphs of = sin, = cos and = tan below. 1 1 90 180 70 360 90 180 70 360 90 180 70 360-1 -1 = sin = cos = tan For trig graphs, how soon the graph repeats itself horizontall is known as the period, and half of the vertical height is known as the amplitude. Function Period Amplitude = sin = cos = tan For the graphs of: = a sinb + c and = a cosb + c : Eample 7: Sketch the graphs of: a = amplitude b = waves in 360 c = vertical shift = a tanb + c: a) sin b) = 5cos + 3 b = waves in 180 c = vertical shift 90 180 70 360 90 180 70 360 c) = -3sin3 90 180 70 360 Page 17 of 107

Compound Angles A compound angle is one containing two parts, e.g. ( 60). The graphs of compound angles can be thought of as the trig version of = f ( a), i.e. shifted left or right b a units. Eample 8: n the aes opposite, sketch: a) = sin 90 180 70 360 b) = sin ( 45) Page 18 of 107

Transformation of Graphs We have seen how the graph of = sin() is different to that of = sin(), and how = differs from = ( 1). The si operations below are used to transform the graph of a function: = f() + a = f( + a) = f() = f() (-, 4) (-, 4) -6 4-6 4 (, -) (, -) = f() + a is obtained b sliding = f(): = f( + a) is obtained b sliding = f(): Verticall upwards if a > 0 Horizontall left if a > 0 Verticall downwards if a < 0 Horizontall right if a < 0 = k f() = f(k) = f() = f() (-, 4) (-, 4) -6 4-6 4 (, -) (, -) = k f() is obtained b verticall: = f(k) is obtained b horizontall: stretching = f() if k > 1 compressing = f() if k > 1 compressing = f() if 0 < k < 1 stretching = f() if 0 < k < 1 = - f() = f(-) = f() = f() (-, 4) (-, 4) -6 4-6 4 (, -) (, -) = - f() is obtained b reflecting = f(): in the - ais = f(-) is obtained b reflecting = f(): in the - ais Page 19 of 107

Multiple Transformations ften, are asked to perform more than one transformation on a graph. Where appropriate, alwas leave sliding verticall to last. Eample 9: Part of the graph of = f() is shown. n separate diagrams, sketch: a) = f(-) + b) = 1 f( + 1) -3 (1, 1) = f() (-, -4) Past Paper Eample: The diagram shows a sketch of the function = f( ). To the diagram, add the graphs of: a) = f() b) = 1 f(). (-4, 8) (, 8) Page 0 of 107

Quadratic Functions Finding the Equation of a Quadratic Function From Its Graph: = k( a)( b) If the graph of a quadratic function has roots at = -1 and = 5, a reasonable guess at its equation would be = 4 5, i.e. from = ( + 1)( 5). However, as the diagram shows, there are man parabolas which pass through these points, all of which belong to the famil of functions = k ( + 1) ( 5). To find the equation of the original function, we need the roots and one other point on the curve (to allow us to determine the value of k). Eample 1: State the equation of the graph below in the form = a + b + c. 6 1 3 Completing the Square (Revision) The diagram shows the graphs of two quadratic functions. If the graph of = is shifted q units to the right, followed b r units up, then the graph of = ( q) + r is obtained. As the turning point of = is (0, 0), it follows that the new curve has a turning point at (q, r). A quadratic equation written as = p ( - q) + r is said to be in the completed square form. = = ( q) + r (q, r) (0, 0) Eample : (i) Write the following in the form = ( + q) + r and find the minimum value of. (ii) Hence state the minimum value of and the corresponding value of. a) = + 6 + 10 b) = - 3 + 1 Page 1 of 107

Completing the Square when the Coefficient 1 Eample 3: Write = 3 + 1 + 5 in the form = p( + q) + r. Eample 4: Write = 5 + 1 in the form = p - ( + q). Eample 5: a) Write = - 10 + 8 in the form = ( + p) + q. b) Hence find the maimum value of 18-10 8 Solving Quadratic Equations via Completing the Square Quadratic equations which do not easil factorise can be solved in two was: (i) completing the square, or (ii) using the quadratic formula. In fact, both methods are essentiall the same, as the quadratic formula is obtained b solving = a + b + c via completing the square. Eample 6: State the eact values of the roots of the equation - 4 + 1 = 0 b: a) using the quadratic formula b) completing the square Page of 107

