Production of this 2015 edition, containing corrections and minor revisions of the 2008 edition, was funded by the sigma Network.

Transcription

1

2 About the HELM Project HELM (Helping Engineers Learn Mathematics) materials were the outcome of a three- ear curriculum development project undertaken b a consortium of five English universities led b Loughborough Universit, funded b the Higher Education Funding Council for England under the Fund for the Development of Teaching and Learning for the period October 00 September 005, with additional transferabilit funding October 005 September 006. HELM aims to enhance the mathematical education of engineering undergraduates through fleible learning resources, mainl these Workbooks. HELM learning resources were produced primaril b teams of writers at si universities: Hull, Loughborough, Manchester, Newcastle, Reading, Sunderland. HELM gratefull acknowledges the valuable support of colleagues at the following universities and colleges involved in the critical reading, trialling, enhancement and revision of the learning materials: Aston, Bournemouth & Poole College, Cambridge, Cit, Glamorgan, Glasgow, Glasgow Caledonian, Glenrothes Institute of Applied Technolog, Harper Adams, Hertfordshire, Leicester, Liverpool, London Metropolitan, Mora College, Northumbria, Nottingham, Nottingham Trent, Oford Brookes, Plmouth, Portsmouth, Queens Belfast, Robert Gordon, Roal Forest of Dean College, Salford, Sligo Institute of Technolog, Southampton, Southampton Institute, Surre, Teesside, Ulster, Universit of Wales Institute Cardiff, West Kingswa College (London), West Notts College. HELM Contacts: Post: HELM, Mathematics Education Centre, Loughborough Universit, Loughborough, LE11 3TU. helm@lboro.ac.uk Web: HELM Workbooks List 1 Basic Algebra 6 Functions of a Comple Variable Basic Functions 7 Multiple Integration 3 Equations, Inequalities & Partial Fractions 8 Differential Vector Calculus 4 Trigonometr 9 Integral Vector Calculus 5 Functions and Modelling 30 Introduction to Numerical Methods 6 Eponential and Logarithmic Functions 31 Numerical Methods of Approimation 7 Matrices 3 Numerical Initial Value Problems 8 Matri Solution of Equations 33 Numerical Boundar Value Problems 9 Vectors 34 Modelling Motion 10 Comple Numbers 35 Sets and Probabilit 11 Differentiation 36 Descriptive Statistics 1 Applications of Differentiation 37 Discrete Probabilit Distributions 13 Integration 38 Continuous Probabilit Distributions 14 Applications of Integration 1 39 The Normal Distribution 15 Applications of Integration 40 Sampling Distributions and Estimation 16 Sequences and Series 41 Hpothesis Testing 17 Conics and Polar Coordinates 4 Goodness of Fit and Contingenc Tables 18 Functions of Several Variables 43 Regression and Correlation 19 Differential Equations 44 Analsis of Variance 0 Laplace Transforms 45 Non- parametric Statistics 1 z- Transforms 46 Reliabilit and Qualit Control Eigenvalues and Eigenvectors 47 Mathematics and Phsics Miscellan 3 Fourier Series 48 Engineering Case Stud 4 Fourier Transforms 49 Student s Guide 5 Partial Differential Equations 50 Tutor s Guide Copright Loughborough Universit, 015 Production of this 015 edition, containing corrections and minor revisions of the 008 edition, was funded b the sigma Network.

3 Contents 17 Conics and Polar Coordinates 17.1 Conic Sections 17. Polar Coordinates Parametric Curves 33 Learning outcomes In this Workbook ou will learn about some of the most important curves in the whole of mathematics - the conic sections: the ellipse, the parabola and the hperbola. You will learn how to recognise these curves and how to describe them in Cartesian and in polar form. In the final block ou will learn how to describe cruves using a parametric approach and, in particular, how the conic sections are described in parametric form.

4 Conic Sections 17.1 Introduction The conic sections (or conics) - the ellipse, the parabola and the hperbola - pla an important role both in mathematics and in the application of mathematics to engineering. In this Section we look in detail at the equations of the conics in both standard form and general form. Although there are various was that can be used to define a conic, we concentrate in this Section on defining conics using Cartesian coordinates (, ). However, at the end of this Section we eamine an alternative wa to obtain the conics. Prerequisites Before starting this Section ou should... Learning Outcomes On completion ou should be able to... be able to factorise simple algebraic epressions be able to change the subject in simple algebraic equations be able to complete the square in quadratic epressions understand how conics are obtained as curves of intersection of a double-cone with a plane state the standard form of the equations of the ellipse, the parabola and the hperbola classif quadratic epressions in, in terms of conics HELM (015): Workbook 17: Conics and Polar Coordinates

5 1. The ellipse, parabola and hperbola Mathematicians, engineers and scientists encounter numerous functions in their work: polnomials, trigonometric and hperbolic functions amongst them. However, throughout the histor of science one group of functions, the conics, arise time and time again not onl in the development of mathematical theor but also in practical applications. The conics were first studied b the Greek mathematician Apollonius more than 00 ears BC. Essentiall, the conics form that class of curves which are obtained when a double cone is intersected b a plane. There are three main tpes: the ellipse, the parabola and the hperbola. From the ellipse we obtain the circle as a special case, and from the hperbola we obtain the rectangular hperbola as a special case. These curves are illustrated in the Figures 1 and. cone-ais plane of intersection generator lines Circle: obtained b intersection of a plane perpendicular to the cone-ais with cone. (A degenerate case is a single point.) As the plane of intersection tilts the other conics are obtained: Ellipse: obtained b a plane, which is not perpendicular to the cone-ais, but cutting the cone in a closed curve. Various ellipses are obtained as the plane continues to rotate. Figure 1: Circle and ellipse HELM (015): Section 17.1: Conic Sections 3

