MEI Core 2. Sequences and series. Section 1: Definitions and Notation

Transcription

1 Notes and Eamples MEI Core Sequences and series Section : Definitions and Notation In this section you will learn definitions and notation involving sequences and series, and some different ways in which sequences and series can be generated. These notes contain subsections on Types of sequence Sequences defined deductively Sequences defined inductively Types of sequence A sequence is a set of numbers in a given order. These numbers may form an algebraic pattern. You need to make sure that you understand what is meant by: an arithmetic sequence (see page 6) a geometric sequence (see pages 6-4) an oscillating sequence (see page 64) a periodic sequence (see pages 64-5) A series is the sum of the terms of a sequence. You need to be familiar with the notation for a series ( is pronounced sigma and is the Greek capital S): 0 a k means the series a + a + a + + a 0. Sequences which follow a pattern can be defined algebraically in one of two ways: deductively or inductively. Sequences defined deductively A deductive definition gives a direct formula for the kth term of the sequence in terms of k. The terms of the sequence can be found by substituting the numbers,, for k. Eample A sequence is defined deductively by ak k (i) Write down the first five terms of the sequence. MEI, 6/0/09 /

2 MEI C Sequences Section Notes and Eamples (ii) Find the 0 th term of the sequence. (iii) Find 5 a k (i) Substituting k =, k =, k = 5 into the epression sequence,, 6,, (ii) Substituting k = 0 a0 0 (iii) a 6 k 40 ak k gives the For further eamples on sequences defined deductively, look at the Flash resources Finding sequences from general terms and Finding general terms. You can also look at further eamples using the sigma notation using the Flash resources Sigma notation () and Sigma notation (). Sequences defined inductively An inductive definition tells you how to find a term in a sequence from the previous term. The definition must also include the value of the first term of the sequence. You can then find the second term from the first term, the third term from the second term, and so on. Eample A sequence is defined inductively as a k a k, a 0 (i) Write down the first si terms of the sequence. (ii) Find 6 a k (i) Each term is found by doubling the previous term and adding. The first term is 0. a a 0 a 0 a a MEI, 6/0/09 /

3 MEI C Sequences Section Notes and Eamples a a a 7 4 a a6 a5 5 The first si terms are 0,,, 7, 5, (ii) 6 a k 57 For further eamples on sequences defined inductively, look at the Flash resources Sequences from recurrence relations and Finding recurrence relations. MEI, 6/0/09 /

4 MEI Core Sequences and series Section : Arithmetic sequences and series Notes and Eamples In this section you learn about arithmetic sequences and series (sometimes called arithmetic progressions or A.P.s), which were introduced in the last section. These notes contain subsections on Formulae for arithmetic series Worked eamples Harder eamples Formulae for arithmetic series All you need to know to answer a question on arithmetic sequence are the two general formulae for any arithmetic sequence: the formula for a k, the kth term of the sequence a a ( k ) d k the formula for S n, the sum of the first n terms of the sequence Sn n a n d [ ( ) ] where a is the first term of the sequence and d is the common difference (the difference between successive terms). S n, the sum of the first n terms of the sequence, can also be calculated as Sn n[first term + last term] To solve a problem, all you need to do is to substitute the information given in the question into the appropriate formula, and solve the resulting equation. Sometimes this may involve solving a quadratic equation or simultaneous equations. The Arithmetic series spreadsheet shows you graphs of the terms of arithmetic series and the sum of n terms. Try varying the values of a and d, looking in particular at what happens when d is negative. MEI, 6/0/09 /4

5 MEI C Sequences Section Notes and Eamples Worked eamples Eample shows a straightforward application of the formulae. Eample An arithmetic sequence which has 6 terms starts, 5, 8 (i) Find the last term. (ii) Find the sum of the terms of the sequence. (i) The first term, a, is and the common difference, d, is. Use the formula ak a ( k ) d To find the 6 th term, substitute k = 6, a = and d = 6 th term = + (6 ) = + 5 = 47 (ii) Use the formula Sn n[ a ( n ) d]. Substitute as before: Sum 6[ (6 ) ] 8[4 5] To see further eamples, use the Flash resources nth terms of an AP and Sum of an AP. For practice in finding a particular term of an arithmetic series, as in Eample (i), try the interactive questions Finding terms in arithmetic series. For practice in finding the sum of an arithmetic series, as in Eample (ii), try the interactive questions Finding the sum of an arithmetic series. Eample An arithmetic series has first term and the sum of the first 0 terms is 88. Find the common difference. Sn n a n d d 0 6 9d 60 90d 90d 8 d. Substituting n = 0, a =, and S n = 88 MEI, 6/0/09 /4

6 MEI C Sequences Section Notes and Eamples For practice in questions like the one above, try the interactive questions Finding the common difference in an A.P. Eample An arithmetic series has common difference -0.5 and the sum of the first 5 terms is 50. Find the first term. Sn n a n d 50 5 a a a 8 a 40 a 0 Substituting n = 5, d = -0.5, and S n = 50 For practice in questions like the one above, try the interactive resource Finding the first term of an A.P. Harder eamples Eample 4 involves solving simultaneous equations. Eample 4 The 5 th term of an arithmetic sequence is 4 and the 9 th term is 4. (i) Find the first term and the common difference. (ii) The last term of the sequence is 6. How many terms are in the sequence? (i) Using the formula ak a ( k ) d For the 5 th term: 4 = a + (5 )d 4 = a + 4d For the 9 th term: 4 = a + (9 )d 4 = a + 8d 4 = a + 4d 4 = a + 8d Subtracting: 0 = -4d d = -5 Substituting d = -5 into either equation gives a = 44. As the 9 th term is smaller than the 5 th term, you would epect the common difference to be negative. (ii) Substituting a = 44 and d = -5 into the formula a a ( k ) d : For the last term: -6 = 44 + (k )-5 k MEI, 6/0/09 /4

7 MEI C Sequences Section Notes and Eamples -80 = -5(k ) 6 = k k = 7 There are 7 terms. In the net eample, you are given the sum of n terms and you need to find the value of n. You need to solve a quadratic equation to find n. Eample 5 The sum of the terms of an arithmetic sequence with first term 5 and common difference 6 is 66. How many terms are there in the sequence? Substituting a = 5 and d = 6 into the formula Sn n[ a ( n ) d] gives 66 n[ 5 ( n) 6] 66 n(0 6n 6) 66 n(4 6 n) 66 nn n n 66 0 (n 44)( n4) 0 Since n must be positive, n = 4 The sequence has 4 terms. For practice in questions like the one above, try the interactive resource Finding the number of terms in an A.P. MEI, 6/0/09 4/4

8 MEI Core Sequences and series Section : Geometric sequences and series Notes and Eamples These notes contain subsections on Formulae for geometric sequences Worked eamples Finding the number of terms Formulae for geometric sequences In the section on Arithmetic sequences, you used two basic formulae to solve problems: the formula for the kth term in an arithmetic sequence, and the formula for the sum of the first n terms of the sequence. When you are working with geometric sequences, you need to use the equivalent formulae, and you may also need to use a third formula, for the sum to infinity of the sequence. The formula for the kth term of a geometric sequence is ak ar k where a is the first term of the sequence and r is the common ratio (the number that each term is multiplied by to obtain the net term). The formula for the sum of the first n terms of a geometric sequence is S n n a( r ) r or S n n ar ( ) r These two formulae are equivalent. If r is less than, then it is easier to use the left-hand formula, and if r is greater than, it is easier to use the righthand version. However, either will give you the right answer! The formula for the sum to infinity of a geometric sequence is a S for < r < r The restriction < r < is very important, as it is only for these values of r that the series converges (i.e. the terms become numerically smaller and smaller). If r >, the terms become larger and larger and so the sum of the terms becomes larger and larger, so it makes no sense to talk about the sum to infinity. If r < -, the terms become numerically larger and larger, but alternate between positive and negative, so the sum of the terms becomes MEI, 8/08/09 /6

9 MEI C Sequences Section Notes and Eamples alternately large and small, and again it makes no sense to talk about the sum to infinity. In both the cases r > and r < -, the series diverges. You can eplore geometric series using the interactive Geometric series spreadsheet. Try varying the values of a and r. Notice in particular how the terms oscillate when r is negative. Worked eamples Eample A geometric sequence has first term 4 and common ratio ½. (i) Find the 6 th term. (ii) Find the sum of the first 0 terms. (iii) Find the sum to infinity of the terms of the sequence. k (i) a ar k 6 th term = Substituting a = 4, r = ½ and k = 6 (ii) S S n n a( r ) r 4( 0 ) 8( 0 ) Substituting a = 4, r = ½ and n = 0 (iii) S a r 4 8 Substituting a = 4, and r = ½ You can see additional eamples using the Flash resources nth terms of a GP, Sum of a GP, and Sum of an infinite GP. For practice in questions like Eample (i), try the interactive resource Finding terms in a geometric series. For practice in questions like Eample (iii), try the interactive resource Finding the sum of a geometric series. Eample A geometric series has common ratio 0.5 and the sum of the first 5 terms is Find the first term. MEI, 8/08/09 /6

