Mathematics. Mathematics 1. hsn.uk.net. Higher HSN21000

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1 Higher Mathematics UNIT Mathematics HSN000 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still Notes. For more details about the copright on these notes, please see

2 Higher Mathematics Unit Mathematics Contents Straight Lines The Distance Between Points The Midpoint Formula Gradients 4 4 Collinearit 6 5 Gradients of Perpendicular Lines 6 6 The Equation of a Straight Line 7 7 Medians 9 8 Altitudes 0 9 Medians and Altitudes in Isosceles Triangles 0 0 Perpendicular Bisectors Intersection of Lines Functions and Graphs 5 Set Theor 5 Functions 6 Inverse Functions 9 4 Eponential Functions 0 5 Introduction to Logarithms 0 6 Radians 7 Eact Values 8 Trigonometric Functions 9 Graph Transformations Differentiation 8 Introduction to Differentiation 8 Finding the Derivative 9 Differentiating with Respect to ther Variables 4 Rates of Change 4 5 Equations of Tangents 5 6 Increasing and Decreasing Curves 8 7 Stationar Points 8 8 Determining the Nature of Stationar Points 9 9 Curve Sketching 4 0 Closed Intervals 4 Graphs of Derived Functions 45 ptimisation 45 Sequences 47 Introduction to Sequences 47 Linear Recurrence Relations 48 Divergence and Convergence 49 4 The Limit of a Sequence 50 5 Solving Recurrence Relations to find a and b 5 - ii - HSN000

3 Higher Mathematics Unit Mathematics UTCME Straight Lines The Distance Between Points Points on Horizontal or Vertical Lines It is relativel straightforward to work out the distance between two points which lie on a line parallel to the - or -ais. (, ) (, ) In the diagram to the left, the points (, ) and (, ) lie on a line parallel to the -ais, i.e. =. The distance between the points is simpl the difference in the -coordinates, i.e. where >. (, ) In the diagram to the left, the points (, ) and (, ) lie on a line parallel to the -ais, i.e. =. (, ) The distance between the points is simpl the difference in the -coordinates, i.e. where >. EXAMPLE. Calculate the distance between the points ( 7, ) and ( 6, ). The distance is 6 ( 7) = = units Page HSN000

4 Higher Mathematics Unit Mathematics The Distance Formula The distance formula gives us a method for working out the length of the straight line between an two points. It is based on Pthagoras s Theorem., The distance d between the points (, ) and (, ) EXAMPLES. A is the point (, 4) and The length is is d = + units ( ) + ( ) = ( ( ) ) + ( 4) = 5 + ( ) = = 4 units B,. Calculate the length of the line AB.. Calculate the distance between the points 5 The distance is ( ) + ( ) ( 5 ) ( ) = = = + 4 = = = = (, ) units d 4, and (, ). Page HSN000

5 Higher Mathematics Unit Mathematics The Midpoint Formula The point half-wa between two points is called the midpoint. It is calculated as follows: The midpoint M between (, ) and (, ) EXAMPLES is M + +,. Calculate the midpoint of the points (, 4) and The midpoint is + +, ( 4, ) =, = 7, 8.. In the diagram below, A ( 9, ) lies on the circumference of the circle with centre C( 7, ), and the line AB is the diameter of the circle. Find the coordinates of B. B Note Simpl writing The midpoint is (4, ) would be acceptable in an eam Since C is the centre of the circle and AB is the diameter, C is the midpoint of AB. Using the midpoint formula, we have: = ( 7, ), where B is the point (, ) B comparing - and -coordinates, we have: = 7 = 9 + = 4 + = 4 = 5 So B is the point ( 5, 6 ). A C = 6 Page HSN000

6 Higher Mathematics Unit Mathematics Gradients Consider a straight line passing through the points (, ) and (, ) The gradient m of the line through (, ) and (, ) change in vertical height is: m = = for change in horizontal distance pposite Also, since tanθ = = Adjacent we obtain: m = tanθ where θ is the angle between the line and the positive direction of the -ais. Note As a result of the above definitions: lines with positive gradients slope up, from left to right (, ) θ θ (, ) : lines with negative gradients slope down, from left to right positive direction lines parallel to the -ais have a gradient of zero lines parallel to the -ais have an undefined gradient We ma also use the fact that: Lines are parallel the have the same gradient. Note A B means: If A is true, then B is true and If B is true, then A is true Page 4 HSN000

7 Higher Mathematics Unit Mathematics EXAMPLES. Calculate the gradient of the straight line shown in the diagram below. m = tanθ = tan = 0 6 (to d.p.). Find the size of angle θ shown in the diagram below. θ m = 5 Using m = tanθ, the angle which the line makes with the positive direction of the -ais is: tan ( m) = tan ( 5) = So θ = = (to d.p.). Find the angle that the line joining P(, ) and Q (, 7 ) makes with the positive direction of the -ais. 7 + The line has gradient m = = = + And m = tanθ tanθ = θ = tan = 7 57 (to d.p.) Page 5 HSN000