6 cone-ais generator line P Parabola: obtained when the plane is parallel to the generator of the cone. Different parabolas are obtained as the point P moves along a generator. cone-ais generator line Hperbola: obtained when the plane intersects both parts of the cone. The rectangular hperbola is obtained when the plane is parallel to the cone-ais. (A degenerate case is two straight lines.) Figure : Parabola and hperbola The ellipse We are all aware that the paths followed b the planets around the sun are elliptical. However, more generall the ellipse occurs in man areas of engineering. The standard form of an ellipse is shown in Figure 3. minor-ais b major-ais a e a ae ae a a e directri foci b directri Figure 3 4 HELM (015): Workbook 17: Conics and Polar Coordinates

7 If a > b (as in Figure 1) then the -ais is called the major-ais and the -ais is called the minorais. On the other hand if b > a then the -ais is called the major-ais and the -ais is then the minor-ais. Two points, inside the ellipse are of importance; these are the foci. If a > b these are located at coordinate positions ±ae (or at ±be if b > a) on the major-ais, with e, called the eccentricit, given b e = 1 b a (b < a) or b e = 1 a b (a < b) The foci of an ellipse have the propert that if light ras are emitted from one focus then on reflection at the elliptic curve the pass through at the other focus. Ke Point 1 The standard Cartesian equation of the ellipse with its centre at the origin is a + b = 1 This ellipse has intercepts on the -ais at = ±a and on the -ais at ±b. The curve is also smmetrical about both aes. The curve reduces to a circle in the special case in which a = b. Eample 1 (a) Sketch the ellipse 4 + = 1 (b) Find the eccentricit e 9 (c) Locate the positions of the foci. Solution (a) We can calculate the values of as changes from 0 to : From this table of values, and using the smmetr of the curve, a sketch can be drawn (see Figure 4). Here b = 3 and a = so the -ais is the major ais and the -ais is the minor ais. Here b = 3 and a = so the -ais is the major ais and the -ais is the minor ais. (b) e = 1 a /b = 1 4/9 = 5/9 e = 5/3 (c) Since b > a and be = 5, the foci are located at ± 5 on the -ais. HELM (015): Section 17.1: Conic Sections 5

8 Solution (contd.) 5 3 foci 5 3 Figure 4 Ke Point 1 gives the equation of the ellipse with its centre at the origin. If the centre of the ellipse has coordinates (α, β) and still has its aes parallel to the - and -aes the standard equation becomes ( α) ( β) + = 1. a b Task Consider the points A and B with Cartesian coordinates (c, 0) and ( c, 0) respectivel. A curve has the propert that for ever point P on it the sum of the distances P A and P B is a constant (which we will call a). Derive the Cartesian form of the equation of the curve and show that it is an ellipse. P (, ) B c O c A Your solution 6 HELM (015): Workbook 17: Conics and Polar Coordinates

9 Answer We use Pthagoras s theorem to work out the distances P A and P B: Let R 1 = P B = [( + c) + ] 1/ and let R = P A = [(c ) + ] 1/ We now take the given equation R 1 + R = a and multipl both sides b R 1 R. The quantit R 1 R on the left is calculated to be 4c, and a(r 1 R ) is on the right. We thus obtain a pair of equations: R 1 + R = a and R 1 R = c a Adding these equations together gives R 1 = a + c a and squaring this equation gives + c + c + = a + c + c Simplifing: a (1 c a ) + = a c whence a + (a c ) = 1 This is the standard equation of an ellipse if we set b = a c, which is the traditional equation which relates the two semi-ais lengths a and b to the distance c of the foci from the centre of the ellipse. The foci A and B have optical properties; a beam of light travelling from A along AP and undergoing a mirror reflection from the ellipse at P will return along the path P B to the other focus B. The circle The circle is a special case of the ellipse; it occurs when a = b = r so the equation becomes r + r = 1 or, more commonl + = r Here, the centre of the circle is located at the origin (0, 0) and the radius of the circle is r. If the centre of the circle at a point (α, β) then the equation takes the form: ( α) + ( β) = r Ke Point The equation of a circle with centre at (α, β) and radius r is ( α) + ( β) = r HELM (015): Section 17.1: Conic Sections 7

10 Task Write down the equations of the five circles (A to E) below: circle E 1.5 circle A circle B circle C circle D Your solution Answer A ( 1) + ( 1) = 1 B ( 3) + ( 1) = 1 C ( + 0.5) + ( + ) = 1 D ( ) + ( + ) = (0.5) E ( + 0.5) + (.5) = 1 8 HELM (015): Workbook 17: Conics and Polar Coordinates

11 Eample Show that the epression = 0 represents the equation of a circle. Find its centre and radius. Solution We shall see later how to recognise this as the equation of a circle simpl b eamination of the coefficients of the quadratic terms, and. However, in the present eample we will use the process of completing the square, for and for, to show that the epression can be written in standard form. Now Also, ( 1) 1 and + 6 ( + 3) 9. Hence we can write ( 1) 1 + ( + 3) = 0 or, taking the free constants to the right-hand side: ( 1) + ( + 3) = 4. B comparing this with the standard form we conclude this represents the equation of a circle with centre at (1, 3) and radius. Task Find the centre and radius of each of the following circles: (a) = 1 (b) = 0 Your solution Answer (a) centre: (, 3) radius 1 (b) centre: ( 1, 0) radius /. HELM (015): Section 17.1: Conic Sections 9