10 MEI C Sequences Section Notes and Eamples n a( r ) Sn r 5 a a a 40 The first term is 40. Substituting r = 0.5, S n = 77.5 and n = 5 For practice in questions like Eample, try the interactive resource Finding the first term of a finite G.P. In Eample, the common ratio is negative. Be careful when you are using your calculator to work out powers of a negative number you will probably need to use brackets. If you type in ^8 then the calculator will probably work out 8 and then apply a negative. So you need to type in (-)^8. Remember that even powers of a negative number are positive, and odd powers are negative, and check that your answer is sensible. Eample A geometric sequence has rd term and 6 th term 96. (i) Find the first term and the common ratio. (ii) Find the sum of the first 8 terms. (i) rd term = ar ar 6 th 5 5 term = ar ar 96 Dividing the second equation by the first gives ar 4a a The first term is and the common ratio is. r 8 r = (ii) S S n 8 n ar ( ) r 8 (( ) ) (56 ) 55 MEI, 8/08/09 /6

11 MEI C Sequences Section Notes and Eamples In the net eample, you have to use the formula for the sum of an infinite geometric series. Eample 4 (i) The sum of an infinite geometric series with common ratio -0.5 is 6.4. Find the first term. (ii) The sum of an infinite geometric series with first term is 5. Find the common ratio. a (i) S r a a.5 a The first term is 8. (ii) a S r 5 r r r The common ratio is 0.4. For practice in questions like Eample 4(i), try the interactive resource Finding the first term of an infinite G.P. For practice in questions like Eample 4(ii), try the interactive resource Finding the common ratio in an infinite G.P. Finding the number of terms In Eamples 5 and 6, you have to find the value of n, which appears as an inde. If you have covered logarithms, you can use these to find n. If not, you can still solve the problem, by using trial and improvement. In Eample 5, it is quite simple to use trial and improvement. In Eample 6, both methods are shown. Eample 5 The sum of a geometric series with first term and common ratio - is 5. Find the number of terms in the series. MEI, 8/08/09 4/6

12 MEI C Sequences Section Notes and Eamples n a( r ) Sn r ( ) 5 ( ) ( ) 5 n 5 ( ) n ( ) 5 By trial and improvement n = 9. There are 9 terms in the series. n n For practice in questions like the one above, try the interactive resource Finding the number of terms in a finite G.P. In the net eample you have to find the number of terms required to eceed a given sum. Eample 6 Tom saves money every year. The first year he saves 00. Each year he increases the amount he saves by 0%. After how many years do Tom s savings first eceed 000 (ecluding any interest he has earned)? The amount Tom saves each year forms a geometric sequence with a = 00 and r =.. The sum of Tom s savings after n years is given by the formula n 00(. ) Sn. n 00(. ) 0. n 000(. ) When this amount eceeds 000 n 000(. ) 000 n. n. MEI, 8/08/09 5/6

13 MEI C Sequences Section Notes and Eamples Solve the inequality either using logarithms or by trial and improvement Using logarithms n. n log. log n log. log n log log. n 7.7 Using trial and improvement The savings first eceed 000 after 8 years. For practice in questions like the one above, try the interactive resource Number of terms in a G.P to eceed a given sum. MEI, 8/08/09 6/6

14 MEI Core Differentiation Section : Introduction to differentiation Notes and Eamples These notes include sub-sections on: What is differentiation? Investigating gradients Rules for finding derivatives Differentiating from first principles Finding tangents and normals to curves What is differentiation? In this section, you will be studying the relationship between the position of a point on a curve and the gradient of the curve. Straight lines are, by definition, lines of constant gradient. Curves, on the other hand, have varying gradient the gradient depends on whereabouts you are on the curve. Differentiation is the process of finding the gradient at any point on a curve from the equation of the curve. Differentiation, together with its reverse process, called integration, form the branch of mathematics called calculus. The discovery of calculus (Made in the 7 th century by Isaac Newton in England and, independently, by Gottfried von Liebnitz in Germany) was one of the most significant advances in the history of mathematics and science, and was crucial to unlocking the mathematical basis of our planetary system. Differentiation is the process of finding the gradient function, or derivative, or derived function. Given an equation for y in terms of, the gradient function or derivative is written d y, and gives the gradient of the curve in terms of. d Investigating gradients You can look at a demonstration of how the gradients of chords approach the gradient of a tangent using the Flash resource The limit of the gradient of a chord on y = ². This looks at the eample on page 9. You can see another eample using the Flash resource The limit of the gradient of a chord on y = ³. To investigate the pattern in the value of the gradient at different points on a curve, use the Flash resources Gradient of a curve at a point on y = ² and Gradient of a curve at a point on y = ³. MEI, 4/05/0 /7

15 MEI C Differentiation Section Notes and Eamples Rules for finding derivatives. If y is a polynomial function (made up of powers of ), the following rules will enable you to find the derivative d y d : The derivative of n is n n, The derivative of k n is kn n, The derivative of a constant is zero. The derivative of a sum (or difference) is the sum (or difference) of the derivatives Eample Differentiate y = The derivative of is = 6 The derivative of 5 is 5 = 0 The derivative of 4 is 0. So d y d = 6 0. You can see further eamples using the Flash resources Basic differentiation: individual powers of and Basic differentiation: sums of powers of. For practice in questions like the one above, try the interactive questions Basic differentiation. For additional practice, try the Differentiation Dominoes activity. Cut out the dominoes and match each epression (on the blue right side of a domino) with its derivative (on the yellow left side of a domino). The dominoes should form a single loop. The net eample involves an epression which is the product of two functions. You cannot differentiate this by differentiating each function separately and then multiplying the results, i.e. the derivative of a product of two functions is not the product of the derivatives! So with eamples involving brackets, you will need to multiply out the brackets first. (There is a rule for differentiating the product of two functions, but you do not need to know this yet.) Eample (i) Find the derivative of y = ( )( + ). (ii) Hence find the gradient of the curve at the point (, 0) (iii) Find the coordinates of the points where the gradient is zero. MEI, 4/05/0 /7

16 MEI C Differentiation Section Notes and Eamples (i) y = ( )( + ) = + dy = 4 + d (ii) (iii) Substituting = into the gradient function, dy 4 d 5 so the gradient of the curve at (, 0) is 5. The gradient of the curve is zero when d y d 4 + = 0 ( )( ) = 0 = and =. When =, y = When =, y = ( )( + ) =. So the points on the curve with gradient zero are (, ) and Now calculate the y coordinates for these values of., 50 7 The points where the gradient is zero are called the turning points or stationary points of the curve. You will look at such points in more detail in Section. The net eample involves the quotient of two functions (i.e. one function divided by another). As with products the derivative of a quotient is not the quotient of the derivatives. You need to divide the fraction first. Eample Differentiate y = = dy d = 6 MEI, 4/05/0 /7

17 MEI C Differentiation Section Notes and Eamples Eample 4 Sketch the graph of the gradient function of the curve shown below. y Gradient positive So derivative positive - - dy d Gradient zero at these points, so derivative is zero Gradient negative So derivative negative Differentiating from first principles The notation d y for the derivative of a function comes from differentiation d from first principles. To find the gradient function of a function f, you find the gradient of the chord which joins the point (, f()) to another point ( +, f( + )), where is very small. When you have simplified this as much as possible, you then let = 0, and the chord becomes a tangent to the graph. Eample 8. in the tetbook illustrates this method for the function y. Here is a further eample. MEI, 4/05/0 4/7

18 MEI C Differentiation Section Notes and Eamples Eample 5 Differentiate y = from first principles. y + y = ( + ) ( + ) and y = Subtracting gives Dividing by : y = ( + ) ( + ) + δy δ = ( δ ) ( δ ) δ δ (δ ) (δ ) δ = δ δ (δ ) (δ ) δ = δ = (δ ) (δ ) Now letting 0, δ y δ d y d =. Notice that this is the same answer you would obtain by applying the rules to find d y d. By differentiating from first principles you can see that the rules do indeed give the correct gradient functions (derivatives). Finding Tangents and Normals to Curves The gradient of a tangent to a curve at a particular point is the same as the gradient of the curve at that point. So to find the equation of a tangent to a curve, you first need to find the gradient m of the curve via differentiation. You can then substitute m and the coordinates (, y ) of the point on the curve into the formula: y y = m( ). You can look at eamples of tangents to a quadratic curve using the Flash resource Tangent at a point on a quadratic curve. Eample 6 Find the equation of the tangent to the curve y = at the point with - coordinate. MEI, 4/05/0 5/7