8 Higher Mathematics Unit Mathematics 4 Collinearit Points which lie on the same straight line are said to be collinear. For three collinear points, the gradient between two of the points is equal to the gradient between two other points and the share a common point. m = m = m = m = m = m = This test for collinearit can onl be used in two dimensions. EXAMPLE not collinear No common point Equal gradient Show that the points P( 6, ), Q ( 0, ) and R ( 8, 6 ) are collinear. ( ) 6 mpq = mqr = 0 ( 6) 8 0 = = = 6 = not collinear Common point Unequal gradients 4 8 Since mpq = mqr and Q is a common point, P, Q and R are collinear 5 Gradients of Perpendicular Lines Two lines at right-angles to each other are said to be perpendicular. For perpendicular lines with gradients m and m : m = m collinear Common point Equal gradients The simple rule is, if ou know the gradient of one of the lines, then the gradient of the other is calculated b inverting the gradient and changing the sign. For eample: If m = then m = Note that this rule cannot be used if the line is parallel to the - or -ais. If a line is parallel to the -ais ( m = 0), then the perpendicular line is parallel to the -ais it has an undefined gradient. If a line is parallel to the -ais then the perpendicular line is parallel to the -ais. Page 6 HSN000

9 Higher Mathematics Unit Mathematics EXAMPLES. If T is the point (, ) and S is ( 4, 5), find the gradient of the line perpendicular to ST. m ST 5 ( ) = 4 = so m = 7 (since mst m = ). Triangle MP has vertices M(, 9), ( 0, 0 ) and P(, 4 ). Show that the triangle is right angled. Sketch: M(, 9) ( 0, 0) P(, 4) m M 9 0 = 0 = m MP 9 4 = = 5 5 = m P 4 0 = 0 = Since mm mp =, M is perpendicular to P which means MP is right-angled at. 6 The Equation of a Straight Line To work out the equation of a straight line, we need to know two things: the gradient of the line, and a point which lies on the line. The equation of a straight line is given b: b = m( a) where m is the gradient and ( a, b ) is a point on the line. Notice that if we have a point ( 0, c ) the -ais intercept then the equation becomes = m + c. You should alread be familiar with this form. It is good practice to rearrange the equation of a straight line into the form: a + b + c = 0 where a is positive Page 7 HSN000

10 Higher Mathematics Unit Mathematics Lines Parallel to Aes If a line is parallel to the -ais (i.e. m = 0 ), its equation is = c = c c If a line is parallel to the -ais (i.e. m is undefined), its equation is = k = k k EXAMPLES. Find the equation of the line with gradient (, 4). b = m( a) ( ) 4 = ( ) + = = 5 5 = 0 passing through the point. Find the equation of the line passing through A (, ) and B(,). To work out the equation, we must first find the gradient of the line AB: mab = = = ( ) 5 Now we have a gradient, and can use this with one of the given points: b = m( a) = 5 using A, and mab = = 5 = = 0. Find the equation of the line passing through ( ) and ( ) 5, 4 The gradient is undefined since the -coordinates are equal. So the equation of the line is = 5. 5, 5. Page 8 HSN000

11 Higher Mathematics Unit Mathematics 7 Medians A median is a line through a verte of a triangle and the midpoint of the opposite side. A BM is a median of ABC M is the midpoint of AC The standard process for finding the equation of a median is shown below. EXAMPLE M B Triangle ABC has vertices A ( 4, 9) B( 0, ) and C( 4, 4)., Find the equation of the median from A. C Start with a sketch: C M B Step Calculate the midpoint of the relevant line. Step Calculate the gradient of the line between the midpoint and the opposite verte. Step Find the equation using this gradient and either of the two points used in Step it is advisable to use the point given in the question. Using B( 0, ) and C( 4, 4) : ( 4) M =, 4, = = ( 7, ) Using A ( 4, 9) and M( 7, ) m AM A = ( 9) = = 7 4 Using A ( 4, 9) 8 : and m 8 AD = : b = m( a) 8 ( ) + 9 = 4 ( ) + 7 = 8 = = 0 Page 9 HSN000

12 Higher Mathematics Unit Mathematics 8 Altitudes An altitude is a line through a verte or a triangle, perpendicular to the opposite side. BD is an altitude of ABC A C D The standard process for finding the equation of an altitude is shown below. EXAMPLE B Triangle ABC has vertices A (, 5), B( 4, ) and C( 7, ). Find the equation of the altitude from A. Start with a sketch: D C B Step Calculate the gradient of the line which is perpendicular to the altitude. Step Calculate the gradient of the altitude using m =. m Step Find the equation using this gradient and the point that the altitude passes through. Using B( 4, ) and C( 7, ) m BC = = = 7 4 Using mbc mad = : m AD = : Using A (, 5) and m AD = : b = m( a) + 5 = ( ) + 8 = 0 A = Medians and Altitudes in Isosceles Triangles In isosceles triangles, the altitude and median through one of the vertices are the same line. In equilateral triangles, this is the case for all three vertices. A B Page 0 HSN000

13 Higher Mathematics Unit Mathematics 0 Perpendicular Bisectors A perpendicular bisector is a line which cuts through the midpoint of another line at right-angles. A D B In both cases, CD is the perpendicular bisector of AB C A The standard process for finding the equation of a perpendicular bisector is shown below. C D B EXAMPLE A is the point (,) and B is the point ( 4, 7 ). Find the equation of the perpendicular bisector of AB. Step Calculate the midpoint of the line being bisected. Step Calculate the gradient of the line used in Step, then find the gradient of its perpendicular bisector using m =. m Step Find the equation of the perpendicular bisector using the point from Step and the gradient from Step. Start with a sketch: A Using A (,) and B( 4, 7 ) : 4 7 Midpoint + + AB, = = (, 4) Using A (,) and B( 4, 7 ) : 7 mab = 4 ( ) = 6 6 = m = since mab = m Using (, 4 ) and m = : b = m( a) 4 = ( ) = = = 0 B Page HSN000

