EDEXCEL NATIONAL CERTIFICATE UNIT 4 MATHEMATICS FOR TECHNICIANS OUTCOME 4 - CALCULUS

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1 EDEXCEL NATIONAL CERTIFICATE UNIT 4 MATHEMATICS FOR TECHNICIANS OUTCOME 4 - CALCULUS TUTORIAL 1 - DIFFERENTIATION Use the elemetary rules of calculus arithmetic to solve problems that ivolve differetiatio ad itegratio of simple algebraic ad trigoometric fuctios. CONTENT CALCULUS ARITHMETIC, DIFFERENTIATION AND INTEGRATION Differetiatio: itroductio to the differetial coefficiet, graphical aalogy of differetial coefficiet, gradiet at a poit; Liebitz otatio for differetial coefficiet; differetiatio of simple polyomial, expoetial fuctios ad siusoidal fuctios, rate of chage applied to simple fuctios Itegratio: itegratio as the reverse of differetiatio, basic rules of itegratio for simple polyomial fuctios, expoetial fuctios ad siusoidal fuctios, idefiite itegrals, defiite itegrals of simple polyomial fuctios; cocept of the itegral as a summatio device. O completio of this tutorial you should be able to do the followig. Explai differetial coefficiets. Apply Newto s rules of differetiatio to basic fuctios. Solve basic egieerig problems ivolvig differetiatio. Defie higher differetial coefficiets. Evaluate higher order differetial coefficiet. D.J.Du 1

2 1. GRADIENTS AND FINITE CHANGES Remember that the symbol (Capital Delta) meas a fiite chage i somethig. Here are some examples. Temperature chage T T T 1 Chage i time t t t 1 Chage i Agle θ θ θ 1 Chage i distace x x x 1 Chage i velocity v v v 1 The symbol δ (lower case delta) meas a small but fiite chage i somethig such as δt, δt, δθ, δx, δv ad so o. Cosider the followig. The distace moved by a object is directly proportioal to time t as show o the graph. Velocity Chage i distace/chage i time. v x/ t This would be the same for a small chage. v δx/δt x/ t The ratio x/ t is the same as the ratio δx/δt ad the ratio is the gradiet of the straight lie. TANGENT OF A CURVE The graph shows a curve that is a plot of x f(t) t The f(t) meas a fuctio of time ad this idea is ofte used istead of writig the variable itself. If we stuck a pi i the paper at poit P ad moved a straight edge aroud util it touched the pi, the straight edge would be the taget at P. This could be a solid triagle as show ad we must keep the other edges horizotal ad vertical. The vertical ad horizotal sides of the triagle give us the gradiet of the taget. Gradiet x/ t CHORD OF A CURVE We could draw a lie ad measure the gradiet but aother way is to draw a chord betwee two poits either side of P. The chord jois poits A ad B which are approximately equally spaced from P. The gradiet would be approximately x/ t but this time we could evaluate x ad t mathematically. D.J.Du

3 WORKED EXAMPLE No.1 Estimate the gradiet of the curve f(t) t at the poit t 6 by fidig the chord betwee t 8 ad t 4 At t 4, f(t) 4 64 At t 8, f(t) 8 51 x t Gradiet of the chord is 448/8 11 Gradiet is approximately 11 The closer we make the poits A ad B to P the more accurate the estimate will be. WORKED EXAMPLE No. Repeat the last questio takig A at t 5 ad B at t 7. At t 5, f(t) 5 15 At t 7, f(t) 7 4 x t 7-5 Gradiet of the chord is 18/ 109 Gradiet is approximately 109 If we took values closer to poit P we should start to use δx ad δt to idicate that the icremets are small. If we made it really small, such that the values got very close to 0 the gradiet would be very accurate ad the ratio δx/δt would have a defiite value. We ca actually do this. If we take the chord betwee poit P ad aother poit B the gradiet would be evaluated as follows. δx f(t+δt) f(t) δx f(t + δt) - f(t) δx (t + δt) - (t) Gradiet ad δt δt δt δt You eed to be able to multiply out the brackets but here is the result whe you do it. ( δt) ( δt) + t δt + t ( δt) δx ( δt) + t + t ( δt) δx t + + t δt + t ( δt) t δt δt δt δt δx δx Now if we brig B so close to P that δt 0 we have t ad at t δt δt Whe the value of δ is made zero, we replace it with 'd' so that the gradiet is 108 We might use the expressio "as δt 0, 108 " This is the value at the limit as δt becomes zero ad we also use the expressio to mea "i the limit as δt teds to zero" D.J.Du

