18 EVEN MORE CALCULUS

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1 8 EVEN MORE CALCULUS Chapter 8 Even More Calculus Objectives After stuing this chapter you shoul be able to ifferentiate an integrate basic trigonometric functions; unerstan how to calculate rates of change; be able to integrate using substitutions; be able to integrate by parts; be able to formulate an solve simple first orer ifferential equations. 8.0 Introuction In the earlier chapters 8,, an 4 you met the concept of ifferentiation an integration. You shoul be familiar with the ifferentiation an integration of functions such as n, e an ln. In this chapter the range of functions is epane to inclue the ifferentiation an integration of trigonometric functions an two important methos of integration, namely substitution an by parts, are ealt with. Finally, there is an introuction to the important topic of ifferential equations. 8. Derivatives of trig functions Intuitive ieas The erivative of a general function f() is given by f ( )= lim h 0 f( + h) f h You are going to obtain the erivatives of sin, cos, an tan. Remember that the erivative of a function at a given point is given by the graient of that function at the given point. 35

2 Suppose you look at a sketch of the graph sin an sketch in the tangents for values of of: This is shown opposite. 3π, π, π,0,π,π,3π y = sin - These tangent values when plotte against will look like the sketch opposite. y This suggests that the graph of the function for the tangent may be something like the sketch opposite. This sketch suggests that the erivative of sin might be cos, but this illustration is certainly not a proof. Activity By consiering the graph of cos, with tangents rawn at key points, show that the erivative of cos might be sin. Using a calculator or computer The erivative of sin is given by ( sin )= lim sin + h h 0 sin h Using a calculator or computer you can try some specific eamples - using a set value for an making h small. Why is it not possible to take the actual value when h = 0? 35

3 For instance, setting = 0. an h = , sin( ) sin = = = = cos So, approimately sin( ) sin cos( 0.) Activity Either calculate sin( + h) sin h for other values of an very small values of h; or write a computer programme to evaluate sin( + h) sin h for various values of an very small values of h. In each case show that sin( + h) sin cos h Geometric proof It is also possible to use the unit circle to emonstrate that, for eample, ( sin )=cos 353

4 In the iagram, shown opposite, h is small, an A B = sin( + h) AB = sin A A = h (raius angle) So sin( + h) sin = AP O A' P A h B' B Now assume that when h is very small the arc AA is approimately straight. So for the 'triangle' A AP, A' h sin( + h) sin h AP AA =cos P A So, approimately, if h is very small sin( + h) sin cos h Now, taking the limit as h 0, gives ( sin )=cos Activity 3 By consiering the same unit circle iagram, show that cos( + h) cos sin h an, taking the limit as h 0, ( cos )= sin To fin the erivative of tan, formally you can use the quotient rule, u v = uv vu u = u, v = v v, 354

5 giving This gives tan = = = sin cos sin cos sin cos ( cos ) cos cos sin sin ( cos ) = ( cos ) + sin cos = ( cos ) ( tan ) = ( sec ) Activity 4 Show that ( cot )= cosec So the three crucial results are ( sin )=cos or y = sin = cos ( cos )= sin or y = cos = sin ( tan )=sec or y = tan = sec an, inverting these results cos =sin + c sin = cos + c sec =tan + c 355

6 Why is it that we cannot fin ( tan ) when =± π,±3 π,±5 π, K? To ifferentiate y = sin k, where k is a constant, you can efine a new variable z by z = k, so that y = sin k = sin z, giving But = cosz. z = z z (function of a function) an since z = k then So giving z = k = ( cos z) k = k cosk. Eercise 8A. Fin the erivative of (a) sin (b) sin (c) sin00 () cos3. Fin cosk an tan k 4. Fin ( cosec ) an ( sec ). Hence etermine sec tan. 3. Differentiate (a) sin ( ) (b) sin cos (c) cos 3 ( ) () tan 356

