Chapter 2 Overview: Anti-Derivatives. As noted in the introduction, Calculus is essentially comprised of four operations.

Transcription

1 Chapter Overview: Anti-Derivatives As noted in the introduction, Calculus is essentially comprised of four operations. Limits Derivatives Indefinite Integrals (or Anti-Derivatives) Definite Integrals There are two kinds of Integrals--the Definite Integral and the Indefinite Integral. The Definite Integral was eplored first as a way to determine the area bounded by a curve rather than bounded by a polygon. We know, from Geometry, how to find the eact area of various polygons, but geometry never considered figures where one or more sides is not made of a line segment. Here we want to consider a figure where one side is the curve y=f() and the other sides are the -ais and the lines = a and = b. y=f() a b As we can see above, the area can be approimated by rectangles whose height is the y value of the equation and whose width we will call Δ. The more rectangles we make, the better the approimation. The area of each rectangle would be n f Δ and the total area of n rectangles would be f ( i ) Δ. If we could 67 i=

2 make an infinite number of rectangles (which would be infinitely thin), we would have the eact area. The rectangles can be drawn several ways--with the left side at the height of the curve (as drawn above), with the right side at the curve, with the rectangle straddling the curve, or even with rectangles of different widths. But once they become infinitely thin, it will not matter how they were drawn--they will have no width and a height equal to the y-value of the curve. We can make an infinite number of rectangles mathematically by taking the Limit as n approaches infinity, or n Lim f ( i ) Δ. n i= This limit is rewritten as the Definite Integral: b a f b is the "upper bound" and a is the "lower bound," and would not mean much if it were not for the following rule. The symbol comes from the 7th century S and stands for sum. For a long time in the mathematical world we did not know that integrals and derivatives were connected. In the mid-600s Scottish mathematician James Gregory published the first proof of what is now called the Fundamental Theorem of Calculus, changing the math world forever. The Indefinite Integral is often referred to as the Anti-Derivative, because, as an operation, it and the Derivative are inverse operations (just as squares and square roots, or eponential and log functions). In this chapter, we will consider how to reverse the differentiation process. In the net chapter, we will eplore the definite integral. 68

3 .: Anti-Derivatives--the Power Rule As we have seen, we can deduce things about a function if its derivative is know. It would be valuable to have a formal process to determine the original function from its derivative accurately. The process is called Anti-differentiation, or Integration. Symbol: ( f () ) ="the integral of f of, d-" The is called the differential. For now, we will just treat it as part of the integral symbol. It tells us the independent variable of the function (usually, but not always, ) and, in a sense, is where the increase in the eponent comes from. It does have meaning on its own, but we will eplore that later. Looking at the integral as an anti-derivative, that is, as an operation that reverses the derivative, we should be able to figure out the basic process. Remember: d [ n ] = n n and D [constant] is always 0 (or, multiply the power in front and subtract one from the power). If we are starting with the derivative and want to reverse the process, the power must increase by one and we should divide by the new power. Also, we do not know, from the derivative, if the original function had a constant that became zero, let alone what the constant was. ( n ) The Anti-Power Rule = n+ + c for n n + The "+ c" is to account for any constant that might have been there before the derivative was taken. NB. This Rule will not work if n = -, because it would require that we divide by zero. But we know from the Derivative Rules what 69

4 yields ( or ) as the derivative--ln. So we can complete the anti-power Rule as: ( n ) The Anti-Power Rule = n+ + c if n n + = Ln + c Since D [f()+g()] = D [f()] + D [g()] and D [cn] = cd [n], then ( f + g ) = f c( f ) = c f + g These allows us to integrate a polynomial by integrating each term separately. OBJECTIVES Find the anti-derivative of a polynomial. Integrate functions involving Transcendental operations. Use Integration to solve rectilinear motion problems. E ( ) = c = c 70

5 E = c = Ln c = Ln + + c E + + = + = Ln + c = + Ln + c Integrals of products and quotients can be done easily IF they can be turned into a polynomial. + E + ( + )( +) d = = c = c 7

6 Eample 5 is called an initial value problem. It has an ordered pair (or initial value pair) that allows us to solve for c. E 5 f ' = 6 +. Find f () if f (0) =. f () = 6 + = + + c f (0) = 0 ( 0) + 0 c = + c = f () = + + E 6 The acceleration of a particle is described by a( t) = t + 8t +. Find the distance equation for (t) if v(0) = and (0) =. ( ) ( ) v(t) = a t dt = t + 8t + dt = t + t + t + c = 0 + ( 0) + ( 0) + c = c v(t) = t + t + t + (t) = v t dt = t + t + t + dt = t + t + t + t + c = ( 0 ) + ( 0 ) + ( 0 ) + ( 0) + c = c ( t) = t + t + t + t + 7