12 Engineering Eample 1 A circle-cutting machine Introduction A cutting machine creates circular holes in a piece of sheet-metal b starting at the centre of the circle and cutting its wa outwards until a hole of the correct radius eists. However, prior to cutting, the circle is characterised b three points on its circumference, rather than b its centre and radius. Therefore, it is necessar to be able to find the centre and radius of a circle given three points that it passes through. Problem in words Given three points on the circumference of a circle, find its centre and radius (a) for three general points (b) (i) for ( 6, 5), ( 3, 6) and (, 1) (ii) for ( 0.7, 0.6), (5.9, 1.4) and (0.8,.8) where coordinates are in cm. Mathematical statement of problem A circle passes through the three points. Find the centre ( 0, 0 ) and radius R of this circle when the three circumferential points are (a) ( 1, 1 ), (, ) and ( 3, 3 ) (b) (i) ( 6, 5), ( 3, 6) and (, 1) (ii) ( 0.7, 0.6), (5.9, 1.4) and (0.8,.8) Measurements are in centimetres; give answers correct to decimal places. Mathematical analsis (a) The equation of a circle with centre at ( 0, 0 ) and radius R is ( 0 ) + ( 0 ) = R and, if this passes through the 3 points ( 1, 1 ), (, ) and ( 3, 3 ) then ( 1 0 ) + ( 1 0 ) = R (1) ( 0 ) + ( 0 ) = R () ( 3 0 ) + ( 3 0 ) = R (3) Eliminating the R term between (1) and () gives ( 1 0 ) + ( 1 0 ) = ( 0 ) + ( 0 ) so that = (4) 10 HELM (015): Workbook 17: Conics and Polar Coordinates

13 Similarl, eliminating R between (1) and (3) gives = (5) Re-arranging (4) and (5) gives a sstem of two equations in 0 and 0. ( 1 ) 0 + ( 1 ) 0 = (6) ( 3 1 ) 0 + ( 3 1 ) 0 = (7) Multipling (6) b ( 3 1 ), and multipling (7) b ( 1 ), subtracting and re-arranging gives 0 = 1 ( ) (3 1 )( + ) + ( 1 )( 3 + 3) + ( 3 )( 1 + 1) (8) while a similar procedure gives 0 = 1 ( ) (1 3 )( + ) + ( 1 )( 3 + 3) + ( 3 )( 1 + 1) (9) Knowing 0 and 0, the radius R can be found from R = ( 1 0 ) + ( 1 0 ) (10) (or alternativel using and (or 3 and 3 ) as appropriate). Equations (8), (9) and (10) can now be used to analse the two particular circles above. (i) Here 1 = 6 cm, 1 = 5 cm, = 3 cm, = 6 cm, 3 = cm and 3 = 1 cm, so that = = 0 and = 61 + = = 5 From (8) 0 = 1 ( ) ( 1) = 0 while (9) gives ( ) = 3 = 0 = = 1 0 The radius can be found from (10) R = ( 6 ( 3)) + (5 1) = 5 = 5 so that the circle has centre at ( 3, 1) and a radius of 5 cm. HELM (015): Section 17.1: Conic Sections 11

14 (ii) Now 1 = 0.7 cm, 1 = 0.6 cm, = 5.9 cm, = 1.4 cm, 3 = 0.8 cm and 3 =.8 cm, so that and = = = = = 8.48 so from (8) 0 = 1 ( ) and from (9) 0 = 1 ( ) and from (10) = = = = R = ( ) + ( ) = = so that, to d.p., the circle has centre at (.71, 0.07) and a radius of 3.45 cm. Mathematical comment Note that the epression appears in the denominator for both 0 and 0. If this epression is equal to zero, the calculation will break down. Geometricall, this corresponds to the three points being in a straight line so that no circle can be drawn, or not all points being distinct so no unique circle is defined. 1 HELM (015): Workbook 17: Conics and Polar Coordinates

15 Engineering Eample The web-flange junction Introduction In problems of torsion, the torsion constant, J, which is a function of the shape and structure of the element under consideration, is an important quantit. A common beam section is the thick I-section shown here, for which the torsion constant is given b flange J = J 1 + J + αd 4 where the J 1 and J terms refer to the flanges and web respectivel, and the D 4 term refers to the web-flange junction. In fact [ tf α = min, t ] ) w ( rtf t w t f flange where t f and t w are the thicknesses of the flange and web respectivel, and r is the radius of the concave circle element between the flange and the web. D is the diameter of the circle of the web-flange junction. D r t w tf web As D occurs in the form D 4, the torsion constant is ver sensitive to it. Calculation of D is therefore a crucial part of the calculation of J. Problem in words Find D, the diameter of the circle within the web flange junction as a function of the other dimensions of the structural element. Mathematical statement of problem (a) Find D, the diameter of the circle, in terms of t f and t w (the thicknesses of the flange and the web respectivel) in the case where r = 0. When t f = 3cm and t w = cm, find D. (b) For r 0, find D in terms of t f, t w and r. In the special case where t f = 3 cm, t w = cm and r = 0.4 cm, find D. HELM (015): Section 17.1: Conic Sections 13

16 Mathematical analsis (a) Consider a co-ordinate sstem based on the midpoint of the outer surface of the flange. R A The centre of the circle will lie at (0, R) where R is the radius of the circle, i.e. R = D/. The equation of the circle is + ( + R) = R (1) In addition, the circle passes through the corner at point A (t w /, t f ), so ( ) tw + ( t f + R) = R () On epanding giving so that t w 4 + t f Rt f + R = R Rt f = t w 4 + t f R = (t w/4) + t f t f = t w + t f 8t f D = R = t w 4t f + t f (3) Setting t f = 3 cm, t w = cm gives D = + 3 = 3.33 cm HELM (015): Workbook 17: Conics and Polar Coordinates