7 8. Rates of change Chapter 8 Even More Calculus From your earlier work on ifferentiation, you will recognise that the rate of change of a variable, say f, is given by f, where t enotes time. Often though, f is a function of another variable, say, so that you nee to use the 'function of a function' rule, f = f We will see in the net two eamples, how this result is use. Eample The raius of a circular oil slick is increasing at the rate of m/s. Fin the rate at which the area of the slick is increasing when its raius is (a) 0 m, (b) 50 m. Solution Now A = πr giving A = ( π r ) = ( π r ) r (using 'function of a function') r = π r r = 4π r, since r = m/s. (a) (b) When r = 0 m, A = 40π 5.7m /s. When r = 50 m, A = 00π 68.3m /s. Eample A spherical balloon is blown up so that its volume increases at a constant rate of 0 cm 3 /s. Fin the rate of increase of the raius when the volume of the balloon is 000 cm

8 Solution Now v = 4 3 π r3 giving v = 4 r 3 π r3 = 4π r r. r But giving v = 0 0 = 4π r r. Thus r = 5 π r. When the volume is 000 cm the raius can be calculate from 000 = 4 3 π r3 r = 750 π 3 r = 5 π 750 π Eercise 8B. The raius of a circular isc is increasing at a constant rate of cm/s. Fin the rate at which the area is increasing when the raius is 0 cm.. A spherical balloon is inflate by gas being pumpe at a constant rate of 00 cm 3 /s. What is the rate of increase of the surface area of the balloon when its raius is 00 cm? 3. The volume of water in a container is given by e cm 3 where is the epth of the water in the container. Fin the rate of increase of volume when the epth is 0. cm an increasing at a rate of 0. cm s. Also etermine when the rate of increase of volume is stationary. (AEB) 4. The area of the region between two concentric circles of raii an y ( > y) is enote by A. Given that is increasing at the rate of 3 ms, y is increasing at the rate of 4 ms, an when t = 0, = 5 m an y = m, fin (a) the rate of increase of A when t = 0 ; (b) the ratio of to y when A begins to ecrease; (c) the time at which A is zero. area A y 358

9 8.3 Integration by substitution Chapter 8 Even More Calculus This metho is the equivalent to using a change of variable when ifferentiating composite functions. Activity 5 Differentiate (a) ( ) (b) ( 3 + ) 5 (c) ( 4 7) 6 () ( 4 7) n an use the above results to fin (a) (c) 5 3 (b) + ( 3 )5 () 3 (Check your results by ifferentiation.) In general, if y = f then = f( ). If we let = gu for some function g of a new variable u, then y becomes a function of u an u = u u = f( ) u = f( g( u) ) u y = f gu ( ) u u 359

10 Eample Fin 3 using a suitable substitution. Solution If u = 3 then u becomes u + = 3 an =. Hence the integral 3 ( 3 ) u u = u + 3 u u 3 = 9 u 3 + u u = 9 5 u u ( 3 )+c = ( + )5 3( 3 )3 +c = 45 ( )5 7 ( +c )3 Eample Fin Solution You can rewrite this integral using the substitution u = ( 3 + ) u = 6 The limits on u for the integration are etermine from = 0 u = = u = 5 360

11 Hence = ( 3 + ) 5 = = u 3 u 6 = 4 u4 5 = [ ] = u u 3 6 u = Eercise 8C. Fin the following integrals, using the suggeste substitution. (a) (b) 4 3 u = 3 3 ( u = ) (c) cos sin 4 ( u = sin ) () e + e ( u = + e ). Using the substitution = 4sin θ or otherwise, show that 0 4 = π 3. Using the substitution = 3tanθ, evaluate (AEB) 36

