New Material Section 1: Functions and Geometry occurring in engineering 1. Plotting Functions: Using appropriate software to plot the graph of a function Linear f(x) = mx+c Quadratic f(x) = Px +Qx+R Cubic f(x) = Ax 3 +Bx +Cx+D Expontential f(x) = Ae kx Expontential f(x) =A(1- e kx ) Log f(x) = ln(x) Log f(x) = ln(x+6) Modulus Trig Trig Trig f(x) = x f(x) = Asin(kx) f(x) = Acos(kx) f(x) = Asin(kx+g) ( See Lab exercises and quiz sheet) 1
Quiz sheet: Which of the following functions best match the plots given: y = 5cos(3x), y = e x, y =5sin(3x), y = ln(x+5), y = 40 (1-e -.8x ), y = 5sin(x), y= x+1, y = x, y = 5cos(x), y = 100e -0.08t sin(6t) y = e -3x 60 40 0 0-1 -0 0 1 3 4-40 6 5 4 3 1 0-90 -60-30-1-0 30 60 90 10 150 180-3 -4-5 -6 6 5 4 3 1 0-5 -4-3 - -1 0 1 3 4 5 5 0 15 10 5 0-4 -3 - -1-5 0 1 3 4-10 -15-0 -5
6 4 0-180 -90 0 90 180 70 - -4 450 400 350 300 50 00 150 100 50 0-1 0 1 3 4-6 5 4 3 1 0-10 -5 0-1 5 10 15 0 5 30 35 40 45 - -3 8 7 6 5 4 3 1 0-4 -3 - -1-1 0-1 3-3 -4-5 -6 9 8 7 6 5 4 3 1 0-1 0 1 3 4 5. Alteration of functional description as a result of a transformation in the plane. How a graphical translation can alter a functional description How a reflection in either axis can alter a functional description How a scaling transformation can alter a functional description Obtain the inverse of a function by a pictorial representation, graphically or algebraically ( See notes on quadratics below and the lab exploration at the end of the graph exercises) 3
Quadratics graphs translations, minimum and maximum values We will now see that if one can sketch of the graph of y x and one can draw any quadratic once you know how to write it down properly Graph of y x First we tabulate some values x -.5 - -1.5-1 -0.5 0 0.5 1 1.5.5 y x 6.5 4.5 1 0.5 0 0.5 1.5 4 6.5 and draw the graph for x from.5 to +.5, 7 6 5 4 3 1 0-5 -4-3 - -1-1 0 1 3 4 5 - Notice that the graph is symmetric about the y axis, and as y x 0, the graph is above the x axis. Notice also that the minimum y = 0 occurs when x = 0. Now, what does the graph of y x look like? In this case, the minimum value of y is again 0, but it occurs when x =. The graph is also symmetric either side of the vertical line at x = (not the y axis as before). Infact, y x is exactly the same graph as below shows, y x, but moved or translated to x =, as the table x -0.5 0 0.5 1 1.5.5 3 3.5 4 4.5 6.5 4.5 1 0.5 0 0.5 1.5 4 6.5 y x So now add the graph to our original plot: 7 6 5 4 3 1 0-5 -4-3 - -1-1 0 1 3 4 5-4
eg1. Without working out any values, do a quick sketch of the following, 1. y x 1. x y 3. y x 4. y x 1 10 9 8 7 6 5 4 3 1 0-6 -5-4 -3 - -1-1 0 1 3 4 5 6 - -3-4 -5-6 -7-8 -9-10 Exercise: If y = -(x+5) can be described as y = x flipped and moved 5 units to the left describe the graphs above in terms of what happens to y = x. 5
Now you have the hang of it, complete the table below and draw a quick sketch of y x, X -3 - -1 0 1 3 y = x 9 4 1 0 1 4 9 y = x - What do you notice? 10 9 8 7 6 5 4 3 1 0-6 -5-4 -3 - -1-1 0 1 3 4 5 6 - -3-4 -5-6 -7-8 -9-10 In this case, the minimum value of y is, and it again occurs at x = 0. To draw the graph, you are drawing the graph of y x, but lowering it by. Exercise: Write a sentence in English to describe the following functions in terms of what happens to an original y = x plot. (1) y = x + 3 () y = x - 3 (3) y = -x + (4) y = -x -1 6
eg. Sketch the graphs of, 1. y x 1. x 3 y 3. x 1 y 4. y x 1 1 10 9 8 7 6 5 4 3 1 0-6 -5-4 -3 - -1-1 0 1 3 4 5 6 - -3-4 -5-6 -7-8 -9-10 Exercise: Write a sentence in English to describe the functions above in terms of what happens to an original y = x plot. 7