4 . DIFFERENTIAL COEFFICIENT- THE DERIVATIVE Summarizig what we have just studied - I geeral if we have a fuctio y f(x) ad we wish to fid the gradiet at a poit P, we take a poit close to it ad form a expressio for the gradiet of the chord ad this is: δy f(x + δx) f(x) where f(x)is the value of y at P at δx δx f(x + δx) is the value of y at P + δx If we reduce the value of δx to zero we get a limitig ratio that is the true gradiet of the curve at poit P ad this ratio is called the DIFFERENTIAL COEFFICIENT or DERIVATIVE δy As δx 0 δx WORKED EXAMPLE No. Usig the method just described, fid the exact gradiet of y f(x) x at the poit x 4 δy f(x + δx) - f(x) (x + δx) - f(x) x + ( δx) + xδx - x δx δx δx δx I the limit as δx 0 x Put t 4 ad 8 ( ) δx + x δx δx + x δx THE DIFFERENTIAL COEFFICIENT FOR f(x) x We seem to be o the way to fidig the solutio for this because we have alrea solved f(x) x ad f(t) t.usig the same method, the differetial coefficiet is : δy f(x + δx) - f(x) i the limit whe δx 0 Puttig f(x) x we have δx δx (x + δx) - x as δx 0 δx Rearrage ito this form (x + δx) δx x 1 + x δx δx x (1+ ) - x (1 ) -1 x + as δx 0 x x as δx 0 δx δx Multiplyig out ay bracket to a power higher tha 1 will produce the result show ext. δx δx x x x as δx 0 The 1 will disappear. δx All the missig terms cotai higher powers tha 1. D.J.Du 4

5 δx δx x x x as δx 0 δx Divide out by δx 1 δx 1 x as δx 0 x x δx Put δx 0 ad all the terms iside the square bracket will be zero except the first x 1 x x This is a geeral solutio for the differetial coefficiet of y f(x) x The otatios used i this sectio were the work of the Great Mathematicia Liebitz ad more ca be foud o this at the followig Web Site ad Newto is credited with developig this work at the same time as Liebitz but it is Newto is regarded as the father of Calculus.. NEWTON S METHOD.1 DIFFERENTIATION OF AN ALGEBRAIC EXPRESSION The equatio x a t / is a example of a algebraic equatio. I geeral we use x ad y ad a geeral equatio may be writte as y ax where a is a costat ad is a power or idex. The rule for differetiatig is : / ax (-1) or ax (-1) Note that itegratig returs the equatio back to its origial form.. DIFFERENTIATING A CONSTANT. Cosider the equatio y a x. Whe 0 this becomes y a x 0 a (the costat). (Remember that aythig to the power of zero is uity). Usig the rule for differetiatio / ax 0-1 a (0)x -1 0 The costat disappears whe itegrated. This explais why, whe you do itegratio without limits, you must add o a costat that might or might ot have bee preset before you differetiated. It is importat to remember that: A costat disappears whe differetiated. D.J.Du 5

6 WORKED EXAMPLE No. 4 Differetiate the fuctio x t / with respect to t ad evaluate it whe t 4. t x ()()t 1 t Puttig t we fid 6 WORKED EXAMPLE No. 5 Differetiate the fuctio y 4 + x with respect to y ad evaluate it whe y 5. y 4 + x 0 + x 1 x Puttig y 5 we fid 10 WORKED EXAMPLE No. 6 Differetiate the fuctio z y 4 with respect to y ad evaluate it whe y. 4 dz z y ()(4)x 4 1 8y Puttig y we fid dz 64 WORKED EXAMPLE No. 7 Differetiate the fuctio p q + q 5 +5 with respect to q ad evaluate it whe q. p q + q dp ()()q dq 1 dp puttig q we get dq + (5)()q q + 15q 4 WORKED EXAMPLE No. 8 The equatio likig distace ad time is x 4t + ½ at where a is the acceleratio. Fid the velocity at time t 4 secods give a 1.5 m/s. x 4t + ½ at velocity v / 4 + at 4 +(1.5)(4) 10 m/s D.J.Du 6

7 SELF ASSESSMENT EXERCISE No.1 1. Fid the gradiet of the fuctio y x - 5x 7 whe x (Aswer -8). Fid the gradiet of the fuctio p q + q + 4q ad evaluate whe q (Aswer 58). Fid the gradiet of the fuctio u v + 4v 4 ad evaluate whe v 5 (Aswer 00) SELF ASSESSMENT EXERCISE No. 1. The electric charge eterig a capacitor is related to time by the equatio q t. Determie the curret (i dq/) after 5 secods. (0 Amp). The agle θ radias tured by a wheel after t secods from the start of measuremet is foud to be related to time by the equatio θ ω 1 t + ½ αt ω 1 is the iitial agular velocity ( rad/s) ad α is the agular acceleratio (0.5 rad/s ). Determie the agular velocity (ω dθ/) 8 secods from the start. (6 rad/s) D.J.Du 7

