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.

6 In fact this is a form of quadratic equation. 4 2. 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.

7 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. 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.

8 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. 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?

9 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. 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.

10 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. 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.