7 E 7 The acceleration of a particle is described by a( t) = t 6t +. Find the distance equation for (t) if v() = 0 and () =. ( ) v(t) = a t dt = t 6t + dt = t t + t + c 0 = + + c 5 = c v(t) = t t + t 5 ( ) (t) = v t dt = t t + t 5 dt = t t + t 5t + c = 6 = c c ( t) = t t + t 5t + 6 7

8 The proof of the Transcendental Integral Rules can be left to a more formal Calculus course. But since the integral is the inverse of the derivative, the discovery of the rules should be obvious from looking at the comparable derivative rules. Derivative Rules d sin u = ( cos u) du d cos u = ( sin u) du d tan u = ( sec u) du d eu = du eu d au = au Lna du d sin- u = u D u d cos- u = u D u d tan- u = u + D u d csc u d sec u d cot u d Ln u = u d Log a u = = ( csc u cot u) du = ( sec u tan u) du = ( csc u) du du u Ln a du d csc- u = u u D u d sec- u = u u D u d cot- u = u + D u 7

9 Transcendental Integral Rules ( cos u ) du = sin u + c ( sin u ) du = cos u + c ( sec u ) du = tan u + c ( e u ) ( a u ) du = e u + c du = au Ln a + c ( csc u cot u ) du = csc u + c ( sec u tan u ) du = sec u + c ( csc u ) du = cot u + c u du = Ln u + c du u = sin u + C du + u = tan u + C du u u = sec u + C Note that there are only three integrals that yield inverse trig functions where there were si inverse trig derivatives. This is because the other three rules derivative rules are just the negatives of the first three. As we will see later, these three rules are simplified versions of more general rules, but for now we will stick with the three. E 8 E 9 ( sin + cos ) ( sin + cos ) = ( sin ) + ( cos ) = cos + sin + c ( e + + csc ) ( e + + csc ) = ( e ) + + ( csc ) = e + cot + c 75

10 Trig Inverse Integral Rules du a u = sin u a + C du u + a = a tan u a + C du u u a = u a sec a + C E 0 If u + u + = tan u + C E If = sec ( sec + tan ), find y if y( 0) = 0. ( ) y = sec sec + tan = sec + sec tan 0 = tan 0 + sec 0 + c 0 = 0 ++ c c = = tan + sec + c y = tan + sec 76

11 . Homework Set A Perform the Anti-differentiation ( + 5) ( + 8) ( +) 0. ( ) Solve the initial value problems.. f ' = 6 +. Find f() if f(0) =.. f ' = + +. Find f() if f() = f ' = ( ) ( +). Find f() if f() =. 6. The acceleration of a particle is described by a( t) = 6t t + 8. Find the distance equation for (t) if v() = and () =. 7. The acceleration of a particle is described by a t = t t +. Find the distance equation for (t) if v(0) = and (0) =. 77

12 . Homework Set B Perform the Anti-differentiation.. ( ) y t + 7t + 9 dt t + 7. ( t +) ( t + 7)dt

13 .: Integration by Substitution--the Chain Rule The other three derivative rules--the Product, Quotient and Chain Rules--are a little more complicated to reverse than the Power Rule. This is because they yield a more complicated function as a derivative, one which usually has several algebraic simplifications. The Integral of a Rational Function is particularly difficult to unravel because, as we saw, a Rational derivative can be obtained by differentiating a composite function with a Log or a radical, or by differentiating another rational function. Reversing the Product Rule is as complicated, though for other reasons. We will leave both these subjects for a traditional Calculus Class. The Chain Rule is another matter. Composite functions are among the most pervasive situations in math. Though not as simple at reverse as the Power Rule, the overwhelming importance of this rule makes it imperative that we address it here. Remember: The Chain Rule: d f ( g() ) = f ' ( g() ) g'() The derivative of a composite turns into a product of a composite and a noncomposite. So if we have a product to integrate, it might be that the product came from the Chain Rule. The integration is not done by a formula so much as a process that might or might not work. We make an educated guess and hope it works out. You will learn other processes in Calculus for when it does not work. Integration by Substitution (The Unchain Rule) 0) Notice that you are trying to integrate a product. ) Identify the inside function of the composite and call it u. ) Find du from u. ) If necessary, multiply a constant inside the integral to create du, and balance it by multiplying the reciprocal of that constant outside the integral. (See EX ) ) Substitute u and du into the equation. 5) Perform the integration by Anti-Power (or Transcendental Rules, in net section.) 6) Resubstitute the -equivalent for u. 79

