1627 simultaneous equations and intersections
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- The document is a demo disc for teaching A-Level maths modules C1 and C2, containing sample slides from 51 presentations covering topics like roots, surds, differentiation, integration, and trigonometry. - The slides demonstrate key concepts, worked examples, and exercises for students, with explanations of important results and places where students should take notes or do exercises independently. - The disc is intended to help both students learn and teachers deliver content for the AS core maths modules.
1627 simultaneous equations and intersections
- 1. Demo DiscDemo Disc ““Teach A Level Maths”Teach A Level Maths” Vol. 1: AS Core ModulesVol. 1: AS Core Modules © Christine Crisp
- 2. Explanation of Clip-art images An important result, example or summary that students might want to note. It would be a good idea for students to check they can use their calculators correctly to get the result shown. An exercise for students to do without help.
- 3. 38: The Graph of tanθ 43: Quadratic Trig Equations 29: The Binomial Expansion 33: Geometric series – Sum to Infinity 25: Definite Integration 13: Stationary Points 11: The Rule for Differentiation 9: Linear and Quadratic Inequalities 8: Simultaneous Equations and Intersections 6: Roots, Surds and Discriminant The slides that follow are samples from the 51 presentations that make up the work for the AS core modules C1 and C2. 18: Circle Problems 46: Indices and Laws of Logarithms 26: Definite Integration and Areas
- 4. 6: Roots, Surds and Discriminant Demo version note: Students have already met the discriminant in solving quadratic equations. On the following slide the calculation is shown and the link is made with the graph of the quadratic function.
- 5. For the equation . . .0742 =+− xx . . . the discriminant acb 42 − 12−= There are no real roots as the function is never equal to zero 2816 −= The Discriminant of a Quadratic Function If we try to solve , we get0742 =+− xx 2 124 −± =x The square of any real number is positive so there are no real solutions to 12− 742 +−= xxy 0< Roots, Surds and Discriminant
- 6. 8: Simultaneous Equations and Intersections Demo version note: The following slide shows an example of solving a linear and a quadratic equation simultaneously. The discriminant ( met in presentation 6 ) is revised and the solution to the equations is interpreted graphically.
- 7. 14 −−= xy 32 += xy e.g. 2 14 −−= xy )(2−−−−−− 32 += xy )(1−−−−−− Eliminate y: 1432 −−=+ xx The discriminant, 0)4)(1(444 22 =−=− acb 0442 =++⇒ xx 0)2)(2( =++⇒ xx (twice)2−=⇒ x The quadratic equation has equal roots. The line is a tangent to the curve. 72 =⇒−= yx 0442 =++ xxSolving Simultaneous Equations and Intersections
- 8. 9: Linear and Quadratic Inequalities Demo version note: Students are shown how to solve quadratic inequalities using earlier work on sketching the quadratic function. The following slide shows one of the two types of solutions that arise. The notepad icon indicates that this is an important example that students may want to copy.
- 9. 542 −−= xxy 542 −−= xxy Solution: e.g.2 Find the values of x that satisfy 0542 ≥−− xx 0542 =−− xx 0)1)(5( =+−⇒ xx 5=⇒ x or 1−=x 1−≤⇒ x There are 2 sets of values of x Find the zeros of where)(xf 54)( 2 −−= xxxf 542 −− xx is greater than or equal to 0 above the x-axis 5≥xor These represent 2 separate intervals and CANNOT be combined Linear and Quadratic Inequalities
- 10. 11: The Rule for Differentiation Demo version note: In this presentation, the rule for differentiation of a polynomial is developed by pattern spotting, working initially with the familiar quadratic function. A later presentation outlines the theory of differentiation.
- 11. ),( 42Tangent at 2 xy = (2, 4)x The Gradient at a point on a Curve Definition: The gradient at a point on a curve equals the gradient of the tangent at that point. e.g. 3 12 The gradient of the tangent at (2, 4) is 43 12 ==m So, the gradient of the curve at (2, 4) is 4 The Rule for Differentiation
- 12. 13: Stationary Points Demo version note: Stationary points are defined and the students practice solving equations to find them, using cubic functions, before going on to use the 2nd derivative to determine the nature of the points. The work is extended to other functions in a later presentation.
- 13. xxxy 93 23 −−= 0= dx dy The stationary points of a curve are the points where the gradient is zero A local maximum A local minimum x x The word local is usually omitted and the points called maximum and minimum points. e.g. Stationary Points
- 14. 18: Circle Problems Demo version note: The specifications require students to know 3 properties of circles. Students are reminded of each and the worked examples, using them to solve problems, emphasise the need to draw diagrams.