17 (b) Again using a co-ordinate sstem based on the mid-point of the outer surface of the flange, consider now the case r 0. R r B Point B (t w / + r, t f r) lies, not on the circle described b (1), but on the slightl larger circle with the same centre, and radius R + r. The equation of this circle is + ( + R) = (R + r) (4) Putting the co-ordinates of point B into equation (4) gives ( ) tw + r + ( t f r + R) = (R + r) (5) which, on epanding gives t w 4 + t wr + r + t f + r + R + t f r t f R rr = R + rr + r Cancelling and gathering terms gives so that t w 4 + t wr + r + t f + t f r = 4rR + t f R = R (r + t f ) R = D = (t w/4) + t w r + r + t f + t fr (r + t f ) so D = t w + 4t w r + 4r + 4t f + 8t fr (8r + 4t f ) (6) Now putting t f = 3 cm, t w = cm and r = 0.4 cm makes D = + (4 0.4) + (4 0.4 ) + (4 3 ) + ( ) (8 0.4) + (4 3) = = 3.5 cm Interpretation Note that setting r = 0 in Equation (6) recovers the special case of r = 0 given b equation (3). The value of D is now available to be used in calculations of the torsion constant, J. HELM (015): Section 17.1: Conic Sections 15

18 The parabola The standard form of the parabola is shown in Figure 5. Here the -ais is the line of smmetr of the parabola. directri focus a a Figure 5 Ke Point 3 The standard equation of the parabola with focus at (a, 0) is = 4a It can be shown that light ras parallel to the -ais will, on reflection from the parabolic curve, come together at the focus. This is an important propert and is used in the construction of some kinds of telescopes, satellite dishes and car headlights. Task Sketch the curve = 8. Find the position of the focus and confirm its lightfocusing propert. Your solution 16 HELM (015): Workbook 17: Conics and Polar Coordinates

19 Answer This is a standard parabola ( = 4a) with a =. Thus the focus is located at coordinate position (, 0). θ θ focus If our sketch is sufficientl accurate ou should find that light-ras (lines) parallel to the -ais when reflected off the parabolic surface pass through the focus. (Draw a tangent at the point of reflection and ensure that the angle of incidence (θ sa) is the same as the angle of reflection.) B changing the equation of the parabola slightl we can change the position of the parabola along the -ais. See Figure 6. = 4a( + 1) = 4a = 4a( 3) 1 3 Figure 6: Parabola = 4a( b) with verte at = b We can also have parabolas where the -ais is the line of smmetr (see Figure 7). In this case the standard equation is = 4a or = 4a focus a Figure 7 HELM (015): Section 17.1: Conic Sections 17

20 Task Sketch the curves = and = ( 3). Your solution Answer = ( 3) 3 = The focus of the parabola = 4a( b) is located at coordinate position (a + b, 0). Changing the value of a changes the conveit of the parabola (see Figure 8). = 3 = = Figure 8 18 HELM (015): Workbook 17: Conics and Polar Coordinates

21 The hperbola The standard form of the hperbola is shown in Figure 9(a). This has standard equation a b = 1 The eccentricit, e, is defined b e = 1 + b a (e > 1) asmptotes focus b focus focus focus ae a a ae b (a) (b) Figure 9 Note the change in sign compared to the equivalent epressions for the ellipse. The lines = ± b a are asmptotes to the hperbola (these are the lines to which each branch of the hperbola approach as ± ). If light is emitted from one focus then on hitting the hperbolic curve it is reflected in such a wa as to appear to be coming from the other focus. See Figure 9(b). The hperbola has fewer uses in applications than the other conic sections and so we will not dwell here on its properties. Ke Point 4 The standard equation of the hperbola with foci at (±ae, 0) is a b = 1 with eccentricit e given b e = 1 + b a (e > 1) HELM (015): Section 17.1: Conic Sections 19

22 General conics The conics we have considered above - the ellipse, the parabola and the hperbola - have all been presented in standard form:- their aes are parallel to either the - or -ais. However, conics ma be rotated to an angle with respect to the aes: the clearl remain conics, but what equations do the have? It can be shown that the equation of an conic, can be described b the quadratic epression A + B + C + D + E + F = 0 where A, B, C, D, E, F are constants. If not all of A, B, C are zero (and F is a suitable number) the graph of this equation is (i) an ellipse if B < 4AC (circle if A = C and B = 0) (ii) a parabola if B = 4AC (iii) a hperbola if B > 4AC Eample 3 Classif each of the following equations as ellipse, parabola or hperbola: (a) = 0 (b) = 0 (c) = 6 (d) = 0 Solution (a) Here A = 1, B =, C = 3 B < 4AC. This is an ellipse. (b) Here A = 1, B =, C = 1 B = 4AC. This is a parabola. (c) Here A =, B = 1, C = B < 4AC also A = C but B 0. This is an ellipse. (d) Here A = 3, B = 0, C = 3 B < 4AC. Also A = C and B = 0. This is a circle. 0 HELM (015): Workbook 17: Conics and Polar Coordinates