14 This is one of those mathematical processes that makes little sense when first seen. But after seeing several eamples, the meaning suddenly becomes clear. Be Patient. OBJECTIVE Use the Unchain Rule to integrate composite, product epressions. 0 E + 5 ( + 5) 0 is the composite function. u = = u 0 du = u + c du = E + 5 ( + 5) is the composite function. So u = = ( + 5) ( ) = u du = u + c + 5 du = 80

15 E (( + ) + 5 ) + 5 is the composite function. So u = + 5 = ( + ) du = + (( + ) + 5 ) = + 5 ( + ) = u du u = du = u c = ( 5 + 5) 5 + c E + 5 ( ) + 5 ( ) u = du = ( + 5) = = u - du = u + c = c 8

16 Of course, the Unchain Rule will apply to the transcendental functions quite well. E 5 ( sin 5 ) ( sin 5 ) = sin = 5 sin u du u = 5 du = 5 = ( 5 cos u ) + c E 6 ( sin 6 cos ) ( sin 6 cos ) = ( u6 ) du = 7 u7 + c = cos 5 + c 5 u = sin du = cos = 7 sin7 + c E 7 5 sin 6 ( 5 sin 6 ) = 6 u = 6 du = 6 5 ( sin 6 ) 6 5 = 6 sin u du = 6 cos u + c = 6 cos 6 + c 8

17 E 8 ( cot csc ) ( cot csc ) = cot u = cot du = -csc ( csc ) du = u = u + c = cot + c E 9 cos u = = du = - = cos = ( cos ) = cos u du = sin u + c = sin + c 8

18 E 0 e + e + d = + e = eu du = eu + c u = + du = = e + + c E u = du = = = du u = sin u + c = sin + c 8

19 E ( e + sin5 ) ( e + sin5 ) = ( e ) + ( ) + ( sin5 ) = e + 5 sin5 5 u = du = u = 5 du = 5 = eu du + 5 sinu du = eu + ( 5 cosu )+ c = e cos5 + c 85

20 . Homework Set A Perform the Anti-differentiation ( 5 + ) ( sin + e ) 0. sec ( )+ ln. cos 5. sin 7 + ( ) ( ). sec. 5. tan sec e Ln ( +) + 0. sin Ln cos sin e. cot csc. sin cos 86

21 .. + cos sin. Homework Set B ( ) t t 5 + dt. 0m +5 dm. m + m + ( + ) 5 5. ( s +) 5 ds 6. 5t t + dt 7. m m + 8 dm 8. ( 8 +) v 5 v dv 0. ( 5 sin + e ) cos + sin. sec + e. sec. e csc e sec( ln )tan( ln ) e + sec e csce cote 8. ( e ) ( e ) 87

22 9. sec y tan 5 y. Verify that your integration is correct by taking the derivative of your answer. 0. cosθe sinθ + θ θ + dθ. Verify that your integration is correct by taking the derivative of your answer. tan( t ) dt. tsec t. Verify that your integration is correct by taking the derivative of your answer.. sin. te 5t + dt. ( e y +) 5. sec tan 6. sin( t )cos 5 ( t )dt 7. cos e sin 8. tanθ ln( secθ )dθ 9. ( e y + y 7cosy) 0. sin + cos sec + e π. e t sec e t dt. 8lnm m dm. sin y ycos y 88

23 . Separable Differential Equations Vocabulary: Differential Equation (differential equation) an equation that contains an unknown function and one or more of its derivatives. General Solution The solution obtained from solving a differential equation. It still has the +C in it. Initial Condition Constraint placed on a differential equation; sometimes called an initial value. Particular Solution Solution obtained from solving a differential equation when an initial condition allows you to solve for C. Separable Differential Equation A differential equation in which all terms with y s can be moved to the left side of an equals sign ( = ), and in which all terms with s can be moved to the right side of an equals sign ( = ), by multiplication and division only. OBJECTIVES Given a separable differential equation, find the general solution. Given a separable differential equation and an initial condition, find a particular solution. E Find the general solution to the differential equation = y. Start here. = y y = y = Integrate both sides. Separate all the y terms to the left side of the equation and all of the terms to the right side of the equation. y = + C You only need C on one side of the equation and we put it on the side containing the. 89