14 Higher Mathematics Unit Mathematics Intersection of Lines Man problems involve lines which intersect (cross each other). nce we have equations for the lines, there are three was of calculating the point of intersection using simultaneous equations. We will demonstrate the three methods b finding the point of intersection of the lines with equations = + 5 and =. B elimination This should be a familiar method, and can be used in all cases. = + 5 = ( ) = + 5 = + = 8 = 9 Substitute = 9 into: = = 9 + = So the lines intersect at the point (, 9 ). B equating This method can be used when both equations have a common - or - coefficient. In this case, both equations have an -coefficient of one. Make the subject of both equations: = 5 = + Equate: 5 = + = 8 = 9 Substitute = 9 into: = = 9 + = So the lines intersect at the point (, 9 ). Page HSN000

15 Higher Mathematics Unit Mathematics B substitution This method can be used when one equation has an - or -coefficient of one (i.e. just an or with no multiplier). Substitute = into: = + 5 ( ) = = + 5 = 4 Substitute = into: = = = 9 = So the lines intersect at the point (, 9 ). EXAMPLE Triangle PQR has vertices P(, ), Q (, ) and R ( 9, ). The medians PS and QT intersect at M. Q M S (a) Find the equations of PS and QT. (b) Hence find the coordinates of M. P (a) To work out the equation of PS, we need a point and the gradient: Find the coordinates of S, the midpoint of QR + 9 S =, = ( 5, ) Find the gradient of PS using P(, ) and S( 5, ) ( ) m PS = = 5 6 = T R Find the equation of PS using P(, ) and m PS = b = m( a) + = ( + ) ( ) + 9 = + = 8 8 = 0 Page HSN000

16 Higher Mathematics Unit Mathematics To work out the equation of QT, we need a point and the gradient: Find the coordinates of T, the midpoint of PR + 9 T =, = 4, Find the gradient of QT using T 4, Q, and m QT = = 4 (b) To find M, equate: 8 = = 5 = Hence M is 5,. 4 Substitute = into: = 8 = 8 = 5 Find the equation of QT using Q (, ) and m 4 QT = b = m a = 4 ( ) ( ) = = = 0 Page 4 HSN000

17 Higher Mathematics Unit Mathematics UTCME Functions and Graphs Set Theor In order to stud functions and graphs, we use set theor. This requires some standard smbols and terms, which ou should become familiar with. set element subset a collection of objects (usuall numbers) (or member) an object which is part of a set a set which is part of another set { 5, 6, 7, 8 } is a set belongs to; is a member of 6 { 5, 6, 7, 8} does not belong to; is not a member of 7 is an element of { 5, 6, 7, 8 } { 5, 6 } is a subset of { 5, 6, 7, 8 } 4 { 5, 6, 7, 8} { } or the empt set a special set with no members Standard Sets There are common sets of numbers which have their own smbols. Note that numbers can belong to more than one set. N natural numbers counting numbers, i.e. N = {,,, 4, 5,...} W whole numbers natural numbers including zero, i.e. W = { 0,,,, 4,...} Z integers positive and negative whole numbers, i.e. Z = {,,, 0,,, } Q rational numbers can be written as a fraction of integers, e.g. 4,, 0 5, R real numbers all points on the number line, e.g. 6,,,, 0 5 Page 5 HSN000

18 Higher Mathematics Unit Mathematics Notice that N is a subset of W, which is a subset of Z, which is a subset of Q, which is a subset of R. These relationships between the standard sets are illustrated in the Venn diagram below. R Q Z W N EXAMPLE List all the numbers in the set P = { N :< < 5}. P contains natural numbers which are strictl greater than and strictl less than 5, so: P = {,, 4} Note In Set Theor, a colon ( : ) means such that Functions A function is a rule which connects a set of numbers to another set. The set of starting numbers is called the domain and the resulting set is called the range. f f ( ) domain range A function is usuall denoted b a lower case letter (e.g. f or g ) and is defined using a formula of the form f ( ) =. Restrictions on the Domain The domain of a function must be defined such that the function can be evaluated for all elements of the domain. In other words, the domain can onl contain numbers which give answers when worked through the function. Division b Zero It is impossible to divide b zero, so in functions involving fractions, the domain must eclude numbers which would give a denominator (bottom line) of zero. Page 6 HSN000

19 Higher Mathematics Unit Mathematics For eample, the function defined b: f ( ) = 5 cannot have 5 in its domain, since this would make the denominator equal to zero. The domain of f ma be epressed formall as { R : 5}. This is read as all belonging to the real set such that does not equal five. Even Roots Using real numbers, we cannot evaluate an even root (i.e. square root, fourth root etc) of a negative number. So an functions involving even roots must eclude numbers which would give a negative number under the root. For eample, the function defined b: f ( ) = 7 must have 7 0. Solving for gives 7, so the domain of f can be epressed formall as { : } EXAMPLE R A function g is defined b g ( ) =. + 4 Define a suitable domain for g. We cannot divide b zero, so 4 Identifing the Range R.. So the domain is { : 4} Some functions cannot produce certain values so these are not in the range. For eample: f ( ) = does not produce negative values, since an number squared is either positive or zero. Looking at the graphs of functions makes identifing the range more straightforward. = f ( ) If we consider the graph of = f ( ) (shown to the left) it is clear that there are no negative -values. The range can be stated formall as f ( ) 0. Page 7 HSN000