23 Task Classif each of the following conics: (a) = 0 (b) = 0 (c) = 0 (d) + 3 = 0 (e) = 7 Your solution Answer (a) A = 1, B =, C = 3 B > 4AC hperbola (b) A =, B = 1, C = 1 B > 4AC hperbola (c) A = 4, B = 0, C = 0 B = 4AC parabola (d) A = 1, B = 1, C = 1 B < 4AC, A = C, B 0 ellipse (e) A =, B = 0, C = B < 4AC, A = C and B = 0 circle HELM (015): Section 17.1: Conic Sections 1

24 Eercises 1. The equation = 0 represents an ellipse. Find its centre, the semi-major and semi-minor aes and the coordinate positions of the foci.. Find the equation of a circle of radius 3 which has its centre at ( 1,.) 3. Find the centre and radius of the circle + 5 = 0 4. Find the position of the focus of the parabola + 3 = 0 5. Classif each of the following conics (a) + 3 = 0 (b) = 0 (c) + 1 = 0 (d) 4 4 = 0 (e) = 0 6. An asteroid has an elliptical orbit around the Sun. The major ais is of length km. If the distance between the foci is km find the equation of the orbit. Answers 1. centre: (, 3), semi-major 3, semi-minor, foci: (, 3 ± 5). ( + 1) + (.) = 9 3. centre: (1, 1) radius 7 4. = ( 3) a = 1, b = 3. Hence focus is at coordinate position (4, 0). 5. (a) parabola with verte ( 1, 4) (b) ellipse (c) hperbola (d) hperbola (e) ellipse with centre (, 3) = HELM (015): Workbook 17: Conics and Polar Coordinates

25 Polar Coordinates 17. Introduction In this Section we etend the use of polar coordinates. These were first introduced in.8. The were also used in the discussion on comple numbers in 10.. We shall eamine the application of polars to the description of curves, particularl conics. Some curves, spirals for eample, which are ver difficult to describe in terms of Cartesian coordinates (, ) are relativel easil defined in polars [r, θ]. Prerequisites Before starting this Section ou should... Learning Outcomes On completion ou should be able to... HELM (015): Section 17.: Polar Coordinates be familiar with Cartesian coordinates be familiar with trigonometric functions and how to manipulate then be able to simplif algebraic epressions and manipulate algebraic fractions understand how Cartesian coordinates and polar coordinates are related find the polar form of a curve given in Cartesian form recognise some conics given in polar form 3

26 1. Polar Coordinates In this Section we consider the application of polar coordinates to the description of curves; in particular, to conics. If the Cartesian coordinates of a point P are (, ) then P can be located on a Cartesian plane as indicated in Figure 10. P P r O (a) O θ (b) Figure 10 However, the same point P can be located b using polar coordinates r, θ where r is the distance of P from the origin and θ is the angle, measured anti-clockwise, that the line OP makes when measured from the positive -direction. See Figure 10(b). In this Section we shall denote the polar coordinates of a point b using square brackets. From Figure 10 it is clear that Cartesian and polar coordinates are directl related. The relations are noted in the following Ke Point. Ke Point 5 If (, ) are the Cartesian coordinates and [r, θ] the polar coordinates of a point P then and, equivalentl, = r cos θ r = + + = r sin θ tan θ = From these relations we see that it is a straightforward matter to calculate (, ) given [r, θ]. However, some care is needed (particularl with the determination of θ) if we want to calculate [r, θ] from (, ). 4 HELM (015): Workbook 17: Conics and Polar Coordinates

27 Eample 4 On a Cartesian plane locate points P, Q, R, S which have their locations specified b polar coordinates [, π ], [, 3π ], [3, π 6 ], [, π] respectivel. Solution P R S 30 o 3 Q Figure 11 Task Two points P, Q have polar coordinates [3, π 3 ] and [, 5π ] respectivel. B locating 6 these points on a Cartesian plane find their equivalent Cartesian coordinates. Your solution Answer Q 3 P P : (3 cos π 3, 3 sin π 3 ) (3, 3 3 ) π/6 π/3 Q : ( cos π 6, sin π 6 ) ( 3, 1) HELM (015): Section 17.: Polar Coordinates 5

28 The polar coordinates of a point are not unique. So, the polar coordinates [a, θ] and [a, φ] represent the same point in the Cartesian plane provided θ and φ differ b an integer multiple of π. See Figure 1. a P a P a P θ θ +π θ + kπ Figure 1 For eample, the polar coordinates [, π 3 ], [, 7π 3 ], [, 5π 3 Cartesian plane. ] all represent the same point in the Ke Point 6 B convention, we measure the positive angle θ in an anti-clockwise direction. The angle θ is interpreted as the angle θ measured in a clockwise direction. θ -θ Figure 13 Eercises 1. The Cartesian coordinates of P, Q are (1, 1) and ( 1, 3). What are their equivalent polar coordinates?. Locate the points P, Q, R with polar coordinates [1, π 3 ], [, 7π 3 10π ], [, ]. What do ou notice? 3 6 HELM (015): Workbook 17: Conics and Polar Coordinates

