3 Which is faster, going up or coming down?

Transcription

1 MATH 35, by T. Lakoba, University of Vermont 18 3 Wic is faster, going up or coming down? Suppose you trow a ball into te air. If one neglects te air resistance, ten it takes te same time for te ball to reac te igest point of its trajectory and to fall down from tat point to te Eart. Now, if one does take te air resistance into account, wic way will it take te ball longer to go, up or down? It can be sown, bot rigorously and at te intuitive level, tat it will take longer for te ball to fall tan to rise. Tis olds for any law of air resistance, as long as te motion of te ball is strictly vertical i.e., one-dimensional) and te air resistance depends only on te ball s velocity. For details see te article by F. Brauer posted online. In tis lecture, we will answer te above question in a muc restricted setting, namely: wen te air resistance is very small compared to te gravity) and, moreover, wen its magnitude is a simple linear or quadratic function of te velocity. Te reason we will study tis restricted setting is tat we will illustrate some basic steps of te perturbative approac to solving algebraic and differential equations. We will also see some applications of te Taylor series expansion of functions. 3.1 Te exact model Let a ball wit mass m be moving vertically up or down), and its initial velocity pointing up) be v 0. We assume tat te only two forces acting on te ball are te gravity and te air resistance, wit te latter being directed oppositely to te direction of te ball s motion. Projecting Newton s Second Law m a = F on te y-axis, we ave te following equations see te figure above). Going up, v > 0: m dv dt = mg F air ; Going down: v < 0: m dv dt = mg + F air. 3.1a) 3.1b)

2 MATH 35, by T. Lakoba, University of Vermont 19 We begin by taking F air to be proportional to te first power of te velocity, i.e. F air = D v. 3.) It is known tat for small bodies e.g., a tennis ball) wit not very large velocities e.g., falling from te roof of a few-story building), 3.) is a good model. On te oter and, for larger objects aving larger velocities e.g., a skydiver or a paracutist), te air resistance is proportional to te square of te velocity: F air = Dv. 3.3) See Sections and 3 of te article by L. Long and H. Weiss posted online, if you want to see some pysical arguments beind tis. For now, we focus on model 3.). In more detail, it can be written as: Substitution of tis into 3.1) yields: v > 0 F air = D v > 0) v < 0 F air = D v) > 0). m dv dt = mg Dv 3.4) for bot te upward and downward motions. It is customary to nondimensionalize equations. For 3.4), we do tis in two steps. First, dividing by m yields dv dt = g D m v. Second, let us introduce a new variable τ = gt. Ten, using te Cain Rule, we ave: dv dt = dv dτ dτ dt = dv dτ g. Substituting tis into te previous equation and cancelling by g, we obtain: dv dτ = 1 D mg v. Finally, we use te notation introduced in Lecture : Ten te nondimensional form of Eq. 3.4) is: dv dτ v. v = 1 v, 3.5) were D/mg). Note tat due to te cange of variables from t to τ, te usual equation dy dt = v now as te form ẏ = 1 v. 3.6) g Te initial conditions for 3.5) and 3.6) are: v0) = v 0, y0) = y )

3 MATH 35, by T. Lakoba, University of Vermont 0 In te specific case we are considering, y 0 = 0.) Wen te air resistance is absent = 0 in 3.5)), we get te usual formulae from 3.5), 3.6), and 3.7): v = v 0 τ 3.8a) y = y v 0 τ τ ). g 3.8b) Verify tat tey are equivalent to te form you are used to.) Now let us consider te case 0 > 0). Equation 3.5) can be solved by separation of variables: dv = 1 v dτ dv = τ + C C = const) 1 + v 1 ln1 + v) = τ + C verify). Using te first initial condition for v in 3.7), we find C: Finally, we solve for v to obtain verify): C = 1 ln1 + v 0). v = v0 )e τ 1 ). 3.9) Using now 3.6) and te second initial condition in 3.7), we obtain verify): y = y v 0 ) 1 e τ g ) τ. 3.10) Question: Answer: We ave obtained an answer. How do we know tat we aven t made a mistake? Use sanity ceck verify if te limiting cases make sense. Tere are two limiting cases: 1 very large) and 1 very small). In te former case, we expect tat te ball will go up by a small amount only wy?), and in te latter case we expect tat te answer is close to tat given by 3.8). 1 For any fixed τ i.e. wen τ is in no way related to ), 3.10) yields: y 1 = y g 1 + v 0 1 e } {{} v 0 τ ) } {{ } 1 0 τ y g v 0 τ), i.e., indeed, te elevation of te ball is very small. Before we consider te oter limiting case, let us introduce a new notation. Suppose ε is a small number: ε 1. Ten one says tat a function fε) is Oε) if fε) lim ε 0 ε = const 0.