3. Completing the square - technique Sometimes it is necessary to change a quadratic expression so that it of the form: x a b where a and b are Real numbers. This process is called completing the square and is done by adding and subtracting half the coefficient of the x term into the expression and tidying up using the factors Key is to concentrate on the the end of the expression: x x xy y xy y x y x y x and x terms and put to one side any constant at This balances out to zero so effectively you have not changed the expression, just done some cosmetic surgery x px p x px p This tidies up to a complete square x p p The sign chosen is the sign of the x coefficient This is the numerical side effect of completing the square Effectively all the terms involving x are now tucked away inside a bracket squared 8
Example Complete the square for the following quadratic expression x writing it in the form x a b where a and b are Real numbers: 14x 5 by Solution We will effectively forget about the 5 until the last line. Now concentrate on working with the first portion to complete the square: 14 14 x 14x 5 x 14x 5 This tidies up to a complete square 7 7 5 x The sign chosen is the sign of the x coefficient This is the numerical side effect of completing the square x 7 44 It is important to realise that if x 14 x 5and x 7 44 are evaluated for any value of x they will produce the same answer. They are in effect the same expression just dressed in different guises 9
Exercises Complete the square for each of the following quadratic expressions by writing them in the form x a b 1. x 4x 1 where a and b are Real numbers:. s 10s 38 3. x 7x 9 4. t 3t 10 5. x 4x 15 Completing the square will have many applications such as quickly sketching any Quadratics ( see lab tasks) 10
Applying completing to quickly sketching any quadratic function (See also lab tasks) Example On the same plot sketch the functions: y = x y = (x + 5) y = (x + 5) + 0 y = (x + 5) - 10 Solution 80 70 60 50 40 30 0 10 0-11 -10-9 -8-7 -6-5 -4-3 - -10 0 1 3 4 5 6 7 8 9 10-0 Example By completing the square write x - 10x + 33 in the form x a b b are Real numbers. Hence sketch the function Solution where a and 11
Exercise (i) Sketch y = x. Describe the relationship between the graphs of y = x and y = x a b where a and b are Real numbers. (ii) By completing the square write 6x 7 and b are Real numbers. Hence sketch the curve y x 6x 7 x in the form x a b where a Example: Write down the equation of the quadratics shown below: 1
4. Quadratic Expression and Conics In dimensions, any expression for the form, f(x, y) = Ax +By + Cxy + Dx + Ey + F where A, B, C, D, E and F are numbers (with not both A and B are equal to 0) is called a quadratic expression. Any expression of the form, Ax + By + Cxy + Dx + Ey + F = 0. (*) Is called a quadratic equation. It represents a restriction on the values x and y are take, and if and (x, y) co-ordinates satisfying (*) are plotted in D you get a curve. These types of curves are called conics as they can represent algebraically the conic sections which are the shapes produced by the intersection of a plane and a cone.. Ellipse Circle Parabola Hyperbola There are 4 standard conics and we will look at each one in turn: Circle Ellipse Hyperbola Parabola In each case we will look at the equation of the conic, sketching them, their standard forms: Circle (x x 0 ) + (y y 0 ) = r ( x x0 ) ( y y0 ) Ellipse 1 a b ( x x0 ) ( y y0 ) Hyperbola 1 a b Parabola ( x x ) 4 py 0 13
