Learning outcomes EDEXCEL NATIONAL CERTIFICATE UNIT 4 MATHEMATICS FOR TECHNICIANS OUTCOME 1 TUTORIAL 2 - LINEAR EQUATIONS AND GRAPHS On completion of this unit a learner should: 1 Know how to use algebraic methods 2 Be able to use trigonometric methods and standard formula to determine areas and volumes 3 Be able to use statistical methods to display data 4 Know how to use elementary calculus techniques. OUTCOME 1- Know how to use algebraic methods Indices and logarithms: laws of indices (a m x a n = a, = a m+n m a m n, a, (a m ) n = a mn ) n a laws of logarithms (log A + log B = log AB, log A n A = n log A, log A log B log ) B e.g. common logarithms (base 10), natural logarithms (base e), exponential growth and decay Linear equations and straight line graphs: linear equations e.g. y = mx + c; straight line graph (coordinates on a pair of labelled Cartesian axes, positive or negative gradient, intercept, plot of a straight line); experimental data e.g. Ohm s law, pair of simultaneous linear equations in two unknowns Factorisation and quadratics: multiply expressions in brackets by a number, symbol or by another expression in a bracket; by extraction of a common factor e.g. ax + ay, a(x + 2) + b(x + 2); by grouping e.g. ax - ay + bx - by, quadratic expressions e.g. a 2 + 2ab + b 2 ; roots of an equation e.g. quadratic equations with real roots by factorisation, and by the use of formula D.J.Dunn www.freestudy.co.uk 1
DIRECTLY PROPORTIONAL RELATIONSHIPS In Engineering and Science, the relationship between two quantities is often DIRECTLY PROPORTIONAL and when one is plotted against the other, a straight line graph is produced. The symbol means directly proportional to For example if y is directly proportional to x we write this as y x To get an equation, we replace ' ' with '= m' hence y = m x or y/x = m For example a mechanical spring usually behaves such that the change in length x is directly proportional to the applied force F. F x = k x where k is the spring constant or constant of proportionality. With no force the spring has its normal length. Suppose that a force of 1 N stretches the spring 2 mm. Being directly proportional it follows that 2 N will stretch it 4 mm and 3 N 6 mm and so on. The graph shows this. Because the graph goes through the zero origin we see that what ever the force the ratio F/x is 0.5 N/mm and this is constant k and also the gradient of the graph. Figure 1 The mathematical law is simply F = 0.5 x and this is an example of a LINEAR EQUATION In general linear equations take the form y = m x y is the quantity plotted vertically x is the quantity plotted horizontally m is the gradient (constant of proportionality) simply found as m = y/x but this is only true if the graph passes through the point 0,0 INTERCEPTS Consider the graph that relates the Fahrenheit and Celsius temperature scales. This may be used to convert from one to the other. 212 o F corresponds to 100 o C so one point on the graph goes there. 0 o C corresponds to 32 o F so another point goes there. Joint the two points with a straight line and we have the complete graph. We can work out that for every change of 1 o C we have a corresponding change of (9/5) o F. Figure 2 If we want to use a formula instead of the graph we can work out that o F = o C x 9/5 + 32 9/5 is the gradient and is simply found as the ratio: (212-32) 100 = 1.8 or 9/5 Note that the graph intercepts the vertical axis at 32 and this is called the intercept. A complete LINEAR EQUATION has the form of y = mx + C m is the gradient C the intercept with the y axis. m is the constant of proportionality as well as the gradient of the graph. D.J.Dunn www.freestudy.co.uk 2
