VIEWS: 129 PAGES: 14 CATEGORY: Teachers POSTED ON: 11/4/2009 Public Domain
Math 223 VECTOR CALCULUS SOLUTION FOR MIDTERM EXAM – I (09/29) September 29 (Mon), 2008 Instructor: Yasuyuki Kachi Line #: 32702. [I] (20pts) In R3 , with the coordinate system x, y, z , (1) the deﬁning equation of the plane Π passing through P = 1, 0, −1 , and having the normal vector n = 3, 6, 10 , is 3 x + 6 y + 10 z = −7. (2) Let Q = 0, −1, 0 . Then − −→ b = PQ = 0 − 1, −1 − 0, 0− −1 = − 1, −1, 1 . (3) The distance between the plane Π and the point Q is n ·b 3, 6, 10 · − 1, −1, 1 projn b = = √ n 32 + 62 + 102 1 = √ . 145 [II] (20pts) (1) We may rewrite the equation which is originally written in cylindrical coordinate system r, θ, z as 1 (∗) z = , sin 2θ 1 using Cartesian coordinate system x, y, z , as follows. First, recall sin 2θ = 2 cos θ sin θ double angle formula . Accordingly, we may “artiﬁcially” rewrite the right-hand side of the given equation (∗) as 1 1 = sin 2θ 2 · cos θ sin θ r2 = . 2 · r cos θ r sin θ Recall that the conversion formulas between the two coordinate systems are x = r cos θ, y = r sin θ, z = z. Also recall that a part of the reverse conversion formulas is x2 + y 2 = r2 . Hence we may rewrite the right-hand side of (∗), using x, y, z, as x2 + y 2 . 2xy In sum, the equation (∗) becomes x2 + y 2 z = , 2xy or the same 2xyz = x2 + y 2 . 2 (2) We may prove that the surface in R3 as in (1) is ruled , as follows: Let z0 be an arbitrary real number such that z0 > 1. Substitute z = z0 1 into the equation z = , thus obtain sin 2θ 1 z0 = , sin 2θ or, the same to say, 1 sin 2θ = . z0 1 Since < 1, there are four θs in 0 ≤ θ < 2 π. We may conﬁrm z0 this by drawing the graph of u = sin 2θ in the θ, u coordinate plane, and use the horizontal line test. For an arbitrary choice of θ = θ1 satisfying the highlighted equation, there is another θ = θ2 satisfying the highlighted equation, such that θ1 − θ2 = π . In other words, we may list the complete set of solutions to the highlighted equation as θ1 , θ2 , θ3 , θ4 , in such a way θ1 − θ2 = π, and θ3 − θ4 = π hold. Accordingly, the half-lines θ = θ1 and θ = θ2 make a pair, and the half-lines θ = θ3 and θ = θ4 make a pair, and each of the two pairs of half-lines makes a ﬂat angle, thus makes a whole line. To conclude, the intersection of the surface with the plane z = z0 is the union of two lines. This proves that the surface is ruled. 3 [III] (20pts) We make the triplet of vectors a, b, c in R3 below an orthonormal frame by ﬁlling in appropriate real numbers in the blanks: 25 144 a = , − , , 169 169 25 b = , , , 169 √ 60 2 119 c = − , , − . 169 169 We rely on the classiﬁcation of orthonormal frames in R3 : a = a2 + b2 − c2 − d2 , 2 bc + ad , 2 bd − ac , b = 2 bc − ad , a2 − b2 + c2 − d2 , 2 cd + ab , c = 2 bd + ac , 2 cd − ab , a2 − b2 − c2 + d2 , where a, b, c, d ∈ R such that a2 + b2 + c2 + d2 = 1. First, observe the ﬁrst entry of a + the second entry of b + the third entry of c = 3 a2 − b2 − c2 − d2 = 4 a2 − a2 + b2 + c2 + d2 = 4 a2 − 1 since a2 + b2 + c2 + d2 = 1 . This equals 25 25 119 69 + − = − . 