Solução_Calculo_Stewart_6e

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F. TX.10 CHAPTER 9 REVIEW ¤ 405 (b) x and y are constant ⇒ x 0 =0and y 0 =0 ⇒ 0=0.4x(1 − 0.000005x) − 0.002xy 0=−0.2y +0.000008xy ⇒ 0=0.4x[(1 − 0.000005x) − 0.005y] 0=y(−0.2+0.000008x) The second equation is true if y =0or x = 0.2 0.000008 =25,000. Ify =0in the first equation, then either x =0 or x = 1 0.000005 =200,000. Ifx =25,000, then0=0.4(25,000)[(1 − 0.000005 · 25,000) − 0.005y] ⇒ 0=10,000[(1 − 0.125) − 0.005y] ⇒ 0 = 8750 − 50y ⇒ y =175. Case (i): y =0, x =0: Zero populations Case (ii): y =0, x = 200,000: In the absence of birds, the insect population is always 200,000. Case (iii): x =25,000, y =175: The predator/prey interaction balances and the populations are stable. (c) The populations of the birds and insects fluctuate (d) around 175 and 25,000, respectively, and eventually stabilize at those values. 25. (a) d 2 y dx 2 = k 1+ 2 dy . dx Setting z = dy dz ,weget dx dx = k √ 1+z 2 ⇒ dz √ = kdx. Using Formula 25 gives 1+z 2 ln z + √ 1+z 2 = kx + c ⇒ z + √ 1+z 2 = Ce kx [where C = e c ] ⇒ √ 1+z 2 = Ce kx − z ⇒ 1+z 2 = C 2 e 2kx − 2Ce kx z + z 2 ⇒ 2Ce kx z = C 2 e 2kx − 1 ⇒ z = C 2 ekx − 1 2C e−kx .Now dy dx = C 2 ekx − 1 2C e−kx ⇒ y = C 2k ekx + 1 2Ck e−kx + C 0 . From the diagram in the text, we see that y(0) = a and y(±b) =h. a = y(0) = C 2k + 1 2Ck + C 0 ⇒ C 0 = a − C 2k − 1 2Ck ⇒ y = C 2k (ekx − 1) + 1 2Ck (e−kx − 1) + a. Fromh = y(±b), wefind h = C 2k (ekb − 1) + 1 2Ck (e−kb − 1) + a and h = C 2k (e−kb − 1) + 1 2Ck (ekb − 1) + a. Subtracting the second equation from the first, we get 0= C e kb − e −kb k 2 − 1 e kb − e −kb = 1 Ck 2 k C − 1 sinh kb. C Now k>0 and b>0, sosinh kb > 0 and C = ±1. IfC =1,then y = 1 2k (ekx − 1) + 1 2k (e−kx − 1) + a = 1 e kx + e −kx − 1 k 2 k + a = a + 1 (cosh kx − 1). IfC = −1, k then y = − 1 2k (ekx − 1) − 1 2k (e−kx − 1) + a = −1 k e kx + e −kx + 1 2 k + a = a − 1 (cosh kx − 1). k Since k>0, cosh kx ≥ 1,andy ≥ a,weconcludethatC =1and y = a + 1 (cosh kx − 1),where k
F. 406 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS TX.10 h = y(b) =a + 1 (cosh kb − 1). Sincecosh(kb) =cosh(−kb), there is no further information to extract from the k condition that y(b) =y(−b). However, we could replace a with the expression h − 1 (cosh kb − 1), obtaining k y = h + 1 (cosh kx − cosh kb). It would be better still to keep a in the expression for y, and use the expression for h to k solve for k in terms of a, b,andh. That would enable us to express y in terms of x and the given parameters a, b, andh. Sadly, it is not possible to solve for k in closed form. That would have to be done by numerical methods when specific parameter values are given. (b) The length of the cable is L = b 1+(dy/dx)2 dx = b −b −b =2 (1/k)sinhkx =(2/k)sinhkb b 0 1+sinh 2 kx dx = b −b cosh kx dx =2 b 0 cosh kx dx
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