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We already know that r = 31/2, so f = Arccos [(61/2/2)/31/2] = Arccos (21/2/2) = p /4 Our spherical ordered triple, listing the coordinates in the order P = (q,f,r), is therefore P = (3p /4,p /4,31/2) This is the original spherical angle in the challenge from the chapter text We ve worked the problem out in both directions without running into any trouble, so we can be confident that we didn t make any errors either way
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APPENDIX
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Worked-Out Solutions to Exercises: 11-19
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These solutions do not necessarily represent the only ways the chapter-end problems can be figured out If you think you can solve a particular problem in a quicker or better way than you see here, by all means go ahead! But always check your work to be sure your alternative answer is correct
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1 The domain of the relation shown in Fig 11-10 is set X We ve been told that the relation never maps any element of set X into more than one element of set Y Set Y contains no elements outside the co-domain Therefore, the relation is an injection The illustration shows that the relation maps elements X completely onto set Y, so the relation is a surjection Because the relation is both an injection and a surjection, it s a bijection by definition In this example, the range happens to be the same as the codomain That s not true of all relations This relation is a function, because no element in the domain maps to more than one element in the range 2 Every positive integer y in set Y (the range) has infinitely many rational numbers x from set X (the domain) assigned to it For example, if we take the integer y = 5 in set Y, it can correspond to any rational x in set X such that 4 < x 5 The relation is clearly not one-to-one, so it s not an injection For any positive integer y in set Y, we can find at least one positive rational x in set X that maps to it, so the relation is a surjection The relation is not a bijection; it would have to be both an injection and a surjection to qualify for that status If we take any positive rational number x in the domain X, we can never map it to more than one positive integer y in the range Y Therefore, our relation is a function of x 3 This relation, like the one described in Problem 2, is not one-to-one, so it isn t an injection For any positive rational number y in set Y, we can find a positive integer x in set X that maps to it, so we have a surjection The relation is not a bijection, because it isn t both an injection and a surjection If we take any positive integer x in the domain X,
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Figure B-1
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Illustration for the solution to Problem 4 in Chap 11
we can map it to infinitely many positive rationals y in the range Y Therefore, this relation is not a function of x 4 A relation whose graph is a circle or ellipse in the Cartesian xy plane can never be a function of x, because such a graph always fails the vertical-line test Figure B-1 shows several examples 5 A relation whose graph is a circle or ellipse in the polar qr plane is a function of qr if the origin is inside the circle or ellipse Figure B-2A shows a simple example in which a circle is centered at the origin in the polar plane When we graph this relation the Cartesian way as shown in Fig B-2B, we get a straight, horizontal line that passes the vertical-line test 6 We ve been given the functions f (x) = x + 2 and g (x) = 3 Their sums are ( f + g)(x) = f (x) + g (x) = (x + 2) + 3 = x + 5
516 Worked-Out Solutions to Exercises: 11-19
Figure B-2
Illustration for the solution to Problem 5 in Chap 11 At A, each radial division represents 1 unit At B, the divisions are as labeled
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