Solving Quadratic Inequations Quadratic inequations are easil solved b making a sketch of the equivalent quadratic function, and determining the regions above or below the ais. Eample 7: Find the values of for which: a) 7 + 6 > 0 b) 7 + 6 < 0 First, sketch = 7 + 6 Roots of Quadratic Equations and The Discriminant (Revision) For = a + b + c, b 4ac is known as the discriminant. b 4ac > 0 gives real, unequal roots b 4ac = 0 gives real, equal roots b 4ac < 0 gives N real roots If b 4ac gives a perfect square, the roots are RATINAL If b 4ac does NT give a perfect square, the roots are IRRATINAL (i.e. surds) Eample 8: Determine the nature of the roots of the equation 4( 3) = 9 Eample 9: Find the value(s) of p given that + 4 + p = 0 has real roots. Eample 10: Find the value(s) of r given that + (r - 3) + r = 0 has no real roots. Page 3 of 107

Tangents to Curves: Using the Discriminant To find the points of contact between a line and a curve, we make the curve and line equations equal (i.e. make = ) to obtain a quadratic equation, and solve to find the -coordinates. B finding the discriminant of this quadratic equation, we can work out how man points of contact there are between the line and the curve. There are 3 options: Two points of contact ne point of contact No points of contact distinct roots Equal roots No real roots b 4ac > 0 b 4ac = 0 b 4ac < 0 The most common use for this technique is to show that a line is a tangent to a curve Eample 11: Show that the line = 3 13 is a tangent to the curve = 7 +1, and find the coordinates of the point of contact. Eample 1: Find two values of m such that = m - 7 is a tangent to = + 3 Page 4 of 107

Past Paper Eample 1: Epress + 1 + 1 in the form a( + b) + c. Past Paper Eample : Given that + p + p + 6 = 0 has no real roots, find the range of values for p. Past Paper Eample 3: Show that the roots of (k ) 3k + k = - are alwas real. Page 5 of 107

The Circle If we draw, suitable to relative aes, a circle, radius r, centred on the origin, then the distance from the centre of an point P (, ) could be determined to be d. As the shape is a circle, then this distance is equal to the radius. It therefore follows that: r P (, ) Since r, then r Therefore, The equation + = r describes a circle with centre (0, 0) and radius r Eample 1: Write down the centre and radius of each circle. a) + = 64 b) + = 361 c) + 3 = 5 Eample : State where the points (-, 7), (6, -8) and (5, 9) lie in relation to the circle + = 100. Circles with Centres Not at the rigin The radius in the above circle is the distance between (, ) and the origin, i.e. r ( -0) ( -0). If we move the centre to the P (, ) point (a, b), then r ( -a) ( -b). Squaring both sides, we can now also sa that: r The equation ( a) + ( b) = r describes a circle with centre (a, b) and radius r Eample 3: Write down the centre and radius of each circle. C (a, b) a) ( 1) + ( + 3) = 4 b) ( + 9) + ( - ) = 0 c) ( 5) + = 400 Page 6 of 107

Eample 4: A is the point (4, 9) and B is the point (-, 1). Find the equation of the circle for which AB is the diameter. Eample 5: Points P, Q and R have coordinates (-10, ), (5, 7) and (6, 4) respectivel. a) Show that triangle PQR is right angled at Q. b) Hence find the equation of the circle passing through points P, Q and R. The General Equation of a Circle For the circle described in Eample 3a, we could epand the brackets and simplif to obtain the equation + + 6 + 6 = 0, which would also describe a circle with centre (1, -3) and radius. For + + g + f + c = 0, Therefore, the circle described b ( + g) + ( + f) = - c + + g + f + c = 0 ( + g + g ) + ( + f + f ) = g + f - c ( + g) + ( + f ) = (g + f c) has centre (-g, -f ) and r = g f c Eample 6: Find the centre and radius of the circle with equation + 4 + 8 5 = 0 Eample 7: State wh the equation + 4 4 + 15 = 0 does not represent a circle. Eample 8: State the range of values of c such that the equation + 4 + 6 + c = 0 describes a circle. Eample 9: Find the equation of the circle concentric with + + 6-54 = 0 but with radius half its size. Page 7 of 107

Intersection of Lines and Circles As with parabolas, there are three possibilities when a line and a circle come into contact, and we can eamine the roots of a rearranged quadratic equation to determine which has occurred. However: We CANNT make the circle and line equations equal to each other: the line equation must be substituted INT the circle equation to obtain our quadratic equation! Two points of contact ne point of contact No points of contact distinct roots Equal roots No real roots b 4ac > 0 b 4ac = 0 b 4ac < 0 As with parabolas, the most common use of this technique is to show tangenc. Eample 10: Find the coordinates of the points of intersection of the line = 1 and the circle + 1 + 7 = 0. Eample 11: Show that the line = 3 + 10 is a tangent to the circle + 8 4 0 = 0 and establish the coordinates of the point of contact. Page 8 of 107

