How Differential Equations Arise. Newton s Second Law of Motion

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1 page 1 CHAPTER 1 Frst-Order Dfferental Equatons Among all of the mathematcal dscplnes the theory of dfferental equatons s the most mportant. It furnshes the explanaton of all those elementary manfestatons of nature whch nvolve tme. Sophus Le 1.1 How Dfferental Equatons Arse In ths secton we wll ntroduce the dea of a dfferental equaton through the mathematcal formulaton of a varety of problems. We then use these problems throughout the chapter to llustrate the applcablty of the technques ntroduced. Newton s Second Law of Moton Newton s second law of moton states that, for an object of constant mass m, the sum of the appled forces actng on the object s equal to the mass of the object multpled by the acceleraton of the object. If the object s movng n one dmenson under the nfluence of a force F, then the mathematcal statement of ths law s m dv = F, (1.1.1) where v(t) denotes the velocty of the object at tme t. We let y(t) denote the dsplacement of the object at tme t. Then, usng the fact that velocty and dsplacement are related va v=, we can wrte (1.1.1) as m d 2y = F. 2 (1.1.2) Ths s an example of a dfferental equaton, so called because t nvolves dervatves of the unknown functon y(t). 1

2 page 2 2 CHAPTER 1 Frst-Order Dfferental Equatons mg Postve y-drecton Fgure 1.1.1: Object fallng under the nfluence of gravty. Gravtatonal Force: As a specfc example, consder the case of an object fallng freely under the nfluence of gravty (see Fgure 1.1.1). In ths case the only force actng on the object s F = mg, where g denotes the (constant) acceleraton due to gravty. Choosng the postve y-drecton as downward, t follows from Equaton (1.1.2) that the moton of the object s governed by the dfferental equaton or equvalently, m d2 y = mg, (1.1.3) 2 d 2 y 2 = g. Snce g s a (postve) constant, we can ntegrate ths equaton to determne y(t). Performng one ntegraton yelds = gt + c 1, where c 1 s an arbtrary ntegraton constant. Integratng once more wth respect to t, we obtan y(t) = 1 2 gt2 + c 1 t + c 2, (1.1.4) where c 2 s a second ntegraton constant. We see that the dfferental equaton has an nfnte number of solutons parameterzed by the constants c 1 and c 2. In order to unquely specfy the moton, we must augment the dfferental equaton wth ntal conons that specfy the ntal poston and ntal velocty of the object. For example, f the object s released at t = 0 from y = y 0 wth a velocty v 0, then, n adon to the dfferental equaton, we have the ntal conons y(0) = y 0, (0) = v 0. (1.1.5) These conons must be mposed on the soluton (1.1.4) n order to determne the values of c 1 and c 2 that correspond to the partcular problem under nvestgaton. Settng t = 0 n (1.1.4) and usng the frst ntal conon from (1.1.5), we fnd that y 0 = c 2. Substtutng ths nto Equaton (1.1.4), we get y(t) = 1 2 gt2 + c 1 t + y 0. (1.1.6) In order to mpose the second ntal conon from (1.1.5), we frst dfferentate Equaton (1.1.6) to obtan = gt + c 1. Consequently the second ntal conon n (1.1.5) requres c 1 = v 0.

3 page How Dfferental Equatons Arse 3 From (1.1.6), t follows that the poston of the object at tme t s y(t) = 1 2 gt2 + v 0 t + y 0. The dfferental equaton (1.1.3) together wth the ntal conons (1.1.5) s an example of an ntal-value problem. Sprng Force: As a second applcaton of Newton s law of moton, consder the sprng mass system depcted n Fgure 1.1.2, where, for smplcty, we are neglectng frctonal and external forces. In ths case, the only force actng on the mass s the restorng force (or sprng force), F s, due to the dsplacement of the sprng from ts equlbrum (unstretched) poston. We use Hooke s law to model ths force: y 0 Mass n ts equlbrum poston y(t) Postve y-drecton Fgure 1.1.2: A smple harmonc oscllator. Hooke s Law: The restorng force of a sprng s drectly proportonal to the dsplacement of the sprng from ts equlbrum poston and s drected toward the equlbrum poston. If y(t) denotes the dsplacement of the sprng from ts equlbrum poston at tme t (see Fgure 1.1.2), then accordng to Hooke s law, the restorng force s F s = ky, where k s a postve constant called the sprng constant. Consequently, Newton s second law of moton mples that the moton of the sprng mass system s governed by the dfferental equaton m d2 y 2 = ky, whch we wrte n the equvalent form d 2 y 2 + ω2 y = 0, (1.1.7) where ω = k/m. At present we cannot solve ths dfferental equaton. However, we leave t as an exercse (Problem 7) to verfy by drect substtuton that y(t) = A cos(ωt φ) s a soluton to the dfferental equaton (1.1.7), where A and φ are constants (determned from the ntal conons for the problem). We see that the resultng moton s perodc wth ampltude A. Ths s consstent wth what we mght expect physcally, snce no frctonal forces or external forces are actng on the system. Ths type of moton s referred to as smple harmonc moton, and the physcal system s called a smple harmonc oscllator.