29 Answer 1. π/3 7π/4 (1, 1) [, 7π/4] ( 1, 3) [, π/3]. All these points lie on a straight line through the origin.. Simple curves in polar coordinates We are used to describing the equations of curves in Cartesian variables,. Thus + = 1 represents a circle, centre the origin, and of radius 1, and = is the equation of a parabola whose ais is the -ais and with verte located at the origin. (In colloquial terms the verte is the sharp end of a conic.) We can convert these equations into polar form b using the relations = r cos θ, = r sin θ. Eample 5 Find the polar coordinate form of (a) the circle + = 1 (b) the parabola =. Solution (a) Using = r cos θ, = r sin θ in the epression + = 1 we have (r cos θ) + (r sin θ) = 1 or r (cos θ + sin θ) = 1 giving r = 1. We simplif this to r = 1 (since r = 1 is invalid being a negative distance). Of course we might have guessed this answer since the relation r = 1 states that ever point on the curve is a constant distance 1 awa from the origin. (b) Repeating the approach used in (a) for = we obtain: r sin θ = (r cos θ) i.e. r sin θ r cos θ = 0 Therefore r(sin θ r cos θ) = 0. Either r = 0 (which is a single point, the origin, and is clearl not a parabola) or sin θ r cos θ = 0 giving, finall r = 1 tan θ sec θ. This is the polar equation of this particular parabola, =. HELM (015): Section 17.: Polar Coordinates 7

30 Task Sketch the curves (a) = cos (b) = π 3 (c) = Your solution Answer π/3 (a) (b) (c) Task Sketch the curve r = cos θ. First complete the table of values. Enter values to d.p. and work in radians: Your solution π θ 0 6 r π 6 3π 6 4π 6 5π 6 6π 6 Answer π π 3π 4π 5π 6π θ r You will see that the values of θ for invalid). Now sketch the curve: Your solution π < θ < 3π give rise to negative values of r (and hence 8 HELM (015): Workbook 17: Conics and Polar Coordinates

31 Answer ( ) 1 circle: centre, 0, radius 1. 1 Task Sketch the curve θ = π/3. Your solution Answer Radial line passing through the origin at angle π 3 to the positive -ais. Task Sketch the curve r = θ. Your solution Answer HELM (015): Section 17.: Polar Coordinates 9

32 3. Standard conics in polar coordinates In the previous Section we merel stated the standard equations of the conics using Cartesian coordinates. Here we consider an alternative definition of a conic and use this different approach to obtain the equations of the standard conics in polar form. Consider a straight line = d (this will be the directri of the conic) and let e be the eccentricit of the conic (e is a positive real number). It can be shown that the set of points P in the (, ) plane which satisf the condition distance of P from origin perpendicular distance from P to the line = e is a conic with eccentricit e. In particular, it is an ellipse if e < 1, a parabola if e = 1 and a hperbola if e > 1. See Figure 14. r cos θ d d O r θ P Figure 14 We can obtain the polar coordinate form of this conic in a straightforward manner. If P has polar coordinates [r, θ] then the relation above gives r = e or r = e(d + r cos θ) d + r cos θ Thus, solving for r: r = ed 1 e cos θ This is the equation of the conic. In all of these conics it can be shown that one of the foci is located at the origin. See Figure 15 in which the pertinent details of the conics are highlighted. e<1 e =1 e>1 [ ed ] [ ed ] 1+e,π 1 e, 0 [ ] d,π [ ed ] 1+e,π Figure HELM (015): Workbook 17: Conics and Polar Coordinates

33 Task Sketch the ellipse r = 4 cos θ and locate the coordinates of its vertices. Your solution Answer Here Then r = de = 4 cos θ = 1 1 cos θ so e = 1 de 1 + e = 3 = 4 3 and de 1 e = 1 = 4 4/3 4 HELM (015): Section 17.: Polar Coordinates 31

34 1. Sketch the polar curves (a) r = Eercises 1 1 cos θ (b) r = e θ (c) r =. Find the polar form of the following curves given in Cartesian form: (a) = 1 + (b) = 1 3. Find the Cartesian form of the following curves given in polar form Answers 1. (a) r = sin θ + cos θ (b) r = 3 cos θ Do ou recognise these equations? (a) parabola e =1,d =1 6 3 cos θ. 1/ (b) decreasing spiral (c) r = 1 1 cos 3 θ ellipse since e = 1 < 1. 3 Also de = 3/ 3. (a) r sin θ = 1 + r cos θ r = cos θ cos θ = 1 1 cos θ (b) r cos θ sin θ = 1 r = cosec θ 3. (a) r(sin θ + cos θ) = + = which is a straight line 3 (b) r = 3 cos θ + = + = 3 + in standard form: ( 3 ) + = 9 ( ) 3. i.e. a circle, centre 4, 0 with radius 3 3 HELM (015): Workbook 17: Conics and Polar Coordinates

35 Parametric Curves 17.3 Introduction In this Section we eamine et another wa of defining curves - the parametric description. We shall see that this is, in some was, far more useful than either the Cartesian description or the polar form. Although we shall onl stud planar curves (curves ling in a plane) the parametric description can be easil generalised to the description of spatial curves which twist and turn in three dimensional space. Prerequisites Before starting this Section ou should... Learning Outcomes On completion ou should be able to... HELM (015): Section 17.3: Parametric Curves be familiar with Cartesian coordinates be familiar with trigonometric and hperbolic functions and be able to manipulate them be able to differentiate simple functions be able to locate turning points and distinguish between maima and minima. sketch planar curves given in parametric form understand how the same curve can be described using different parameterisations recognise some conics given in parametric form 33