24 y = + C + y = C Multiply both sides by. Note: C is still a constant, so we ll continue to note it just by C. This equation should seem familiar. It s the family of circles centered at the origin with radius C. y = ± C Isolate y. Usually, if it s possible, we will solve our equation for y so our solution can be written as y = ± C. Also to note, we could check our solution by taking the derivative of our solution. + y = C + y = 0 = y. Steps to Solving a Differential Equation:. Separate the variables. Note: Leave constants on the right side of the equation.. Integrate both sides of the equation. Note: only write the +C on the right side of the equation. Here s why you will have a constant on either of your equation when you integrate both sides of your differential Equation, but you would wind up subtracting one from the other eventually, and two constants subtracted from one another is still a constant, so we only write +C on one side of the equation.. Solve for y, if possible. If you integrate and get a natural log in your result, solve for y. If there is no natural log, solve for C. Note: e ln y = y because e raised to any power is automatically positive, so the absolute values are not necessary.. Plug in the initial condition, if you are given one, and solve for C. Note: Solve for C immediately if the left integral does not result in a Ln. Simplify before solving if there is a Ln. 90

25 E Find the general solution to the differential equation dm dt = mt. dm dt = mt Start here. dm m = tdt Separate all the m terms to the left side of the equation and all of the t terms to the right side of the equation. dm = tdt m Integrate both sides. ln m = t + C You only need C on one side of the equation and we put it on the side containing the. e ln m = e t +C e both sides of the equation to solve for y. m = e t e C m = Ke t e C is still a constant, so we will just note it as K. E Find the particular solution to y = y y, given y =. = ( ) y y = ( ) ln y = + C y = e +C = e e C = Ke = = Ke 0 y K = y = e 9

26 E Solve the differential equation = y + e y This is the general solution. = y + e y ( y + e y ) = ( ) y + e y = C y + e y 5 = C E 5 Find the particular solution to dr dt = t sint r given that r( 0) =. dr dt = t sint r rdr = ( t sint)dt r = t + cost + C = 0 + cos0 + C C = 7 r = t + cost +7 r = ± t + cost + 7 r = t + cost + 7 Why do we only need the positive r? What would have happened if you had solved for r before plugging in the initial condition? 9

27 r = t + cost + C r = ± t + cost + C = ± 0 + cos0 + C C = 7 r = t + cost + 7 Again, you can check your solution by taking the derivative of your solution. E 6 AP 998 BC # 9

28 . Homework Set A Solve the differential equation.. = y. = y. ( +) = y. = y + 5. = y 6. dt = sec t ye 5y 7. = e 8. sec y tan y y = + Find the solution of the differential equation that satisfies the given initial condition = y ; y 0 = 5 0. = y ; y = 0. = ( y +); y() = 0. du dt = t + sec t ; u(0) = u = y ; y 0 = = ( +) ( y); y = = y + y ; y ( 0 ) = = y ysin ; y( 0) = 5e 7. Solve the initial-value problem y' = sin sin y ; y(0) = π, and graph the solution. 8. Solve the equation e - y y'+ cos = 0 and graph several members of the family of solutions. How does the solution curve change as the constant C varies? 9

29 . Homework Set B Solve the differential equation.. = y. y' = y sin. dv = + v + t + tv dt. dt = t y y + 5. dθ dr = + r θ Find the solution of the differential equation that satisfies the given initial condition. 6. = ycos + y ; y ( 0 ) = 7. Find an equation of the curve that satisfies = y and whose y-intercept is Solve the initial-value problem y' = cos siny ; y π = π, and graph the solution. 95

30 .: Integration by Substitution--Back Substitution Sometimes when applying the Chain Rule, the other factor is not the du, or there are etra s that must be replaced with some form of u. The method of choosing u to equal the inside of the composite function remains the same, but there is more substitution necessary. This is best understood in an eample: E : + u = + = u du = + = = + ( u )u du = u 5 u du = u 7 u c ( ) 7 ( 5 + ) 5 + c = 7 + E : + u = = u + du = 96

31 ( +) u du ( +) = u + = ( u + )u du = u + u du + u + c 5 = u5 = ( 5 ) 5 + ( ) + c Integration by Back Substitution (The Unchain Rule) 0) Notice that you are trying to integrate a product. ) Identify the inside function of the composite and call it u. ) Find du from u. ) If necessary, multiply a constant inside the integral to create du, and balance it by multiplying the reciprocal of that constant outside the integral. (See EX ) a) Identify any etra s. b) Isolate in the scratch work ) Substitute u and du as well as the etra s into the equation. 5) Perform the integration by Anti-Power (or Transcendental Rules, in net section.) 6) Resubstitute the -equivalent for u. 97

