MATHEMATICS FOR ENGINEERING DIFFERENTIATION TUTORIAL 1 - BASIC DIFFERENTIATION

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1 MATHEMATICS FOR ENGINEERING DIFFERENTIATION TUTORIAL 1 - BASIC DIFFERENTIATION Tis tutorial is essential pre-requisite material for anyone stuing mecanical engineering. Tis tutorial uses te principle of learning by example. Te approac is practical rater tan purely matematical and may be too simple for tose wo prefer pure mats. Calculus is usually divided up into two parts, integration and differentiation. Eac is te reverse process of te oter. Integration is covered in tutorial 1. On completion of tis tutorial you sould be able to do te following. Explain differential coefficients. Apply Newton s rules of differentiation to basic functions. Solve basic engineering problems involving differentiation. Define iger differential coefficients. Evaluate iger order differential coefficient. D.J.Dunn 1

2 1. DIFFERENTIAL COEFFICIENTS DIFFERENTIATION Differentiation is te reverse process of integration but we will start tis section by first defining a differential coefficient. Remember tat te symbol means a finite cange in someting. Here are some examples. Temperature cange T = T T 1 Cange in time t = t t 1 Cange in Angle = 1 Cange in distance x = x x 1 Cange in velocity v = v v 1 Te symbol means a small but finite cange in someting suc as T, t,, x, v and so on. Consider te following. Te distance moved by an object is directly proportional to time t as sown on te grap. Figure 1 Velocity = Cange in distance/cange in time. v = x/ t Tis would be te same for a small cange. v = x/ t = x/ t Te ratio x/ t is te same as te ratio x/ t and te ratio is te gradient of te straigt line. INFINITESIMALLY SMALL CHANGES d Te symbol d is used to denote a cange tat is infinitesimally small. On our grap te ratios are all te same and equal to te velocity. Tis value is te same at any point on a straigtline grap. v = / = x/ t = x/ t. Tis ratio olds true even wen te canges approac zero. D.J.Dunn

3 Now consider te case wen acceleration occurs and te velocity is canging. Te grap of x against t is an upwards curve. Figure We can no longer say v = x/ t but v = x/ t migt give a reasonable approximation if it is possible to measure te small canges. Te result of evaluating v = x/ t would give te velocity at te time te measurements were made. At some oter time te value would be different. If we were able to take our measurements over smaller and smaller intervals, te velocity would become te instantaneous velocity. Wen te value of t tends to zero we would write t 0. We would say tat in te limit, as t 0, te ratio x/ t becomes te differential coefficient (te true ratio) and we write it as /. It sould be obvious now tat te differential coefficient is te rate of cange of one variable wit anoter and is also te gradient of te grap at a given point. v = / gives te precise velocity at an instant in time but of course we could not find it by measuring and. Newton s calculus metod allows us to find tese differential coefficients if we ave a matematical equation linking te two variables. Wen a bo accelerates at a m/s te formula relating distance and time is x = a t /. Te velocity is te ratio / and it may be found at any moment in time by applying Newton s rules for differentiation. Remember Differentiation gives te gradient of te function. D.J.Dunn

4 . NEWTON S METHOD.1 DIFFERENTIATION OF AN ALGEBRAIC EXPRESSION Te equation x = a t / is an example of an algebraic equation. In general we use x and y and a general equation may be written as y = Cx n were C is a constant and n is a power or index. Te rule for differentiating is : / = Cnx (n-1) or = Cnx (n-1) Note tat integrating returns te equation back to its original form. DERIVATION For tose wo want to know were tis comes from ere is te derivation. Consider to points A and B on a curve of y = Cx n Join AB wit a straigt line and te gradient of te line is approximately te gradient at point P. Te closer A and B are te more accurate tis becomes. Gradient at P BC/AC is te difference between te value of x at B and A. Gradient f x B f x A = lim f(x A + ) f(x) A 0 = lim 0 Cx n 1 + x If we expand 1 + x n Cx n f(x) B f(x) A = lim 0 C(x + ) n Cx n = lim 0 Cx n 1 + x = lim 0 n we get a series 1 + n x + a x n 1 + b x + c x 4.. = lim 0 Cx n 1 + n x + a x + b x + c x = lim 0 Cx n n x + a x + b x + c x 4.. divide out te = lim n 0 Cxn x + a + b x x + Put = 0 and = Cnxn 1 D.J.Dunn 4

5 . DIFFERENTIATING A CONSTANT. Consider te equation y = a x n. Wen n = 0 tis becomes y = a x 0 = a (te constant). (Remember tat anyting to te power of zero is unity). Using te rule for differentiation / = anx 0-1 = a (0)x -1 = 0 Te constant disappears wen integrated. Tis explains wy, wen you do integration witout limits, you must add on a constant tat migt or migt not ave been present before you differentiated. It is important to remember tat: A constant disappears wen differentiated. WORKED EXAMPLE No.1 Differentiate te function x = t / wit respect to t and evaluate it wen t =. t x ()()t 1 t Putting t we find 6 WORKED EXAMPLE No. Differentiate te function y = 4 + x wit respect to y and evaluate it wen y = 5. y 4 x 1 0 x x Putting y 5 we find 10 D.J.Dunn 5