36 1. Parametric curves Here we eplore the use of a parameter t in the description of curves. We shall see that it has some advantages over the more usual Cartesian description. We start with a simple eample. Eample 6 Plot the curve = cos t }{{ = 3 sin t } parametric equations of the curve 0 t π }{{} parameter range Solution The approach to sketching the curve is straightforward. We simpl give the parameter t various values as it ranges through 0 π and, for each value of t, calculate corresponding values of (, ) which are then plotted on a Cartesian plane. The value of t and the corresponding values of, are recorded in the following table: t 0 π 0 π 0 3π 0 4π Plotting the (, ) coordinates gives the curve in Figure 16. 5π 0 6π 0 7π 0 8π 0 9π 0 10π 0 3 t = π t = 7π 0 t = 3π 0 3π 0 t = 0 Figure 16 The curve in Figure 16 resembles part of an ellipse. This can be verified b eliminating t from the parametric equations to obtain an epression involving, onl. If we divide the first parametric equation b and the second b 3, square both and add we obtain ( ) + ( 3 ) = cos t + sin t 1 i.e = 1 which we easil recognise as an ellipse whose major-ais is the -ais. Also, as t ranges from 0 π = cos t decreases from 0, and = 3 sin t increases from 0 3. We conclude that the 34 HELM (015): Workbook 17: Conics and Polar Coordinates

37 parametric equations = cos t, = 3 sin t together with the parametric range 0 t π describe that part of the ellipse 4 + = 1 in the positive quadrant. On the curve in Figure 16 we have 9 used an arrow to indicate the direction that we move along the curve as t increases from its initial value 0. Task Plot the curve = t + 1 = t 3 0 t 1 Do ou recognise this curve as a conic section? First construct a table of (, ) values as t ranges from 0 1: Your solution t Answer t Now plot the points on a Cartesian plane: Your solution Answer t = 1 3 t = 0.75 t = 0.5 t = 0 t = 0.5 Now eliminate the t-variable from = t + 1, = t 3 to obtain the form of the curve: HELM (015): Section 17.3: Parametric Curves 35

38 Your solution Answer = 4 1 which is the equation of a parabola. Eample 7 Sketch the curve = t + 1 = t t 1 Solution This is ver similar to the previous Task (ecept for t 4 replacing t in the epression for and t replacing t in the epression for ). The corresponding table of values is t t = 1 3 t = 0 t = 0.5 t = 0.5 t = 0.75 Figure 17 We see that this is identical to the curve drawn previousl. This is confirmed b eliminating the t-parameter from the epressions defining,. Here t = 1 so = ( 1) 3 which is the same as obtained in the last Task. The main difference is that particular values of t locate (in general) different (, ) points on the curve for the two parametric representations. We conclude that a given curve in the plane can have man (in fact infinitel man) parametric descriptions. 36 HELM (015): Workbook 17: Conics and Polar Coordinates

39 Task Show that the two parametric representations below describe the same curve. (a) = cos t = sin t 0 t π (b) = t = 1 t 0 t 1 Eliminate t from the parametric equations in (a): Your solution Answer + = cos t + sin t = 1 Eliminate t from the parametric equations in (b): Your solution Answer = 1 = 1 or + = 1 What do ou conclude? Your solution Answer Both parametric descriptions represent (part of) a circle centred at the origin of radius 1.. General parametric form We will assume that an curve in the plane ma be written in parametric form: = g(t) = h(t) }{{} parametric equations of the curve t 0 t t 1 }{{} parameter range in which g(t), h(t) are given functions of t and the parameter t ranges over the values t 0 t 1. As we give values to t within this range then corresponding values of, are calculated from = g(t), = h(t) which can then be plotted on an plane. In derivatives d dt 1.3, we discovered how to obtain the derivative d d d d = d dt d dt d and. We found dt and d d = ( d d dt dt d dt ) d dt from a knowledge of the parametric ( ) 3 d dt HELM (015): Section 17.3: Parametric Curves 37

40 Note that derivatives with respect to the parameter t are often denoted b a dot: d dt ẋ so that d dt ẏ d dt ẍ etc d d = ẏ d ẋÿ ẏẍ and = ẋ d ẋ 3 Knowledge of the derivative is sometimes useful in curve sketching. Eample 8 Sketch the curve = t 3 + 3t + t = 3 t t 3 t 1. Solution = t 3 + 3t + t = t(t + )(t + 1) = 3 t t = (t + 3)(t 1) so that = 0 when t = 0, 1, and = 0 when t = 3, 1. We calculate the values of, at various values of t: t We see that t = and t = 0 give rise to the same coordinate values for (, ). This represents a double-point in the curve which is one where the curve crosses itself. Now d dt = 3t + 6t +, d dt = t d d = (1 + t) 3t + 6t + so there is a turning point when t = 1. The reader is urged to calculate d and to show that d this is negative when t = 1 (i.e. at = 0, = 4) indicating a maimum when. (The reader should check that vertical tangents occur at t = 0.43 and t = 1.47, to d.p.) We can now make a reasonable sketch of the curve: t = 0.5 t = 1 t = 1.5 t =, 0(double point) t =.5 t = 0.5 t increasing t = 3 6 t = 1 6 Figure HELM (015): Workbook 17: Conics and Polar Coordinates

41 3. Standard forms of conic sections in parametric form We have seen above that, given a curve in the plane, there is no unique wa of representing it in parametric form. However, for some commonl occurring curves, particularl the conics, there are accepted standard parametric equations. The parabola The standard parametric equations for a parabola are: = at = at Clearl, we have t = ( ) and b eliminating t we get = a or = 4a which we recognise a 4a as the standard Cartesian description of a parabola. As an illustration, Figure 19 shows the curve with a = and 1 t.3 t = t = 1 t = t = 1 Figure 19 The ellipse Here, the standard equations are = a cos t = b sin t Again, eliminating t (dividing the first equation b a, the second b b, squaring and adding) we have ( ) ( ) + = cos t + sin t 1 or, in more familiar form: a b a + b = 1. If we choose the range for t as 0 t 7π 4 the following segment of the ellipse is obtained. t = π t = 3π b t = π 4 4 3π 4 π 4 t = π t = 0 a t = 5π 4 t = 7π 4 3 t = π Figure 0 Here we note that (ecept in the special case when a = b, giving a circle) the parameter t is not the angle that the radial line makes with the the positive -ais. HELM (015): Section 17.3: Parametric Curves 39