32 E : + u = = u + du = ( + ) ( ) = ( ( u + )+ )u du = ( u + 5)u du = u 5 + 5u du = u u5 5 + c = ( 6 ) 6 + ( ) 5 + c E : + + There are two ways to approach this problem. One could use polynomial long division to simplify before integrating: Then + = u = + du = = ( )+ 8 du u = = + 8ln u + c + 8ln + + c An alternative would be to make the denominator u and use back-substitution: 98

33 E (again): = u + u u = + = u du = + = u u + + du u = u u + 8 du u = u + 8 u du = u = + u + 8ln u + c ( + )+ 8ln + + c The answer looks different from what the answer was the first time, but, with FOILing and adding like terms, it can be shown that these are the same answer. As a rule of thumb, doing algebra simplification before Calculus will generally make the problem shorter and more simple. In this case, Polynomial Long Division made the problem easier than Back-Substitution. 99

34 . Homework Set ( + ln ) ln

35 .5: Powers of Trig Functions: Sine and Cosine Another instance of Back Substitution involves the Trig Functions and the Pythagorean Identities. As we saw in previous sections, since d! cos = ( sin!) and d! sin! = ( cos ), one of these functions can serve as the du when the other serves as u. But what about when there are higher eponents involved? In general, what about: OBJECTIVE! sin n!!cos m!! Use the Integration by Substitution to integrate integrands involving Sine and Cosine. There are two cases of integration of this kind of integrand, depending on the powers m and n. Case. The easier (and more common case on the AP test) is when either m or n is an odd number. One of whichever function has the odd power will be the du and the rest of those functions can convert to the other trig function by means of the Pythagorean Identities. Remember: sin θ + cos θ = cos θ = sin θ sin θ = cos θ 0

36 E sin cos Since cos has the odd power, u = sin, du = cos, and cos = sin = u. sin cos = sin cos cos cos du = sin sin = u u = ( u u 6 ) du = 5 u5 7 u7 + c = 5 sin5 7 sin7 + c E sin 5!cos!! Since sin has the odd power, u = cos, du = sin, and sin = cos = u. sin 5 cos = sin cos ( sin ) cos ( sin ) cos sin = sin = cos = ( u ) u du = ( u + u )u du = ( u u + u 6 )du = u 5 u5 + 7 u7 + c = cos + 5 cos5 7 cos7 + c 0

37 If both powers are odd, either function can serve as u. Because of the negative sign, it is usually deemed easier to choose u = sin. E tan At first, this does not appear to be a sin/cos integral, but a basic substitution reveals it is: tan = sin cos = u du = ln cos + c = ln cos + c = ln sec + c This gives us two more integral rules: tanu du = ln secu + c cotu du = ln sinu + c The last two trig integrals, secu du and cscu du, seem to be of the same kind as these two, but proving these by sin and cos is difficult. We will give the rules here and prove them in later. secu du = ln secu + tanu + c cscu du = ln cscu cotu + c 0

38 Case. The more difficult situation is when both powers are even. In this case, variations on the half angle argument rules come into play. sin = ( cos) and cos! = + cos! E cos cos = ( + cos) = = ( + cos) ( + cosu)du = u + sinu + c = ( ) + sin + c = + sin + c This eample leads to two more integral equations that are helpful to know: cos u!du =! u + sinu + c sin u du = u sinu + c 0

39 E 5 sin cos! = ( cos) ( + cos)! + cos = 8 cos + cos! = 8 ( cos cos + cos ) u = du = = 8 ( cos cos + cos ) = 6 ( cosu cos u + cos u)!du = 6 ( )du cosu 6 du 6 ( cos u)du + 6 ( cos u)du =! 6 u 6 sinu 6 u sinu + cos u 6!cosudu + c = 6 u 6 sin!u 6 = 6 u 6 sinu u + 6 sinu + = 6 u 6 sinu u + 6 sinu + u sin!u + 6 ( sin!u)!cos!u!du + c v 6 dv + c 6 v v + c 6 sin sin + c =! 8 6 sin sin + 6 sin 8 sin + c = ( 6 ) sin ( 6 )+ sin+ 6 = sin 8 sin + c v = sinu dv = cosu 05

40 .5 Homework: Perform the Anti-differentiation.. sin cos. sin cos 5. sin cos 7. sin 5 cos 6 5. sin cos 5 6. sin 5 cos 5 7. sin cos 8. sin cos 06