4 page 4 4 CHAPTER 1 Frst-Order Dfferental Equatons Newton s Law of Coolng We now buld a mathematcal model descrbng the coolng (or heatng) of an object. Suppose that we brng an object nto a room. If the temperature of the object s hotter than that of the room, then the object wll begn to cool. Further, we mght expect that the major factor governng the rate at whch the object cools s the temperature dfference between t and the room. Newton s Law of Coolng: The rate of change of temperature of an object s proportonal to the temperature dfference between the object and ts surroundng medum. To formulate ths law mathematcally, we let T(t)denote the temperature of the object at tme t, and let T m (t) denote the temperature of the surroundng medum. Newton s law of coolng can then be expressed as the dfferental equaton dt = k(t T m ), (1.1.8) where k s a constant. The mnus sgn n front of the constant k s traonal. It ensures that k wll always be postve. 1 After we stu Secton 1.4, t wll be easy to show that, when T m s constant, the soluton to ths dfferental equaton s T(t)= T m + ce kt, (1.1.9) where c s a constant (see also Problem 12). Newton s law of coolng therefore predcts that as t approaches nfnty (t ), the temperature of the object approaches that of the surroundng medum (T T m ). Ths s certanly consstent wth our everyday experence (see Fgure 1.1.3). T(t) T(t) T 0 Object that s coolng T m T m t T 0 Object that s heatng t Fgure 1.1.3: Accordng to Newton s law of coolng, the temperature of an object approaches room temperature exponentally. The Orthogonal Trajectory Problem Next we consder a geometrc problem that has many nterestng and mportant applcatons. Suppose F(x,y,c) = 0 (1.1.10) 1 If T>T m, then the object wll cool, so that dt/ < 0. Hence, from Equaton (1.1.8), k must be postve. Smlarly, f T<T m, then dt/ > 0, and once more Equaton (1.1.8) mples that k must be postve.

5 page How Dfferental Equatons Arse 5 defnes a famly of curves n the xy-plane, where the constant c labels the dfferent curves. For nstance, the equaton x 2 + y 2 c = 0 descrbes a famly of concentrc crcles wth center at the orgn, whereas x 2 + y c = 0 descrbes a famly of parabolas that are vertcal shfts of the standard parabola y = x 2. We assume that every curve n the famly F(x,y,c) = 0 has a well-defned tangent lne at each pont. Assocated wth ths famly s a second famly of curves, say, G(x,y,k)= 0, (1.1.11) y x wth the property that whenever a curve from the famly (1.1.10) ntersects a curve from the famly (1.1.11), t does so at rght angles. 2 We say that the curves n the famly (1.1.11) are orthogonal trajectores of the famly (1.1.10), and vce versa. For example, from elementary geometry, t follows that the lnes y = kx n the famly G(x,y,k)= y kx = 0 are orthogonal trajectores of the famly of concentrc crcles x 2 + y 2 = c 2. (See Fgure ) Orthogonal trajectores arse n varous applcatons. For example, a famly of curves and ts orthogonal trajectores can be used to defne an orthogonal coordnate system n the xy-plane. In Fgure the famles x 2 + y 2 = c 2 and y = kx are the coordnate curves of a polar coordnate system (that s, the curves r = constant and θ = constant, respectvely). In physcs, the lnes of electrc force of a statc confguraton are the orthogonal trajectores of the famly of equpotental curves. As a fnal example, f we consder a two-dmensonal heated plate, then the heat energy flows along the orthogonal trajectores to the constant-temperature curves (sotherms). Fgure 1.1.4: The famly of curves x 2 + y 2 = c 2 and the orthogonal trajectores y = kx. Statement of the Problem: Gven the equaton of a famly of curves, fnd the equaton of the famly of orthogonal trajectores. Mathematcal Formulaton: We recall that curves that ntersect at rght angles satsfy the followng: The product of the slopes 3 at the pont of ntersecton s 1. Thus f the gven famly F(x,y,c) = 0 has slope m 1 = f(x,y) at the pont (x, y), then the slope of the famly of orthogonal trajectores G(x,y,k) = 0sm 2 = 1/f (x, y), and therefore the dfferental equaton that determnes the orthogonal trajectores s dx = 1 f(x,y). 2 That s, the tangent lnes to each curve are perpendcular at any pont of ntersecton. 3 By the slope of a curve at a gven pont, we mean the slope of the tangent lne to the curve at that pont.

