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MATH 31BH HOMEWORK 8 SOLUTIONS Problem 3.1.2 x 2 2 2 The set M = ∈ R : x + x + y = 2 is a smooth curve. y Proof. Consider the function F : R2 → R defined by x F = x + x2 + y 2 − 2. y x Now M is exactly the set of points for which F = 0. y Notice that [DF ] = [1 + 2x, 2y]. This matrix is onto so long as not both entries are equal to 0. This occurs if and only if x = − 12 and y = 0. But if x = − 12 and y = 0, then 2 1 1 1 2 2 + (0)2 = − 6= 2, x+x +y =− + − 2 2 4 so this point does not lie on M . It follows that [DF ] is onto for every point in M , and therefore by Theorem 3.1.10 M is a 1-dimensional smooth manifold. Problem 3.1.8 (a) Let Xa = {(x, y, z) ∈ R3 : x − y 2 = a} and Yb = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = b}. Let F (x, y, z) = x − y 2 − a. Then [DF ] = [1, −2y, 0], Notice that this matrix is onto regardless of the values of x, y, z, or a. Therefore, Xa is a smooth surface. Next, let G(x, y, z) = x2 + y 2 + z 2 − b. Then [DG] = [2x.2y, 2z], which is onto for every point in R3 except for (0, 0, 0). But if b 6= 0, this point does not belong to Yb . By Theorem 3.1.10 we conclude that Yb is a smooth surface for b 6= 0. When b = 0, notice that Yb = {(0, 0, 0)}. So Y0 is also a smooth manifold. 1 2 MATH 31BH HOMEWORK 8 SOLUTIONS (b) We now ask when Xa ∩ Yb is a smooth manifold. Define x − y2 − a F(x, y, z) = . x2 + y 2 + z 2 − b Then 1 −2y 0 DF] = . 2x 2y 2z We may reduce this matrix to row echelon form as 1 −2y 0 . 0 2y − 4xy 2z This matrix is surjective except for when 2y − 4xy = 0 and 2z = 0. That is, the matrix is surjective for all points except those in 1 {(x, 0, 0) : x ∈ R} ∪ , y, 0 : y ∈ R . 2 Now notice that if a point in Xa ∩ Yb may have the form (x, 0, 0) if and only if a2 = b. Similarly, one finds that a point in Xa ∩ Yb will have the form 12 , y, 0 if and only if a + b = 43 . Therefore Xa ∩ Yb is a smooth curve if a2 6= b and a + b 6= 34 . Problem 3.1.12 Let X ⊂ R3 be the set of midpoints of segments joining a point of the curve C1 of equation y = x2 , z = 0 to a point of the curve C2 of equation z = y 2 , x = 0. (a) For all real t, C1 is parameterized by the function γ1 defined by t γ1 (t) = t2 . 0 Similarly, C2 is parameterized by the function γ2 defined by 0 γ2 (s) = s s2 for all real s. (b) We may parameterize X by choosing any point on C1 and any point on C2 and calculating their midpoint. That is, X is parameterized by t γ1 (t) + γ2 (s) t22+s . γ(s, t) = = 2 2 s2 2 MATH 31BH HOMEWORK 8 SOLUTIONS 3 x 2 2 (c) If y is a point in X, then x = 2t , y = t 2+s , and z = s2 for some real t and s. z The first equation tells us that t = 2x. Using this information in the second equation, we see that s = 2y − t2 = 2y − 4x2 . Therefore, from the last equation, we find that points in X satisfy 2 1 2y − 4x2 , z= 2 or z = 2y 2 − 8x2 y + 8x4 . (d) Let f (x, y) = 2y 2 − 8x2 y + 8x4 . Then f is a C 1 mapping, and X is defined by the surface z = f (x, y). Therefore, X is a smooth surface. Problem 3.2.10 (a) cos(t) Let C be the helicoid parameterized by γ(t) = sin(t) . t The tangent line to C at the point γ(t) is given by r(s) = γ 0 (t)s + γ(t), so − sin(t) cos(t) −s sin(t) + cos(t) r(s) = cos(t) s + sin(t) = s cos(t) + sin(t) . 1 t s+t If we allow t to vary, we get the parameterization for the union X of all tangent lines to C: −s sin(t) + cos(t) τ (s, t) = s cos(t) + sin(t) . s+t Problem 3.2.6 Let X ⊆ R4 be the set of points such that x21 + x22 − x23 − x24 = 0 and x1 + 2x2 + 3x3 + 4x4 = 4. 1 0 (a) Let p = 1. Let 0 x21 + x22 − x23 − x24 F(x1 , x2 , x3 , x4 ) = . x1 + 2x2 + 3x3 + 4x4 − 4 Then F(x) = 0 for every x ∈ X, and 2x1 2x2 −2x3 −2x4 [DF] = . 