41 .6: Powers of Trig Functions: Secant and Tangent As with sin and cos, Tan and Sec work together in a Pythagorean Identity and Csc and Cot work together in a Pythagorean Identity. So, we can consider integrals of these forms to be cases of Integration by Substitution. sec n tan m or csc n cot m Remember: tan θ + = sec θ sec θ = tan θ sec θ tan θ = + cot θ = csc θ csc θ = cot θ csc θ cot θ = = ( sec u) du du d tan u d cot u = csc u d csc u d sec u = ( csc u cot u) du = ( sec u tan u) du OBJECTIVE Use the Integration by Substitution to integrate integrands involving Secant and Tangent or Cosecant and Cotangent. There are three cases of integration involving these kinds of integrand, depending on the powers m and n. Case. When the Secant s (or Cosecant s) power is even, then u = tan du = sec and tan += sec or u + = sec 07

42 E sec!tan!! sec tan = sec tan sec tan sec u du = + tan = + u = ( u + u ) du = u + 5 u5 + c = tan + 5 tan5 + c Case. When the Tangent s (or Cotangent s) power is odd, then E csc!cot!! u = sec du = ( sec tan ) and tan = sec or u = tan Because cot has an odd power, it will be part of du and u = csc du = ( csc cot ) cot = u 08

43 csc cot = csc cot ( csc cot ) = u ( u )du = ( u u )du = 5 u5 + u + c = 5 csc5 + csc + c If both cases are present (that is, the tangent power is odd and the secant power is even), either function can serve as u. Case. When the Tangent s (or Cotangent s) power is even AND the Secant s (or Cosecant s) power is odd, then The problem is not an Integration-by-Substitution problem. It is an Integration-by-Parts problem. We will need to wait until Chapter 8 to do them. 09

44 .6 Homework: Perform the Anti-differentiation.. sec tan 5. sec 6 tan. sec 5 tan 7. sec tan 6 5. sec 6 tan 6. csc cot 5 7. csc cot 5 8. csc 7 cot 5 9. csc cot 0. csc cot 0

45 .7 Slope Fields This section is very much an AP-driven section. Their philosophy is that math should be understood and eplained algebraically, graphically, numerically and verbally. The topic of differential equations fits nicely into this paradigm. Vocabulary: Slope field Given any function f, a slope field is drawn by taking evenly spaced points on the Cartesian coordinate system (usually points having integer coordinates) and, at each point, drawing a small line with the slope of the function. There are four ways that the AP Eam usually approaches Slope Fields:. Draw a Slope Field (free response). Sketch the solution to a Slope Field (free response). Identify the differential equation for a Slope Field (multiple choice). Identify the solution equation to a Slope Field (multiple choice) NB The first two are usually two parts of the same problem Objectives: Given a differential equation, sketch its slope field. Given a slope field, sketch a particular solution curve. Given a slope field, determine the family of functions to which the solution curves belong. Given a slope field, determine the differential equation that it represents. Steps to Sketching a Slope Field:. Determine the grid of points for which you need to sketch (many times the points are given).. Pick your first point. Note its and y coordinate. Plug these numbers into the differential equation. This is the slope at that point.. Find that point on the graph. Make a little line (or dash) at that point whose slope represents the slope that you found in Step.. Repeat this process for all the points needed.

46 Additional Steps to Sketching a Solution Curve for an Initial Condition Problem: 5. Start by finding the initial condition point on the graph. 6. Follow the marks from the slope field in either direction from your initial condition. E Sketch the slope field for = y at the points indicated. Note that, wherever y =, = 0. So, the segments at y = will be horizontal. Also, where = 0, = dne. So, there are not segments on the -ais. We can then plug the numerical values of the points into = y the slant of the line segments. to determine (, ) = (, 0) = (, ) = etc. (, ) = (, 0) = (, ) =

47 E Sketch the slope field for = at the points indicated and sketch the particular solution given the initial condition of,. y y 5 5

48 E Sketch the slope field for = at the points indicated and sketch the particular solution given the initial condition of(, ).. We first need to solve this equation for. = y and sketch 5 What are some trends you notice? Solve this differential equation analytically. What connection can you make between the analytic solution and the graphical solution from the slope field? Solve the equation analytically for the initial condition of(,).