Eample 1: Find the equations of the tangents to the circle + = 9 from the point (0, 5). 5 Tangents to Circles at Given Points Remember: at the point of contact, the radius and tangent meet at 90 (i.e., the are perpendicular). To find a tangent at a given point: Find the centre of the circle Find the gradient of the radius (joining C and the given point) Find the gradient of the tangent (flip and make negative) Sub the gradient and the original point into b = m ( a) C P Eample 13: Find the equation of the tangent to + 14 + 6 87 = 0 at the point (-, 5). Page 9 of 107

Past Paper Eample 1: A circle has centre C (-, 3) and passes through point P (1, 6). a) Find the equation of the circle. P(1, 6) C(-, 3) b) PQ is a diameter of the circle. Find the equation of the tangent to this circle at Q. Q Past Paper Eample : a) Show that the line with equation = 3 is a tangent to the circle with equation + + 14 + 4 19 = 0 and state the coordinates of P, the point of contact. b) Relative to a suitable set of coordinate aes, the diagram opposite shows the circle from a) and a second smaller circle with centre C. The line = 3 is a common tangent at the point P. The radius of the larger circle is three times the radius of the smaller circle. Find the equation of the smaller circle. C P Page 30 of 107

Past Paper Eample 3: Given that the equation + p 4p + 3p + = 0 represents a circle, determine the range of values of p. Past Paper Eample 4: Circle P has equation 8 10 9 0. Circle Q has centre (-, -1) and radius. a) i) Show that the radius of circle P is 4. ii) Hence show that circles P and Q touch. b) Find the equation of the tangent to circle Q at the point (-4, 1) Page 31 of 107

Calculus 1: Differentiation In the chapter on straight lines, we saw that the gradient of a line is a measure of how quickl it increases (or decreases) at a constant rate. This is eas to see for linear functions, but what about quadratic, cubic and higher functions? As these functions produce curved graphs, the do not increase or decrease at a constant rate. For a function f(), the rate of change at an point on the function can be found b measuring the gradient of a tangent to the curve at that point. The rate of change at an point of a function is called the derived function or the derivative. Finding the rate of change of a function at a given point is part of a branch of maths known as calculus. For function f() or the graph = f(), the derivative is written as: f'() ( f dash ) R d ( d b d ) d Derivative = Rate of Change of the Function = Gradient of the Tangent to the Curve The Derivative of f() = a n Eample 1: Find the derivative of f() = To find the derivative of a function: 1. Make sure it s written in the form = a n. Multipl b the power 3. Decrease the power b one Eample : f() = 3. Find f ( ). This means: At = 1, the gradient of the tangent to 3 = At = -, the gradient of the tangent to 3 = If f() = a n, then f () = na n-1 The DE rivative DE creases the power! To find the derivative of f(): f() MUST be written in the form f() = a n Rewrite to eliminate fractions b using negative indices Rewrite to eliminate roots b using fractional indices Revision from National 5 Eample 3: Write with negative indices: Eample 4: Write in inde form: a) b) 1 5 4 c) 5 3 a) b) 3 c) 3 7 Page 3 of 107

Eample 5: For each function, find the derivative. a) f() = 35 b) g() = - -3 ( 0) c) p() = 1 ( > 0) d) = 1 5 + 3 + 9 e) = 3 1 ( > 0) f) = ( 0) Eample 6: Find the rate of change of each function: 5 3 6 3 a) f ( ) b) c) 3 f ( ) 5 3 3 Points to note: Number terms disappear (e.g. if f() = 5, f () = 0) terms leave their coefficient (e.g. if f() = 135, f () = 135) Give our answer back in the same form as the question Equation of a Tangent to a Curve Eample 7: Find the equation of the tangent to the curve = 15 when = 4. To find the equation of a tangent to a curve: Find the point of contact (sub the value of into the equation to find ) d Find d d Find m b substituting into d Use b = m ( a) Page 33 of 107

Eample 8: a) Find the gradient of the tangent to the curve = 3 at the point where = 3 7. b) Find the other point on the curve where the tangent has the same gradient. Eample 9: Find the point of contact of the tangent to the curve with equation gradient of the tangent is 9. 7 3 when the Stationar Points and their Nature d An point on a curve where the tangent is horizontal (i.e. the gradient or = 0) is commonl known as d a stationar point. There are four tpes of stationar point: f() f() f() f() Minimum Turning Point Maimum Turning Point Rising Point of Inflection Falling Point of Inflection To locate the position of stationar points, we find the derivative, make it equal zero, and solve for. To determine their tpe (or nature), we must use a nature table. Page 34 of 107

