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This is solution manual to help with Advanced Engineering Mathematics course at Bengal Engineering and Science University. It includes: Aperture, Streamlines, Stagnation, Points, Poisson, Integral, Formula, Potentials, Variables, Cauchy, Riemann, Equations

Typology: Exercises

2011/2012

1 / 10

Download Part 31 Numerics in General-Advanced Engineering Mathematics-Solution Manual and more Exercises Engineering Mathematics in PDF only on Docsity! 14. w arccosh z implies z x iy cosh w cos iw sin (iw 1_2). Along with an interchange of the roles of the z- and w-planes, this reduces the present problem to the consideration of the sine function in Sec. 17.4 (compare with Fig. 388). We now have the hyperbolas 1 where c is different from the zeros of sine and cosine, and as limiting cases the y-axis and the two portions of the aperture. 16. Team Project. (b) We have F(z) ln z ln z arg z. K 2 iK 2 iK 2 y2 cos2 c x2 sin2 c Instructor’s Manual 295 x y Section 18.4. Problem 8 x y Section 18.4. Problem 10 im18.qxd 9/21/05 1:09 PM Page 295 docsity.com Hence the streamlines are circles ln z const, thus z const. The formula also shows the increase of the potential (x, y) arg z under an increase of arg z by 2, as asserted in (b). (d) F1(z) ln (z a) (source). F2(z) ln (z a) (sink). The minus sign has the consequence that the flow is directed radially inward toward the sink because the velocity vector V is V F(z) • • • . For instance, at z a i (above the sink), V , which is directed vertically downward, that is, in the direction of the sink at a. (e) The addition gives F(z) z ln z x arg z i(y ln x2 y2) . Hence the streamlines are (x, y) Im F(z) y ln x2 y2 const. In both flows that we have added, z 1 is a streamline; hence the same is true for the flow obtained by the addition. Depending on the magnitude of K, we may distinguish among three types of flow having either two or one or no stagnation points on the cylinder wall. The speed is V F(z) F(z) j(1 ) j . We first note that V * 1 as z * ; actually, V * 1, that is, for points at a great distance from the cylinder the flow is nearly parallel and uniform. The stagnation points are the solutions of the equation V 0, that is, (A) z2 z 1 0. iK 2 iK 2z 1 z2 K 2 y x2 y2 K 2 y x2 y2 K 2 x x2 y2 iK 2 1 z i 2 x a iy (x a)2 y2 1 2 1 x iy a 1 2 1 z a 1 2 1 2 1 2 K 2 K 2 296 Instructor’s Manual im18.qxd 9/21/05 1:09 PM Page 296 docsity.com 1 0 2 0 [(3 cos )2 (8 sin )2]r dr d 20 [(3 cos )2 (8 sin )2] d [(9 0 ) (64 0 )] 2 55. 8. Set r 0. 10. Team Project. (a) (i) Polar coordinates show that F(z) z2 assumes its maximum 52 and its minimum 20 at the boundary points 6 4i and 4 2i, respectively, and at no interior point. (ii) Use the fact that e3z e3x is monotone. (iii) At z i we obtain the maximum F(z) [sin2 x sinh2 y]1/2 sinh 1 1.1752 and at z 0 the minimum 0. (b) cos z2 cos2 x sinh2 y (Sec. 13.6) shows that, for instance, in the unit disk the maximum 1 sinh2 1 1.5431 is taken at z i. (c) F(z) is not analytic. (d) The extension is simple. Since the interior R of C is simply connected, Theorem 3 applies. The maximum of F(z) is assumed on C, by Theorem 3, and if F(z) had no zeros inside C, then, by Theorem 3, it would follow that F(z) would also have its minimum on C, so that F(z) would be constant, contrary to our assumption. This proves the assertion. The fact that F(z) const implies F(z) const for any analytic function F(z) was shown in Example 3, Sec. 13.4. 