19 MEI C Differentiation Section Notes and Eamples y = d y d = 6. dy At the point with -coordinate, 6 d The gradient of the tangent is therefore. y = y = To find the gradient of the tangent, first differentiate the equation of the curve. Find the y-coordinate of the point where = by substituting into the equation of the curve. y y m( ) y ( ) ( ) y y 4 This is the required equation of the tangent. Use the formula for the equation of a line with m =, = and y = The normal to a curve is the line perpendicular to the tangent. Remember that the gradient of a line perpendicular to a line with gradient m is m, where m. m You can look at eamples of normals to a quadratic curve using the Flash resource Normal at a point on a quadratic curve. Eample 7 Show that the normal to the curve y = at the point (, ) passes through the origin. y = d y = 4. d dy At the point with coordinate, 4 d Gradient of the tangent = so gradient of the normal = y y m( ) y ( ) ( ) y y This line passes through the origin. First, find the gradient of the tangent by differentiating y. Use the formula for the equation of a line with m = -, = and y = MEI, 4/05/0 6/7

20 MEI C Differentiation Section Notes and Eamples You may find the Mathcentre video Tangents and normals useful. Note that this video may contain eamples which use some differentiation techniques that you have not yet met. For practice in questions like Eamples 6 and 7, try the interactive questions Finding tangents and normals. Eample 8 A curve has equation y. (i) Find the -coordinates of the points on the curve with gradient 6. (ii) Find the -coordinates of the points on the curve for which the gradient of the normal is. dy y d (i) Gradient = ( 5)( ) 0 or (ii) Gradient of normal = gradient of curve = Gradient = 5 0 ()( ) 0 or For practice in questions like the one above, try the interactive questions Finding points on curves, given gradients. MEI, 4/05/0 7/7

21 MEI Core Differentiation Section : Applications of differentiation Notes and Eamples These notes contain sub-sections on: Turning points Second derivatives Maimum and minimum problems Turning Points Points on a curve where the tangent is horizontal are called stationary points, or turning points. dy At these points, the gradient of the curve is zero, so d = 0. Stationary points are classified into three different types: Local maimum The gradient is positive to the left, zero at the point, and negative to the right. gradient positive gradient zero gradient negative Local minimum The gradient is negative to the left, zero at the point, and positive to the right. Stationary point of inflection gradient negative gradient zero gradient positive The gradient goes from positive to zero to positive or negative to zero to negative. gradient positive gradient zero gradient positive gradient negative gradient zero gradient negative MEI, 0//0 /6

22 MEI C Differentiation Section Notes and Eamples To distinguish between these, you can either: Test the value of the derivative either side of the stationary point, or Use the second derivative test - see later. Eample Find the stationary points on the curve y =, investigate their nature, and sketch the curve. y = dy d =. = 0 = = = or Step : Differentiate the function. Step : Solve d y 0 d When =, y = () () = () = ; When =, y = =. So the stationary points are (, ) and (, ). Step : Calculate the ycoordinates for these values of (called the stationary values). Step 4: Use a table to investigate the sign of d y d for values of either side of the stationary values dy d -ve +ve -ve So (, ) is a local minimum and (, ) is a local maimum The curve crosses the y ais when = 0. When = 0, y = 0. Step 5: Find where the curve cuts the aes It crosses the ais when y = 0. When y = 0, = 0 ( ) = 0, = 0, and. MEI, 0//0 /6

23 MEI C Differentiation Section Notes and Eamples y (, ) 0 Step 6: Sketch the curve. Make sure your sketch includes the coordinates of the intercepts and the turning points. (, ) You may like to look at the PowerPoint animation of this eample. For etra practice in finding stationary points, use the Flash resources Stationary points on quadratic curves and Stationary points on cubic curves. For each problem, try to find and identify the stationary points yourself, then check your answer and look at the graph. Second Derivatives If you differentiate a derivative, you get the second derivative. If you start with an equation for y in terms of, the first derivative is d y (you say: dee y by d dee ) and the second derivative is written squared ) d d y (you say: dee two y by dee Eample Given that y =, find d d y. y = dy d = d y = 6 d For some more eamples, look at the Flash resource Second derivatives. One important application of second derivatives is in the motion of a particle. If you study Mechanics you will learn more about this. If you start with MEI, 0//0 /6

24 MEI C Differentiation Section Notes and Eamples displacement s as a function of time t, then the first derivative d s gives the rate dt d s of change of displacement, or the velocity, and the second derivative dt gives the rate of change of velocity, or acceleration. Eample The displacement s metres of a particle from a point P at time t seconds is given by the equation: s = t 4t. Calculate the velocity and the acceleration of the particle after seconds have elapsed. s = t 4t v = d s = t 4 dt d s a = = 6t dt So when t =, v = 4 = 8 m s a = 6 = m s The second derivative measures the rate of change of the first derivative. We can use this fact to investigate the nature of turning points: Maimum points d y If < 0, the gradient function d y is decreasing. d d At a maimum point, the gradient goes from + to 0 to, in other words is decreasing. gradient negative So d y d < 0 the turning point is a maimum. gradient positive gradient zero Minimum points d y If > 0, the gradient function d y is increasing. d d At a minimum point, the gradient goes from to 0 to +, in other words is increasing. gradient negative gradient zero gradient positive So d y d >0 the turning point is a minimum. MEI, 0//0 4/6

25 MEI C Differentiation Section Notes and Eamples Points of inflection d y You might think that when = 0, the turning point d is a stationary point of inflection, and it is certainly true that if the stationary point is an inflection, then the value of the second derivative at the point is indeed zero. gradient positive gradient zero However, the converse statement is not true: dy d y You cannot conclude that if 0 and 0, the point is a stationary d d inflection. gradient positive In this case, you need to use the table method to investigate the turning point, or consider what you know about the graph of the function. For eample, 4 consider y. You may find the Mathcentre video Maima and minima useful. Note that this video may contain eamples which use some differentiation techniques that you have not yet met. Eample 4 Check the nature of the turning points of y = using the second derivative test. y = dy d = dy d = 0 when = 0 = or When =, y = ; when =, y = The stationary points are (, ) and (-, -) d y = 6 d d y When =, = 6 < 0 maimum d d y When =, = 6 > 0 minimum. d (, ) is a maimum point and (-, -) is a minimum point. MEI, 0//0 5/6

26 MEI C Differentiation Section Notes and Eamples Maimum and Minimum Problems One important immediate application of differentiation is to problems that involve maimising or minimising a variable quantity. Eample 5 A rectangular sheet of metal of length m and width m has squares cut from each corner. The sides are then folded up to form an open topped bo. Find the maimum possible volume of the bo. Step : Draw a diagram and use it to help you to formulate the problem mathematically. Call the side length of the squares cut out. What are the length, width and depth of the bo in terms of? Length = width = Depth = The volume V m of the bo is given by V = ( ). Step : Find the maimum volume by differentiating. The maimum volume will occur when d V 0 d. Before differentiating, epand the brackets. V ( ) dv = 8 + d Now put d V 0, and solve for : d 8 + = 0 ( )( 6) = 0 = or = 6. When When 6,, V 0. This must be the minimum. V. This must be the maimum So the maimum possible volume of the bo is m. 7 MEI, 0//0 6/6

27 MEI Core Integration Section : Introduction to integration Notes and Eamples These notes contain subsections on: Reversing differentiation The rule for integrating n Finding the arbitrary constant An application: the motion of a particle Reversing differentiation Integration is the reverse of differentiation. If you are given an epression for dy, and you want to find an epression for y, you need to use integration. d This is sometimes called solving a differential equation. Remember that when you integrate, you must always add an arbitrary constant (see the tetbook for the eplanation of this). Eample shows how you can integrate a function by thinking about what function you would need to differentiate to obtain the given function. Eample Find the general solution of the following differential equations: dy (i) d (ii) dy 6 d (iii) d y t dt df (iv) d (v) ds 4 t dt (vi) d u 5 dv dy 4 (i) y 4 c d 4 The derivative of is 4. 4 So integrating 4 gives. Therefore integrating gives 4 4. (ii) dy d y c The derivative of is So integrating 7 gives. 6 7 Therefore integrating gives 7. MEI, 0/0/ /5

28 MEI C Integration Section Notes and Eamples (iii) dy dt t y t c t is t. The derivative of So integrating t gives t. Therefore integrating t gives t. (iv) (v) (vi) df d ds d f c t t s t c du 5 u 5v c dv The derivative of is. So integrating gives. Therefore integrating gives t, and integrating gives 5t. t. 5 4 The derivative of t is 4 So integrating 5t gives t Therefore integrating t gives t 5, and integrating 4 t gives 5 5 t. The derivative of v is. So integrating gives v. Therefore integrating 5 gives 5v. The rule for integrating n The method for integrating any polynomial function can be summed up as: n n Integrating, where n is a positive integer, gives n n Integrating k, where n is a positive integer and k is a constant, n k gives n You can integrate the sum of any number of such functions by simply integrating one term at a time. Eample Integrate each of the following gradient functions. dy (i) d MEI, 0/0/ /5