20 Higher Mathematics Unit Mathematics EXAMPLE. A function f is defined b f ( ) = sin for R. Identif its range. Sketching the graph of = f ( ) shows that sin onl produces values from to inclusive. = sin This can be written as f ( ). Composite Functions Functions can be combined to give a composite function. If we have two functions defined b f ( ) and g ( ), then f g ( ) and g f ( ) define composite functions. In most cases f g ( ) g f. f f ( ) g h g( f ) or h EXAMPLES. Defined on suitable domains, f ( ) = and g ( ) =. Find: (a) f (b) f g ( ) (c) g f ( ) (a) f = = 4 = ( ) (b) f g ( ) f = ( ) = ( ) (c) g f ( ) g = 4. Functions f ( ) = + and g ( ) = are defined on suitable domains. Find formulae for h ( ) = f g ( ) and k ( ) = g f ( ). h ( ) = f g ( ) = f ( ) = + k ( ) = g f ( ) = g + = + Page 8 HSN000

21 Higher Mathematics Unit Mathematics Inverse Functions Defined on suitable domains, all the functions we will meet have an inverse function, which reverses the effect of the function. If we have a function defined b f ( ), its inverse is usuall denoted f ( ). If a number is worked through a function f then a function g, and the result is the same as ou started with, i.e. g f ( ) =, then f and g are inverses. + For eample, f ( ) = 4 and g ( ) = are inverse functions. If we 4 work the number through f and then the result through g, we will get back again. + f = 4 g = 4 = = Graphs of Inverses If we have the graph of a function, then we can find the graph of its inverse b reflecting in the line =. For eample, the diagrams below show the graphs of two functions and their inverses. = f ( ) = f f ( ) g = g ( ) = = f ( ) = g ( ) Page 9 HSN000

22 Higher Mathematics Unit Mathematics 4 Eponential Functions An eponential function is one in the form f ( ) = a where a, R and a > 0. This is known as an eponential function to the base a; is referred to as the power, inde or eponent. 0 Notice that when = 0, f ( ) = a =. Also when =, f ( ) = a = a. Hence the graph of an eponential alwas passes through ( 0, ) and (, a ) : = a, a > = a, 0 < a < (, a) (, a) EXAMPLE Sketch the curve with equation = 6. The curve passes through ( 0, ) and (, 6 ). = 6 (, 6) 5 Introduction to Logarithms Until now, we have onl been able to solve problems involving eponentials when we know the inde, and have to find the base. For eample, we can 6 solve k = 5 b taking sith roots to get k = 6 5. But what if we know the base and have to find the inde? To solve 6 k = 5 for k, we need to find the power of 6 which gives 5. To save writing this each time, we use the notation k = log6 5, read as log to the base 6 of 5. In general: log a is the power of a which gives The properties of logarithms will be covered in Unit utcome. Page 0 HSN000

23 Higher Mathematics Unit Mathematics Logarithmic Functions A logarithmic function is one in the form f ( ) = log a where a, > 0. Logarithmic functions are inverses of eponentials, so to find the graph of = log a, we can reflect the graph of = a in the line =. = log a ( a,) The graph of a logarithmic function alwas passes through (, 0 ) and ( a,). EXAMPLE Sketch the curve with equation = log6. The curve passes through (, 0 ) and ( 6, ). = log 6 ( 6,) 6 Radians Degrees are not the onl units used to measure angles. The radian (RAD on the calculator) is a measurement also used. Degrees and radians are related such that: π radians = 80 The other equivalences that ou should be familiar with are: 0 = π 6 radians 45 = π radians 60 = π 4 radians 90 = π radians 5 = π 4 radians 60 = π radians Converting between degrees and radians is straightforward. To convert from radians to degrees, multipl b 80 and divide b π. Degrees To convert from degrees to radians, multipl b π and divide b 80. For eample, 50 = 50 π 5 80 = 8π radians. 80 π 80 π Radians Page HSN000

24 Higher Mathematics Unit Mathematics 7 Eact Values The following eact values must be known. You can do this b either memorising the two triangles involved, or memorising the table. DEG RAD sin cos tan π 6 45 π 4 60 π 90 π 0 8 Trigonometric Functions Periodic functions have a repeating pattern in their graphs. The length of the smallest repeating pattern in the -direction is called the minimum period. If the repeating pattern has a minimum and maimum value, then half of the difference between the minimum and maimum is called the amplitude. ma. value amplitude min. value minimum period The three basic trigonometric functions (sine, cosine, and tangent) are periodic, and have graphs as shown below. = sin = cos = tan Period = 60 = π radians Amplitude = Period = 60 = π radians Amplitude = Period = 80 =π radians Amplitude is undefined Page HSN000

25 Higher Mathematics Unit Mathematics 9 Graph Transformations The graphs below represent two functions. ne is a cubic and the other is a sine wave, focusing on the region between 0 and 60. There are three different things we can do to the graphs. Translation ( p, q) = g A translation moves ever point on a graph a fied distance in the same direction. The shape of the graph does not change = sin Translation parallel to the -ais f ( ) + a moves the graph of f ( ) up or down. The graph is moved up if a is positive, and down if a is negative. (,) a is positive ( p, q + ) = g ( ) + (,) = sin + (, ) a is negative ( p, q ) = g ( ) (, ) = sin Page HSN000