49 y 5 E Recall from the last section we looked at =. Below is its slope field. y y 5 What was its analytical solution (from last section)? Again, note the connection between the analytical solution that comes from separating the variables and the graphical solution that comes from the slope field. 5

50 Why even bother with Slope Fields? Why use slope fields when we can just separate variables and solve the differential equation analytically. Well, since you asked, sometimes we can t solve differential equations analytically. What if = y or = e? For the first differential equation, we can t separate the variables. For the second differential equation, there is no analytical antiderivative. But slope fields give us a graphical solution to a differential equation even is we cannot find an analytical solution. This just means that even though we don t have an actual equation for a solution, we still know what the solution looks like. Finding the differential equation for a Slope Field. ) Horizantal Dashes = 0 ) Dashes \ < 0 ) Dashes / > 0 ) All Dashes in any column // to each other has no y 5) All Dashes in any row // to each other has no 6

51 E 5 Match the slope fields with their differential equations. (a) = y (b) = e (c) = cos (d) = y y y 5 5 I II y y 5 5 III IV I and III have all slopes in a column parallel therefore there is no y in those equations, namely, a and c. I also appears to have horizontal asymptotes. Therefore, I = (a) and III = (c). II has positive slopes in QI and QIII and negative slopes in QII and QIV. Also, the slope = 0 at = 0 and the slope does not eist at y = 0, therefore, II = (d) That leaves IV = (b) 7

52 To find the solution to a differential equation, we need to remember what we learned last year about the general shapes of the graphs of various equations. Summary of Graphs from PreCalculus Polynomials: y = a n n + a n n... General Rational Functions: y = A n n + A n n... B m m + B m m.... y y y

53 y Radical Functions: y = n a n n a 0 : y 0 y Eponential Functions: Logarithmic Functions: y y 5 5 Trigonometric Functions: y = sin y = cos y = tan 9

54 y = csc y = sec y = cot Finding the solution equation entails looking at the pattern of the slopes and matching it against the graphs we know. E 6 Match the slope fields with their solution curve. (a) y = (b) y = + (c) y = (d) y = (e) y = (f) y = (g) y = ln (h) y = tan (i) y = sin (j) y = cos y y y I II III y y y IV V VI 0

55 y y 5 5 VII VIII y y 5 5 IX X I forms parabolas, so the only equation that could be is (c). II forms horizontal lines, so that is (a). III has a vertical asymptote at = 0, so it could be (e) or (g). But the slopes go down as you move toward = 0, therefore, III is (g). Etc.

56 .7 Homework Set A. A slope field for the differential equation y' = y y is shown. (a) Sketch the graphs of the solutions that satisfy the given initial conditions. (i) y(0) = (ii) y(0) = - (iii) y(0) = - (iv) y(0) = y 5 Match the differential equation with its slope field (labeled I-IV). Give reasons for your answer.. = y y. = y. = y y 5. = y 5 5 I II y y 5 5 III IV

57 6. Use the slope field labeled I (for eercises -5) to sketch the graphs of the solutions that satisfy the given initial conditions. (a) y(0) = (b) y(0) = 0 (c) y(0) = - 7. Sketch a slope field for the differential equation. Then use it to sketch three solution curves. y' = + y 8. Sketch the slope field of the differential equation. Then use it to sketch a solution curve that passes through the given point. y' = y ; (,0) 9(a). A slope field for the differential equation y' = y( y- ) ( y- ) is shown. Sketch the graphs of the solutions that satisfy the given initial conditions. (i) y(0) = -0. (ii) y(0) = (iii) y(0) = (iv) y(0) =. y 5 9(b). If the initial condition is y(0) = c, for what values of c is lim t y(t) finite?

58 0. Which of the following differential equations corresponds to the slope field shown in the figure below? a) = y b) = y c) = y d) = e) = + y. Which of the following differential equations corresponds to the slope field shown in the figure below? a) = y b) = 5y c) = 0 y d) = y e) = y

59 . Which of the following differential equations corresponds to the slope field shown in the figure below? a) = y b) = 5y c) = 0 y d) = y e) = y. Which of the following equations might be the solution to the slope field shown in the figure below? a. y = b. y = cos c. y = sec d. = y e. = y 5

60 . Which of the following equations might be the solution to the slope field shown in the figure below? a. y = b. y = cos c. y = sec d. = y e. = y 5. Which of the following equations might be the solution to the slope field shown in the figure below? a. y = b. y = cos c. y = sec d. = y e. = y 6

61 6. Which of the slope field shown below corresponds to = y? a) b) c) d) e) 7

62 7. Which of the slope field shown below corresponds to = y? a) b) c) d) 8

63 8. Which of the slope field shown below corresponds to y = e? a) b) c) d) 9

64 8. Which of the slope field shown below corresponds to y = sec? a) b) c) d).7 Homework Set B Draw the slope fields for the intervals, and y, 0

