12/13/2019 FINAL PRACTICE 4 Math 21a, Fall 2019 Name: MWF 9 Oliver Knill MWF 9 Arnav Tripathy • Print your name in the above box and check MWF 9 Tina Torkaman your section. MWF 10:30 Jameel Al-Aidroos • Do not detach pages or unstaple the packet. MWF 10:30 Karl Winsor • Please write neatly. Answers which are illeg- ible for the grader cannot be given credit. MWF 10:30 Drew Zemke • MWF 12 Stepan Paul Show your work. Except for problems 1-3, we need to see details of your computation. MWF 12 Hunter Spink • No notes, books, calculators, computers, or MWF 12 Nathan Yang other electronic aids can be allowed. MWF 1:30 Fabian Gundlach • You have 180 minutes time to complete your MWF 1:30 Flor Orosz-Hunziker work. MWF 3 Waqar Ali-Shah 1 20 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 10 12 10 13 10 14 10 Total: 150 Problem 1) True/False questions (20 points). No justifications are needed. 1) T F The parametrization ~r(u; v) = [cos(u); sin(u); v] describes a cone. Solution: It is a cylinder. 2) T F The vectors ~v = [2; 1; 5] and ~w = [2; 1; −1] are perpendicular. Solution: Their dot product is zero. RR Let E be a solid region with boundary surface S. If F~ · dS~ = 0, then 3) T F S div(F~ )(x; y; z) = 0 everywhere inside E. Solution: There could be a cancellation of divergence like for F~ = [x2; 0; 0] with the unit sphere S. RR If div(F~ )(x; y; z) = 0 for all (x; y; z) then F~ ·dS~ = 0 for any closed surface 4) T F S S. Solution: This follows from the divergence theorem. If F~ is a conservative vector field in space, then F~ is has zero divergence 5) T F everywhere. Solution: The vector field F~ (x; y; z) = [x; 0; 0] is conservative but has divergence 1 everywhere. ~ ~ ~ − ~ ~ RRIf F; G areRR two vector fields for which F G = curl(H), then 6) T F ~ · ~ ~ · ~ S F dS = S G dS for any closed surface S. Solution: When adding two vector fields, the corresponding flux is added. Since the flux of the curl of a vector field through any closed surface is zero (follows from Stokes or from the divergence theorem), the flux of F~ and G~ is the same. The linearization of f(x; y) = x2 + y3 − x at (2; 1) is L(x; y) = 4 + 4(x − 7) T F 2) + 3(y − 1). Solution: Almost. It is just the constant which is wrong. The correct linearization is L(x; y) = 3 + 4(x − 2) + 3(y − 1). RR The volume of the solid E is [x; 2x+z; x−y]·dS~, where S is the surface 8) T F S of the solid E. Solution: Since the divergence of the vector field is 1, this follows from the divergence theorem. The vector field F~ (x; y; z) = [x+y; x−y; 3] has zero curl and zero divergence 9) T F everywhere. Solution: yes, both are zero. If F~ is a vector field, then the flux of the vector field curl(curl(F )) through 10) T F a sphere x2 + y2 + z2 = 1 is zero. Solution: By the divergence theorem and using the fact that the vector field is curl(G) for some other vector field G = curl(F ). If the vector field F~ has zero curl everywhere then the flux of F~ through 11) T F any closed surface S is zero. Solution: It would follows from the divergence theorem if F~ were incompressible. It would also follow that the line integral along any closed curve is zero. The equation div(grad(f)) = 0 is an example of a partial differential equa- 12) T F tion for an unknown function f(x; y; z). Solution: If we write it out, it is. 13) T F The vector (~i +~j) × (~i −~j) is the zero vector if ~i = [1; 0; 0] and ~j = [0; 1; 0]. Solution: Either compute directly [1; 1; 0] × [1; −1; 0] = [0; 0; −2] or foil out in your head to get the result is −2~i × ~j which is equal to −2~k. If f is maximal under the constraint g = c, then the angle between rf(x; y) 14) T F and rg(x; y) is zero. Solution: It is zero or π. The second case is also possible. Let L be the line x = y; z = 0 in the plane Σ : z = 0 and let P be a point. 