Eample 10: Find the stationar points of the curve = 3 1 + 18 and determine their nature. Page 35 of 107

Increasing & Decreasing Functions d if > 0, then is increasing d For an curve, d if < 0, then is decreasing d d if = 0, then is stationar d If a function is alwas increasing (or decreasing), it is said to be strictl increasing (or decreasing). Eample 11: State whether the function f() = 3 5 + is increasing, decreasing or stationar when: a) = 0 Eample 1: Show algebraicall that the function f() = 3 6 + 1 5 is never decreasing. b) = 1 c) = Eample 13: Find the intervals in which the function f() = 3 6 + 5 is increasing and decreasing. Page 36 of 107

Curve Sketching To accuratel sketch and annotate the curve obtained from a function, we must consider: Eample 14: Sketch and annotate full = 3 (4 ) 1. and intercepts. Stationar points and their nature Page 37 of 107

Closed Intervals Sometimes, we ma onl be interested in a small section of the curve of a function. To find the maimum and minimum values of a function in a given interval, we find stationar points as normal, but we also need to consider the value of the function at the ends of the interval. Eample 15: Find the greatest and least values of = 3 1 on the interval -3 1. Note: In a closed interval. The maimum and minimum values of a function occur either at a Stationar Point within the interval or at the end point of the interval. Page 38 of 107

Differentiation in Contet: ptimisation Differentiation can be used to find the maimum or minimum values of things which happen in real life. Finding the maimum or minimum value of a sstem is called optimisation. Eample 16: A carton is in the shape of a cuboid with a rectangular base and a volume of 3888cm 3. The surface area of the carton can be represented b the formula A() = 4 + Find the value of such that the surface area is a minimum. 583. In eams, optimisation questions almost alwas consist of two parts: part one asks ou to show that a situation can be described using an algebraic formula or equation, whilst part two asks ou to use the given formula to find a maimum or minimum value b differentiation. Leave part 1 of an optimisation question until the end of the eam (if ou have time), as the are almost alwas (i) more difficult than finding the stationar point and (ii) worth fewer marks. Remember that part is just a well-disguised find the minimum/maimum turning point of this function question! Eample 17: A square piece of card of side 30cm has a square of side cm cut from each corner. An open bo is formed b turning up the sides. a) Show that the volume, V, of the bo ma be epressed as 900 10 4 3 Page 39 of 107

b) Find the maimum volume of the bo. Eample 18: An architect has designed a new open-plan office building using two identical parabolic support beams spaced 5m apart as shown below. The front beam, relative to suitable aes, has the equation = 7. The inhabited part of the building is to take the shape of a cuboid. = 7 a) B considering the point P in the corner of the front face of the building, show that the area of this face is given b A() = 54 3. P b) Find the maimum volume of the inhabited section of the building. Page 40 of 107

= f() Graph of the Derived Function From the graph of = f(), we can obtain the graph of = f () b considering its stationar points. n the graph of = f (), the -coordinate comes from the derivative of = f(). 1. Draw a set of aes directl under a cop of = f().. Locate the stationar points. 3. At SP s, f () = 0, so the coordinate of f () = 0 on the new graph. 4. Where f() is increasing, f () is above the ais. 5. Where f() is decreasing, f () is below the ais. 6. Draw a smooth curve which fits this information. Eample 19: For the graphs below. Sketch the corresponding derived graphs of = f () a) = f() Page 41 of 107

b) = f() Past Paper Eample 1: A curve has equation = 4 4 3 + 3. Find the position and nature of its stationar points. Page 4 of 107

Past Paper Eample : Find the equation of the two tangents to the curve = 3 3 1 + 0 which are parallel to the line 48 = 5. Past Paper Eample 3: An open water tank, in the shape of a triangular prism, has a capacit of 108 litres. The tank is to be lined on the inside in order to make it watertight. The triangular cross-section of the tank is right-angled and isosceles, with equal sides of length cm. The tank has a length of L cm. L a) Show that the surface area to be lined, A cm, is given b A ( ) = + 43000 b) Find the minimum surface area of the tank. Page 43 of 107

Past Paper Eample 4: A function is defined on the domain 0 3 b f() = 3 4 + 6. Determine the maimum and minimum values of f. Page 44 of 107