- 15. e.g.2 The centre of a circle is at the point C (-1, 2). The radius is 3. Find the length of the tangents from the point P ( 3, 0). xC (-1, 2) Solution: 2 12 2 12 )()( yyxxd −+−= P (3,0) x Method: Sketch! • Find CP and use Pythagoras’ theorem for triangle CPA A 222 ACPCAP −= 11920 =−=⇒ AP tangent tangent • Use 1 tangent and join the radius. The required length is AP. 22 )20())1(3( −+−−=⇒ CP Circle Problems 20416 =+=⇒ CP 20 11 3
- 16. 25: Definite Integration Demo version note: The next slide shows a typical summary. The clip-art notepad indicates to students that they may want to take a note.
- 17. SUMMARY Find the indefinite integral but omit C Draw square brackets and hang the limits on the end Replace x with • the top limit • the bottom limit Subtract and evaluate The method for evaluating the definite integral is: Definite Integration
- 18. 26: Definite Integration and Areas Demo version note: The presentations are frequently broken up with short exercises. The next slide shows the solution to part of a harder exercise on finding areas. The students had been asked to find the points of intersection of the line and curve, sketch the graph and find the enclosed area.
- 19. 2+= xy 2 4 xy −= ,02 =⇒−= yx Area of the triangle ⇒ 2−=x 1=xor Substitute in :2+= xy 31 =⇒= yx Area under the curve 1 2 31 2 2 3 44 −− −=−= ∫ x xdxx 9= 33 2 1 ××= (b) ; 2+= xy2 4 xy −= 022 =−+ xx⇒ 0)1)(2( =−+ xx⇒ Shaded area = area under curve – area of triangle 2 9 = 2 9 = Definite Integration and Areas 2 42 xx −=+⇒
- 20. 29: The Binomial Expansion Demo version note: The following short exercise on Pascal’s triangle appears near the start of the development of the Binomial Expansion. Answers or full solutions are given to all exercises.
- 21. Exercise Find the coefficients in the expansion of 6 )( ba + Solution: We need 7 rows 1 2 1 1 3 3 1 1 1 1 1 4 6 4 1 1 5 10 110 5 1 6 15 120 15 6Coefficients The Binomial Expansion
- 22. 33: Geometric series – Sum to Infinity Demo version note: The students are shown an example to illustrate the general idea of a sum to infinity. A more formal discussion follows with worked examples and exercises.
- 23. Suppose we have a 2 metre length of string . . . . . . which we cut in half We leave one half alone and cut the 2nd in half again m1 m1 m1 m2 1 . . . and again cut the last piece in half m1 m2 1 m4 1 m4 1 m2 1 Geometric series – Sum to Infinity
- 24. 38: The Graph of tanθ Demo version note: The next slide shows part way through the development of the graph of using and . θtan=y θsin=y θcos=y
- 25. θsin=y θ y θcos=y θ y x The graphs of and for areθsin θcos 3600 ≤≤ θ x x 190sin = 090cos ∞= This line, where is not defined is called an asymptote. θtan θtan=y Dividing by zero gives infinity so is not defined when .θtan 90=θ The Graph of tanθ
- 26. 43: Quadratic Trig Equations Demo version note: By the time students meet quadratic trig equations they have practised using both degrees and radians.
- 27. e.g. 3 Solve the equation for the interval , giving exact answers.πθ 20 ≤≤ 02cos3cos2 2 =−+ θθ 2 1cos =θ 2cos −=θ⇒ or 0232 2 =−+ cc Factorising: 0)2)(12( =+− cc 22 1 −== cc or⇒ The graph of . . .θcos=y Solution: Let . Then,θcos=c shows that always lies between -1 and +1 so, has no solutions for . θcos 2cos −=θ θ 0 π π2 θ 1 y θcos=y-1 Quadratic Trig Equations
- 28. θcos=y-1 0 π π2 θ 1 y Principal Solution: 3 60 π θ == Solving for . 2 1cos =θ πθ 20 ≤≤ 50⋅=y 3 π 3 5π Ans: 3 5 , 3 ππ θ = Quadratic Trig Equations
- 29. 46: Indices and Laws of Logarithms Demo version note: The approach to solving the equation started with a = 10 and b an integer power of 10. The word logarithm has been introduced and here the students are shown how to use their calculators to solve when x is not an integer. The calculator icon indicates that students should check the calculation. ba x =
- 30. A logarithm is just an index. To solve an equation where the index is unknown, we can use logarithms. e.g. Solve the equation giving the answer correct to 3 significant figures. 410 =x x is the logarithm of 4 with a base of 10 4log410 10=⇒= xx We write In general if bx =10 then bx 10log= = logindex ( 3 s.f. )6020⋅= Indices and Laws of Logarithms
- 31. Full version available from:- Chartwell-Yorke Ltd. 114 High Street, Belmont Village, Bolton, Lancashire, BL7 8AL England, tel (+44) (0)1204 811001, fax (+44) (0)1204 811008 info@chartwellyorke.com www.chartwellyorke.com