4 MATH 35, by T. Lakoba, University of Vermont 1 For example: ε + 100ε = Oε), Similarly, one defines Oε ), Oε 3 ), etc: sin ε = Oε), 15ε + 4ε = Oε). Oε n ) lim ε 0 ε n = a nonzero number. Te following aritmetic rules apply to te O-notation: Oε) + Oε) = Oε) Oε) Oε) = Oε) const Oε) = Oε) Oε) ± Oε ) = Oε) ε Oε) = Oε ) 1 Oε) ε = O1). Te generalization of tese rules to oter Oε n ) is obvious. Tus, returning to our analysis, we can say tat wen 1, ) 1 y y 0 = O. Now let us consider te oter limiting case. 1 Recall tat we want to confirm tat in te limit 0, Eq. 3.10) reduces to Eq. 3.8b). To tis end, use te Maclaurin series for e x, e x = 1 + x 1! + x! + x3 3! +..., to expand te expression in te large parenteses in 3.10) up to O ) you will see wy in a moment): 1 + v 0 ) 1 1 τ + τ + O 3 )) τ = 1 + v 0 ) τ ) + O ) τ = τ τ + O ) + v 0 τ + O ) + O 3 ) τ = v 0 τ τ ) + O ). Substituting tis back into 3.10), we find: y = y v 0 τ τ ) g ) + O ) = y v 0 τ τ ) + O). g

5 MATH 35, by T. Lakoba, University of Vermont Tus, indeed, te O1)-term in te above expression coincides wit 3.8b). Tis indicates tat our answer 3.10) is, most likely, correct. To conclude tis section, let us answer te question posed in te title of tis lecture, for model 3.5) and ence for its solution 3.9), 3.10)). It is easy to find te time, τ m, needed for te ball to reac te maximum elevation: just set v = 0 in 3.9). Ten verify): τ m = 1 ln1 + v 0). 3.11) However, it is not possible to find analytically te time τ 1 wen te ball its te ground, because it is not possible to solve analytically te transcendental equation 3.10) wose l..s. is set to 0. Noneteless, we can still answer our question by evaluating y m ). Indeed, if y m ) > 0, ten te ball is still in te air wen τ = τ m + τ m, i.e. going down is slower tan going up. If, on te oter and, y m ) < 0, ten te ball as already it te ground before τ = τ m + τ m, so tat in tis case we would conclude tat going down is faster. So, we compute setting y 0 = 0): y m ) = v 0 ) 1 ) e τm ) τ m g use 3.11) = v0 g v 0 ) ) ) ln1 + v 0). Let us denote 1 + v 0 x> 1). Ten y m ) = 1 g x 1 x ln x) verify). Te r..s. of tis expression is a function of x, fx): fx) = x 1 ln x. x It is easy to see tat: f1) = 1 1 ln 1 = 0, and f x) = x x = 1 x) 1 > 0, for x > 1. Terefore, fx) increases, and so fx) > 0 for x > 1. Tus, y m ) > 0, and ence it takes te ball longer to fall down tan to go up. 3. Perturbative treatment of model 3.5) Te perturbative treatment of te above model can be motivated by two different observations. First, we note tat expression 3.10) is rater cumbersome. In practice, te air resistance is quite small, and so all we really need is te first-order correction to te equations of motion 3.8) witout te air resistance. To obtain suc a correction, we need to expand 3.10) keeping iger orders of tan we did above wen considering te limiting case 1.