6 WORKED EXAMPLE No. Differentiate te function z = y 4 wit respect to y and evaluate it wen y =. z y dz 4 ()(4)x 4 1 8y Putting y we find dz 16 WORKED EXAMPLE No.4 Differentiate te function p = q + q 5 +5 wit respect to q and evaluate it wen q =. p q q dp ()()q dq 5 1 puttingq we get 5 (5)()q q 15q dp dq 4 WORKED EXAMPLE No.5 Te equation linking distance and time is x = 4t + ½ at were a is te acceleration. Find te velocity at time t = 4 seconds given a = 1.5 m/s. x = 4t + ½ at velocity = v = / = 4 + at = 4 +(1.5)(4) = 10 m/s D.J.Dunn 6

7 SELF ASSESSMENT EXERCISE No.1 1. Find te gradient of te function y = x - 5x 7 wen x = (Answer -8). Find te gradient of te function p = q + q + 4q and evaluate wen q = (Answer 1). Find te gradient of te function u = v + 4v 4 and evaluate wen v = 5 (Answer 00) SELF ASSESSMENT EXERCISE No. 1. Te electric carge entering a capacitor is related to time by te equation q = t. Determine te current (i = dq/) after 5 seconds. (0 Amp). Te angle radians turned by a weel after t seconds from te start of measurement is found to be related to time by te equation = 1 t + ½ t 1 is te initial angular velocity ( rad/s) and is te angular acceleration (0.5 rad/s ). Determine te angular velocity ( = d /) 8 seconds from te start. (6 rad/s) D.J.Dunn 7

8 . OTHER STANDARD FUNCTIONS For oter common functions te differential coefficients may be found from te look up table below. y sin(ax) y cos(ax) y tan(ax) y ln(ax) y ae kx acos(ax) asin(ax) a atan(ax) 1 1 x x kx ake WORKED EXAMPLE No.6 Te distance moved by a mass oscillating on a spring is given by te equation: x = 5 cos (8 t) mm. Find te distance and velocity after 0.1 seconds. At 0.1 seconds x = 5 cos (0.8) =.48 mm v = / = -40 sin (8t) = -40 sin (0.8) = mm/s Note tat your calculator must be in radian mode wen looking up sine and cosine values. WORKED EXAMPLE No.7 Te distance moved by a mass is related to time by te equation : x = 0e 0.5t mm. Find te distance and velocity after 0. seconds. At 0. seconds x = 0e 0.5t = 0e 0.1 =.1 mm v = / = (0)(0.5) e 0.5t = 10 e 0.1 = mm/s D.J.Dunn 8

9 SELF ASSESSMENT EXERCISE No. 1. If te current flowing in a circuit is related to time by te formula i = 4sin(t), find te rate of cange of current after 0. seconds. (9.9 A/s). Te voltage across a capacitor C wen it is being discarged troug a resistance R is related to time by te equation v = 4e -t/t were T = is a time constant and T = RC. Find te voltage and rate of cange of voltage after 0. seconds given R = 10 k and C = 0 F. (1.47 V and -7.6 V/s ). Te voltage across a capacitor C wen it is being carged troug a resistance R is related to time by te equation v = 4-4e -t/t were T = is a time constant and T = RC. Find te voltage and rate of cange of voltage after 0. seconds given R = 10 k and C = 0 F. (.58 V and 7.6 V/s ) 4. Te distance moved by a mass is related to time by te equation : x = 17e 0.t mm. Find te distance and velocity after 0.4 seconds. (19.17 mm and 5.75 mm/s) 5. Te angle turned by a simple pendulum is given by te equation: = 0.05 sin (6 t) mm. Find te angle and angular velocity (d /) after 0. seconds. ( radian and rad/s) D.J.Dunn 9

10 . HIGHER ORDER DIFFERENTIALS Consider te function y = x. Te grap looks like tis. Figure Te gradient of te grap at any point is / =x. Tis may be evaluated for any value of x. If we plot / against x we get te following grap. Figure 4 Tis grap is also a curve. We may differentiate again to find te gradient at any point. Tis is te gradient of te gradient. We write it as follows. d d y 6x Te grap is a straigt line as sown wit a gradient of 6 at all points. If we differentiate again we get Figure 5 d y d d y 6 D.J.Dunn 10

11 WORKED EXAMPLE No.8 Te distance moved by a bo (in metres) wit uniform acceleration is given by s = 5t. Find te distance moved, velocity and acceleration after 1 seconds. distance s 5t 70 m ds velocity v 10t 10 m/s dv d s acceleration 10 m/s WORKED EXAMPLE No.9 Te distance moved by an oscillating bo is related to time by te function: x = 1. sin(t) mm. Find te distance moved, velocity and acceleration after 0. seconds. distance x 1.sin(t) 1.sin (0.6) (1.)(.5646) mm velocity v ()(1.)cos(t) (.4)cos(0.6) mm/s dv d x acceleration ()()(1.)sin(t) - 4.8sin(0.6) -.71mm/s D.J.Dunn 11

12 SELF ASSESSMENT EXERCISE No.4 1. Evaluate te first and second derivative of te function p = 8e -0.t wen t =. (Answers and 0.15). Te motion of a mecanism is described by te equation x = 50 Cos(0.5t) mm. Calculate te distance, velocity and acceleration after 0. seconds. (Answers mm, -.74 mm/s and -1.6 mm/s ). Evaluate te first and second derivatives of te function z = x 4 + x + x - 5 wen x = 4 (Answers 658 and 456) 4. Te motion of a bo is described te equation x = Asin( t) were x is te distance moved and t is te variable time. Sow by successive differentiation and a substitution tat te acceleration is given by a = - x. D.J.Dunn 1