Circle: The general equation of a circle is, (x x 0 ) + (y y 0 ) = r (*) where (x 0, y 0 ) is the centre, and r the radius. We have already studied this and just wish to look at the application of completing the square to finding the centre and radius of a circle given its equation in non-standard form Example Find the centre and radius of x + y + 4x 6y + 9 = 0. Hence sketch the circle. Solution We need to put this into the centre form (*) above to do this we use Completing the Square on the x and y terms. Sketch: So that, x y 4x ( x ) 6x ( y 3) ( x ) ( y 3) 3 ( x ) 4 ( y 3) ( x ) ( y 3) 4 9 9 0 4 9 The equation of a circle, centre (-, 3) and radius. Exercises Find the centre and radius of: 1. x y x y 7 0 14
. x y 4x 8y 16 0 3. x y 6x 8 0 15
The Ellipse This has the form, x a b y 1 a, b > 0 centre (0,0) 1 1 [In (*) A =, B, C = 0, D = 0, E = 0, F = -1] a b The curve looks like this, b - a a Translation of the axes Suppose you have an ellipse centred at (x 0, y 0 ), The equation of the ellipse with centre (x 0, y 0 ) is, - b y 0 ( x x ) ( y y ) a b 0 0 1 x 0 Eg. ( x 1) ( y ) sketch the ellipse 1 4 9 Solution Compare with standard form : Then a = 4 => a = and b = 9 => b = 3 ( x x ) ( y y ) a b 0 0 1 Centre of ellipse (x 0, y 0 ) = (1, ) 16
Eg. ( y ) Sketch the ellipse ( x 3) 1 4 Solution ( y ) ( x 3) 1` => 4 Compare with standard form : Then a = 1 => a = 1 And b = 4 => b = ( x3) ( y ) 1 1 4 ( x x0) ( y y0) 1 a b Centre of ellipse (x 0, y 0 ) = (-3, ) Sketch: Eg. Sketch the ellipse 4(x + 1) + y = 16 Solution 17
Eg. Write the ellipse, ellipse. Solution x y x y 4 3 51 0 in standard form. Hence sketch the 18
Exercise. Write the ellipse 3x y 6x 4y 1 0 in standard form Solution Group x and y terms together 3x 6x y 4y 1 0 Then, 3x 6x y 4y 1 0 3x 6x y 4y 1 0 3 x x y y 1 0 3 x x (1) (1) y y (1) (1) 1 0 x y 3 1 1 1 1 1 0 So we have, x y x y x y 3 1 3 1 1 0 3 1 1 6 6 1 6 1 6 3 6 x 1 y 1 1 3 Which is an ellipse with centre (-1, -1) and a, b 3 19
5. Cubics and Quartics Plotting Cubics 1. recognise the graphs of y = x 3 and y = -x 3. recognise the main features of the graph of Quartics y= ax 3 + bx +cx +d recognise the main features of the graphs of quartic polynomials y= ax 4 +bx 3 + cx +dx +e See lab tasks 6.0 Absolute Value Function The absolute or modulus of x : x is the distance the number x is from the origin on a number diagram (line ) or It makes the number positive Diagram: 5 5 13 13 See also lab tasks. 0
7.0 Exponential Functions Any function of the form, f ( x) e kx where k is any positive or negative number, is called an exponential function. We have already seen, exp( x) Example Consider f ( x) e x (i.e. k = -1) e x i.e. k = 1. (i) (ii) Check e x means 1 e. x Graph of e x for x from 3 to +3. Solution (i) Check this, (a)on your calculator, type in, then do e x (b)on your calculator, type in, do e x, then do 1 x. The answers to (a) and (b) should be the same should be the same. (ii) Graph of e x, x from 3 to +3. Note that this is just a mirror image of e x across the y-axis. Growth/Decay The number k is called the growth rate of the function if k>0 The number k is called the decay rate if k<0. 1
Example Check you can complete the following table e x e x e 3x e 3.5x e 10x x =1.7E+00 7.39E+00.01E+01 3.31E+01.0E+04 x =15 3.7E+06 1.07E+13 3.49E+19 6.3E+ 1.39E+65 What do you notice? Graphs of e, e and e x x 3x Note that as k gets larger, the graphs get steeper. What will graphs of e, e and e x x 3 x look like? -3 - -1 1 3 x