6 MATH 35, by T. Lakoba, University of Vermont 3 Second, note tat te air resistance could ave been given by a more complicated function of v tan 3.), in wic case it would not be possible to obtain an exact analytical solution for te counterpart of model 3.5). Yet, as long as te air resistance is small, we could ope to find an approximate solution of tat model as being a perturbation of te solution 3.8) witout te air resistance. Below we illustrate bot approaces, starting wit te first one mentioned above. Let us seek a representation of solution 3.10) as: y) = y 0) ) + y 1) ) + O ), were y 0) ) is te resistance-free solution 3.8b) and y 1) is te first-order correction to it. To find y 1), we repeat te calculations done before Eq. 3.11), but keep one more power of : y) = v 0 ) 1 1 τ + ) τ 3 τ 3 + O 4 )) 6 τ g = v 0 ) τ ) ) + τ 3 + O 3 ) τ g 6 = 1 τ + τ 3 + O 3 ) + v 0 τ v 0 τ g 6 = 1 v 0 τ τ ) + v 0τ g + τ ) ) 3 + O 3 ) 6 = 1 v 0 τ τ ) + v 0τ g + τ ) ) 3 + O ). 6 Tus, we ave obtained tat y) = 1 v 0 τ τ ) + g ) + O 3 ) τ v 0τ + τ ) ) 3 + O ). 3.1) 6 Question: Wen is tis perturbative solution valid? Answer: Wen te correction term is muc smaller tan te resistance-free term, i.e. v 0τ + τ ) 3 6 v 0τ τ. 3.13) For 3.13) to old, it suffices tat eac term on te l..s. be smaller tan eac term on te r..s.. Tis occurs wen verify) v 0 1, τ 1, and τ v ) Let us sow tat all tree of tese inequalities are equivalent. Indeed, we are interested in te times wen te ball is in te air, i.e. τ < τ m. One can alternatively write tis as τ τ m, were te symbol means equals in te order of magnitude sense. E.g., 1 or 1 3, i.e. tis new notation allows us to ignore a factor of order in our formulae. 1 Next, we estimate τ m from 3.11). Using te Maclaurin series for ln1 + x): ln1 + x) = x x + x3 3..., ln1 + x) = x + Ox ) for x 1, 1 A legitimate question to ask would be: Is 1 10? An answer depends on particular circumstances. E.g., 1 10 if we compare bot tese numbers wit 1000, but 1 10 if we compare tem wit 0.

7 MATH 35, by T. Lakoba, University of Vermont 4 we see tat v 0 1 ln1 + v 0 ) = v 0 + O ). From te last expression and 3.11), it follows tat wic finally leads to v 0 1 τ m = v 0 + O), τ v 0. Wit te above estimate, it is now clear tat all te tree strong inequalities in 3.14) are equivalent. Now, let us explore te second venue, described at te beginning of tis section. Consider model 3.5) were te term v 1. Note tat tis is precisely te condition under wic te perturbative solution is valid; owever, in tis case, it arises from purely pysical consideration tat te air resistance be small compared to te gravity. Let us seek te solution v) in te form: v = v 0) + v 1) + v ) +..., 3.15) were v 0), v 1), v ), etc. do not depend on. Substituting 3.15) into 3.5) we obtain: v 0) + v 1) + O ) = 1 v 0) + v 1) + O ) ). Let us now collect te terms at te like powers of : at 0 : v 0) = 1. Tis is te equation for te resistance-free case, as expected. Wit te initial condition from 3.7), we ave: v 0) = v 0 τ wic, of course, is 3.8a)). Next, at 1 : v 1) = v 0) see te equation above) v 1) = v 0 + τ v 1) = v 1) = 0) v 0 τ + τ = v 0 τ + τ. Here we ave again used te initial condition 3.7), wic implies tat since v = 0) = v 0 and since we ave taken v 0) 0) = v 0, ten v n) 0) = 0 for n = 1,,.... Substituting te expressions for v 0) and v 1) into 3.15), we find: v = v 0 τ) v 0 τ τ ) + O ), 3.16a)

8 MATH 35, by T. Lakoba, University of Vermont 5 and ence using y 0 = 0 and omitting te O )-term): y = 1 v 0 τ τ ) g v0 τ τ 3 6 )). 3.16b) Tis is te same as 3.1), as it sould be. Tus, we ave sown tat te same perturbative solution can be obtained by two independent approaces: by Taylor-expanding te exact solution and by perturbatively solving te model, Eq. 3.5). In most practical cases, wen te exact solution is not available, te second approac may be te only one tat can give an approximate solution. To conclude tis section, let us find te perturbative expressions for te times of going up and down, and tereby confirm our earlier conclusion tat going down takes longer. First, from 3.16a), we find te time of going up as te particular value of τ wen v = 0: 0 = v 0 τ m v 0 τ m + τ m, 3.17) were we ave omitted te O )-term. Tis is a quadratic equation for τ and can be solved exactly. However, a muc easier approac is to seek te solution τ m in te form similar to 3.15): τ m = τ m 0) + τ m 1) + O ). 3.18) Substituting 3.18) into 3.17), we obtain: 0 = v 0 m 0) + τ m 1) + O )) v 0 m 0) + τ m 1) + O )) + 0) m + τ m 1) + O )). Collecting terms at like powers of : at 0 : at 1 : Tus, 0 = τ 1) m v 0 τ 0) m + 0 = v 0 τ 0) m τ 0) m = v 0. 0) m ) τ 1) m = v 0 τ 0) m + 0) m ) = v 0. τ m = v 0 v 0 + O ). 3.19) Verify tat tis agrees wit te first two terms of te expansion of 3.11) wen 1 use te Maclaurin series stated after Eq. 3.14)). Now let us use te same metod to find te time, τ, wen te ball its te ground. Substituting into 3.16b) wit y = 0 an expansion we find, omitting O ) terms: τ = τ 0) + τ 1) + O ), 0 = v 0 0) + τ 1) ) 1 0) + τ 1) ) [ v0 0) + τ 1) ) 1 ] 0) + τ 1) 6 )3