65 . = y. = cos π y. 5. = y. = y 6. = = ln y 7. = y 8. = ey 9. Given the differential equation, = + y a. On the ais system provided, sketch the slope field for the at all points plotted on the graph. b. If the solution curve passes through the point (0, ), sketch the solution curve on the same set of aes as your slope field. C. Find the equation for the solution curve of = y, given that y( 0) = y

66 0. Given the differential equation, = y sin π a. On the ais system provided, sketch the slope field for the at all points plotted on the graph. b. There is some value of c such that the horizontal line y = c satisfies the differential equation. Find the value of c. c. Find the equation for the solution curve, given that y = y. Given the differential equation, = y a. On the ais system provided, sketch the slope field for the at all points plotted on the graph. b. If the solution curve passes through the point (0, ), sketch the solution curve on the same set of aes as your slope field. c. Find the equation for the solution curve, given that y 0 =

67 y

68 Anti-Derivative Chapter Test. a. ( ) + c b. ( ) + c c. ( ) + c d. ( ) + c e. ( ) + c. 5 + e (5Ln) a. 5 Ln5 + 5 e5ln + c b. 5 Ln5 + 5 e5ln + c c. 5 Ln5 + e5ln 5 + c d. 5 Ln c e. 5 Ln c y 5. Shown above is the slope field for which of the following differential equations? (A) = y (B) = y (C) = (D) = (E) y = y

69 is shown in the. The slope field for a differential equation = f, y figure. Which of the following statements are true? I. All solution curves have the same slope for a given value. II. The solution curve has a horizontal asymptote as approaches. III. is undefined at =. y 5 (A) I. only (B) II. Only (C) I. and II. Only (D) II. And III. Only (E) None 5. Which of the following could be a solution curve for the slope field of? y π π π π (A) y = tan (B) y = sec (C) y = ln (D) y = sec tan (E) y = e 5

70 The curve is symmetrical about y ais At =, the solution curve has ngent with a slope of. The solution curve has a rizontal tangent at y = The slope field for a differential equation = f, y Which of the following statements are true? is shown in the figure. y 5 (A) I. only (B) II. only (C) II. and III. only (D) I. and III. only (E) I., II., and III e e. sin cos 5.. Find f, if f " = ( + ), f ' 0 = and f π = π. 5. a( t) = sin( t) describes the acceleration of a particle. Find ( t)if v π = and π =. 6

71 6. Sketch a slope field for the differential equation y' = + y on the aes provided. Then use it to sketch three solution curves. y 7

72 . Homework. + + c.. + c c c C c c c ln + c c c. + c. f = f = f = ( t) = t t + t t + 7. t = t t + t + t + 8

73 . Homework: c ( ) 5 + c c. ( 5 ) 5 + c c + c c c cos + e + c tan + ln + c 5 sin 5 + c. cos ( 7 +) + c 7 tan( ) + c. cos + c 5. 5 tan5 + c 6. Ln + c 9

74 e6 + c 8. csc + c ( Ln +) + c 0. e + c. cot + c. sin + c. tan + c. + c. Homework. y = k. y = + C. y = k +. y = C tan + C 5. y = 6. y = ± 5 ln 5 tan + C 7. y = 7. y = ± e + C 8. y = sec + + C 9. y = y = +. y = π. y = e e tan. y = tan( ). y = e 5. u = t + tant y = cos ( cos ) 0

75 8. y = ln sin + C. Homework:. 8 ( ) ( 5 )5 + c c ( + )+ 7 ln + + c ( 8 +) ( 6 +) 5 + c 5 ( + ln ) ( + ln ) + c 6. 6 ( ) + ( 5 ) 5 + c ( + ) 5 ( + ) + 8 ( + ) + c c.5 Homework:. cos + 5 cos5 + c. 5 sin5 7 sin7 + 9 sin9 + c. sin 5 sin5 + 7 sin7 9 sin9 + c. 7 cos7 + 9 cos9 cos + c

76 5. 6 cos6 + c 6. 6 sin6 sin8 + 0 sin0 + c 7. 8 sin + c sin + 8 sin + c

77 .6 Homework:. 6 tan6 + c. 9 tan9 + 7 tan7 + 5 tan5 + c. sec sec9 + 7 sec7 5 sec5 + c. 7 tan7 + c 5. 8 tan8 + tan6 + tan + c 6. 6 cot6 + c 7. 7 csc7 + 5 csc5 csc + c 8. csc + 9 csc9 7 csc7 + c 9. 5 cot5 7 cot7 + c 0. cot cot + c.7 Homework.. IV. I. III 5. II

78 B. E. B. A. E 5. B 6. A 7. D 8. B 9. D