Footnotes on expontential functions 1. notice that e x is never negative. in fact, e x doesn t equal 0 either i..e there is no x for which e x = 0. 3. The function e x is called the exponential function and is e to the power of x. 4. The number e = e 1 is approximately.718. 5. In fact the number.7188188.. occurs a lot in nature and so is given a special symbol the letter e. 6. Some laws of growth and decay in Nature featuring e: 0 Linear expansion l l e Change in electrical resistance with temperature R R e Newtons Law of Cooling T T e t 1 Discharge of a capacitor q Qe CR t Decay of current in an inductive circuit i Ie 0 R L t Growth of current in an capacitive circuit i I 1 e CR t 7. Finally the exponential function arises in any mechanical problem involving damping or friction. Example The flow of water into a fresh water tank is controlled by a ballcock valve. After filling from empty, the volume at time t (minutes) is given. by, V ( t) 100 (1 e 0 t ) where V is in litres. Plot a graph of V(t) against t for t = 0, 4, 8, 1, 16, 0 mins. Solution: t( mins) 0 4 8 1 16 0 V(t) 0 55.0671 79.81035 90.98 95.9378 98.16844 F HG 1 T T I KJ As you can see, this is an increasing function it is one-to-one. 3
8.0 Logarithmic Functions There occurs in engineering applications a range of functions known as log functions. The most common are log 10 ( log to the base 10) and log e( x ) ln( x ) (the natural log function also known as log to the base e). The natural log function is the one which occurs the most and we have already met it as the inverse of the e x function and we will concentrate mainly on this: The natural log function Before we used your calculator to graph, g( x) log e( x ) ln( x ) ( Note log e or ln on a calculator, not log 10 ). x 0.000001 0.1 0.5 1 3 10 1000 ln(x) Notice that ln is increasing, so it is one-to-one (and vice-versa). Notes on the domain of ln 1 Note that if x 0 then ln) will give an error message (try not working out ln(-1) or ln(0)). This means ln(x) is not sensible for x 0 or, in other words, domain of the function ln is all x such that x 0 4
Moving Functions Example Based on the function y = ln(x) graph the functions (i) y= ln (x+3) (ii) y = ln(x-3) (iii) y = ln(x) 5 Solution Example Based on the function 0. t y 100( 1e ) graph the functions (i) (ii) (iii) y 100 1 e y 100 1 e Solution 0. ( t ) ( ) 0. ( t ) ( ) y e 0. ( ) 100( 1 t ) 40 5
8.0 Graphing trigonometrical functions( waves, signals) There are three basic trigonometrical functions: y = sin t, y= cos t and y = tan t. y = sin t t ( degrees) 0 30 60 90 10 150 180 10 40 70 300 330 360 sin t 0 0.500 0.866 1.000 0.866 0.500 0-0.500-0.866-1 -0.866-0.500 0 There are three basic trigonometrical functions: y = sin(t), y = cos(t) and y = tan(t). From now on we will assume that the variable t is in radians think of t as time and you can t go far wrong! The maximum vertical displacement is 1. y = sin(t) 4 parts:up from 0, down to 0, down from 0, up to 0 sin(t) is the height of the point P as it rotates round anticlockwise from the positive horizontal axis. This height is positive above the horizontal and negative below it, giving the graph of height against t above. This pattern repreats itself every π as the height is back where it started every π. The pattern also repeats if you wind around clockwise. 6
The maximum vertical displacement is 1. y = cos(t) 4 parts:down to 0, down from 0, up to 0, up from 0 cos(t) is the base length of the point P as it rotates round anticlockwise from the positive horizontal axis. This length is positive to the right of the vertical and negative to the left, giving the graph of length against t above. This pattern repreats itself every π as the length is back where it started every π. The pattern also repeats if you wind around clockwise. y = tan(t) Graph from to is shown. tan(t) is the ratio sin( t) cos( t) Think about what this means in terms of the height and length above. What is this ratio when t is or? 7