WORKED EXAMPLE No.1 Convert 60 o C into Fahrenheit o F = o C x 9/5 + 32 = 60 x 9/5 + 32 = 108 + 32 = 140 o F WORKED EXAMPLE No.2 The graph shows the relationship between the pressure p of a gas plotted vertically and the temperature θ of a gas plotted horizontally when the volume is kept constant. Deduce the law relating them. Calculate the temperature at which the pressure is zero. Figure 3 The vertical axis is p and the horizontal is θ so the graph has a law p = mθ + C The gradient m is found from the two marked points. Vertical change = 210 100 = 110 kpa Horizontal change = 300 0 = 300 o C m = 110/300 = 0.367 kpa/ o C The intercept with the vertical axis is at 100 kpa so C = 100 kpa The law is p = 0.367θ + 100 where p is in kpa and θ is in o C Now put p = 0 and find θ 0 = 0.367θ + 100 0.367θ = -100 θ = -100/0.367 = -272.5 o C D.J.Dunn www.freestudy.co.uk 3
WORKED EXAMPLE No.3 The graph shows the relationship between a variable y plotted vertically and x plotted horizontally. Deduce the law relating them. Figure 4 The law is y = mx + C The gradient m is found by choosing any two points such as (17.5, 30) and (0, -15). The vertical change in y is 30 (-15) = 45 and the horizontal change in x is 17.5 0 = 17.5 m = 45/17.5 = 90/35 = 2.571 The intercept with the y axis is C = -15. The law is y = (90/35)x 15 or 2.571x - 15 Note 90/35 is an exact number and 2.571 is rounded off. Check if it works by choosing any value of x say 10. y = (2.571)(10) 15 = 10.71 and this is correct when checked on the graph. WORKED EXAMPLE No.4 The graph shows the relationship between a variable y plotted vertically and x plotted horizontally. Deduce the law relating them. Figure 5 The law is y = mx + C The gradient m is found by choosing any two points such as (-2, 30) and (7, -10). The vertical change in y is -10 (30) = -40 and the horizontal change in x is 7 - (-2) = 9 m = -40/9 = -4.444 and the intercept with the y axis is C = 21. The law is y = -(40/9)x + 21 or -4.444x + 21 Note 90/35 is an exact number and 2.571 is rounded off. Check if it works by choosing any value of x say 0. y = (4.444) (0) + 21 = 21 and this is correct when checked on the graph. Note that when the line slopes down to the right y is decreasing as x increases and the gradient is always negative. D.J.Dunn www.freestudy.co.uk 4
2. INVERSELY PROPORTIONAL The 1/x button on your calculator is called the inverse button. If you enter 2 and press the 1/x button you get the answer 0.5 because it evaluates ½. Inverse means 1 divided by For example inverse y means 1/y Inverse A means 1/A and so on. Consider the equation x y = C Rearrange and we get y = C/x or y = C (1/x) 1/x is the inverse of y. If we have an equation x 1/y it means that x is inversely proportional to y. An example of this is Boyle s law used in physics pv = C. In this law p is pressure and V is volume and the relationship is true when the temperature is constant. If we rearrange the law into p = C/V we may say that pressure is inversely proportional to volume or p = C (1/V). If we plot p against (1/V) we will get a straight line passing through the origin and C (the constant) is the gradient of the graph. WORKED EXAMPLE No.5 Determine the law relating p and V from the graph. Determine the pressure when V = 0.5 m 3. Figure 6 The law is p = C(1/V) and C is the gradient 1000/3 Hence p = (1000/3)(1/V) and pv = 1000/3 = 333.3 Nm When V = 0.5 1/V = 2 p = 333.3 (2) = 666.7 N/m 2 and the graph bears this out. D.J.Dunn www.freestudy.co.uk 5
SELF ASSESSMENT EXERCISE No.1 1. Determine the law that relates the variables for the following graphs. (a) Figure 7 a and b (Answers y = 2.286x + 2 and y = -20.833x + 250) (b) 2. Find the relationship for y and x from the graph. (Answer y = 57.1(1/x) + 10) Figure 8 3. Sketch the graph of the following relationships. (a) y = 0.4x 2 (b) y = -3x -2 (c) s = -2u + 4 D.J.Dunn www.freestudy.co.uk 6