1 2 3 4 4 MATH 31BH HOMEWORK 8 SOLUTIONS Now we have 2 0 −2 0 [DF(p)] = , 1 2 3 4 which is onto. So by the implicit function theorem, X is a smooth manifold in some neighborhood of p. (b) The tangent space to X at p is spanned by the null space of [DF(p)]. We have 2 0 −2 0 1 0 −1 0 [DF(p)] = → , 1 2 3 4 0 1 2 2 1 0 −2 −2 , . so the tangent space to X at p is span 1 0 0 1 (c) We notice that every pair of columns of [DF(p)] except for columns 2 and 4 are linearly independent. Therefore, by the implicit function theorem, each pair of variables besides x2 and x4 may be written as a function of the other two in some neighborhood of p. (d) We claim that X is a smooth manifold. Indeed, recall 2x1 2x2 −2x3 −2x4 1 2 3 4 [DF] = → . 1 2 3 4 0 2x2 − 4x1 −2x3 − 6x1 −2x4 − 8x1 The only way this matrix is not onto is if 2x2 = 4x1 , −2x3 = 6x1 , and −2x4 = 8x1 . But plugging these values into the first equation that defines X tells us that x1 = 0. Therefore, x1 = x2 = x3 = x4 = 0. But this point does not satisfy the second equation defining X, and therefore the point does not belong to X. It follows that [DF] is onto for every point in X, and thus by Theorem 3.1.10 X is a smooth manifold. Problem 3.1 (a) The set X ⊆ R3 of equation x3 + xy 2 + yz 2 + z 3 = 4 is a smooth surace. Proof. Let F (x, y, z) = x3 + xy 2 + yz 2 + z 3 − 4. Then F (x, y, z) = 0 if and only if (x, y, z) ∈ X. Moreover, [DF ] = 3x2 + y 2 2xy + z 2 2yz + 3z 2 . Now this matrix is not onto if and only if all of its entries are 0. Notice that the first entry is 0 if and only if x = y = 0. In this case, the second and third entries will simultaneously be 0 if and only if z = 0. Therefore this matrix is onto for all points except (0, 0, 0). But notice that (0, 0, 0) 6∈ X, because it does not satisfy the equation defining X. Thus, by Theorem 3.1.10, X is a smooth surface. MATH 31BH HOMEWORK 8 SOLUTIONS 5 (b) We have [DF (1, 1, 1)] = 4 3 5 . It follows that the tangent plane to X at (1, 1, 1) has equation 4(x − 1) + 3(y − 1) + 5(z − 1) = 0 and the tangent space to X at (1, 1, 1) has the form 4ẋ + 3ẏ + 5ż = 0. Problem 3.4 x 0 0 Consider the space X of triples p = 0 , q = y , r = 0 such that y 6= 0 and 0 0 z the segments p, q and q, r form an angle of π/4. (a) We notice that the angle between these segments is the same as the angle between ~ and qr: ~ vectors qp x 0 ~ = −y and qr ~ = −y . qp 0 z Now the cosine of this angle is just since cos(π/4) = √12 we assert that ~ qr ~ qp· ~ ~ . ||qp|||| qr|| Since we desire the angle to be π/4 and y2 1 √ =p . 2 (x2 + y 2 )(y 2 + z 2 ) Therefore any point in X will satisfy f (x, y, z) = 0 for f (x, y, z) = p y2 (x2 + y 2 )(y 2 + z2) 1 −√ . 2 (b) Notice that (1)2 1 f (0, 1, 1) = p −√ 2 (02 + 12 )(12 + 12 ) 1 1 =p −√ 2 (1)(2) = 0. We also have [Df ] = [−xy 2 ((y 2 + z 2 )(x2 + y 2 ))−3/2 (y 2 + z 2 ), 1 2y((x2 + y 2 )(y 2 + z 2 ))−1/2 − y 2 ((x2 + y 2 )(y 2 + z 2 ))−3/2 (4y 3 + 2z 2 y + 2x2 y), 2 −zy 2 ((y 2 + x2 )(z 2 + y 2 ))−3/2 (y 2 + x2 )] so √ √ # 2 2 [Df (0, 1, 1)] = 0, ,− . 4 4 " 6 MATH 31BH HOMEWORK 8 SOLUTIONS Since the derivative is onto at this point, it follows that X is a smooth manifold in some neighborhood of (0, 1, 1). (c) It is true that in some neighborhood of (0, 1, 1) that X is locally the graph of a function expressing z in terms of x and y. This follows from the implicit function theorem, since the third column of the derivative of f at (0, 1, 1) is nonzero and hence invertible. (d) The tangent plane to X at (0, 1, 1) is given by √ √ 2 2 (y − 1) − (z − 1) = 0 4 4 and the tangent space to X at this point is given by √ √ 2 2 ẏ − ż = 0. 4 4 Problem 3.7.16 Let f (x, y, z) = x + y + z. We wish to find the critical points of f constrained to the surface parameterized by sin(uv) + u g(u, v) = u + v . uv Constrained to this surface, our function takes on the values f (u, v) = (sin(uv) + u) + (u + v) + uv = sin(uv) + 2u + v + uv for all real u and v. Therefore, we wish to find the critical points of this function of two variables. We have fu = v cos(uv) + 2 + v and fv = u cos(uv) + 1 + u, so we seek to solve the system of equations ( v cos(uv) + 2 + v = 0 u cos(uv) + 1 + u = 0. By setting equal the lefthand sides of these two equations, we find that v = 2u. Substituting this into, say, the second equation tells us that any critical point (u, v) will satisfy u cos(2u2 ) + u + 1 = 0.