169 169 169 169 In short, 69 4 a2 − 1 = − . 169 4 From this it follows 1 69 a2 = 1− 4 169 25 = . 169 Next, observe 2 b2 − 2 c2 = the ﬁrst entry of a − the second entry of b 25 25 = − 169 169 = 0. Hence b2 = c2 . Accordingly, 25 a 2 − d2 = the ﬁrst entry of a = . 169 25 Since we had a2 = , it follows d = 0. Now we substitute d = 0, 169 and also b2 = c2 , into the above expression of the triplet a , b , c : a = a2 , 2 bc, − 2 ac , b = 2 bc, a2 , 2 ab , c = 2 ac, − 2 ab, a2 − 2 b2 . In particular, √ 60 2 the third entry of a = − the ﬁrst entry of c = , 169 and also 144 the ﬁrst entry of b = the second entry of a = − . 169 5 The latter also reads 144 2 bc = − < 0. 169 This and b2 = c2 prove c = −b. Hence √ 60 2 the third entry of b = the third entry of a = , 169 and √ 60 2 the second entry of c = − the third entry of b = − . 169 To conclude, √ 25 144 60 2 a = − , 169 , 169 , 169 √ 144 25 60 2 b = − , 169 , 169 , 169 √ √ 60 2 60 2 119 c = − − − . 169 , 169 , 169 Note : More generally, if 2 2 a = cos α , − sin α , , 2 b = , cos α , , 1 c = − √ sin 2α , , cos 2α 2 6 form an orthonormal frame, then they are 2 2 1 a = cos α , − sin α , √ sin 2α , 2 2 2 1 b = − sin α , cos α , √ sin 2α , 2 1 1 c = − √ sin 2α , − √ sin 2α , cos 2α . 2 2 [IV] (20pts) (1) For three vectors a = a1 , a2 , a3 , b = b1 , b2 , b3 , c = c1 , c2 , c3 in R3 , their scalar triple product a × b · c coincides with the following quantity: a1 a2 a3 b1 b2 b3 = a1 b2 c3 + a2 b3 c1 + a3 b1 c2 c1 c2 c3 − a3 b2 c1 − a1 b3 c2 − a2 b1 c3 . This last quantity is called the determinant . The determinant possesses the following property: a1 a2 a3 c1 c2 c3 b1 b2 b3 = a1 a2 a3 , c1 c2 c3 b1 b2 b3 which can be rewritten in terms of the scalar triple product of vectors in R3 , as a ×b ·c = c ×a · b. 7 (2) For three vectors a , b and c in R3 , their vector triple product a ×b ×c coincides with a · c b − b · c a. (3) We may prove the identity a ·c b ·c a ×b · c ×d = a ·d b ·d for vectors a , b , c and d in R3 , as follows: a ×b · c ×d = c ×d ×a ·b by the formula in (1) = c · a d − d · a c ·b by the formula in (2) = c · a d · b − d · a c · b = a · c b · d − a · d b · c a ·c b ·c = . a ·d b ·d Note 1 : We may use (3) to prove 2 2 2 2 a ×b = a b − a ·b , for vectors a and b in R3 , as follows: In the formula obtained in (3), substitute c with a , and substitute d with b , and obtain 8 a ·a b ·a a ×b · a ×b = . a ·b b ·b This is the same as 2 2 a a ·b a ×b = , 2 a ·b b or 2 2 2 2 a ×b = a b − a ·b . Note 2 : An instant application of the identity in “Note 1” above is the following Cauchy–Schwarz inequality for vectors in R3 : a ·b ≤ a b . [V] (20pts) Consider the equation 2/5 2/5 −5/2 (#) r = cos θ + sin θ , written in the polar coordinate system . (1a) Substitute (#) into x = r cos θ, and y = r sin θ each: cos θ x = 5/2 , 2/5 2/5 cos θ + sin θ sin θ y = 5/2 . 