29 MEI C Integration Section Notes and Eamples (ii) (iii) dy 4 5 d dy ( )( ) d dy (i) d 4 y 4 c dy (ii) 4 5 d 4 5 y c dy (iii) ( )( ) d 6 y c 6 Remember the arbitrary constant You can see further eamples using the Flash resources Basic indefinite integration: single powers of and Basic indefinite integration: sums of powers of. Note that these resources use notation for integration which is not used in the tetbook until a little later: for now you just need to know that f ( )d means integrate the function f(). You can also practise integration using the interactive questions Indefinite integration of polynomials. Note that w.r.t. means with respect to and just means that is the variable that you are using. In these questions, is not always the variable being used, but you integrate in eactly the same way whatever letter is used. You could also try the Indefinite integrals puzzle. This also uses the notation f ( )d to mean integrate the function f(). The Calculus Basic puzzle includes both differentiation and integration, using positive integer powers of. Finding the arbitrary constant If you are given additional information, you can find the value of the arbitrary constant by substituting the given information. This is sometimes called finding the particular solution of a differential equation. The net eample shows how this is done. MEI, 0/0/ /5

30 MEI C Integration Section Notes and Eamples Eample The gradient of a curve at any point (, y) is given by d y d = ( + ). The curve passes through the point (, 5). Find the equation of the curve. dy d = ( + ) = + 4 Integrating: y c 4 c 4 When =, y = 5 5 = c c = 5 = 5 6 So the equation of the curve is y = Just as with differentiating, you need to epand the brackets first Substitute the given values of and y You can see more eamples like the one above using the Flash resource Reverse of differentiation. An application: the motion of a particle One very useful application of integration and differentiation is in the relationships between the displacement, velocity and acceleration of a particle. You are not required to know these relationships for this component, but if you study Mechanics you may have already looked at them. The velocity, v, of a particle is the derivative of its displacement, s, with respect to time, t. ds v dt So you can find the displacement by integrating the velocity. The acceleration, a, of a particle is the derivative of its velocity with respect to time. dv a dt So you can find the velocity by integrating the acceleration. Eample 4 The velocity of a particle t seconds after leaving a point P is given by: v = t + Find the displacement of the particle from the point P after seconds. MEI, 0/0/ 4/5

31 MEI C Integration Section Notes and Eamples The velocity is the derivative of the displacement with respect to t. So: ds dt = t + Integrate to find s in terms of t. s = t t c The particle passes through P when t = 0, so substituting t = 0 and s = 0: 0 = 0 0 c c = 0. s = t t To find the displacement after seconds, substitute t = into this equation: s = = 6.5 m. MEI, 0/0/ 5/5

32 MEI Core Integration Section : Finding the area under a curve Notes and Eamples These notes contain subsections on: Indefinite integration: formal notation Definite integration The definite integral as an area The trapezium rule Indefinite integration: formal notation This section introduces the formal notation for integration. In Section you were given epressions for d y and asked to find an epression for y. d So you would write: dy y c d. Using the formal notation, you would write this as: d c You would read this as the integral of with respect to The eample below is the same as Eample from section, but is epressed in formal notation. Eample Integrate each of the following functions. (i) (ii) 4 5 (iii) ( )( ) Remember the arbitrary constant (i) 4 d 4 (ii) d (iii) c c ( )( ) d 6 d 6 c MEI, 05/0/ /6

33 MEI C Integration Section Notes and Eamples Definite integration The definite integral from a to b of a function f(), which is written as b a f ( )d, is found as follows: Integrate f() suppose we call the integral g() Write the integral in square brackets, with the limits on the right hand side: g( ) b a Work out the value of g() with = a and = b, and subtract: b g( ) g( b) g( a). a Eample Evaluate ( )d. d 4 8 Watch the signs here For additional eamples like this one, look at the Flash resource Definite integration. For practice in eamples like the one above, try the interactive resource Definite integration of polynomials. You could also try the Definite integrals puzzle. The definite integral as an area b The definite integral f ( )d calculates the area a between the curve y = f() and the -ais. f() The formal proof of this result is given on pages 4 44 of the tetbook. a b Note that you are not epected to be able to reproduce this formal proof you need only have a general understanding of the ideas involved. MEI, 05/0/ /6

34 MEI C Integration Section Notes and Eamples If the curve is above the -ais, so that the value of y is positive, the definite integral works out to be positive. However, if the curve is below the -ais, so that y is negative, the integral works out to be negative. It is important that you have a sketch of the curve before you use integration to work out areas like this. If the curve crosses over the -ais, then the integral subtracts any negative area from a positive area, and therefore calculates the difference of the two areas above and below the -ais. Eample (i) Show that ( 4 )d = 0, and interpret the result. 0 (ii) Find the total area enclosed by the curve, the -ais and the line =. : (i) ( 4 )d Since the integral is zero, there must be equal areas above and below the ais. (ii) Sketch the graph: The curve crosses the -ais when 4 = 0 ( 4) = 0 = 0 or = 4 Red area = 4 0 ( 4 )d This integral works out to be negative because the curve is below the -ais. MEI, 05/0/ /6

35 MEI C Integration Section Notes and Eamples Blue area = ( 4 )d So the total area enclosed = 7 7 square units. You can test yourself using the Flash resources Areas under a curve above the -ais and Areas above and below the -ais. You can move the curve around to get different functions, and move the limits of the area to be found. Use integration to work out the area and then check your answer. You may also find the Mathcentre video Finding areas using integration useful. Note that this video may contain eamples which use some integration techniques that you have not yet met. The work in the tetbook on finding the area between two curves and the area between the curve and the y-ais, are outside the syllabus. However if you are interested in this etension work you might like to look at the Flash resource Area between a line and a curve. The trapezium rule Splitting the area into bits gives a method for calculating the approimate area under a curve. Instead of using rectangles, it is more accurate to use trapezia: O y 0 y y y y n a h b If the trapezia have width h, and the y-coordinates are y, y,, y n, then the total area A is: MEI, 05/0/ 4/6

36 MEI C Integration Section Notes and Eamples h A y y y y y 0 n... n where ( b a) h This result is derived on pages 6-6 of the tetbook.. n Eample 4 (i) Use the trapezium rule with: (a) strips and (b) 6 strips, to find the approimate value of d. (ii) 0 Given that the eact value of this integral is 4, find the relative error in each case, and eplain why the trapezium rule is an under-estimate of the true value. (i) (a) With strips of width h = : A (b) With 6 strips of width h = 0.5: A (ii) (a) relative error = (b) relative error = 0 y y = = 0.00 From the graph of the function, it is clear that the trapezia lie underneath the curve, so the trapezium rule gives an under-estimate MEI, 05/0/ 5/6

37 MEI C Integration Section Notes and Eamples You can see more eamples like this one using the Flash resource The trapezium rule. You can also use the Geogebra resource The trapezium rule to see the result of applying the trapezium rule with different numbers of strips. MEI, 05/0/ 6/6

38 MEI Core Trigonometry Section : Trigonometric Functions and Identities Notes and Eamples In this section you learn about the trigonometric functions and some trigonometric identities. These notes contain sub-sections on: The trigonometric functions for angles between 0 and 90 Common values of sin θ, cos θ and tan θ Trigonometric functions for angles of any size Trigonometric identities Graphs of trigonometric functions The trigonometric functions for angles between 0 and 90 From GCSE, you know that for a right-angled triangle the trigonometric functions are defined as: opposite You can also write: hypotenuse θ adjacent adjacent sin(90 θ) and hypotenuse opposite sin θ hypotenuse adjacent cosθ hypotenuse opposite tan θ adjacent opposite cos(90 θ) hypotenuse So: cos θ sin(90 θ) and sin θ cos(90 θ) For eample: cos 70 = sin 0 sin 5 = cos 75 MEI, 7/05/0 /8

39 MEI C Trigonometry Section Notes and Eamples Common values of sin θ, cos θ and tan θ You should learn the values of sin θ, cos θ and tan θ for θ = 0, 0, 45, 60 and 90. θ cos θ sin θ 0 tan θ 0 undefined 0 You can see how these values are found on page 7 of the tetbook. If you find it hard to remember the values for 0, 45 and 60 at first, you can sketch the relevant triangles. B B 0 45 A 60 D C A 45 C and then you can simply read off the value you need. You can test yourself on these values using the Flash resource Standard values of trig ratios. Try to remember the eact value of each trig ratio, and click on the bo to check your answer. In the eample below you need to substitute eact values into an epression. Eample Show that sin 0cos 0 sin 0 and cos0 MEI, 7/05/0 /8