26 Higher Mathematics Unit Mathematics Translation parallel to the -ais f ( + a) moves the graph of f ( ) left or right. The graph is moved left if a is positive, and right if a is negative. a is positive a is negative = g ( + ) = g ( ) ( p, q) ( p +, q) 5 = sin( + 90 ) = sin( 90 ) Reflection When reflecting, the graph is flipped about one of the aes. It is important to appl this transformation before an translation. Reflection in the -ais f ( ) reflects the graph of f ( ) in the -ais. = g ( ) ( p, q) = sin Reflection in the -ais f ( ) reflects the graph of f ( ) in the -ais. ( p, q) = g ( ) = sin( ) From the graphs, sin( ) = sin Page 4 HSN000

27 Higher Mathematics Unit Mathematics Scaling Scaling verticall kf ( ) scales the graph of f ( ) in the vertical direction. The -coordinate of each point on the graph is multiplied b k, roots are unaffected. These eamples consider positive k. k > 0 < k < ( p, q) = g ( ) ( p, q) = g ( ) = sin Negative k causes the same scaling, but the graph must then be reflected in the -ais: = g ( ) = sin ( p, q) Page 5 HSN000

28 Higher Mathematics Unit Mathematics Scaling horizontall f ( k ) scales the graph of f ( ) in the horizontal direction. The coordinates of the -ais intercept sta the same. The eamples below consider positive k. k > 0 < k < = g ( ) = g ( p, q ) ( p, q) 6 = sin = sin Negative k causes the same scaling, but the graph must then be reflected in the -ais: ( p, q) = g ( ) Page 6 HSN000

29 Higher Mathematics Unit Mathematics EXAMPLES. The graph of = f ( ) is shown below. = f ( ) 0 (, 4) Sketch the graph of = f ( ). 5 Reflect in the -ais, then shift down b : = f ( ) ( 5, ) 0 (, 6). Sketch the graph of = 5cos( ) where = 5cos( ) Page 7 HSN000

30 Higher Mathematics Unit Mathematics UTCME Differentiation Introduction to Differentiation Differentiation belongs to a branch of Mathematics called calculus an area which we have not covered before. Calculus provides a tool for solving problems involving motion (e.g. the orbits of planets or flight path of a rocket). To introduce this utcome, we will look at two distance/time graphs which describe the journe of a bugg. distance (m) distance (m) time (s) time (s) The first graph implies that the bugg could travel at a constant speed, then change instantl to a different speed. This is impossible the speed actuall changes graduall, as shown in the second graph. Instantaneous speed is the speed of an object at an eact moment in time. It is also known as the rate of change of distance with respect to time. Problems that involve rates of change are studied in differential calculus. The calculation of instantaneous speed can be a ver long process but differentiation provides a quick method. Page 8 HSN000

31 Higher Mathematics Unit Mathematics Finding the Derivative The basic rule for differentiating f ( ) = with respect to is: If n n f ( ) = then f ( ) = n where n R Stated simpl: the power (n) multiplies to the front of the term, and the power lowers b one ( n ). EXAMPLES 4. A function f is defined for R b f ( ) =. Find f ( ). f ( ) = 4. Differentiate f ( ) = with respect to. f ( ) = 4 For an epression in the form =, the derivative with respect to is epressed as d d. EXAMPLE. Differentiate d = d 4 = with respect to. To find the derivative of an epression in with respect to, the notation d is used. d EXAMPLE 4. Find the derivative of d d = Preparing to differentiate with respect to. It is important that before ou differentiate, all brackets are multiplied out, and there are no fractions with an term in the denominator (bottom line), for eample: = = = n 4 5 = = 5 Page 9 HSN000

32 Higher Mathematics Unit Mathematics EXAMPLES. Differentiate with respect to. d d ( ) = = =. Find the derivative of = with respect to. = d = d = Note It is good practice to tid up our answer Terms with a coefficient The rule for differentiating f ( ) = a with respect to is as follows: If n n f ( ) = a then f ( ) = an where n R, a is a constant Stated simpl: the power (n) multiplies to the front of the term, then the power lowers b one ( n ). EXAMPLES. A function f is defined for R b f ( ) =. Find f ( ). f ( ) = 6. Differentiate = 4 with respect to. d = 8 d 8 =. Differentiate d d ( ) = = 4 with respect to. n Page 0 HSN000

33 Higher Mathematics Unit Mathematics 4. Given =, find d d. = d = d 4 = 4 Differentiating more than one term The following rule allows us to differentiate epressions with several terms: If f ( ) = g ( ) + h ( ) then f ( ) = g ( ) + h ( ) Stated simpl: differentiate each term separatel. EXAMPLES. A function f is defined for R b f ( ) = + 5. Find f ( ). f ( ) = Differentiate d d = = with respect to Note The derivative of an term (e.g.,, 0 ) is alwas a constant. For eample: d ( 6 ) = 6 d d ( d ) = The derivative of a constant (e.g., 0, π ) is alwas zero. For eample: d d = d 0 0 d = Page HSN000

34 Higher Mathematics Unit Mathematics Differentiating more comple epressions We will now consider more comple eamples were we will have to use several of the rules we have met. EXAMPLES. Find d when d = +. = ( )( + ) = + = d = d 6 6. A function f is defined for R, 0 b f ( ) = +. Find f ( ). 5 f = f ( ) = = 5. Differentiate d d = = = 5 5 with respect to. 4. Differentiate = with respect to. f = = f ( ) = Page HSN000