Periodic Functions an introduction The three basic trigonometrical functions: y = sin x, y= cos x and y = tan x are examples of periodic functions. Periodic functions are functions whose graph is made up of a simple pattern repeated. Period The minimum length in the horizontal direction of the basic pattern in the graph is called the period. The period of y = sin t is degrees or radians The period of y = cos t is degrees or radians The period of y = tan t is degrees or radians The amplitude of y = A sin t and y = A cos t The maximum vertical displacement is called the amplitude of the wave. As we have seen the maximum size of any sin or cos function is 1 so the amplitude of y = A sin t and y = A cos t is A. Examples State the amplitude and hence sketch the functions given over one period using the same axes: 1 y = 3 sin x y = 4.7 sin x 3 y = 0.8 sin x x 4 y sin 3 State the amplitude and hence sketch the functions given over one period: 5 y = cos x x 6 y cos 8
The functions y = A sin t and y = A cos t y = sin t The function y = sin x completes a full cycle as x varies from 0 to 360. In the function y = sin t as the x is doubled before the sin is taken then y = sin t will complete two full cycles as x varies from 0 to 360and so will complete a full cycle as x varies from 0 to 180 i.e a full cycle is completed in 360 degrees. 1-1 0 90 18 7 36 Conclusion Amplitude does not effect the periodicity of trigonometric functions and so y = A sin t completes a full cycle in 360 radians. (Similarly for y = A cos xt) Summary(Period) degrees or k The functions y = A sin t and y = A cos t have a period 360o or radians Frequency of a wave The frequency, f, of a wave is the number of cycles completed in 1 second. It is measured in hertz (Hz). One hertz is one cycle per second. Consider y = A sin x period radians => one cycle completed in seconds cycles completed in seconds cycles completed in 1second Therefore the frequency is Note that Examples f T 1 f 1 State the number of cycles completed in 360 by the following: x (i) y =3 sin 4x (ii) y= 4 cos 3x (iii) y 1 x sin (iv) y F 3 5cos I H 4 K 9
Technique for Sketching Engineering Waves To sketch y = A sin ( ( t ) 1 Find the amplitude and period of the wave, Calculate quarters of the period and mark these out on the horizontal axis 3 Mark out the vertical axis + and A. 4 Sketch y Asin( t) over one period, Similarly for y Acos( t) State the amplitude and the period of the following functions and hence sketch them: (i) y = sin 3t (ii) y= 4 cos 4 t (iii) y t 1 cos 4 F 3t (iv) y H G I 6sin K J 4 (v)the graph shows the output from an oscilloscope attached to an accelerometer sensor on a piston. Estimate the amplitude, period and frequency of motion of the piston. 4 3 1 0-1 - -3-4 Acceleration ((10 6 m s -1 ) Time t 0 0.0005 0.001 0.0015 0.00 0.005 0.003 30
Waves of the form y = A sin ( t + ) or y = A cos (t +) We now wish to consider waves of the form y = A sin( t + ) and y = A cos( t + ). Introducing the ( called a phase angle)has the effect of moving the wave either to the left or the right, but does not effect either the amplitude or the period of the wave. To see the effect we are going to approach this in terms of moving the waves form y = A sin( t ) and y = A cos( t ) Figure 1 shows the graphs of y = 3 sin( t ) And y = 3 sin( t +1) Figure shows the graphs of y = 3 sin( t ) and y = 3 sin( t -1 ). So y = 3sin( t +1 ) is y = 3 sin( t ) moved to the left by 0.5 and y = 3sin( t +1 is y = 3 sin( t ) moved to the right by 0.5 But we know already this!!! Why Because we know that y = 3sin( t +1 ) can be written as y = 3sin( (t + 0.5) ) which is the graph of y = 3 sin( t ) moved to the left by 0.5 as we have replaced t with t + 0.5. Similarly we know already that y = 3sin( t - 1 ) can be written as y = 3sin( (t - 0.5) ) is the graph of y = 3 sin( t ) moved to the right by 0.5 as we have replaced t with t - 0.5. The phase of y Asin( t ) is radians The time displacement d of the wave is seconds 31