2/5 2/5 cos θ + sin θ 9 (1b) x2/5 + y 2/5 2/5 2/5 cos θ sin θ = 2/5 2/5 + 2/5 2/5 cos θ + sin θ cos θ + sin θ 2/5 2/5 cos θ + sin θ = 2/5 2/5 = 1. cos θ + sin θ (2) Know that the equation (#) in the previous page is modiﬁed as 1 −6/5 −3/5 −3/5 (##) r cos θ sin θ 1 − r2 5 −1/5 −1/5 1/5 1/5 = r −2/5 cos θ sin θ − r 2/5 cos θ sin θ . Substitute x = r cos θ, and y = r sin θ in (##): 1 5 5 1 5 1 5 1 − x2 − y 2 −1 −1 x−3/5 y −3/5 = x y − x y . 5 (3) Take the 5-th power of the both sides of the identity in (2): 1 5 x−3 y −3 1 − x2 − y 2 55 −3 5 −3 5 3 5 3 5 = x−1 y −1 − x y − 5 x y − x y −1 5 −1 5 1 5 1 5 + 10 x y − x y = x−1 y −1 − x y − 5 x−2/5 y −2/5 x−1/5 y −1/5 − x1/5 y 1/5 2 + 5 x1/5 y 1/5 x−1/5 y −1/5 − x1/5 y 1/5 . 10 (4) Substitute the result of (2) into the identity you obtained in (3): 1 5 x−3 y −3 1 − x2 − y 2 55 = x−1 y −1 − x y 1 − 5 x−2/5 y −2/5 x−3/5 y −3/5 1 − x2 − y 2 5 2 1/5 1/5 1 2 2 + 5 x y x−3/5 y −3/5 1 − x − y . 5 Clean up the right-hand side: 1 5 x−3 y −3 1 − x2 − y 2 55 = x−1 y −1 − x y 1 2 − x−1 y −1 1 − x2 − y 2 + x−1 y −1 1 − x2 − y 2 . 5 Multiply x3 y 3 to the both sides of the identity: 1 5 1 − x2 − y 2 55 = x2 y 2 − x4 y 4 1 2 − x2 y 2 1 − x2 − y 2 + x2 y 2 1 − x2 − y 2 . 5 This last identity can be rewritten as 5 1 − x2 − y 2 − 3125 x2 y 2 − x 4 y 4 2 + 3125 x2 y 2 1 − x2 − y 2 − 625 x2 y 2 1 − x2 − y 2 = 0. The left-hand side is indeed a polynomial in x, and y. (5) The graph of the curve prescribed by the equation in (4) is the asteroid. It has cusps at each of 1, 0 , 0, 1 , − 1, 0 , 0, −1 . 11 [VI] (Extra 20pts) (1) The Jacobi Identity for vectors in R3 is as follows: a × b × c + b × c × a + c × a × b = 0. (2) We may calculate the Lie bracket A, B = AB − BA for matrices 0 −a3 a2 0 −b3 b2 A = a3 0 −a1 and B = b3 0 −b1 , −a2 a1 0 −b2 b1 0 as follows. First, 0 −a3 a2 0 −b3 b2 AB = a3 0 −a1 b3 0 −b1 −a2 a1 0 −b2 b1 0 − a2 b2 − a3 b3 a2 b1 a3 b1 = a1 b2 − a1 b1 − a3 b3 a3 b2 . a1 b3 a2 b3 − a1 b1 − a2 b2 Second, 0 −b3 b2 0 −a3 a2 BA = b3 0 −b1 a3 0 −a1 −b2 b1 0 −a2 a1 0 − a2 b2 − a3 b3 a1 b2 a1 b3 = a2 b1 − a1 b1 − a3 b3 a2 b3 . a3 b1 a3 b2 − a1 b1 − a2 b2 12 Accordingly, A, B = AB − BA 0 a2 b1 − a1 b2 a3 b1 − a1 b3 = a1 b2 − a2 b1 0 a3 b2 − a2 b3 . a1 b3 − a3 b1 a2 b3 − a3 b2 0 (3) We may paraphrase the result of (2) as 0 −c3 c2 A, B = c3 0 −c1 , −c2 c1 0 where a2 a3 a3 a1 a1 a2 c1 = , c2 = , c3 = . b2 b3 b3 b1 b1 b2 Agree c1 , c2 , c3 = a1 , a2 , a3 × b1 , b2 , b3 . (4) By the result of (3), the cross product for vectors in R3 and the Lie bracket for skew-symmetric 3 × 3 matrices are parallel in structure. We may use this fact to prove the Jacobi Identity as in (1), as follows: Clearly it suﬃces to prove A , B , C + B , C , A + C , A , B = O. 13 The left-hand side equals A B − B A C − C A B − B A + B C − C B A − A B C − C B + C A − A C B − B C A − A C . This is simpliﬁed as ABC − BAC − CAB + CBA + BCA − CBA − ABC + ACB + CAB − ACB − BCA + BAC = O, where we have used the associativity law for matrix multiplications: AB C = A BC . 14