35 Higher Mathematics Unit Mathematics 5. Differentiate d d + 6 with respect to = + = = = = 6. Find the derivative of ( ) = + = + d = + 6 d 5 5 = = + with respect to. Remember a a b = b Remember a b a+ b = Differentiating with Respect to ther Variables So far we have differentiated functions and epressions with respect to. However, the rules we have been using still appl if we differentiate with respect to an variable. Variables such as t (for time) are used commonl to model real-life problems. EXAMPLES. Differentiate d d t t t = 6t t with respect to t.. Given A( r ) = πr, find A ( r ). A( r ) = πr A ( r ) = πr When differentiating with respect to a certain variable, all other variables are treated as constants. EXAMPLE. Differentiate d dp p = p with respect to p. Remember π is just a constant Note Since we are differentiating with respect to p, we treat as a constant Page HSN000

36 Higher Mathematics Unit Mathematics 4 Rates of Change The derivative of a function describes its rate of change. This can be evaluated for specific values b substituting them into the derivative. EXAMPLES 5. Given f ( ) = for R, find the rate of change of f when =. f ( ) = f = 0 = 0 8 = 80. If =, calculate the rate of change of when = 8. = d At = 8, = d 8 d 5 = d = = 5 = 96 = = Remember 48 5 a b b a = Displacement, Velocit and Acceleration The velocit ( v ) of an object is defined as the rate of change of displacement ( s ) with respect to time ( t ). That is: v ( t ) = s ( t ) Also, acceleration ( a ) is defined as the rate of change of velocit with respect to time: a( t ) = v ( t ) EXAMPLE. A ball is thrown so that its displacement s after t seconds is given b s ( t ) = t 5t. Find its velocit after seconds. v ( t ) = s ( t ) = 0 t b differentiating s ( t ) = t 5 t with respect to t Substitute t = into v ( t ) : v = 0 = After seconds, the ball has velocit metres per second. 5 5 Page 4 HSN000

37 Higher Mathematics Unit Mathematics 5 Equations of Tangents As we alread know, the gradient of a straight line is constant. We can determine the gradient of a curve, at a particular point, b considering a straight line which touches the curve at the point. This line is called a tangent. tangent The gradient of the tangent to a curve = f ( ) at = a is given b f ( a). This is the same as finding the rate of change at = a. To work out the equation of a tangent we use b = m( a). Therefore we need to know two things about the tangent: A point, of which at least one coordinate will be given. The gradient, which is calculated b differentiating and substituting in the value of at the required point. EXAMPLES. Find the equation of the tangent to the curve with equation at the point (, ). We know the tangent passes through (, ). = To find its equation, we need the gradient at the point where = : = d = d At =, m = = 4 Now we have the point (, ) and the gradient m = 4, so we can find the equation of the tangent: b = m( a) = 4( ) = = 0 Page 5 HSN000

38 Higher Mathematics Unit Mathematics. Find the equation of the tangent to the curve with equation at the point where =. = We need a point on the tangent. Using the given -coordinate, we can find the -coordinate of the point where the tangent meets the curve: = = ( ) ( ) = + = So the point is (, ) We also need the gradient at the point where = : = d = d At =, m = ( ) = Now we have the point (,) and the gradient m =, so: b = m( a) = ( + ) + = 0. A function f is defined for > 0 b f ( ) =. Find the equation of the tangent to the curve = f ( ) at P. P = f ( ) We need a point on the tangent. Using the given -coordinate, we can find the -coordinate of the point where the tangent meets the curve: f ( ) = Remember a = a = b b c c = So the point is (, ) Page 6 HSN000

39 Higher Mathematics Unit Mathematics We also need the gradient at the point where = : f ( ) = f ( ) = = At =, m = 4 = 4 Now we have the point b = m( a) ( ) = 4 = = 0, and the gradient m = 4, so: 4. Find the equation of the tangent to the curve where = 8. = at the point We need a point on the tangent. Using the given -coordinate, we can work out the -coordinate: = 8 = ( ) = 4 So the point is 8, 4 We also need the gradient at the point where = 8: = = d = d = At = 8, m = 8 = = Now we have the point ( 8, 4) and the gradient m =, so: b = m( a) 4 = ( 8) = = 0 Page 7 HSN000

40 Higher Mathematics Unit Mathematics 6 Increasing and Decreasing Curves For the curve with equation = f ( ), if increases as increases, then the curve is said to be increasing. When the curve is increasing, tangents will slope upwards d from left to right (i.e. their gradients are positive) so 0 d >. Similarl, when decreases as increases, the curve is said to be decreasing d and 0 d <. increasing d 0 d > decreasing d 0 d < increasing d 0 d > 7 Stationar Points Some points on a curve ma be neither increasing nor decreasing we sa that the curve is stationar at these points. This means that the gradient of the tangent to the curve is zero at stationar d points, so we can find them b solving f ( ) = 0 or = 0. d The four possible stationar points are: Turning point Horizontal point of inflection Maimum Minimum Rising Falling A stationar point s nature (tpe) is determined b the behaviour of the graph to its left and right. This is done using a nature table. Page 8 HSN000