Technique for Sketching Engineering Waves To sketch y = A sin ( t + ) 1. Find the amplitude, period and time displacement of the wave,. As before sketch y Asin( t) over one period, 3. Work out the time displacement by writing y Asin( ( t d)) 4. Hence by translating the wave y Asin( t) by the time displacement d (to the left for a lead, to the right for a lag) sketch y Asin( t ). Similarly for y Acos( t ) Example Write y 4sin( 3t 1) in the form A sin( (t + d)) Hence describe how you would sketch the curve of y 4sin( 3t 1) having already sketched the curve y 4sin( 3t ) Example Write y sin( 4t 6 ) in the form A sin( (t - d)) Hence describe how you would sketch the curve of y sin( 4t 6 ) having already sketched the curve y sin( 4t ) Example Write y cos( t 0. 5) in the form A cos( (t + d)) Hence describe how you would sketch the curve of y cos( t 0. 5) having already sketched the curve y cos( t ) Example t 4 Write y 3cos 3 in the form A cos( (t + d)) Hence describe how you would sketch the curve of already sketched the curve 3cos t y 3 t 4 y 3cos 3 having 3
Examples Find the amplitude, period, phase angle and time displacement of the following waves. Hence make a rough sketch of one cycle of the graph. 1. x 4sin(t 1) A = T = Phase = Time disp =. x 5cos(t ), 4 A = T = Phase = Time disp = 3. x 55sin(0 t 0.40), A = T = Phase = Time disp = 4. x 40cos( 10t.5 ), A = T = Phase = Time disp = 33
Example: Solve 3sin5t 3, for t from π to π. Solution: Sketch y 3sin5t and the line y= 3 The horizontal line y = cuts the wave at the 10 points A to J. The solutions of the equation are then the 10 corresponding t values on the horizontal axis. Read them off as accurately as you can (remember that π is about 3.14). The period T of this sin wave is 1.56. If you read off the t values for F and 5 G the other t values are these two plus or minus multiples of the Period T. Exercise: Identify the period and amplitude of f(t) = 3sin(t). Using this information plot the function for t from - to on graph paper. Hence plot the function g(t) = 3sin((t+1)) for t from - to on the same graph, clearly labeling which graph is which. Using your graph find all of the solutions of the equation 3sin((t+1)) = 1.8 for t between - and. Exercise: On graph paper, plot the function g(t) = 4cos(t) over 3 periods between - and 4. Hence plot the function f(t) = 4cos(t- ) over 3 periods. Use your diagram to estimate to one decimal place the first four positive values of t which satisfy the equation = 4cos (t-) 34
Introduction 9.0 Differentiation of Functions (derivatives as formulas for slopes of tangents to curves) Differentiation is a method of finding the formula for the slope of the tangent line to a function at a point on the graph of the function. y=f(x) tangent line to function at x 1 y 1 (x 1, y 1 ) x 1 x As you can see, for a general function, the slope will vary as x varies. In other words, the slope is itself a function of x. 1.1 Finding the slope of the Tangent Line A tangent line is a straight line, and from the work in semester 1 the slope of a straight line can be found as follows: 1 Find points on it and get the slope using these i.e. if bx1, y1g and bx, yg are on the line, then the slope is given by, slope of line = F HG y x y x 1 1 I. KJ y y 1 x 1 ( x1, y1 ) ( x, y ) x 35
But with a tangent line there is a problem, you only know one point on a tangent line. The way we get round this is to look at chords of the curve which get closer and closer to the tangent line each chord will have its own slope which can be found. f ( x) y chord 1 chord tangent y 1 x 1 x x By taking x a smaller and smaller distance from x 1 we can find the slope of the limiting chord this is nothing other than the slope of the tangent line at x 1. Example Find the slope of the tangent line to the curve f ( x) x at the point x =. Solution f ( x) x 4 x 36