9 MATH 35, by T. Lakoba, University of Vermont 6 Collecting te coefficients at like powers of : at 0 : at 1 : 0 = v 0 τ 0) 1 0) ) τ 0) = v 0. 0 = v 0 τ 1) 1 [ 0) τ τ 1) v0 0) ) 1 ] 0) 6 )3. IMPORTANT NOTE: Altoug te original equation for τ was nonlinear see 3.16b) wit y = 0), te equation for te correction τ 1) and for all iger-order corrections τ ), τ 3), etc., if we decide to find tem) is linear, and ence can always be solved and yields a unique solution. Continuing, from te above equation we ave: τ 1) v 0 τ 0) and, using te above expression for τ 0) : verify). Tus, ) = v 0 τ 1) = 3 v 0 0) ) 1 0) 6 )3, τ = v 0 3 v 0 + O ). 3.0) From 3.19) and 3.0), te time required for te ball to go down is: τ τ m = v 0 3 v 0 + O ) v 0 + v 0 O ) = v v 0 + O ). 3.1) Comparing 3.1) wit 3.19), we see tat te time to go down is greater tan te time to go up, as was proved in general in Section Model wit quadratic air resistance We will now follow te steps of Sections 3.1 and 3. to analyze te solution of te model wit te air resistance force given by 3.3): v > 0 v < 0 v = 1 v, v = 1 + v 3.a) 3.b) see 3.1) and 3.3)). Note tat pysically, tis model is not applicable to te motion of te ball, but te matematical perturbation approac carries over to it witout canges, and it is tis approac tat we intend to practice in tis lecture.) I will go briefly over te main steps of te solution. You will be asked to supply te missing details in te omework. We ave to analyze 3.a) and 3.b) separately, since tese are different equations. Let us begin wit 3.a). Te solution to 3.a) and 3.7) is given by: arctan v) = τ + arctan v 0 ), 3.3a)

10 MATH 35, by T. Lakoba, University of Vermont 7 y = y g ln cos [ arctan v 0 ) τ ] cosarctan v 0 )). Te exact time to reac te igest point is found from 3.3a): 3.3b) τ m = arctan v 0 ). 3.4) As in Section 3., ere our goal will be to obtain te perturbation-type solution for v, y, and τ m in two ways: by Taylor-expanding te exact solutions 3.3) and 3.4), and by perturbatively solving 3.a). Let us start wit te Taylor expansion. It is, of course, possible to solve 3.3a) for v and Taylor-expand te answer, and also to Taylor-expand 3.3b). However, tis is an awkward approac. It will be muc easier to Taylor-expand 3.3a) witout solving for v, and ten integrate te result wit respect to τ to obtain te approximate answer for y. Te reason wy suc an approac is easier is te same as wy implicit differentiation is sometimes easier tan explicit differentiation. To obtain te approximate solution of 3.3a), substitute tere expansion 3.15). Note tat te entire arguments of bot arctangents are small because 1. Ten, use te Maclaurin expansion of arctan x: to obtain tat arctan x = x x3 3 + Ox5 ), x 1 v = v 0 τ) + v 0τ + v 0 τ τ ) 3 + O ). 3.5a) 3 In doing so, you sould follow te lines of te derivation of Eqs. 3.19) and 3.0). Integrate 3.5a) to obtain wit y 0 = 0): y = 1 [v 0 τ τ g ) + v 0τ + v 0τ 3 3 τ ) ] 4 + O ). 3.5b) 1 Wen are te perturbative expansions 3.5a) and 3.5b) valid? See a similar discussion after Eq. 3.1). Finally, from 3.4), obtain: τ m = v 0 v O ). 3.6) Tus, you ave found te perturbative solution 3.5) and 3.6) by Taylor-expanding te exact solution 3.3) and 3.4). Now follow te approac presented after Eq. 3.15) to re-obtain tese results by te oter metod considered in Section 3.. First, substitute expansion 3.15) into 3.a) to re-obtain 3.5a). Ten, to conclude te treatment of te upward motion of te ball, re-obtain 3.6) starting wit 3.5a). Use expansion 3.18) for τ m and follow te approac presented after tat equation. Now, turn to te downward motion of te ball, described by Eq. 3.b). As before, begin by finding its exact solution. In an implicit form, it is: 1 ln 1 + v 1 = τ + C, v 3.7a)