Calculus : Integration The reverse process to differentiation is known as integration. Differentiation f() f () Integration As it is the opposite of finding the derivative, the function obtained b integration is sometimes called the anti-derivative, but is more commonl known as the integral, and is given the sign. If f() = n, then n d is the integral of n with respect to Indefinite Integrals and the Constant of Integration Consider the three functions a() = 3 + + 5, b() = 3 + 8 and c() = 3 13 + -. 4 In each case, the derivative of the function is the same, i.e. 6 +. This means that (6 )d has more than one answer. Because there is more than one answer, we sa that this is an indefinite 13 integral, and we must include in the answer a constant value C, to represent the 5, -8, etc which 4 we would need to distinguish a() from b() from c ( ) etc. To find the integral of a function, we do the opposite of what we would do to find the derivative: f() Multipl b the old power Decrease the power b 1 f () f() f () n1 a n 1 n In general: a d C (n 1) Eample 1: Find (remember +C ): IN tegration IN creases the power! 1. Write as a n. Increase the power b 1 3. Divide b the new power a) d b) 4t dt c) (3 5 4)d Page 45 of 107

3 4 3 d) dg 4 (g 0) e) g 6 5 3 p dp f) d 3 ( 0) The Definite Integral A definite integral of a function is the difference between the integrals of f( ) at two values of. Suppose we integrate f( ) and get F( ). Then the integral of f( ) when = a would be F(a), and the integral when = b would be F(b). The definite integral of f( ), with respect to, between a and b, is written as: b f()d F(b) F(a) (where b > a) a For eample, the integral of f( ) = 4 between the values = -3 and = 5 is written as 5 ( 4)d and reads the integral from -3 to 5 of 4 with respect to. Eample : Evaluate 3 Note: definite integrals do NT include the constant of integration! 3 b f() a ( 1)d 1 [F(b) C] [F(a) C] F(b) F(a) To find a definite integral: prepare the function for integration integrate as normal, but write inside square brackets with the limits to the right sub each limit into the integral, and subtract the integral with the lower limit from the one with the higher limit Eample 3: Evaluate p 1 p1dp Eample 4: Evaluate 3 ( 1 0 -)d Page 46 of 107

Eample 5: Find the value of g such that g (6 5)d = 6. Area Between a Curve and the ais. In the diagram opposite, the area of the shaded section can be obtained b finding the area under the graph from to b, and subtracting the area from to a. =f() The value of each of these areas can be determined b integrating the function and substituting b or a respectivel. The area enclosed b the curve = f(), the lines = a, = b and the ais is equal to the definite integral of f() between a and b a b i.e. Area = b a f() d Eample 6: For each graph below, (i) write down the integrals which describe the shaded regions (ii) calculate the area of the shaded region a) b) = 6 = 9 NTE: Eample 6b shows that areas UNDER the ais give NEGATIVE values! Page 47 of 107

Eample 7: 7 a) Evaluate ( 6) d b) (i) Sketch below the area described b the 1 7 integral ( 6) d. 1 The answers for 5a and 5b do not match! This is because the area below the ais and the area above cancel each other out (as in 4b, areas below the ais give negative values). To find the area between a curve and the -ais: 1. Determine the limits which describe the sections above and below the ais. Calculate areas separatel 3. Find the total, IGNRING THE NEGATIVE VALUE F THE SECTIN BELW THE AXIS. Eample 8: Determine the area of the regions bounded b the curve = 4 + 3 and the and aes. = 4 + 3 Page 48 of 107

Area Between Two Curves Consider the area bounded b the curves = ( ) and =. =( ) = =( ) = =( ) = 1 3 1 3 1 3 Area = 3 3 d - ( ) d The diagrams above show that the area between the curves is equal to the area between the top function ( ) and the ais MINUS the area between the bottom curve (( ) ) and the ais. a =g( ) =f( ) b 1 The area between the curves = f( ) and = g( ) (which meet at the points where = a and = b ) is given b: where: A b a (f() g())d f() is the TP function and g() is the BTTM b > a 1 Eample 9: Write down the integrals used to determine the areas shown below: a) b) c) = 6 = 5 = 3 = = = 1 To find the area between two curves: 1. Make a sketch (if one has not been given). Find points of intersection (make = and solve) 3. Subtract the bottom function from the top function, PUTTING THE BTTM FUNCTIN IN BRACKETS! 4. Integrate Page 49 of 107