6 page 6 6 CHAPTER 1 Frst-Order Dfferental Equatons Example Determne the equaton of the famly of orthogonal trajectores to the curves wth equaton y 2 = cx. (1.1.12) Soluton: Accordng to the precedng dscusson, the dfferental equaton determnng the orthogonal trajectores s dx = 1 f(x,y), where f(x,y) denotes the slope of the gven famly at the pont (x, y). To determne f(x,y), we dfferentate Equaton (1.1.12) mplctly wth respect to x to obtan 2y = c. (1.1.13) dx We must now elmnate c from the prevous equaton to obtan an expresson that gves the slope at the pont (x, y). From Equaton (1.1.12) we have c = y2 x, whch, when substtuted nto Equaton (1.1.13), yelds dx = y 2x. Consequently, the slope of the gven famly at the pont (x, y) s f(x,y) = y 2x, so that the orthogonal trajectores are obtaned by solvng the dfferental equaton dx = 2x y. A key pont to notce s that we cannot solve ths dfferental equaton by smply ntegratng wth respect to x, snce the functon on the rght-hand sde of the dfferental equaton depends on both x and y. However, multplyng by y, we see that y dx = 2x, or equvalently, ( ) d 1 dx 2 y2 = 2x. Snce the rght-hand sde of ths equaton depends only on x, whereas the term on the left-hand sde s a dervatve wth respect to x, we can ntegrate both sdes of the equaton wth respect to x to obtan whch we wrte as 1 2 y2 = x 2 + c 1, 2x 2 + y 2 = k, (1.1.14)

7 page How Dfferental Equatons Arse 7 y 2x 2 y 2 k y 2 cx x Fgure 1.1.5: The famly of curves y 2 = cx and ts orthogonal trajectores 2x 2 + y 2 = k. where k = 2c 1. We see that the curves n the gven famly (1.1.12) are parabolas, and the orthogonal trajectores (1.1.14) are a famly of ellpses. Ths s llustrated n Fgure Exercses for 1.1 Key Terms Dfferental equaton, Intal conons, Intal-value problem, Newton s second law of moton, Hooke s law, Sprng constant, Smple harmonc moton, Smple harmonc oscllator, Newton s law of coolng, Orthogonal trajectores. Sklls Gven a dfferental equaton, be able to check whether or not a gven functon y = f(x)s ndeed a soluton to the dfferental equaton. Be able to fnd the dstance, velocty, and acceleraton functons for an object movng freely under the nfluence of gravty. Be able to determne the moton of an object n a sprng mass system wth no frctonal or external forces. Be able to descrbe qualtatvely how the temperature of an object changes as a functon of tme accordng to Newton s law of coolng. Be able to fnd the equaton of the orthogonal trajectores to a gven famly of curves. In smple geometrc cases, be prepared to provde rough sketches of some representatve orthogonal trajectores. True-False Revew For Questons 1 11, decde f the gven statement s true or false, and gve a bref justfcaton for your answer. If true, you can quote a relevant defnton or theorem from the text. If false, provde an example, llustraton, or bref explanaton of why the statement s false. 1. A dfferental equaton for a functon y = f(x)must contan the frst dervatve y = f (x). 2. The numercal values y(0) and y (0) accompanyng a dfferental equaton for a functon y = f(x) are called ntal conons of the dfferental equaton. 3. The relatonshp between the velocty and the acceleraton of an object fallng under the nfluence of gravty can be expressed mathematcally as a dfferental equaton. 4. A sketch of the heght of an object fallng freely under the nfluence of gravty as a functon of tme takes the shape of a parabola.