11 MATH 35, by T. Lakoba, University of Vermont 8 were C is te integration constant. At τ = τ m i.e. wen te ball is at te igest point of its trajectory), v = 0. From tis condition, deduce te value of C, by setting v = 0 and τ = τ m in 3.7a). Next, solve 3.7a) for v to obtain: v = 1 1 exp C)) 1 + exp C)). 3.7b) Pysically, wat does te minus sign in front of tis expression tell you?) To conclude solving for v, transform 3.7b) into: v = 1 tan C)). 3.8a) To obtain y), integrate 3.8a) and use te intial condition tat at τ = τ m, te ball is at its igest elevation. Let us denote tis elevation y m. Ten, obtain from 3.8a) tat y = y m 1 [ g ln cos ] C)). 3.8b) Finally, find y m from 3.3b) assuming tat y 0 = 0). Wen you put all tese results togeter, your answer for y) sould be equivalent to y = v g ln 0 τ ). 3.9) cos arctan v0 ) Interestingly enoug, unlike for te model wit te linear in v) air resistance, for te model wit te quadratic in v air resistance, it is possible to find te analytic expression for te time wen te ball its te ground. By setting y = 0 in 3.9), sow tat tis time, τ, is given by: τ = τ m + 1 ln Here you need to make use of te identity 1 + v 0 + v 0 ). 3.30) arccos x = ln x + x 1). Tis concludes finding te exact solution for te going-down case. From tis point on, repeat te steps you did for te going-up case. Tat is, you will first obtain approximate solutions from te exact ones by Taylor-expanding te latter. To start, obtain te Taylor expansions of 3.7a), valid up to terms O) i.e., obtain te expression for te coefficient of te O)-term). Tis sould be done by substituting expansion 3.15) into 3.7a) and ten using te approac of obtaining Eqs. 3.19) and 3.0). Next, integrate te expression for v and obtain an O)-accurate expression for y. Ten derive an O)-accurate expression for τ from 3.30). You sould obtain te following results: v = C) ) 3 C)3 + O ), 3.31a) It is actually easier to obtain 3.31a) by Taylor-expanding te explicit solution 3.8a) tan te implicit solution 3.7a). Te reason tat you are asked to use te more difficult approac ere is tat it is more advanced and also more general. In particular, it does not rely on te possibility to solve an implicit equation, like 3.7a), for an explicit answer.

12 MATH 35, by T. Lakoba, University of Vermont 9 y = y m 1 C) C)4 g 1 τ = τ m + v 0 v3 0 6 ) + O ), 3.31b) + O ) ) ; 3.31c) a tecnical comment about te O )-term in 3.31c) is found in te footnote 3. In 3.31a) and 3.31b), substitute te expression for C wic you ave found earlier. You do not need to Taylor-expand tat expression for C in powers of.) Wen are te expansions 3.31) valid? Tis concludes te step of obtaining approximate solutions from te exact solutions 3.7) 3.30). Finally, re-obtain te approximate solutions 3.31) starting from te differential equation 3.b). To begin, re-obtain 3.31a) by substituting expansion 3.15) into 3.b). Once you ave obtained 3.31a), it can be integrated to yield 3.31b). You do not need to do tis integration ere because you ave already done it above wen obtaining 3.31b) for te first time. Now, to conclude, use te expansion for τ m ) similar to te one found after Eq. 3.19), i.e.: τ τ m = τ 0) + τ 1) + O ), 3.3) to re-obtain 3.31c) from 3.31b). Do tis in two steps. First, express te answer in terms of y m witout expanding y m in powers of. Your answer sould look like 3.3) were τ 0) and τ 1) may depend on y m. Ten, expand te expression for y m in powers of and substitute it into your expression for τ m ) to make it look like 3.31c). In so doing, you sould use a property of logaritms before expanding ln 1 + x in a Maclaurin series. 3 It is possible to sow tat it is actually O ), but tis is more difficult to do; so just sow tat tis term is no greater tan O ).