Eample 10: Find the area enclosed between the curve = 3 5 and the line = = = 3 5 Differential Equations If we know the derivative of a function (e.g. f ' 6 3), we can obtain a formula for the original function b integration. This is called a differential equation, and gives us the function in terms of and C (which we can then evaluate if we have a point on the graph of the function). Eample 11: The gradient of a tangent to the curve of given than the graph of f f passes through the point (-1, -10). is 4 10, Epress in terms of, Page 50 of 107

Past Paper Eample 1: Evaluate 1 9 1 d Past Paper Eample : Find area enclosed between the curves = 1 + 10 and = 1 + 5. = 1 + 10 = 1 + 5 Page 51 of 107

Past Paper Eample 3: The parabola shown in the diagram has equation = 3. The shaded area lies between the lines = 14 and = 4 Calculate the shaded area. = 4 = 14 Straight Line Recurrence Relations The Circle Calculus Applications Unit Topic Checklist (Unit Assessment Topics in Bold) Topic Questions Done? Gradients (inc. m = tan) Eercise 1A, Q 8 10; Eercise 1B, p 4, Q 4, 5 Y/N Perpendicular Gradients Eercise 1D, Q 1 4, 7 Y/N Eercise 1E, Q 1, 3, 7, 8 ( =m + c) Y/N Equations of straight lines Eercise 1F, Q 1, (A + B + C = 0) Y/N Eercise 1G, Q, 3 ( b = m( a)) Y/N Collinearit Eercise 1B, Q 1 3, 9 Y/N Perpendicular bisectors Eercise 1I, Q 1, ; Eercise 1N, Q 5 Y/N Altitudes Eercise 1K, Q 1, 5; Eercise 1N, Q 1 3 Y/N Medians Eercise 1M, Q 1, 3; Eercise 1N, Q 4 Y/N Distance Formula Eercise 1B, Q 1 Y/N Finding terms Eercise 5D, Q 1 3 Y/N Creating & using formulae Eercise 5C, Q 5 11 Y/N Finding a limit Eercise 5H, Q 1 3 Y/N Eercise 5H, Q 4 10; Eercise 5L, p 83, Q, 4 Y/N Solving to find a and b Eercise 5I, Q 1, Y/N Eercise 5I, Q 3, 4 Y/N Circles centred on Eercise 1D, Q 1 3 Y/N ( a) + ( b) = r Eercise 1F, Q 1 3, 10 Y/N General equation Eercise 1H, Q 1, 4, 1 15; Eercise 1M, Q 1, 7 Y/N Intersection of lines & circles Eercise 1J, Q 3 Y/N Tangenc Eercise 1K, Q, 6; Eercise 1M, Q 4, 8 Y/N Equations of tangents Eercise 1L, Q 1 4 Y/N ptimisation Eercise 6Q, Q 1,, 4 Y/N Eercise 6R, Q 1, 5; Eercise 6S, Q 19 Y/N Area under a curve Eercise 9K, Q 1; Eercise 9N, Q 1, 3, 4 Y/N Area between two curves Eercise 9P, Q 1,, 4; Eercise 9R, Q 7, 11 Y/N Differential Equations Eercise 9Q, Q, 3; Eercise 9R, Q 14, 15 Y/N Page 5 of 107

Polnomials A polnomial is an epression with terms of the form a n, where n is a whole number. For eample, 5p 4 3p 3 is a polnomial, but 3p -1 or 3 p are not. The degree of a polnomial is its highest power, e.g. the polnomial above has a degree of 4. The number part of each term is called its coefficient, e.g. the coefficients of p 4, p 3 and p above are 5, - 3 and 0 (as there is no p term!) respectivel (note that 5p 4 would also be a polnomial on its own, with coefficients of zero for all other powers of p). Evaluating Polnomials An eas wa to find out the value of a polnomial function is b using a nested table. Eample 1: Evaluate f(4) for f() = 4 3 3 10 5 + 7. 4-3 -10-5 7 Line up coefficients Eample : Evaluate f(-1) for f() = 3 5 3 + 4. -1 3 0-0 0 4 3 Missing powers have coefficients of zero! Snthetic Division Dividing 67 b 9 gives an answer of 7 remainder 4. We can write this in two was: Eample 3: 67 9 = 7 remainder 4 R 9 7 + 4 = 67 For this problem, 9 is the divisor, 7 is the quotient, and 4 is the remainder (note that if we were dividing 63 b 9, the remainder would be zero, since 9 is a factor of 63). a) Remove brackets and simplif: b) Evaluate f(3) for f() = 3 + 6 39 + 47 ( 3)( + 9 1) + 11 This shows that: 3 + 6 39 + 47 = ( 3)( + 9 1) + 11 R ( 3 + 6 39 + 47) ( 3) = ( + 9 1) remainder 11 Compare the numbers on the bottom row of the nested table in part b) with the coefficients in part a). This shows that we can use nested tables to divide polnomial epressions to give both the quotient and remainder (if one eists). This process is known as snthetic division. Page 53 of 107