40 MEI C Trigonometry Section Notes and Eamples So substituting these values intosin 0cos 0: 4 4 as required. Trigonometric functions for angles of any size The tetbook shows how the definitions for sine, cosine and tangent can be etended to angles of any size using a diagram like the one below. This gives the definitions: y sin θ y cosθ y tan θ y O θ P (, y) y In Activity 0., you are asked to investigate the sign of sin, cos and tan from the definitions above. Try this before looking at the eplanation below. In the first quadrant, lies between 0 and 90. The values of and y are both positive, so the values of sin, cos and tan are all positive. In the second quadrant, lies between 90 and 80. The value of is negative and the value of y is positive, so sin is positive but cos and tan are both negative. In the third quadrant, lies between 80 and 70 (or alternatively, between -90 and -80 ). The values of and y are both negative, so tan is positive but sin and cos are both negative. In the fourth quadrant, lies between 70 and 60 (or alternatively, between 0 and -90 ). The values of is positive and the value of y is negtive, so cos is positive but sin and tan are both negative. This can be summarised in the diagram below. Sine positive All positive S A Tangent positive or, more simply Cosine positive T C MEI, 7/05/0 /8

41 MEI C Trigonometry Section Notes and Eamples This is often known as the CAST diagram: some people remember it using the word CAST (make sure you start the word in the right place!); others use a phrase such as All School Teachers Criticise (which, while rather unfair to the teaching profession, does have the advantage of starting in the first quadrant). The symmetry of the diagram allows you to find relate trig rations of angles in the second, third and fourth quadrant to trig ratios angles in the first quadrant To find a trig ratio of an angle in the second, third or fourth quadrant in terms of the trig ratio of an acute angle, you need to find the equivalent acute angle using symmetry, and then use the CAST diagram to find whether the trig ratio is positive or negative. You may find the Mathcentre video Trigonometric ratios in all quadrants useful to help you understand this work. Trigonometric identities You need to learn the following identities: sin tan cos sin cos e.g. sin 0 tan 0 cos0 An identity is true for all values of θ. Proofs of these two identities are given on page 76 of the tetbook. These identities are demonstrated in the Flash resources Identities: sin / cos = tan and Identities: sin² + cos² =. MEI, 7/05/0 4/8

42 MEI C Trigonometry Section Notes and Eamples In the net eample you need to use the trigonometric identities to rewrite an epression. Eample Show that (sin cos )(sin cos ) sin Working with the LHS and epanding the brackets gives: (sin cos )(sin cos ) sin cos Since sin cos then cos sin Substituting into gives: (sin cos )(sin cos ) sin ( sin ) Simplifying: (sin cos )(sin cos ) sin as required. Graphs of trigonometric functions Now that you can define sin θ, cos θ and tan θ for any value of θ you can draw the graphs of these functions: Graph of y = sin y y = sin Note:. The graph of y = sin has a period of 60. So it repeats every 60.. It has rotational symmetry about the origin.. sin So y = sin lies between and There is a line of symmetry at = 90 and = -90. e.g. sin 5 = sin 75 and sin 5 = sin -45 e.g. sin 0 = 0.5 so sin -0 = -0.5 e.g. sin = sin (80 - ) MEI, 7/05/0 5/8

43 MEI C Trigonometry Section Notes and Eamples You can see how this graph relates to the circle diagram on page of these notes using the Flash resource The graph of y = sin. Graph of y = cos y y = cos e.g. cos 5 = cos 85 and cos 5 = cos -5 Note:. The graph of y = cos has a period of 60. So it repeats every 60.. It has a line of symmetry in the y-ais.. cos So y = cos lies between and -. e.g. cos 0 = cos The graph of y = cos is a translation of the graph y = sin by 90 to the left. In general, cos sin( 90 ) e.g. cos 0 = sin 90 and cos 0 = sin 00 You can see how this graph relates to the circle diagram on page of these notes using the Flash resource The graph of y = cos. MEI, 7/05/0 6/8

44 MEI C Trigonometry Section Notes and Eamples Graph of y = tan y y = tan Note:. The graph of y = tan has a period of 80. So it repeats every 80.. It has rotational symmetry about the origin. e.g.tan 5 = tan 95 and tan 5 = tan -65 e.g. tan 60 = so tan -60 = -. tan So tan can be any value not just those between - and. 4. There are asymptotes at = 90, = 70, = 450 where the value of tan is undefined (since cos = 0 for these values of ) You can see how this graph relates to the circle diagram on page of these notes using the Flash resource The graph of y = tan. You may also find the Mathcentre video Trigonometric functions useful. In the net eample you need to use the symmetries of the trigonometrical graphs. Eample Write down the eact values of (i) tan 0 (ii) sin (-5 ) (iii) cos 00 Eact means leave as a fraction or a surd don t just give all the decimal places from your calculator! (i) y = tan θ has period of 80 so tan 0 tan (0-80 ) = tan (-60 ). MEI, 7/05/0 7/8

45 MEI C Trigonometry Section Notes and Eamples y = tan θ has rotational symmetry about the origin so tan (-60 ) = -tan 60 So tan0 (ii) y = sin has rotational symmetry so sin (-5 ) = -sin 5. y = sin has a line of symmetry at = 90 so sin 5 = sin (80-5 ) = sin 45 So sin (-5 ) = -sin 45 (iii) y = cos has period of 60 so cos 00 = cos (00-60 ) = cos (-60 ). y = cos is symmetrical about the y-ais so cos (-60 ) = cos 60. So cos 00 = cos 60 =. For practice in questions like the one above, try the interactive resource Eact trig function values of standard angles. MEI, 7/05/0 8/8

46 MEI Core Trigonometry Section : Trigonometric equations Notes and Eamples In this section you learn how to solve trigonometric equations. These notes contain subsections on Principal values Solving simple trigonometrical equations More complicated eamples of trigonometrical equations. Principal values There are infinitely many roots to an equation like sin. Your calculator will only give one solution the principal value. You find this by pressing the calculator keys for arcsin 0.5 (or sin or invsin 0.5). Check that you can get the answer of 0. You can find other roots by looking at the symmetry of the appropriate graph. y A second solution in a 60 cycle can be found by 80 - θ y = cos θ y When you use the inverse cosine function, your calculator will always give you an answer from 0 to A second solution in a 60 cycle can be found by 60 - θ MEI, 7/05/0 /5

47 MEI C Trigonometry Section Notes and Eamples y = tan θ 4 y When you use the inverse tan function, your calculator will always give you an answer between -90 and A second solution in a 60 cycle can be found by θ + 80 or θ Alternatively, you can use the quadrant diagram to find other solutions, by thinking about which quadrants the solutions will be in. Solving simple trigonometrical equations Because there are infinitely many solutions to a trigonometric equation you are only ever asked to find a few of them! Any question at this level asking you to solve a trigonometric equation will also give you the interval or range of values in which the solutions must lie, e.g. you might be asked to solve tan for You can only directly solve trigonometric equations like or tan. Here is an eample. sin or cos 4 Eample Solve sin for sin 60 There will be a second solution in the second quadrant = 0 is also a solution. Since y sin has a period of 60 any other solutions can be found by adding/subtracting 60 to these two solutions. So the other solutions are: = -00 and 0-60 = -40 So the values of θ for which sin are -00, -40, 60, 0. MEI, 7/05/0 /5

48 MEI C Trigonometry Section Notes and Eamples You can see some more eamples like the one above using the Flash resource Solving basic trigonometric equations. Look at the graph each time. More complicated trigonometrical equations Any more complicated equations need to be manipulated algebraically before they can be solved. There are a number of techniques you can use:. Rearrange the equation to make cos,sin or tan the subject.. Check to see if the equation factorises to give two (or more) equations which involve just one trigonometric function (see Eample ). If it is a quadratic in either sin θ, cos θ, or tan θ it can either be factorised or solved using the formula for solving quadratic equations (see Eample ).. If the equation involves just sin θ and cos θ (and no powers), check to sin see if you can use the identity tan (see Eample 4). cos 4. If the equation contains a miture of trigonometric functions (e.g. cos θ and sin θ) then you may need to use the identity sin cos to make it a quadratic in either sin θ, cos θ, or tan θ (see Eample 5). Eample Solve cos sin cos 0 for cos sin cos 0can be factorised as there is cos in both terms on the LHS. Factorise: cos (sin ) 0 So either cos 0 or sin 0 It is wrong to divide through by cos because cos 0 90 you lose the solutions to cos = = 70 is also a solution. sin 0 sin This has solutions in the third and fourth quadrants. The solutions are = 0 and 60-0 = 0. So the values of θ for which cos sin cos 0 are 90, 0, 70 and 0. In Eample you need to solve a quadratic equation. Eample Solve cos cos for You can replace cos with to make things simpler! Or factorise straightaway to get: (cos )(cos + ) = 0 and then solve. MEI, 7/05/0 /5