41 Higher Mathematics Unit Mathematics 8 Determining the Nature of Stationar Points To illustrate the method used to find stationar points and determine their nature, we will consider the graph of f ( ) = Step Differentiate the function. f ( ) = Step Find the stationar values b f ( ) = 0 solving f ( ) = = 0 ( ) 6 + = 0 ( 6) ( )( ) = 0 = or = Step Find the stationar points. f = 9 so (, 9 ) is a stat. pt. f = 8 so (, 8 ) is a stat. pt. Step 4 Write the stationar values in the top row of the nature table, with arrows leading in f ( ) and out of them. Graph Step 5 Calculate f ( ) for the values in the table, and record the results. This gives the f ( ) 0 0 gradient at these values, so zeros confirm Graph that stationar points eist here. Step 6 Calculate f ( ) for values slightl lower and higher than the stationar values and record the sign in the second row, e.g.: f ( 0.8) > 0 so enter + in the first cell. Step 7 We can now sketch the graph near the stationar points: + means the graph is increasing and means the graph is decreasing. Step 8 The nature of the stationar points can then be concluded from the sketch. f ( ) Graph f ( ) Graph (, 9 ) is a ma. turning point (, 8 ) is a min. turning point Page 9 HSN000

42 Higher Mathematics Unit Mathematics EXAMPLES. A curve has equation = Find the stationar points on the curve and determine their nature. Given = d = + 9 d d Stationar points eist where 0 d = : + 9 = 0 ( ) 4 + = 0 ( ) 4 + = 0 ( )( ) = 0 = 0 or = 0 = = When =, = = = 0 Therefore the point is (, 0 ) Nature: d d Graph So (, 0 ) is a maimum turning point, (, 4) is a minimum turning point. When =, = = = 4 Therefore the point is (, 4) Page 40 HSN000

43 Higher Mathematics Unit Mathematics. Find the stationar points of 4 Given = 4 d = 8 d d Stationar points eist where 0 d = : 8 = 0 ( ) 6 4 = 0 = 0 or 6 4 = 0 = 0 = When = 0, = 4( 0) ( 0) 4 = 0 Therefore the point is ( 0, 0 ) Nature: 0 d d Graph So ( 0, 0 ) is a rising point of inflection, 7, is a maimum turning point. 8 4 = 4 and determine their nature. When =, = 4 = = Therefore the point is 7 (, ) 8 9 Curve Sketching In order to sketch a curve, we need to first find the following: -ais intercepts (roots) solve = 0 -ais intercept find for 0 = stationar points and their nature. Page 4 HSN000

44 Higher Mathematics Unit Mathematics EXAMPLE Sketch the graph of =. -ais intercept: = 0 = ( 0) ( 0) = 0 Therefore the point is ( 0, 0 ) Given = d = 6 6 d d Stationar points eist where 0 d = : 6 6 = 0 6 ( ) = 0 6 = 0 or = 0 = 0 When = 0, = ( 0) ( 0) = 0 Therefore the point is ( 0, 0 ) = Nature: 0 d d Graph ais intercepts (roots): = 0 = 0 = 0 = 0 ( 0, 0) When =, ( ) = 0 or = = = Therefore the point is (, ) = 0 = (, 0) ( 0, 0 ) is a maimum turning point (, ) is a minimum turning point = (, ) Page 4 HSN000

45 Higher Mathematics Unit Mathematics 0 Closed Intervals Sometimes it is necessar to restrict the part of the graph we are looking at using a closed interval (also called a restricted domain). The maimum and minimum -values can either be at stationar points or at the end points of the closed interval. Below is a sketch of a curve, with the closed interval 6 shaded. maimum minimum EXAMPLE A function f is defined for R b f ( ) = Find the maimum and minimum value of f ( ) where 4. Given f ( ) = f ( ) = Stationar points eist where f ( ) = 0: = 0 ( ) 5 = 0 ( )( + ) = 0 = 0 or + = 0 = = To find coordinates of stationar points: f = = = Therefore the point is (, ) 6 ( ) ( ) ( ) ( ) = ( ) f = = = 46 7 Therefore the point is (, 46 ) 7 Page 4 HSN000

46 Higher Mathematics Unit Mathematics Nature: f ( ) Graph 46 (, ) 7 is a ma. turning point (, ) is a min. turning point Points at etremities of closed interval: f ( ) = ( ) 5( ) 4( ) + = = Therefore the point is (, ) f 4 = = = Therefore the point is ( 4, ) Now we can make a sketch: ( 4, ) (, 46 ) 7 (, ) (, ) The maimum value is which occurs when = 4 The minimum value is which occurs when = Page 44 HSN000

47 Higher Mathematics Unit Mathematics Graphs of Derived Functions n n The derivative of an function is an function the degree lowers b one. For eample the derivative of a cubic ( ) function, is a quadratic ( ) function. When drawing a derived graph: All stationar points of the original function become roots (i.e. lie on the -ais) on the derived graph The sign (+ or ) of the original graph s gradient becomes the value of the derived graph between the roots. For eample, if the gradient between two stationar points of the original graph is negative, then the graph between the roots of the derived graph will lie below the -ais (i.e. it will be negative). Quadratic Cubic Quartic Linear + + Quadratic + + Cubic ptimisation Differentiation can be used to help solve real-life problems. However, often we need to set up an equation before we can differentiate. Page 45 HSN000

