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 www.freestu.co.uk 1
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 www.freestu.co.uk
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 51 64 448 t 8 4 4 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 4 15 18 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 6 108 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 www.freestu.co.uk
. 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 1 + +.... -1 x x x as δx 0 The 1 will disappear. δx All the missig terms cotai higher powers tha 1. D.J.Du www.freestu.co.uk 4
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 http://e.wikipedia.org/wiki/calculus ad http://e.wikipedia.org/wiki/leibiz%7s_otatio_for_differetiatio 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 www.freestu.co.uk 5
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 5 + 5 dp ()()q dq 1 dp puttig q we get 4 + 40 64 dq + (5)()q 5 1 + 0 6q + 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 www.freestu.co.uk 6
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 www.freestu.co.uk 7
. 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) -8.69 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 0.1.1 mm v / (0)(0.5) e 0.5t 10 e 0.1 11.05 mm/s D.J.Du www.freestu.co.uk 8
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. (0.0466 radia ad 0.1087 rad/s) D.J.Du www.freestu.co.uk 9
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 www.freestu.co.uk 10
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) 0.678 mm velocity v ()(1.)cos(t) (.4)cos(0.6) 1.981 mm/s dv d x acceleratio ()()(1.)si(t) - 4.8si(0.6) -.71mm/s D.J.Du www.freestu.co.uk 11
SELF ASSESSMENT EXERCISE No.4 1. Evaluate the first ad secod derivative of the fuctio p 8e -0.t whe t. (Aswers -1.07 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 49.44 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 www.freestu.co.uk 1