12 8.4 Integration by parts Our secon metho of integration is a very important one which is particularly useful for a number of ifferent integrals. You can see how it works by first ifferentiating with respect to, sin + cos Now use your answer to fin the integral of cos. What form will the integral of sin take? Check your answers by ifferentiating. Activity 6 By assuming e = ( a + b)e +c fin the values of the constants a an b so that the result is vali. (Hint: ifferentiate the R.H.S.) By now you shoul be wonering if there is a more precise metho of evaluating integrals of the type above, where the metho requires inspire guesswork to start with. In fact, there is a more formal metho. If u an v are two functions of then from your work on ifferentiation you know that ( uv)=vu + u v an integrating each sie with respect to gives uv = v u + u v u v = uv v u 36

13 Eample Fin cos Solution Here u = an v = cos, so v = sin. This gives cos = sin sin. = sin + cos + c We can use the same metho to fin the integral in Activity 6. Eample Fin e Solution Here u =, v = e, giving v = e an e = e e. = e e +c = e ( )+c Activity 7 (a) Given that cos =sin + cos + c show that sin = cos + sin + cos + k (b) Given that e = e e +c fin (i) e (ii) e 363

14 One of the ifficulties with integration by parts is that of ientifying u an v. For eample, if I = n v then taking u =, = ln will not really work as you might not be able to solve for v. But taking gives v =, an u = n, v = I = n ( ) = n = n 4 + c It is also important to note that some functions which cannot be immeiately integrate can be ealt with by writing the integran as (function) since can be integrate. This is illustrate below. Eample Fin n Solution I = n = n So take u = n, v = to give v = an I = n. = n = n + c 364

15 Finally in this section it shoul be note that we can have limits on integration in the usual way, for eample π sin = cos 0 [ ] 0 π π = ( 0 0)+ cos π = [ sin ] 0 = π 0 0 cos Eercise 8D. Fin n.. By writing cos 3 θ as cos θ cosθ fin cos 3 θ θ an hence show that 4. Fin 3 n( 4) (AEB) 5. Use integration by parts twice to fin cos cos 3 θ θ = Evaluate sin π 8.5 First orer ifferential equations You might well woner why you have stuie integration in such epth. The main reason is the crucial role it plays in solving what are known as ifferential equations. Here you will only be ealing with first orer ifferential equations, which can be epresse, in general, in the form = f, y when y = y is an unknown function of, an f is a given function of an y. 365

16 Eamples of such equations inclue = y = + y = y ( y 0) but first you will see how they are forme. As an eample of the process, consier the problem of moelling the way in which human populations change. An English economist, Thomas Malthus, in his 80 article "An Essay on the Principle of Population" formulate the first moel which attempte to escribe an preict the way in which the magnitue of the human population was changing. In mathematical terms you can summarise his assumptions by introucing the variable N = Nt () to represent the total population, where t is time. In a small time interval, say δ t, Malthus argue that both births an eaths are proportional to the population size an the time interval. Hence in time δt, there will be ( α Nδ t) births an ( β Nδ t) eaths where α, β are positive constants, an so the increase in the population size is given by δ N = α Nδ t β Nδ t = ( α β )Nδt Writing α β = γ,an iviing by δt gives δ N δ t = γ N Taking the limit as δ t 0 results in the ifferential equation N = γ N Ae to this is an initial conition; for eample, N = N 0 at t = 0. This is an eample of a 'variable separable' ifferential equation, since it can be written in the form N = γ N 366

17 an integration with respect to t, N N = γ N = γ N ln N = γ t + c (c constant) N = e γ t+c = e γ t e c N = Ke γ t ( K = e c ) But N = N 0 at t = 0, giving N 0 = Ke 0 = K. Hence the complete solution is N N = N 0 e γ t This solution is illustrate opposite for positive γ. γ < 0 What oes the solution look like when (a) γ = 0 (b) γ is negative? N 0 0 t Activity 8 By ifferentiation check that N = N 0 e γ t satisfies the ifferential equation N = γ N You can see how to use this population moel by applying it to the 790 an 800 USA population figures shown opposite. The time t = 0 correspons to 790, an t = to 800 etc; so N 0 = 3.9 Year USA Population (in millions)