8 page 8 8 CHAPTER 1 Frst-Order Dfferental Equatons 5. Hooke s law states that the restorng force of a sprng s drectly proportonal to the dsplacement of the sprng from ts equlbrum poston and s drected n the drecton of the dsplacement from the equlbrum poston. 6. If room temperature s 70 F, then an object whose temperature s 100 F at a partcular tme cools faster at that tme than an object whose temperature at that tmes90 F. 7. Accordng to Newton s law of coolng, the temperature of an object eventually becomes the same as the temperature of the surroundng medum. 8. A hot cup of coffee that s put nto a cold room cools more n the frst hour than the second hour. 9. At a pont of ntersecton of a curve and one of ts orthogonal trajectores, the slopes of the two curves are recprocals of one another. 10. The famly of orthogonal trajectores for a famly of parallel lnes s another famly of parallel lnes. 11. The famly of orthogonal trajectores for a famly of crcles that are centered at the orgn s another famly of crcles centered at the orgn. Problems 1. An object s released from rest at a heght of 100 meters above the ground. Neglectng frctonal forces, the subsequent moton s governed by the ntal-value problem d 2 y 2 = g, y(0) = 0, (0) = 0, where y(t)denotes the dsplacement of the object from ts ntal poston at tme t. Solve ths ntal-value problem and use your soluton to determne the tme when the object hts the ground. 2. A fve-foot-tall boy tosses a tenns ball straght up from the level of the top of hs head. Neglectng frctonal forces, the subsequent moton s governed by the dfferental equaton d 2 y 2 = g. If the object hts the ground 8 seconds after the boy releases t, fnd (a) the tme when the tenns ball reaches ts maxmum heght. (b) the maxmum heght of the tenns ball. 3. A pyrotechnc rocket s to be launched vertcally upward from the ground. For optmal vewng, the rocket should reach a maxmum heght of 90 meters above the ground. Ignore frctonal forces. (a) How fast must the rocket be launched n order to acheve optmal vewng? (b) Assumng the rocket s launched wth the speed determned n part (a), how long after t s launched wll t reach ts maxmum heght? 4. Repeat Problem 3 under the assumpton that the rocket s launched from a platform 5 meters above the ground. 5. An object thrown vertcally upward wth a speed of 2 m/s from a heght of h meters takes 10 seconds to reach the ground. Set up and solve the ntal-value problem that governs the moton of the object, and determne h. 6. An object released from a heght h meters above the ground wth a vertcal velocty of v 0 m/s hts the ground after t 0 seconds. Neglectng frctonal forces, set up and solve the ntal-value problem governng the moton, and use your soluton to show that v 0 = 1 2t 0 (2h gt 2 0 ). 7. Verfy that y(t) = A cos(ωt φ) s a soluton to the dfferental equaton (1.1.7), where A, ω, and φ are constants wth A and ω nonzero. Determne the constants A and φ (wth φ < π radans) n the partcular case when the ntal conons are 8. Verfy that y(0) = a, (0) = 0. y(t) = c 1 cos ωt + c 2 sn ωt s a soluton to the dfferental equaton (1.1.7). Show that the ampltude of the moton s A = c c Verfy that, for t>0, y(t) = ln t s a soluton to the dfferental equaton 2 ( ) 3 = d 3 y 3.

9 page How Dfferental Equatons Arse Verfy that y(x) = x/(x + 1) s a soluton to the dfferental equaton y + d2 y dx 2 = dx + x3 + 2x 2 3 (1 + x) Verfy that y(x) = e x sn x s a soluton to the dfferental equaton 2y cot x d2 y dx 2 = By wrtng Equaton (1.1.8) n the form 1 T T m dt = k and usng u 1 du = d (ln u), derve (1.1.9). 13. A glass of water whose temperature s 50 Fstaken outsde at noon on a day whose temperature s constant at 70 F. If the water s temperature s 55 Fat2 p.m., do you expect the water s temperature to reach 60 F before 4 p.m. or after 4 p.m.? Use Newton s law of coolng to explan your answer. 14. On a cold wnter day (10 F), an object s brought outsde from a 70 F room. If t takes 40 mnutes for the object to cool from 70 Fto30 F, dd t take more or less than 20 mnutes for the object to reach 50 F? Use Newton s law of coolng to explan your answer. For Problems 15 20, fnd the equaton of the orthogonal trajectores to the gven famly of curves. In each case, sketch some curves from each famly. 15. x 2 + 4y 2 = c. 16. y = c/x. 21. y = mx + c. 22. y = cx m. 23. y 2 + mx 2 = c. 24. y 2 = mx + c. 25. We call a coordnate system (u, v) orthogonal f ts coordnate curves (the two famles of curves u = constant and v = constant) are orthogonal trajectores (for example, a Cartesan coordnate system or a polar coordnate system). Let (u, v) be orthogonal coordnates, where u = x 2 +2y 2, and x and y are Cartesan coordnates. Fnd the Cartesan equaton of the v-coordnate curves, and sketch the (u, v) coordnate system. 26. Any curve wth the property that whenever t ntersects a curve of a gven famly t does so at an angle a π/2 s called an oblque trajectory of the gven famly. (See Fgure ) Let m 1 (equal to tan a 1 ) denote the slope of the requred famly at the pont (x, y), and let m 2 (equal to tan a 2 ) denote the slope of the gven famly. Show that m 1 = m 2 tan a 1 + m 2 tan a. [Hnt: From Fgure 1.1.6, tan a 1 = tan(a 2 a). Thus, the equaton of the famly of oblque trajectores s obtaned by solvng dx = m 2 tan a 1 + m 2 tan a.] m 1 tan a 1 slope of requred famly m 2 tan a 2 slope of gven famly 17. y = cx y = cx y 2 = 2x + c. 20. y = ce x. Curve of requred famly a a 1 a 2 For Problems 21 24, m denotes a fxed nonzero constant, and c s the constant dstngushng the dfferent curves n the gven famly. In each case, fnd the equaton of the orthogonal trajectores. Curve of gven famly Fgure 1.1.6: Oblque trajectores ntersectng at an angle a.