Eample 4: Find the remainder on dividing 3 + 5 b ( + 5). Eample 5: Write 4p 4 + p 3 6p + 3 (p 1) in the form (ap b) Q(p) + R Remainder Theorem and Factor Theorem Considered together, these two theorems allow us to factorise algebraic functions (remember that a factor is a number or term which divides eactl into another, leaving no remainder). If polnomial f() is divided b ( h), then the remainder is f(h) n division of polnomial f() b ( h), if f(h) = 0, then ( h) is a factor of f() In other words, if the result of snthetic division on a polnomial b h is zero, then h is a root of the polnomial, and ( h) is a factor of it. Eample 6: f() = 3 9 + + 1. Eample 7: Factorise full 3 3 + 1 8. a) Show that ( 4) if a factor of f(). b) Hence factorise f() full. Eample 8: Find the value of k for which ( + 3) is a factor of 3 3 + k + 6 Page 54 of 107

Eample 9: Find the values of a and b if ( 3) and ( + 5) are both factors of 3 + a + b 15 Solving Polnomial Equations Polnomial equations are solved in eactl the same wa as we solve quadratic equations: make the right hand side equal to zero, factorise, and solve to find the roots. Eample 10: The graph of the function = 3 7 + 7 + 15 is shown. Find the coordinates of points A, B and C. = 3 7 + 7 + 15 A B C Finding a Function from its Graph This uses eactl the same sstem as that for quadratic graphs, but with more brackets (see page 19). Remember: tangents to the ais have repeated roots! Eample 11: Find an epression for cubic function f(). = f() 5-5 Page 55 of 107

Sketching Polnomial Functions Eample 1: a) Find the and intercepts of the graph of = 4 6 3 + 13 1 + 4. b) Find the position and nature of the stationar points of = 4 6 3 + 13 1 + 4. c) Hence, sketch and annotate the graph of = 4 6 3 + 13 1 + 4. Page 56 of 107

Trigonometr: Addition Formulae and Equations What ou must know from National 5!!! Sine Graph 1 Cosine Graph 1 = sin = cos 0 90 180 70 360 0 90 180 70 360-1 -1 We can use the above graphs to find the values of: sin00 cos 01 sin 90 1 cos 900 sin1800 cos180 1 sin 70 1 cos 700 sin 360 0 cos 3601 sin 0 0 360 0,180,360 cos 0 0 360 90,70 We can use these graphs to solve the following: sin 1 0 360 70 cos 1 0 360 70 sin 1 0 360 90 cos 1 0 360 0,360 90 sin + All + Remember, this means that: sin 160 would be + 180 tan + 70 cos + 0 360 cos 00 would be tan 00 would be + sin 30 would be and so on... Page 57 of 107

Related Angles 180-90 This diagram can be used to find families of related angles. 180 180 + 70 360-0 360 For eample, for = 30. The famil of related angles would be: 30, 150, 10, 330 These angles are related since: sin30 = 0.5 sin150 = 0.5 sin10 = -0.5 sin330 = -0.5 Note: The sine of these angles have the same numerical value. Eample A: sin = 0.43 (0 360) 1 = sin (0.43) = 5 (R.A) = (0 + 5), (180 5) = 5, 155 Eample B: cos = -0.584 (0 360) 1 = cos (0.584) = 54.3 (R.A) = (180 54.3), (180 + 54.3) = 15.7, 34.3 Equations Step 1: Consider 0.43 Step : We know that we can find the other 3 angles in the famil 155, 05, 335 Step 3: We onl want the angles which will give +ve answers for sin. Step 1: Consider 0.584 (ignore ve) Step : We know that we can find the other 3 angles in the famil 15.7, 34.3, 305.7 Step 3: We onl want the angles which will give -ve answers for cos. Radians If we draw a circle and make a sector with an arc of eactl one radius long, then the angle at the centre of the sector is called a radian. r Remember that Circumference = D = r. This means that there are radians in a full circle. 360 = radians 180 = radians r Page 58 of 107