48 Higher Mathematics Unit Mathematics EXAMPLE Small wooden tras, with open tops and square bases, are being designed. The must have a volume of 08 cm. The internal length of one side of the base is cm, and the internal height of the tra is h cm. (a) Show that the total internal surface area A of one tra is given b 4 A = + (b) Find the dimensions of the tra which use the smallest amount of wood. (a) Volume = area of base height = h We are told that the volume is 08 cm, so: Volume = 08 h = h = h Let A be the surface area for a particular value of : A = + 4h 08 We have h =, so: A = = + 4 (b) The smallest amount of wood is used when the surface area is minimised. da 4 = d da Stationar points occur when 0 d = Nature: 6 4 = 0 da d 0 + = 6 Graph = 6 So the minimum surface area occurs when = 6. For this value of : 08 h = = 6 So a length and depth of 6 cm and a height of cm uses the smallest amount of wood. Page 46 HSN000

49 Higher Mathematics Unit Mathematics UTCME 4 Sequences Introduction to Sequences A sequence is a list of numbers which follow a pattern. To save writing out a list each time, we can define a sequence using a rule or formula. Recurrence Relations Recurrence relations are one wa of describing sequences. These give the nth term of a sequence in terms of the ( n ) th term. Consider the eample where we are adding on 4% interest annuall to 000 for a 6 ear period. When adding on 4% to the initial value, we could do the following: Year Interest = 4% of 000 = = 40 Total = = 040 Calculating 040 b this method is the same as calculating 04% of the initial value. In recurrence relations, we use u 0 to stand for the initial value. Therefore the first term of the sequence is u. u u 0 = 000 = 04% of 000 = = 040 The recurrence relation for this sequence can be written as un+ = 04 un with u 0 = 000 where u n is the value after n ears, and u n + is the value after n + ears. Page 47 HSN000

50 Higher Mathematics Unit Mathematics EXAMPLE The value of an endowment polic increases at the rate of 5% per annum. The initial value is (a) Write down a recurrence relation for the polic s value after n ears. (b) Calculate the value of the polic after 4 ears. (a) Let u n be the value of the polic after n ears. So un+ = 05 un with u = (b) u 0 = 7000 u = = 750 u = = u = = u4 = = After 4 ears, the polic is worth Linear Recurrence Relations Linear recurrence relations are of the form un+ = aun + b where a 0, b R. Note To properl define a sequence using a recurrence relation, we must specif the initial value u 0. EXAMPLES. A patient is injected with 56 ml of a drug. Ever 8 hours, % of the drug passes out of his bloodstream. To compensate, a further 5 ml dose is given ever 8 hours. (a) Find a recurrence relation for the amount of drug in his bloodstream. (b) Calculate the amount of drug remaining after 4 hours. (a) Let u n be the amount of drug in his bloodstream after 8n hours. un+ = 0 78u n + 5 with u = 56 0 (b) u 0 = 56 u = = u u = = = = 740 After 4 hours, he will have.74 ml of drug in his bloodstream. Page 48 HSN000

51 Higher Mathematics Unit Mathematics. A sequence is defined b the recurrence relation un+ = 0 6un + 4 with u =. 0 7 Calculate the value of u and the smallest value of n for which u > 9 7. u u u u u u = 7 = = 8 = = 8 9 = = 9 5 The value of u is 9 5 = 9 6 = The smallest value of n for which u > 9 7 is 5 Using a Calculator Using the ANS button on the calculator, we can carr out the above calculation more efficientl. Divergence and Convergence n 7 = 0 6 ANS + 4 = If we plot the graphs of some of the sequences that we have been dealing with, then some similarities will occur. Divergence Sequences defined b recurrence relations in the form un+ = aun + b where a < or a >, will have a graph like this: = = n u n a > Sequences like this will continue to increase or decrease forever. u n a < n n The are said to diverge. Page 49 HSN000

52 Higher Mathematics Unit Mathematics Convergence Sequences defined b recurrence relations in the form un+ = aun + b where < a <, will have a graph like this: u n 0 < a < n Sequences like this tend to a limit. The are said to converge. u n < a < 0 n 4 The Limit of a Sequence For convergent sequences defined b un+ = aun + b with < a <, u n tends to a limit l as n (i.e. as n gets larger and larger) and: b l = where < a < a You will need to know this formula, as it is not given in the eam. EXAMPLE The deer population in a forest is estimated to drop b 7 % each ear. Each ear, 0 deer are introduced to the forest. The initial deer population is 00. (a) How man deer will there be in the forest after ears? (b) What is the long term effect on the population? (a) un+ = 0 97u n + 0 u = 00 u u u 0 = = 05 4 = = = = Therefore there are 5 deer living in the forest after ears. (b) A limit eists, since < 0 97 < b l = where a = 0 97 and b = 0 a 0 = 0 97 = 7 97 (to d.p.) Therefore the number of deer in the forest will settle around 7. Note Whenever ou calculate a limit, ou must state that A limit eists since < a < Page 50 HSN000

53 Higher Mathematics Unit Mathematics 5 Solving Recurrence Relations to find a and b If we know that a sequence is defined b a linear recurrence relation of the form un+ = aun + b, and we know several terms of the sequence, then we can find the values of a and b. This can be done easil b solving the equations simultaneousl. EXAMPLE A sequence is defined b un+ = aun + b with u = 4, u = 6 and u = 04. Find the values of a and b. Find equations for two values of n: u = au + b u = au + b 6 = 4a + b 04 = 6a + b 4a + b = 6 6a b = 04 So a =. 9 and b =. Solve for a b eliminating b: 4a + b = 6 6a b = a = 56 a = a = Substitute a = 9 into equation: 4( 9 ) + b = b = 6 b = b = Page 5 HSN000