x 1 y1 f ( x1) x 1 y 1 4 b slope = y x 3 9 1 5 5.5 6.5 0.5.5 4.5.4 5.76 0.4 1.76 4.4.3 5.9 0.3 1.9 4.3..1.05.01 As can be seen from the table, the slope gets closer and closer to 4 as x gets closer to. We say that the slope at x = is 4, though it might really be thought of as the slope as you get nearer and nearer to x =. Note As stated earlier, the slope will usually vary as x varies, so what is really required is to find the slope at x in terms of x, not just at a single point (i.e. want to find a formula that gives the slope of the tangent to the curve at any point on it). Example Find a formula for the slope of the tangent line to the curve f ( x) x at any point (x, y). Solution Instead of looking to the right of by 1, 05, 0.4 etc., we will look at the right of x by h (a very small number). So take two points on a chord, ( x, f ( x)) ( x, x ) and x h, f ( x h) ( x h,( x h) ) 1 1 4 g b g f ( x) x bx hg x x x h 37
The slope of this chord is, f ( x h) f ( x) x h x bx hg x h x xh h x h xh h h x h This is the slope of the chord shown. Now, we take h smaller and smaller (just as we did at x = ) and so we get the limiting slope: limiting slope of f ( x) x at x is x. Therefore the formula for slope of the tangent to the curve f ( x) x at any point on the curve is slope = x. Notation for the limiting process: We say when talking about the limiting process above that: Or Or Or x + h tends to x as h tends to 0. the limit as h tends to 0 of x + h is x. lim( x h) x h0 x h x as h 0 Example What are the slopes of the tangents to the curve f ( x) x at x =, 3, 5 and 3. Solution From the above slope = x. At x =, slope = x =() = 4 (which agrees with what we saw earlier.) Slopes at x = 3, 5 and 3 are 6, 10 and 6 respectively also. 38
Definition of Derivative The slope of (the tangent line to) f ( x) at x is called the derivative of f with respect to x. It is represented by the symbol, df dx Example If f ( x) x find df df and hence find at the points x =, 5, -3. dx dx Solution So that df dx df at x = or dx ( ) is 4. Similarly, df dx ( 5) 10, df dx ( 3) 6 etc. In Summary: df dx x So the derivative of a function f(x) gives a slope formula for the tangent to the graph of f(x) at x. Example: Find the equations of the tangent and the normal to the curve 4 3 f ( x) x x 3x x at the point (1, 6) Solution 39
Example Find the equations of the tangent and the normal to the curve 3 y x 3x 1x 1 at the point (0, 1) Solution Example Find the equations of the tangent and the normal to the curve f ( t) 3sin(4t ) at the point (1,.77898) Solution FOOTNOTES: 1 Note on use of different letters for variables Notice also that if f ( t) t, then df t! dt So if f ( s) s then df s, if f ( p) p, df p and so on. ds dp Derivative as a Rate of Change In this topic so far, we have concentrated on the derivative of a function as a formula for the slope of the tangent to a curve, but there is another interpretation of the derivative which is very important in practice. The derivative of f at x is also called the rate of change of f(x) with respect to x. It measure how quickly f(x) is changing with respect to x as x changes. Any engineering situation which involves change will be described mathematically using the derivative. Change with respect to time is the most common situation. We will now see how to calculate the derivative of any function efficiently this efficient set of calculation rules is called differential calculus. 40