Eample 1: Convert: a) 90 to radians b) 60 to radians c) 5 to radians 4 11 d) radians to degrees e) radians to degrees f) radians to degrees 4 3 6 Consider the following triangles: Eact Values 60 1 30 A right-angled triangle made b halving an equilateral triangle of side units 1 A right-angled triangle made b halving an square of side 1 unit 1 45 nce we have found the lengths of the missing sides (b Pthagoras Theorem), the following table of values can be constructed: 0 30 45 60 90 0 6 4 3 (180 - ) SIN Positive 90 () ALL Positive Sin 180 Quadrant Quadrant 3 Quadrant 1 Quadrant 4 0 360 Cos TAN Positive CS Positive Tan (180 + ) 70 (360 -) Eample : State the eact values of: 7 a) sin 150 b) tan 315 c) cos 6 Addition Formulae Page 59 of 107

Finding the value of a compound angle is not quite as simple as adding together the values of the component angles, e.g. sin90 sin60 + sin30. The following formulae must be used: sin(a + B) = sinacosb + cosasinb sin(a B) = sinacosb - cosasinb Eample 3: Epand each of the following: a) sin(x + Y) b) sin(q + 3P) Eample 4: Find the eact value of sin75. Eample 5: A and B are acute angles where 1 tana and 5 3 tanb. Find the value of sin(a + B). 4 Eample 6: Epand each of the following: a) sin( - ) b) sinb 3 Eample 7: In the diagram opposite: AC = CD = units, and AB = BC = 1 unit. A Show that sin X is eactl 1. 10 B X C D cos(a + B) and cos (A B) Page 60 of 107

cos(a + B) = cosacosb - sinasinb cos(a - B) = cosacosb + sinasinb Eample 8: Epand the following: a) cos(x Y) b) cos(x + 315) Eample 9: a) Show that 3 4 1 b) Hence find the eact value of cos 1 To summarise: sin A B sin Acos B cos AsinB cos A B cos Acos B sin AsinB Page 61 of 107

Trigonometric Identities NTE: these are important formulae which are not provided in the eam paper formula sheets! sin tan cos Note that due to the second formula, we can also sa that: sin + cos = 1 cos = 1 sin AND sin = 1 cos To prove that an identit is true, we need to show that the epression on the left hand side of the equals sign can be changed into the epression on the right hand side. Eample 10: Prove that: a) 4 4 cos sin cos sin b) sin4 tan3 tan cos cos 3 c) 1 sin 1 tan tan sin cos Page 6 of 107

Double Angle Formulae sina = sin(a + A) cosa = cos(a + A) = = Since cos = 1 sin and sin = 1 cos, we can further epand the formula for cosa: cosa = cos A sin A cosa = cos A sin A = = sina = sinacosa To summarise: = cos A sin A cosa = cos A - 1 = 1 sin A Eample 11: Epress the following using double angle formulae: a) sinx b) sin6y c) cosx (sine version) d) cos8h (cosine version) e) sin5q f) cos (cos and sin version) Page 63 of 107

Eample 1: sin, where is an acute angle. Find the eact values of: 13 a) sin() b) cos() Eample 13: Prove that sin tan 1 cos Solving Comple Trig Equations Trig equations can also often involve (i) powers of sin, cos or tan, and (ii) multiple and/or compound angles. Eample 14: Solve 4cos 3 = 0 for 0 Page 64 of 107

Trig equations can also be written in forms which resemble quadratic equations: to solve these, treat them as such, and solve b factorisation. Eample 15: Solve 6sin sin = 0 for 0 360 If the equation contains a multiple angle term, solve as normal (paing close attention to the range of values of ). Eample 16: Solve 3 tan 135 1 for 0 360 To solve trig equations with combinations of double- and single-angle angle terms: Rewrite the double angle term using the formulae on Page 59 Factorise Solve each factor for When the term is cosx, the version of the double angle formula we use depends on the other terms in the equation: use cos 1 if the other term is cos ; 1 sin if the other term is sin. Eample 17: Solve sin sin = 0, 0 360 Page 65 of 107

Eample 18: Solve cos 7cos = 0, 0 Formulae for cos and sin Rearranging the formulae for cos allows us to obtain the following formulae for cos and sin cos sin 1 1 cos 1 cos 1 Eample 19: Epress each of the following without a squared term: a) cos θ b) sin 3X c) sin ( ) Past Paper Eample 1: In the diagram, DEC = CEB =, and CDE = BEA = 90. CD = 1 unit; DE = 3 units. B writing DEA in terms of, find the eact value of cos(dêa). C 1 D 3 B E A Page 66 of 107