49 MEI C Trigonometry Section Notes and Eamples cos cos is a quadratic equation in cos Rearrange the quadratic: cos cos 0 Let cos = : 0 Factorise: ()( ) 0 or cos or cos cos has no solutions. So we need to solve cos cos 60 There is also a solution in the 4 th quadrant, so = 00 is also a solution. So the values of θ for which cos cos are 60 and 00. sin In the net eample you need to use the identity tan. cos Eample 4 Solve sincos 0 for You need to rearrange the equation. sincos 0 sin Dividing by cos : 0 cos sin Since tan : tan 0 cos tan 6.4 to d.p. There is also a solution in the rd quadrant. So = 4.4 is also a solution. You can safely divide by cos θ because it can t be equal to 0. If it were then sin θ would also have to be 0 and cos θ and sin θ are never both 0 for the same value of θ. So the values of θ for which sincos 0 are 6.4 and 4.4 to d.p. You can see more eamples like this one using the Flash resource Solving trig equations using identities (). In the net eample you need to use the trigonometric identity sin cos. Eample 5 Solve sin sin cos for 0 60 MEI, 7/05/0 4/5

50 MEI C Trigonometry Section Notes and Eamples Rearranging the identity gives: sin cos cos sin Substituting into the equation sin sin cos gives: sin sin sin This is a quadratic in sin. Rearranging: sin sin 0 Rearranging: sin sin 0 This factorises to give: (sin )(sin ) 0 So either: sin 0 or sin 0 sin sin 0 or So the solutions to sin sin cos are 0,50 or 70 You can see more eamples like this one using the Flash resource Solving trig equations using identities (). You may also find the Mathcentre video Trigonometric equations helpful. (Some of the eamples in this video involve the use of radians, which are covered in a later section). MEI, 7/05/0 5/5

51 Notes and Eamples MEI Core Trigonometry Section : Sine and Cosine Rules In this unit you learn about finding an unknown side or angle in any triangle. You will also learn a new formula for finding the area of a triangle. These notes contain subsections on: The sine rule The cosine rule Choosing which rule to use The area of a triangle The sine rule The sine rule: a b c sin A sin B sin C This form is easier to use when finding an unknown side. b A c The sine rule can also be written as: sin A sin B sinc a b c This form is easier to use when finding an unknown angle. C a B Note: When you use the sine rule to find a missing angle, θ, always check whether 80 - θ is a possible solution as well. Eample shows a straightforward application of the sine rule to find an unknown side. Eample Find the side BC in the triangle ABC. c B 0 A 0 cm C MEI, 7/05/0 /6

52 MEI C Trigonometry Section Notes and Eamples The triangle is isosceles so BAC is b By the sine rule: sin A sin B 0 So: sin 75 sin0 0sin 75 sin 0 so = 9. cm (to sig.fig.) You can see more eamples like this using the Flash resource The sine rule. Eample shows a straightforward application of the sine rule to find an unknown angle. Eample A, B and C are three points on a level plane. B is 6 km due west of A. C is 5 km from B and is on a bearing of 85 from A. Find ACB. First draw a diagram: By the sine rule: So: a C 5 km Due west is a bearing of 70, so this angle must be 5. 5 B 6 km c sin A sin C a c sin5 sin C 5 6 6sin5 sin C 5 sin C C = 8. to d.p. Check whether 80 - C is also a solution: = 6.9 to d.p. This also works so ACB is 8. to d.p. or 6.9 to d.p b N A Your diagram doesn t need to be accurate just large enough to show all the information Don t round here! Store the number in your calculator. Angles A and C still add up to less than 80 MEI, 7/05/0 /6

53 MEI C Trigonometry Section Notes and Eamples You can see eamples similar to this using the Geogebra resource The sine rule finding an angle. This resource also shows geometrically what is happening when there is more than one possible solution. The cosine rule The cosine rule: A a b c bc cos A This form is easier to use when finding an unknown side. b c The cosine rule can also be written as: C b c a cos A bc This form is easier to use when finding an unknown angle. a B Eample shows an application of the cosine rule to find an unknown side. Eample Find the side YZ in the triangle XYZ. The cosine rule for this triangle is: So: Z y 7 cm X 95 y z yz cos X z 6 cm Y cos = 9.6 cm to sig. fig. You can see more eamples like this using the Flash resource The cosine rule. Eample 4 shows a straightforward application of the cosine rule to find an unknown angle. MEI, 7/05/0 /6

54 MEI C Trigonometry Section Notes and Eamples Eample 4 Find the angle θ in the triangle ABC. A 5 m b The cosine rule for this triangle is: c 7 m θ B C 4 m a a b c cos C ab cos C 45 cos C 0. C = 0.5 to d.p. You can see eamples similar to this using the Geogebra resource The cosine rule finding an angle. You can test yourself using the interactive questions Finding an angle using the cosine rule. You are given the coordinates of all three vertices of a triangle. Choosing which rule to use Use the sine rule when: you know sides and angle (not between the two sides) and want a nd angle ( rd angle is now obvious!) you know angles and side and want a nd side Use the cosine rule when: you know sides and want any angle you know sides and the angle between them and want the rd side You may find the Mathcentre video The sine and cosine formulae useful. Eample 5 shows how to decide whether to use the sine or the cosine rule. MEI, 7/05/0 4/6

55 MEI C Trigonometry Section Notes and Eamples Eample 5 A ship sails from a port, P, 6 km due East to a lighthouse, L, 6 km away. The ship then sails 0 km on a bearing of 00 to an island, A. Find: (i) The distance AP (ii) The bearing of P from A First draw a diagram: N N A P 6 km a l 0 L p 0 km 0 (i) You know sides and the angle between them so you need the cosine rule. The cosine rule for this triangle is: l a p ap cos l So the distance AP is 4 km. l cos0 l 96 l 4 km (ii) You can now use either the cosine rule or the sine rule to find the angle PAL. The sine rule for this triangle is: sin A sin L a l Check whether sin A sin = 58. is So: also a solution. 6 4 It isn t because the angles 6sin0 in the triangle would total sin A 4 more than 80. sin A 0.7. A =.8 So the bearing is =.8 to d.p. The area of a triangle To find the area of any triangle you can use the rule: Area of triangle ABC sin ab C So you need two sides and the angle between them. MEI, 7/05/0 5/6

56 MEI C Trigonometry Section Notes and Eamples Eample 6 shows how to use this formula. Eample 6 Find the area of triangle ABC from Eample 4. A c 7 m 5 m b θ B C 4 m a In Eample 4, angle C was found to be 0.5 to d.p. Using the formula Area = absin C gives: Area of triangle ABC = 4 5 sin m You can see more eamples like this using the Flash resource Area of a triangle. For practice in these techniques, try the interactive questions The area of a triangle given sides and an angle and The area of a triangle given three sides. In the first one, the angle given may not always be the angle between the two given sides, so always draw a diagram and decide whether you need to use the sine rule or cosine rule first. For the second one, you will need to use the cosine rule to find an angle before finding the area. For an additional challenge, try the etension worksheet Cosy Cubes. MEI, 7/05/0 6/6

57 MEI Core Trigonometry Section 4: Circular Measure Notes and Eamples In this section you learn about radians and circular measure. These notes contain subsections on: Radians Common values of trig functions in radians Trigonometric graphs in radians Solving trigonometric equations using radians Sectors of circles Radians Often mathematicians use radians rather than degrees to measure angles as they sometimes make calculations easier. To convert degrees to radians you multiply by 80 To convert radians to degrees you multiply by 80 In the eample below you need to convert degrees into radians. Eample Convert these degrees to radians: (a) 60 (b) 70 (c) 7 (a) 60 = 60 = 80 (b) 70 = 70 = 80 (c) 7 = 7 =.0 rads 80 Page: You can see further eamples like this one using the Flash resource Converting degrees to radians. MEI, //0 /7

58 MEI C Trigonometry Section 4 Notes and Eamples For etra practice in questions like the one above, try the interactive questions Convert degrees to radians. In the net eample you need to convert radians into degrees. Eample Convert these radians to degrees: 7 (a) (b) (a) = = (b) 7 = 7 80 = (c) 0.65 rads = 0.65 = 7. (c) 0.65 rads You can see further eamples like this one using the Flash resource Converting radians to degrees. For etra practice in questions like the one above, try the interactive resource Convert radians to degrees. Common values of trig functions in radians You should soon be able to remember the radian equivalent of many common angles: 0 = 6 radians 45 = 4 radians 60 = radians 90 = radians 80 = radians 60 = radians. You should normally epress these as fractions of rather than as decimals. Just as when working in degrees, you should be able to recognise the values of sin, cos and tan for common values of. This table reminds you of the common values you should know, with the angles given in radians. MEI, //0 /7

59 MEI C Trigonometry Section 4 Notes and Eamples θ 0 6 cos θ sin θ 0 tan θ undefined You can test yourself on these values using the Flash resource Standard values of trig ratios. Switch to radians, then try to remember the eact value of each trig ratio, and click on the bo to check your answer. Trigonometric graphs in radians You need to be familiar with the graphs of the trig functions when working in radians as well as in degrees. Here are reminders of these graphs. The graph of y = sin θ where θ is in radians: y The graph of y = sin has a period of. So for eample, sin 0. sin(0. ) It has rotational symmetry about the origin. So for eample, since sin 0.5 then sin There is a line of symmetry at and It repeats every radians 80 rad MEI, //0 /7