42 In the stud of the orbits of planets and satellites it is often preferable to use plane polar coordinates (r, θ) to treat the problem. In these coordinates an ellipse has an equation of the form 1 r = A + B cos θ, with A and B positive numbers such that B < A. Not onl is there a difference in the equations on passing from Cartesian to polar coordinates; there is also a change in the origin of coordinates. The polar coordinate equation is using a focus of the ellipse as the origin. In the Cartesian description the foci are two points at +e along the -ais, where e obes the equation e = a b, if we assume that a < b i.e. we choose the long ais of the ellipse as the -ais. This problem gives some practice at algebraic manipulation and also indicates some shortcuts which can be made once the mathematics of the ellipse has been understood. Eample 9 An ellipse is described in plane polar coordinates b the equation 1 r = + cos θ Convert the equation to Cartesian form. [Hint: remember that = r cos θ.] Solution Multipling the given equation b r and then using = r cos θ gives the results 1 = r + so that r = 1 We now square the second equation, remembering that r = +. We now have 4( + ) = (1 ) = 1 + so that = 1 We now recall the method of completing the square, which allows us to set 3 + = 3( + 3 ) 1 9 ) Putting this result into the equation and collecting terms leads to the final result ( ) a + b = 1 with a = 1 3 and b = 3. This is the standard Cartesian form for the equation of an ellipse but we must remember that we started from a polar equation with a focus of the ellipse as origin. The presence of the term in the equation above actuall tells us that the focus being used as origin was a distance of 1 3 to the right of the centre of the ellipse at = 0. The preceding piece of algebra was necessar in order to convince us that the original equation in plane polar coordinates does represent an ellipse. However, now that we are convinced of this we can go back and tr to etract information in a more speed wa from the equation in its original (r, θ) form. 40 HELM (015): Workbook 17: Conics and Polar Coordinates

43 Solution (contd.) Tr setting θ = 0 and θ = π in the equation 1 r = + cos θ We find that at θ = 0 we have r = 1 3 while at θ = π we have r = 1. These r values correspond to the two ends of the ellipse, so the long ais has a total length = 4 3. This tells us that a = 3, eactl as found b our longer algebraic derivation. We can further deduce that the focus acting as origin must be at a distance of 1 from the centre of the ellipse in order to lead to the two r values 3 at θ = 0 and θ = π. If we now use the equation e = a b mentioned earlier then we find that 1 9 = b, so that b =, as obtained b our length algebra. 3 The hperbola The standard equations are = a cosh t = b sinh t. In this case, to eliminate t we use the identit cosh t sinh t 1 giving rise to the equation of the hperbola in Cartesian form: a b = 1. In Figure 1 we have chosen a parameter range 1 t. t = t =0 t = 0.5 t =1 t =0.5 t = 1 t =1.5 Figure 1 To obtain the complete curve the parameter range < t < must be used. These parametric equations onl give the right-hand branch of the hperbola. To obtain the left-hand branch we would use = a cosh t = b sinh t HELM (015): Section 17.3: Parametric Curves 41

44 Eercises 1. In the following sketch the given parametric curves. Also, eliminate the parameter to give the Cartesian equation in and. (a) = t, = t 0 t 1 (b) = t, = t t (c) = t = t 0 < t < 3 (d) = 3 sin πt = 4 cos πt 1 t 0.5. Find the tangent line to the parametric curve = t t t + t at the point where t = For each of the following curves epressed in parametric form obtain epressions for d d and d and use this information to help make a sketch. d (a) = t t, = t 4t (b) = t 3 3t, = t t Answers 1. (a) = t = (b) = 3 1 t = 0 t = 3 (c) = ( + ) = t = 0 (d) = 1 t = 1 t = 0.5 d d. = t + 1 = t 1 dt dt d = t + 1 d t 1 when t = 1 = 0, = d when t = 1 then = 3 d tangent line is = HELM (015): Workbook 17: Conics and Polar Coordinates

45 Answer 3. (a) d d = t 4 = t dt dt d d = = dt dt d d = t 4 t = t t 1 d d = [(t ) (t 4)] 8(t 1) 3 = 1 (t 1) 3 t = 0 t = t = (b) = (t )(t + t + 1) = (t )(t + 1) = (t + 1)(t ) d d = t 1 = 3t 3 dt dt d d = = 6t dt dt d = [(3t 3) (t 1)6t] = 6t + 6t 6 d (3t 3) 3 7(t 1) 3 t = 1, HELM (015): Section 17.3: Parametric Curves 43

46 NOTES

47 Inde for Workbook 17 Asteroid Cartesian coordinates 4 Circle 3, 7, 7, 37 Circle cutting machine 10 Conic - general equation 0, 30 Directri 4, 16 Eccentricit 5, 30 Ellipse 3-6, 30, 31, 34, 39 Focus 4-5, 16, 30 Hperbola 4, 19, 30, 41 Light ras 16 Major ais 4-5 Minor ais 4-5 Orbit Parabola 4, 16-18, 7, 30, 35-36, 39 Parametric curves Polar coordinates 4 Pthagoras theorem 7 Rectangular hperbola 4 Torsion 13 Web-flange junction 13 EXERCISES, 6, 3, 4 ENGINEERING EXAMPLES 1 Circle cutting machine 10 The web-flange junction 13

48