8 . OTHER STANDARD FUNCTIONS For other commo fuctios the differetial coefficiets may be foud from the look up table below. y si(ax) y cos(ax) y ta(ax) y l(ax) y ae kx acos(ax) asi(ax) a + ata(ax) 1 1 x x kx ake WORKED EXAMPLE No. 9 The distace moved by a mass oscillatig o a sprig is give by the equatio: x 5 cos (8 t) mm. Fid the distace ad velocity after 0.1 secods. At 0.1 secods x 5 cos (0.8).48 mm v / -40 si (8t) -40 si (0.8) mm/s Note that your calculator must be i radia mode whe lookig up sie ad cosie values. WORKED EXAMPLE No. 10 The distace moved by a mass is related to time by the equatio : x 0e 0.5t mm. Fid the distace ad velocity after 0. secods. At 0. secods x 0e 0.5t 0e mm v / (0)(0.5) e 0.5t 10 e mm/s D.J.Du 8

9 SELF ASSESSMENT EXERCISE No. 1. If the curret flowig i a circuit is related to time by the formula i 4si(t), fid the rate of chage of curret after 0. secods. (9.9 A/s). The voltage across a capacitor C whe it is beig discharged through a resistace R is related to time by the equatio v 4e -t/t where T is a time costat ad T RC. Fid the voltage ad rate of chage of voltage after 0. secods give R 10 kω ad C 0 µf. (1.47 V ad -7.6 V/s ). The voltage across a capacitor C whe it is beig charged through a resistace R is related to time by the equatio v 4-4e -t/t where T is a time costat ad T RC. Fid the voltage ad rate of chage of voltage after 0. secods give R 10 kω ad C 0 µf. (.58 V ad 7.6 V/s ) 4. The distace moved by a mass is related to time by the equatio : x 17e 0.t mm. Fid the distace ad velocity after 0.4 secods. (19.17 mm ad 5.75 mm/s) 5. The agle tured by a simple pedulum is give by the equatio: θ 0.05 si (6 t) mm. Fid the agle ad agular velocity (dθ/) after 0. secods. ( radia ad rad/s) D.J.Du 9

10 4. HIGHER ORDER DIFFERENTIALS Cosider the fuctio y x. The graph looks like this. The gradiet of the graph at ay poit is / x. This may be evaluated for ay value of x. If we plot / agaist x we get the followig graph. This graph is also a curve. We may differetiate agai to fid the gradiet at ay poit. This is the gradiet of the gradiet. We write it as follows. d d y 6x The graph is a straight lie as show with a gradiet of 6 at all poits. If we differetiate agai we get d y d d y 6 D.J.Du 10

11 WORKED EXAMPLE No. 11 The distace moved by a bo (i metres) with uiform acceleratio is give by s 5t. Fid the distace moved, velocity ad acceleratio after 1 secods. distace s 5t 70 m ds velocity v 10t 10 m/s dv d s acceleratio 10 m/s WORKED EXAMPLE No.1 The distace moved by a oscillatig bo is related to time by the fuctio: x 1. si(t) mm. Fid the distace moved, velocity ad acceleratio after 0. secods. distace x 1.si(t) 1.si (0.6) (1.)(.5646) mm velocity v ()(1.)cos(t) (.4)cos(0.6) mm/s dv d x acceleratio ()()(1.)si(t) - 4.8si(0.6) -.71mm/s D.J.Du 11

12 SELF ASSESSMENT EXERCISE No.4 1. Evaluate the first ad secod derivative of the fuctio p 8e -0.t whe t. (Aswers ad 0.15). The motio of a mechaism is described by the equatio x 50 Cos(0.5t) mm. Calculate the distace, velocity ad acceleratio after 0. secods. (Aswers mm, -.74 mm/s ad -1.6 mm/s ). Evaluate the first ad secod derivatives of the fuctio z x 4 + x + x - 5 whe x 4 (Aswers 658 ad 456) 4. The motio of a bo is described the equatio x Asi(ωt) where x is the distace moved ad t is the variable time. Show by successive differetiatio ad a substitutio that the acceleratio is give by a -ωx. D.J.Du 1