60 MEI C Trigonometry Section 4 Notes and Eamples The graph of y = cos θ where θ is in radians y The graph of y = cos θ has a period of. So for eample, cos 0. cos(0. ) It has line symmetry about the y-ais. So for eample, cos cos The graph of y = tan θ where θ is in radians 4 y 4 The graph of y = tan has a period of. So for eample, tan 0. tan(0. ) It has rotational symmetry about the origin. So for eample, since tan then tan 80 rad 60 Solving trigonometric equations using radians In section you looked at solving trig equations, giving your answers in degrees. The same methods are used when working in radians. Make sure that your calculator is set to Radians. However, remember that you should be able to recognise common values of the trig functions and give the eact angle in radians as a fraction of where appropriate. MEI, //0 4/7

61 MEI C Trigonometry Section 4 Notes and Eamples In this eample you need to find a solution to a trigonometric equation in radians. Eample Solve cos 0.5 for 0 When the range is given in radians then you must find a solution in radians. Page: 5 You know that cos 60 = radians 80 So is one solution. There is another solution in the 4 th quadrant. 5 is also a solution. 5 So the values of θ for which cos 0.5 are and In this eample you need to use a trigonometric identity. Eample 4 Solve sin cos for You can divide by cos since we know that cos 0 as sin and cos are not both equal to 0 for the same value of. sin cos sin So: cos cos cos sin You know tan cos Substituting into : tan tan rads There is also a solution in the rd quadrant π = -.68 rads is also a solution. So the values of θ for which sin cos are -.68 rads and rads. In this eample you need to use the identity sin θcos θ. Eample 5 Solve sin θ cos θfor 0θ π MEI, //0 5/7

62 MEI C Trigonometry Section 4 Notes and Eamples Using the identity sin θcos θ gives: Substituting into sin θ cos θ: cos θ sin θ sin θ ( sin θ) sin θ sin θ This is a quadratic in sinθ which rearranges to: sin θsin θ 0 Factorising: (sin θ)(sin θ) 0 So either: (sin θ ) 0 sin θ θ π or π π π or 6 6 or (sin θ ) 0 sin θ θ π or π π π 7π π π So the solutions to sin θ cos θ are θ, or 6 6 Sectors of circles A sector of a circle is the shape enclosed by an arc of the circle and two radii. A minor sector is a sector which is smaller than a semi-circle. A major sector is a sector which is larger than a semi-circle. major sector minor sector The formula for arc length is: Arc length = r where is in radians. The formula for the area of a sector is: Area of a sector = r where is in radians. In this eample you need to use the formula for arc length. Eample 6 A sector of a circle with radius 6 cm has an arc length of. Find the angle subtended at the centre of the circle. MEI, //0 6/7

63 MEI C Trigonometry Section 4 Notes and Eamples Page: 6 Use arc length = r. So π = 6θ So 6 You can see more eamples like this using the Flash resource Radians: arc length. In the net eample you need to use the formula for sector area. Eample 7 Find the shaded area. A This is called a segment. 5 0 O 5 B Convert 0 to radians: 0 = 0 = 80 Page: 7 Area of segment = Area of sector OAB area of triangle OAB Area of sector = r where θ is in radians. Area of triangle = sin r So area of segment r r sin Area of segment = r ( sin ) 5 sin = 5.4 cm Area of triangle is ½absin C but a and b are equal to the radius of the circle. Make sure your calculator is in radians mode. You can see more eamples like this using the Flash resource Radians: sector area. You may also find the Mathcentre video Radian measure helpful. MEI, //0 7/7

64 MEI Core Trigonometry Section 5: Transformations Notes and Eamples These notes contain sub-sections on: Translations Stretches Solving equations Reflections (etension work) In this section you learn how to sketch transformations of graphs of trigonometrical functions and other functions. This will allow you to sketch graphs like f( ) sin, f( ) tan and f( ) cos( 0 ). Work on translations has already been covered in chapter : this work is revised and stretches and reflections are introduced. You can investigate the effect of translations and stretches of the graphs of trigonometric functions using the Geogebra resource Transformations of trigonometric graphs. Translations In a translation, the graphs of sin, cos or tan are translated (or moved) a certain number of units in the or y direction. A function of the form: f( ) cos( t) f ( ) cos f( ) sin( t) translates of the graphs of f ( ) sin f( ) tan( t) f ( ) tan t by the vector (i.e. t units in the positive direction). 0 In general, any function of the form y = f( t) represents a translation of the graph of t y = f() by the vector 0. If t is negative, giving a function of the form y = f( + t), then of course the translation is in the negative direction, i.e. to the left. A function of the form: f ( ) s cos f ( ) cos f ( ) s sin translates the graphs of f ( ) sin f ( ) s tan f ( ) tan 0 by the vector (i.e. s units in the positive y direction). s MEI, 8/0/0 /7

65 MEI C Trigonometry Section 5 Notes and Eamples In general, any function of the form y = s + f() represents a translation of the graph of 0 y = f() by the vector. s Combining these two transformations: Any function of the form y = s + f( t) represents a translation of the graph of t y = f() by the vector. s This eample shows how to translate a graph. Eample Sketch the graphs of (i) y sin (ii) y cos( 60 ) (i) This is a translation of the graph y sin by unit vertically upwards. y y = + sin y = sin The graph of y = sin has been moved one unit in the positive y direction. (ii) This is a translation of the graph.5 y cos by 60 to the right. y y = cos 0.5 y = cos ( -60 ) The graph of y = cos has been moved 60 units in the positive direction. MEI, 8/0/0 /7

66 MEI C Trigonometry Section 5 Notes and Eamples Stretches In a stretch, the graphs of sin, cos or tan are stretched by a certain scale factor parallel to the or y aes. A function of the form: f ( ) cos( a) f ( ) sin( a) stretches the graphs of f ( ) tan( a) parallel to the ais with scale factor a. f ( ) cos f ( ) sin f ( ) tan In general, a function of the form y = f(a) represents a one-way stretch of the graph of y = f() parallel to the ais with scale factor a. A function of the form: f ( ) a cos f ( ) asin stretches the graphs of f ( ) a tan parallel to the y ais with scale factor a. f ( ) cos f ( ) sin f ( ) tan In general, a function of the form y = af() represents a one-way stretch of the graph of y = f() parallel to the y ais with scale factor a. This eample shows how to stretch a graph. Eample Sketch the graph of y cos This is a one-way stretch parallel to the y ais with scale factor. y y = cos y = cos The y co-ordinate of every point on the original curve needs to be doubled. MEI, 8/0/0 /7

67 MEI C Trigonometry Section 5 Notes and Eamples The net eample shows how to find the equation of a curve. Eample Write down the equation of the following graph: y Compare the graph with that of y = sin : y y = sin The co-ordinate of every point on the graph of y = sin has been halved. Therefore the equation is y = sin. Stretches like these can, of course, be applied to any graphs, not just trigonometric graphs. Look at the Flash resources Horizontal stretches and Vertical stretches to see stretches applied to a polynomial graph. Solving equations In section you solved simple equations like sin 0.5. As you know, almost all equations of this type have two solutions in the range Graphs like the ones above can be very useful in helping to solve equations like sin 0.5. Notice that the graph of y sin repeats itself every 80, whereas the graph of y sin repeats itself every 60. This means that the equation sin 0.5 has four solutions in the range 0 60, as shown in the graph below. MEI, 8/0/0 4/7

68 MEI C Trigonometry Section 5 Notes and Eamples y y sin sin 0.5 0, 50, 90, 50 5, 75, 95, 55 Since four solutions are epected in the range 0 60, add 60 to each of the first two values for to give two more. Halving each value for gives four solutions for in the range 0 60 Eample 4 Solve each of the following equations in the given range: (i) tan for 0 (ii) cos( 40 ) 0.5 for (i) tan 5 9 7,,,,, ,,,,, 4 4 (ii) cos( 40 ) , 00, 40, 660 0, 60, 80, 60 0, 0, 90, 0 There will be si solutions in the range. The range was given in radians, so the solutions must be in radians There will be four solutions in the range. Subtract 40 from each value, then divide by. Reflections This section is etension work. It will be covered in more depth in C. Special cases of one-way stretches occur when the scale factor of the stretch is given by -. A one-way stretch with scale factor - in the y direction results in the graph being reflected in the ais. MEI, 8/0/0 5/7