18 The moel solution is Nt ()=3.9e γ t an γ is etermine by the population value at t = (800). This gives 5.3 = 3.9e γ e γ = γ = ln an Nt ()=3.9e t What oes this moel preict for the population if t becomes large? Is this realistic? Activity 9 Use the moel above to preict the USA population for 80 ( t = ) to 930 ( t = 4). Compare your preiction with the real values given opposite. Can you eplain any iscrepancies in the preiction an real values? The moel gives a reasonably goo fit for quite some time, but eventually preicts a population growing far too quickly. Why oes the moel eventually go wrong? Let us go back to the original assumption an consier what factors have been neglecte. The moel as it stans preicts the growth of a population in an ieal situation where there are no limits (e.g. foo, lan, energy) to continue growth. In 837, Verhulst propose an etension to the Malthusian moel which took into account the 'crowing' factor. In his moel, he assume that the population change was proportional to Year USA Population (in millions) (i) the current population level N (ii) the ratio of unuse population resources to total population resources when it is assume that a maimum population size of N can be sustaine, ( N i.e. N) 368 N

19 Thus i.e. N N ( = γ N N N) N = γ N γ N N When N is small, γ N will be the ominant term on the right han sie, an so we are back with the Malthusian moel; but as N becomes large, the γ N term becomes important. N This is again a first orer ifferential equation, but much more ifficult to solve. *Activity 0 Show, by ifferentiating the epression below, that the Verhulst moel has a solution given by Nt ()= where N = N 0 at t = 0. N + N N 0 e γt Use your graphic calculator, or computer program to illustrate the shape of this function. For eample, take γ = 0.03 an (a) N 0 = N (b) N 0 = 4 N (c) N 0 = N What happens as t? Returning to simple first orer ifferential equations, you will see how to solve them in the following two eamples. Eample For 0, solve = y given that y = when =

20 Solution You can write this as y = an integrating both sies with respect to. y = y = y = + K (K constant) Since y = when = 0, = 0 + K K = an y = + y = + Eample Solve = y given that y = when =. Solution As before y = y = = y lny = ln + C (C constant) 370

21 Writing C = nk, ln y = ln + ln K ln y = ln ( K ) y = K Since y = when =, = K. giving K = an the solution y =. Eercise 8E Solve the following first orer ifferential equations.. 3y = 5 given that y = when =0. = 3y given that y = when = = y given that y = when = = + 5. e = y given that y = 0 when = 0 6. = y = y + y 8. During the initial stages of the growth of yeast cells in a culture, the rate of increase in the number of cells is proportional to the number of cells present. If n = nt () represents the number of cells at time t, show that n = kn for some constant k. If the cells ouble in unit time, etermine k, an solve for n in terms of t an the initial number of cells,n Miscellaneous Eercises. Differentiate (a) sin 3 (b) sec tan (c) tan.. Fin (a) cos 4 sin (b) (c) sin 3 tan sec 3. The raius r of a circular ink spot, t secons after it first appears, is given by + 4t r = + t Calculate (a) the time taken for the raius to ouble its initial value; (b) the rate of increase of the raius in cm s when t = 3 ; (c) the value to which r tens as t tens to infinity. (AEB) 37

22 4. The volume of liqui V cm 3 in a container when the epth is cm is given by 4 V =, > 0 ( + ) (a) Fin V an etermine the value of for which V = 0. (b) Calculate the rate of change of volume when the epth is cm an increasing at a rate of 0.0 cm s, giving your answer in cm 3 s to three significant figures. (AEB) 5. (a) Differentiate ( + 3 ) with respect to. (b) Use the result from (a), or an appropriate substitution, to fin the value of 7. (a) Use integration by parts to fin ln (b) Fin the solution of the ifferential equation = y ln for which y = when =. (AEB) 8. Fin the general solution of the following first orer ifferential equations (a) ( 3) = y (b) y = ln (c) = y Fin the following integrals using the suggeste substitution (a) cos 5 sin ( u = cos ) (b) (c) u = e ( + e ) u = + e 37

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