EDEXCEL NATIONAL CERTIFICATE UNIT 28 …

1 1 EDEXCEL NATIONAL CERTIFICATE unit 28 further mathematics FOR TECHNICIANS OUTCOME 1 ADVANCED GRAPHICAL TECHNIQUES CONTENTS 1 Be able to use advanced graphical techniques Advanced graphical techniques: graphical solution of a pair of simultaneous equations with two unknowns, to find the real roots of a quadratic equation, for the intersection of a linear and a quadratic equation, non-linear laws such as xbayb,axy2by the use of logarithms to reduce laws of type y = axn to straight line form, of a cubic equation such as 2x3 - 7x2 + 3x + 8 = 0 , recording, evaluating and plotting manual, computerised It is assumed that the student has completed the module mathematics FOR TECHNICIANS.

2 2 1. SIMULTANEOUS EQUATIONS - GRAPHICAL SOLUTION Let's remind ourselves what a simultaneous equation is about. If we have a problem with two unknowns (typically x and y), then we need two equations with the same variables in order to solve them. In general simultaneous equations take the form: ax + by = f1(x,y) x + y = f2(x,y) Where a, b, c and d are known coefficients. Remember that 'f ' means "a function of " and to solve x and y we need to have a value for the function. It is easier to follow if we write the equations as: ax + by = C x + y = D This tutorial only deals with two unknown variables.

3 If we had three unknown variables we would need three simultaneous equations and so on. We can rearrange our equation to give : y D x and ay bCx or as x Dy and baxCy We could plot y against x for both equations and determine the point where x and y are the same for both as the point where they cross. WORKED EXAMPLE No. 1 Solve x and y given the following simultaneous equation x + 7y = 39 ..(1) 2x + 3y = 23 ..(2) SOLUTION Rearrange to make y the subject (you could make x the subject). y = (39 x)/7 ..(3) y = (23 2x)/3 ..(4) Plot x against y using both equations and we get: There is only one value of x and y that are the same for both graphs so x = 4 and y = 5 satisfies both equations.

4 3 WORKED EXAMPLE No. 2 Solve x and y given the following simultaneous equation 2x2 - y = (1) 4x - y = 0 ..(2) SOLUTION Rearrange to make y the subject (you could make x the subject). y = (3) y = 4x ..(4) Plot x against y using both equations and we get the graphs shown. We find that x = 2 and y = 8 satisfies both equations but x = 0 and y = 0 is also a solution. WORKED EXAMPLE No. 3 When two bodies travelling towards each other on the same line collide, the resulting velocities v1 and v2 after the collision are represented by the simultaneous equations: v1 - v2 = -6 and 5v1 + 3v2 = -2 Solve the velocities using the graphical method.

5 SOLUTION Rearrange the equations to make v1 the subject. v1 = -6 + v2 and 53v2v21 Plot and we get the graph shown. v1 = m/s and v2 = + m/s SELF ASSESSMENT EXERCISE No. 1 Solve the variables in the following simultaneous equations by plotting suitable graphs. 1. x + y = 7 and 2x + 3y = 19 (Answers x = 2, y = 5) 2. x - y = -9 and 3x + y = 5 (Answers x = -1, y = 8) 3. 3x2 + 2x - y = -4 and 8x - y = -4 (Answers x = 2, y = 20 but 0,0 is also a solution) You might observe from the previous examples that when one graph is a curve of x2, the possibility exists that there are two solutions. In fact this is a form of quadratic equation. 4 2.

6 QUADRATIC EQUATIONS - GRAPHICAL SOLUTION Most relationships in Engineering and Science are anything but proportional and many are quadratic. Quadratic equations have the general form:- y = f(x) = ax2 + bx1 + cx0 This is normally written as :- y = f(x) = ax2 + bx + c a, b, c and d are coefficients and note that x1 = x and xo =1. There are two values of x that satisfy the equations. Depending on the data, these may be solvable as real numbers or not. We will only deal with equations that you can solve at this level. When y = 0, the values of x that satisfy the equation are called the ROOTS of the equation. If y is a non zero number, we only need to subtract y from both sides of the equation to produce a new equation that equates to zero.

7 The roots may be solved by plotting suitable graphs. WORKED EXAMPLE No. 4 Solve the values of x that satisfy the equation : y = 0 = 2x2 - 4x 6 SOLUTION Plot x against y over a suitable range we get the graph shown. From the graph determine the value of x where y = 0. We find that x = -1 and x = 3. It is always a good idea to check that putting these values into the equation produces y = 0. x = -1 y = 2(-1)2- 4(-1) 6 = 2 + 4 6 = 0 x = 3 y = 2(3)2 - 4(3) 6 = 18 12 6 = 0 An alternative method is to rearrange the equation and make x2 the subject. In this case we get 2x2 = (4x + 6)/2 = 2x + 3 If we let f1(x) = x2 and f2(x) = 2x + 3 and plot both functions against x, the point where the graphs cross is the point where the functions are equal and give the answer for x.

8 5 WORKED EXAMPLE No. 5 A closed metal cylindrical canister has a mean radius R and length L. The surface area of the metal used is given by the formula A = 2 (R2 + RL) Given that the area is 1000 mm2 and the length is 20 mm what is the radius? SOLUTION Put the values into the formula. 1000 = 2 (R2 + 20R) Rearrange into standard form. F(R) =0 = 2 (R2 + 20R) - 1000 0 = 2 R2 + 40 R - 1000 Plot the graph. We see that the value of R required is mm. The negative solution is not a possible solution for a real problem. SELF ASSESSMENT EXERCISE No. 2 1. Solve the following equations by plotting a suitable graph.

9 2x2 3x -2 = 0 (Answers x = 2 or ) -x2 + x + 2 = 0 (Answers x = 2 or -1) x2 + = 0 (Answers x = or 2) 3 x2 -20 log(x) = 0 (Answer x = ) 2. A closed metal cylindrical canister has a mean radius R and length L. The surface area of the metal used is given by the formula A = 2 (R2 + RL) Given that the area is 2000 mm2 and the length is 30 mm what is the radius? (Answers mm, the negative answer is ignored as it is not physically possible) 6 3. GRAPHICAL SOLUTION OF CUBIC EQUATIONS Cubic equations have the general form :- y = f(x) = 0 = ax3 + bx2 + cx + d If a solution exists then it may be found graphically in the same way as for a quadratic equation.

10 A cubic equation has three roots. Many cubic plots do not cross the y axis and there may be three, two, one or no real answers that satisfy the equation. WORKED EXAMPLE No. 6 Find the values of x that satisfy the equation 2x3 + 10x2 - 20x 50 = 0 SOLUTION The solution is the roots of f(x) = 2x3 + 10x2 - 20x 50 To solve x we must plot f(x) = y = 2x3 + 10x2 - 20x 50 and see if any value of x produces a result y = 0 The plot is shown. The graph crosses the y axis three times so there are three values of x that give a solution and these are x = -6, x = and Check them out to see if they are correct. x = -6 2x3 + 10x2 - 20x 50 = -2 so the answer is not exact but close.