69 MEI C Trigonometry Section 5 Notes and Eamples A function of the form: f ( ) cos f ( ) sin reflects the graphs of f ( ) tan f ( ) cos f ( ) sin in the ais. f ( ) tan In general, any function of the form y = -f() represents a reflection of the graph of y = f() in the ais. A one-way stretch with scale factor - in the direction results in the graph being reflected in the y ais. A function of the form: f ( ) cos( ) f ( ) sin( ) reflects the graphs of f ( ) tan( ) f ( ) cos f ( ) sin in the y ais. f ( ) tan In general, any function of the form y = f(-) represents a reflection of the graph of y = f() in the y ais. This eample shows how to find the equation of a curve. Eample 5 Write down the equation of the following graph: y This graph is clearly something to do with tan. Comparing the graph with y = tan : MEI, 8/0/0 6/7

70 MEI C Trigonometry Section 5 Notes and Eamples y y = tan You can see that the graph is a reflection of y = tan in the y ais. So the equation is y = -tan MEI, 8/0/0 7/7

71 MEI Core Logarithms and eponentials Section : Introduction to logarithms Notes and Eamples These notes contain subsections on Indices and logarithms The laws of logarithms Eponential functions Solving eponential equations using logarithms An old practical application of logarithms (etension work) Indices and logarithms The important thing to remember about logarithms is that, although they appear to be a new topic, they are simply about writing what you already know about indices in a different way. If you find it difficult to work out the meaning of a statement involving logarithms, it can be simpler to change the statement into the equivalent statement involving indices. log a b a b. To remember this, notice that a is both the base of the logarithm and the base of the inde, and, the logarithm, is the inde. The value of log a b is the answer to the question: What power must I raise a to in order to get b? Eample (i) Find log 4 (ii) Find, where log5 (i) The statement log 4 Since 4, then must be. So log4. is equivalent to 4. (ii) The statement log5 is equivalent to So MEI, 4/05/0 /6

72 MEI C Logs Section Notes and Eamples You can also look at the Flash resource Basic logarithms. For practice in understanding statements involving logarithms, try the interactive questions Bases of logarithms. The laws of logarithms The laws of logarithms are: log log y log y log log y log y n log nlog These can be proved using the laws of indices: It s a worthwhile eercise to try to work through these proofs First convert into inde notation: log log a c a c b c y b c y To prove the first law: Similarly for the second law: c c a b c c a b y c c y ab log y a b c y log y log log y ab logc a b y c c c y log log log y c c c y Using the laws of indices For the third law: log c c a c an n a n a logc n n log a c n log log c c n As the first two laws of indices require the indices to have the same base, then the first two laws of logarithms require the logarithms to have the same base. MEI, 4/05/0 /6

73 MEI C Logs Section Notes and Eamples Eample (i) Write log y z in terms of log, log y and log z. (ii) Write log a log b log c as a single logarithm. y (i) log log log y log z z log log y log z log log y log z (ii) log a log b log c log a log b log c log a (log b log c) log a b c You can also look at the Flash resource Laws of logarithms, and you may find the Mathcentre video Logarithms useful. For some practice in using the laws of logarithms, try the Logarithms puzzle. Eponential functions An eponential function is any function of the form show some different eponential functions. y a. The graphs below y 5 y y.5 MEI, 4/05/0 /6

74 MEI C Logs Section Notes and Eamples Eponential functions are the inverse of logarithm functions: the function y a is the inverse of the function y log a. This relationship is very useful for solving equations, as shown below. You can eplore graphs of this type using the Flash resource The graph of y = ka. Solving eponential equations using logarithms Logarithms are very useful for solving equations where the unknown variable is an inde, such as the equation 0. Many equations are solved using inverse functions, for eample you solve the equation + = 5, in which addition is applied to the unknown variable, by subtracting from each side. Similarly you solve the equation ² = 0 by using the square root function, which is the inverse of the square function. An equation like 0 involves an eponential function of. So to solve this equation, it follows that you need to use the inverse of the eponential function, which is the logarithm function. This is shown in the net eample. Eample Solve the following equations. (i) 0 (ii) 4 (iii) 0. (i) 0 log log0 log log0 log0. log (ii) 4 log log 4 ( )log log 4 log 4 log log 4. log MEI, 4/05/0 4/6

75 (iii) MEI C Logs Section Notes and Eamples 0. log 0. log ( )log 0. log log log 0. log.4 log 0. You can see similar eamples using the Flash resource Solving a^ = b. For practice in this type of problem, try the interactive questions Logarithms: Solving powers. An old practical application of logarithms Before calculators eisted, logarithms were used to make calculations easier. For eample, suppose you had to divide 467 by You could do this by long division, but it would take a long time and the chances of making a mistake would be quite high. So you would apply the second law of logarithms: log ( ) = log 467 log 9675 To do the calculation, you would have to find the log to base 0 of the two numbers, subtract the results, and then find the inverse log of the answer. You would have to find the values of log 467 and log 9675 from a book of tables. Unfortunately most tables would only tell you the values of log for values of between 0 and 99. So you would then use the fact that log 467 log( ) log4.67 log00000 log You would then use the tables to find the value of log 4. (which is as accurate as most tables would give you). This would give a value for log 467 of You would then go through a similar process to find log log 9675 log log Net you would subtract these two logarithms (without a calculator of course!), giving MEI, 4/05/0 5/6

76 MEI C Logs Section Notes and Eamples Now you would have to find the number whose logarithm is Inverse log tables usually give values between and log 4.8log log 0 log.48 So =.48. Most pupils did not understand the theory behind these calculations; they just followed a set of instructions to use the tables of logarithms and work out the calculation. Even after calculators became widely available, it was several years before this technique was removed from eamination syllabuses! Logarithms were also the basis of slide rules, which were also used before calculators eisted to work out calculations quickly. MEI, 4/05/0 6/6

77 MEI Core Logarithms and eponentials Section : Modelling curves using logarithms Notes and Eamples These notes contain subsections on: Modelling curves of the form y = k n Modelling curves of the form y = ka n Modelling curves of the form y k When you collect data from an eperiment, you may want to find a relationship between two variables, such as the speed of a moving object at a particular time, or temperature of an object and its distance from a heat source. You may plot a graph of one variable against another to help find this relationship. However, unless the graph is a straight line, it may be difficult to see the relationship from the graph. The graphs of y, y etc. look quite similar for 0, and as the eperimental data may not be very accurate, it can be impossible to tell with any certainty what would be the best graph to model the data. This is where logarithms can be very useful. If the relationship is of the form n y k, then plotting log y against log gives a straight line graph. y k n log y log k log y log k log log y log k nlog log y nlog log k n n This is of the form y = m + c, so plotting log y against log gives a straight line with gradient n and intercept log k. The value of n is therefore the gradient of the graph, and the value of k is found by taking the intercept of the graph and finding its inverse logarithm (i.e. intercept 0 if you are using logs to base 0). Eample n The relationship between two variables and y is believed to be of the form y k, where k and n are constants. In an eperiment, the following values of and y are recorded y MEI, 6/0/09 /5

78 MEI C Logarithms Section Notes and Eamples Verify that the model constants k and n. y n k is appropriate and find the approimate values of the y k Taking logarithms: log y log k n log y log k log n log y log k nlog If log y is plotted against log, this is the equation of a straight line graph with gradient n and intercept log k. Plot the values of log y against log : n log y log y Since the graph is approimately a straight line, the relationship appropriate model. 0.8 Gradient = n = Intercept log k 0. k 0 y n k is an MEI, 6/0/09 /5

79 MEI C Logarithms Section Notes and Eamples The relationship is approimately y 0.5 For practice in problems similar to this, try the interactive questions Equations from log-log graphs. Modelling curves of the form y ka Similarly, if the relationship is of the eponential form y against gives a straight line graph. y ka, then plotting log y ka log y log ka log y log k log a log y log k log a log y log a log k This is of the form y = m + c, so plotting log y against gives a straight line with gradient log a and intercept log k. The value of a is found by taking the gradient of the graph and finding its gradient inverse logarithm (i.e. 0 if you are using logs to base 0), and the value of k is found by taking the intercept of the graph and finding its inverse intercept logarithm (i.e. 0 if you are using logs to base 0). Eample The relationship between two variables p and q is believed to be of the form where p and q are constants. q p ab, In an eperiment, the following values of p and q are recorded. p q Verify that the model q p ab is appropriate, and estimate the values of a and b. q ab Taking logarithms: log q log ab p log q log a logb p log q log a p logb If log q is plotted against p, this is the equation of a straight line with gradient log b and intercept log a. p MEI, 6/0/09 /5

80 MEI C Logarithms Section Notes and Eamples Plot the values of log q against p: p q log q Since the graph is approimately a straight line, the relationship appropriate model. q p ab is an.5 Gradient = log b = b. 0.5 Intercept = log a = 0.5 a 0. The relationship is approimately.5/. 0.5 q..5 p You do not need to remember the details of what to plot and what to do with the gradient and intercept of the graph all you need to do is to take logs of both sides of the suggested relationship and apply the laws of logarithms to obtain a relationship of the form y = m + c, as shown above for each of the n relationships y k and y ka. Once you have done this, you can see what you need to plot and how to find the values of the constants. MEI, 6/0/09 4/5