SELF ASSESSMENT EXERCISE No.2 1. In the equation y = 6 x what is the constant of proportionality? 2. If a ball is dropped and allowed to fall, it is observed that the velocity v is directly proportional to the time t measured from the moment it was dropped. Write out the equation linking v and t It is observed that the velocity after 1 second is 9.81 m/s. What is the value and units of the constant? 3. It is observed that the volume of a gas V inside a balloon is inversely proportional to the pressure p. Write down the equation linking p and V It is observed that when the volume is 2 m 3 the pressure is 5 kn/m 2. What is constant of proportionality and what are its units? Plot p against V over the range V = 0.1 to 2 cm 3 and show that this is a curve. What would you have to plot in order to get a straight line graph? Plot this straight line graph. D.J.Dunn www.freestudy.co.uk 7
LOGARITHMIC GRAPHS Logarithms may be used to change to simplify functions by changing them into a straight line graph law. Consider the function y = f(x) = Cx n C is a constant and n a power. Except when n = 1, this a curve when plotted. If we take logarithms we find:- log(y) = φ(x) = log(c) + n log(x) The graph of φ(x) is now a straight line law where log(c) is the intercept and n is the gradient. This is most useful in determining the function from experimental data. WORKED EXAMPLE No. 6 The graph shows the results of an experiment in which a variables x and y are recorded and plotted. When log(x) and log(y) are plotted the straight line graph shown is produced. Determine the function f(x). Figure 11 From the straight line graph we have an intercept of 0.7 and a gradient of (3.7 0.7)/1 = 3 φ(x) = 0.7 + 3 log(x) Take antilogs f(x) = 5x 3 D.J.Dunn www.freestudy.co.uk 8
WORKED EXAMPLE No. 7 The graph shows the results of an experiment in which a variables x and y are recorded and plotted as logs. Determine the function f(x). Figure 12 From the straight line graph we have an intercept of 0.845. The gradient is (2.345 0.845)/(-15) = - 0.1 φ(x) = 0.845 0.1 log(x) Take antilogs f(x) = 7x -0.1 SELF ASSESSMENT EXERCISE No. 3 Determine the function f(x) for each of the graphs below, Figure 13 Answers f(x) = 1.5x 2 and f(x) = 0.5x -0.5 D.J.Dunn www.freestudy.co.uk 9
SIMULTANEOUS EQUATIONS - LINEAR Suppose we have two similar equations y = m 1 x + C 1 and y = m 2 x + C 2 The problem is to find a value of x and y (knowing the values of m and C) that is the same for both equations (if they exist). The simplest but perhaps most time consuming method is to plot two graphs and see where they cross. At the crossing point x and y are the same for both. WORKED EXAMPLE No. 8 Solve the values of x and y that are the same for both of the following equations. y = 3x + 4 y = 5x + 2 Plot both graphs for x = 0 to x = 2 We get the following result. The graphs cross at x = 1 and y = 7. We can check this is correct as follows. y = 3(1) + 4 = 7 y = 5(1) + 2 = 7 WORKED EXAMPLE No. 9 Solve x and y given the following simultaneous equation 39 = x + 7y...(1) 23 = 2x + 3y..(2) Rearrange to make y the subject. y = (39 x)/7.(3) y = (23 2x)/3 (4) Plot x against y using both equations. We get the graph shown. We find that x = 4 and y = 5 satisfies both equations. D.J.Dunn www.freestudy.co.uk 10
SELF ASSESSMENT EXERCISE No. 4 1. A resistance thermometer has a resistance R = 101Ω at a temperature θ = 20 o C and 103Ω at 60 o C. The law relating resistance and temperature is R = Ro + αθ where Ro is the resistance at 0 o C and R is the resistance at any other temperature. α is the temperature coefficient of resistance. Find the temperature Ro and α. Ro + 20α = 101.(1) Ro + 60α = 103..(2) Answers α = -2/-40 = 0.05 and Ro = 100Ω 2. Solve x and y given. 3x + 2y = 12.. (1) x + 3y = 11...... (2) Answers x = 2, y = 3 3. What are the values of x and y that satisfies both the following equations. x/7 y/2 = -3.....(1) x/3 + y/4 = 10..(2) Answers x = 21 and y = 12 D.J.Dunn www.freestudy.co.uk 11