15) T F Then d(P; L) ≥ d(P; Σ). Solution: Yes, restricting to a line can make the distance not smaller. d 0 r 0 · 00 16) T F The chain rule assures that dt f(~r (t)) = f(~r (t)) ~r (t). Solution: Yes. It is just the usual chain rule applied to the parametrization ~r 0(t). If K is a plane in space and P is a point not on K, there is a unique point 17) T F Q on K for which the distance d(P; Q) is minimized. Solution: This point Q is the projection of the point onto the plane. Every other point has larger distance as we can draw a triangle which has a right angle at Q. If B~ (t) is the bi-normal vector to an ellipse ~r(t) contained in the plane 18) T F x + y + z = 1, then B~ is parallel to [1; 1; 1]. Solution: It is always the normal vector The parametrized surface ~r(u; v) = [u; v; u2 + v2] is everywhere perpendic- 19) T F ular to the vector field F~ (x; y; z) = [x; y; x2 + y2]. Solution: Since we do not see any theoretical reason why this should be true, lets experiment. At the point [1; 1; 2], the vector field is [1; 1; 2] and the normal vector to the surface is [1; 0; 2] × [0; 1; 2] = [−2; −2; 1]. The dot product is not zero. Assume ~r(t) is a flow line of a vector field F~ = rf. Then ~r 0(t) = ~0 if ~r(t) 20) T F is located at a critical point of f. Solution: By definition ~r0(t) = F~ (~r(t)) = rf(~r(t)). At a critical point, this is zero. Problem 2) (10 points) No justifications are necessary. a) (2 points) Match the following surfaces. There is an exact match. 1 2 3 4 Surface 1-4 x2 − z2 = 0 ~r(u; v) = [u cos(v); u sin(v); uv] ~r(u; v) = [u; u2; v] x2 − y2 = z2 b) (2 points) Match the expressions. There is an exact match. Integral Enter A-D R R Type of integral b d ~ · × Ra c F (~r(u; v)) (~ru ~rv) dudv A line integral b j~r 0(t)j dt B flux integral Ra b ~ · 0 C arc length Ra RF (~r(t)) ~r (t) dt b d j × j D surface area a c ~ru ~rv dudv c) (2 points) Match the solids. There is an exact match. Solid a-d a b c d 0 ≤ x2 − y2 − z ≤ 1 x2 + y2 − z2 ≤ 1; x ≥ 0 x2 + y2 ≤ 3; y2 + z2 ≥ 1; x2 + z2 ≥ 1 x2 + y2 ≤ 1; y2 + z2 ≤ 1; x2 + z2 ≤ 1 d) (2 points) The figures display vector fields in the plane. There is an exact match. I II III IV Field I-IV F~ (x; y) = [0; 2y] F~ (x; y) = [−x; −2y] F~ (x; y) = [−2y; −x] F~ (x; y) = [−2y; 1] e) (2 points) Match the partial differential equations with formulas and functions u(t; x). There is an exact match. Equation 1-3 A-C Formulas Functions 2 2 wave 1 utt = −uxx A u(t; x) = t + t − x 2 2 heat 2 ut = uxx B u(t; x) = t + t + x 2 Laplace 3 utt = uxx C u(t; x) = x + 2t Solution: a) 4,2,1,3 b) B,C,A,D c) b,a,d,c d) IV,II,I,III e) 3,2,1 and B,C,A Problem 3) (10 points) a) (2 points) Mark the statement or statements which have a zero answer. Statement Must be zero The curl of the gradient rf(x; y; z) at (1; 1; 1) The divergence of the curl r × F~ (x; y; z) at (1; 1; 1) The flux of a gradient field rf(x; y; z) through a sphere The dot product of F~ (1; 1; 1) with the curl(F )(1; 1; 1) The divergence of a gradient field rf(x; y; z) at (1; 1; 1) b) (2 points) Two of the following statements do not make sense. Recall that \incompressible" means zero divergence everywhere and that \irrotational" means zero curl everywhere. Statement Makes no sense The discriminant of the gradient field A conservative and incompressible vector field The flux of the gradient of a function through a surface The gradient of the curl of a vector field c) (2 points) Match the following formulas with the geometric object they describe. Fill in the blanks as needed. Geometric object A-D Formula A unit tangent vector to a curve (~ru × ~rv)=j~ru × ~rvj B unit normal vector to a surface rf(x; y)=jrf(x; y)j C unit normal vector to a level curve ~r 0(t)=j~r 0(t)j D unit tangent vector to the surface (~ru + ~rv)=j~ru + ~rvj d) (2 points) All three curves in the figure are oriented counter clockwise.