12. Since z 1, the triangle inequality yields az b a b. The maximum lies on z 1. Write a aei, z ei, b bei. Choose . Then az b jaeii() beij (a b)ei a b. Answer: a b, taken at z ei(). 14. ex eb with equality only at b. Also, cos y 1 with equality only at 0 and 2, and (b, 0) and (b, 2) lie on the boundary. 16. exp (x2 y2) cos 2xy, R: z 1, x 0, y 0. Yes, (u1, v1) (1, 0) is the image of (x1, y1) (1, 0); this is typical. (u1, v1) is found by noting that on the boundary (semicircle), * eu cos (1 u2) increases monotone with u. Similarly for R. SOLUTIONS TO CHAP. 18 REVIEW QUESTIONS AND PROBLEMS, page 775 12. (20/ln 10) ln r 14. (220 ln r) /ln 4; 0 if r 1, the unit circle, which is closer to the inner circle than to the outer, reflecting the convexity of the curve of the logarithm. 16. 10 (12/) Arg z 1 2 1 2 1 2 1 2 1 Instructor’s Manual 299 im18.qxd 9/21/05 1:09 PM Page 299 docsity.com 18. , (x c)2 (y c)2 2c2, circles through the origin with center on y x. 20. 600[1 (2/) Arg z], so that the answer is 900 V and 600 V. 22. T(x, y) 50[1 Arg (z 2)] 24. 43.22°C, which is obtained as follows. We have T(r) a ln r b and at the outer cylinder, (1) T(10) a ln 10 b 20 and from the condition to be achieved (2) T(5) a ln 5 b 30. (1) subtracted from (2) gives a(ln 5 ln 10) 10, a 10/ln 1_2 14.43. From this and (1) b 20 a ln 10 53.22. Hence on the inner cylinder we should have T(2) a ln 2 b 43.22. 26. z/4 4/z 28. 25 (r cos r3 cos 3 r5 cos 5 • • •) 30. 1_3 2 4(r cos 1_4r 2 cos 2 1_9r 3 cos 3 _116r 4 cos 4 • • • ) 1 5 1 3 100 1 1 2c x y x2 y2 300 Instructor’s Manual im18.qxd 9/21/05 1:09 PM Page 300 docsity.com 301 PART E. Numeric Analysis The subdivision into three chapters has been retained. All three chapters have been updated in the light of computer requirements and developments. A list of suppliers of software (with addresses etc.) can be found at the beginning of Part E of the book and another list at the beginning of Part G. CHAPTER 19 Numerics in General Major Changes Updating of this chapter consists of the inclusion of ideas, such as error estimation by halving, changes in Sec. 19.4 on splines, the presentation of adaptive integration and Romberg integration, and further error estimation techniques in integration. SECTION 19.1. Introduction, page 780 Purpose. To familiarize the student with some facts of numerical work in general, regardless of the kind of problem or the choice of method. Main Content, Important Concepts Floating-point representation of numbers, overflow, underflow, Roundoff Concept of algorithm Stability Errors in numerics, their propagation, error estimates Loss of significant digits Short Courses. Mention the roundoff rule and the definitions of error and relative error. SOLUTIONS TO PROBLEM SET 19.1, page 786 2. 0.286403 • 101, 0.112584 • 102, 0.316816 • 105 6. 19.9499, 0.0501; 19.9499, 0.0501256 8. 99.980, 0.020; 99.980, 0.020004 10. Small terms first. (0.0004 0.0004) 1.000 1.001, but (1.000 0.0004) 0.0004 1 (4S) 14. The proof is practically the same as that in the text. With the same notation we get x y (x y) (x x) (y y) 1 2 1 2 1 2. 16. Since x2 2/x1 and 2 is exact, r(x2) r(x1) by Theorem 1b. Since x1 is rounded to 4S, we have (x1) 0.005, hence r(x1) 0.005/39.95. im19.qxd 9/21/05 1:10 PM Page 301 docsity.com