SOLUTIONS TO PRACTICE EXAM FOR FINAL: 1/13/2002 Math 21aName:MWF9 Sasha BravermanMWF10 Ken ChungMWF10 Jake RasmussenMWF10 WeiYang QuiMWF10 Spiro KarigiannisMWF11 Vivek MohtaMWF11 Jake RasmussenMWF12 Ken ChungTTH10 Oliver KnillTTH11 Daniel Goroff• Start by printing your name in the above box and checkyour section in the box to the left.• Try to answer each question on the same page as thequestion is asked. If needed, use the back or next emptypage for work. If you need additional paper, write yourname on it.• Do not detach pages from this exam packet or unstaplethe packet.• Please write neatly. Answers which are illegible for thegrader can not be given credit. Justify your answers.• No notes, books, calculators, computers or other elec-tronic aids are allowed.• You have 180 minutes time to complete your work.1 202 103 104 105 106 107 108 109 1010 1011 1012 10Total: 130Which section-specific problem do youchoose? Check exactly one problem. Onlythis problem can be graded. If you don’tcommit yourself here, the first attemptedproblem will be graded.12a12b12c12dProblem 1) TF questions (20 points) Circle the correct letter. No justifications are needed.T FThe length of the curve r(t) = (sin(t), t4+ t, cos(t)) on t ∈ [0, 1] isthe same as the length of the curve r(t) = (sin(t2), t8+ t2, cos(t2))on [0, 1].True. This is a consequence of the chain rule:Rba|r(s(t))0| dt =Rba|r0(s(t))||s0(t)|dt =Rs(b)s(a)|r0(s)|ds.T FThe parametric surface r(u, v) = (5u −3v, u −v − 1, 5u −v −7) isa plane.True. The coordinate functions r(u, v) = (x(u, v), y(u, v), z(u, v)) are all linear.T FAny function u(x, y) that obeys the differential equation uxx+ ux−uy= 1 has no local maxima.True. If ∇u = (ux, uy) = (0, 0), then uxx= 1 which is incompatible with a local maximum,where uxx> 0 by the second derivative test.T FThe scalar projection of a vector a onto a vector b is the length ofthe vector projection of a onto b.True. By definition.T FIf f(x, y) is a function such that fx−fy= 0 then f is conservative.False. The notion of conservative applies to vector fields and not to functions.T F(u × v) · w = (u × w) ·v for all vectors u, v, w.False. While |(u ×w) ·v| and |(u ×w) ·v| are both equal to the volume of the parallelepipeddetermined by u, v and w, the sign is different. An example: for u = h1, 0, 0i, v = h0, 1, 0i, w =h0, 0, 1i, we have (u × v) · w = 1 and (u ×w) · v = −1.T FThe equation ρ = φ/4 in spherical coordinates is half a cone.False. The equation ρ = φ/4 defines a heart shaped rotational symmetric surface. The surfaceφ = c = const would define half a cone for c ∈ [0, π] ˙T FThe function f(x, y) =(xx2+y2if (x, y) 6= (0, 0)0 if (x, y) = (0, 0)is continuous atevery point in the plane.2False. Taking y = 0, we get f(x, 0) = 1/x which is discontinuous.T FR10Rx01 dydx = 1/2.True. This is the area of half of the unit square.T FLet a and b be two vectors which are perpendicular to a given planeΣ. Then a + b is also perpendicular to Σ.True. If v is a vector in the plane, then a · v = 0 and b · v = 0 then also (a + b) · v = 0.T FIf g(x, t) = f(x −vt) for some function f of one variable f(z) theng satisfies the differential equation gtt− v2gxx= 0.True. Actually one could show that g(x, t) = f(x − vt) + h(x + vt) is the general solution ofthe wave equation gtt− v2gxx= 0.T FIf f(x, y) is a continuous function on R2such thatR RDf dA ≥ 0for any region D then f(x, y) ≥ 0 for all (x, y).True. Assume f(a, b) < 0 at some point (a, b), then f(x, y) < 0 in a small neighborhood D of(a, b) and alsoR RDf dA < 0 contradicting the assumption.T FAssume the two functions f(x, y) and g(x, y) have both the criticalpoint (0, 0) which are saddle points, then f + g has a saddle pointat (0, 0).False. Example f(x, y) = x2− y2/2, g(x, y) = −x2/2 + y2have both a saddle point at (0, 0)but f + g = x2/2 + y2/2 has a minimum at (0, 0).T FIf f(x, y) is a function of two variables and if h(x, y) = f(g(y), g(x)),then hx(x, y) = fy(g(y), g(x))g0(y).False. The correct identity would be hx(x, y) = fy(g(y), g(x))g0(x) according to the chain rule.T FIf we rotate a line around the z axes, we obtain a cylinder.False. The surface could also be a one-sheeted hyperboloid or a cone.T FThe line integral of F(x, y) = (x, y) along an ellipse x2+ 2y2= 1 iszero.True. The curl Qx− Pyof the vector field F(x, y) = (P, Q) is 0. By Green’s theorem, the3line integral is zero. An other way to see this is that F is a gradient field F = ∇f withf(x, y) = (x2+ y2)/2. Therefore F is conservative: the line integral along any closed curve inthe plane is zero.T FIf u(x, y) satisfies the transport equation ux= uy, then the vectorfield F(x, y) = hu(x, y), u(x, y)i is a gradient field.True. F = (P, Q) = (u, u). From ux= uywe get Qx= Pywhich implies that F is a gradientfield.T F3 grad(f) =ddtf(x + t, y + t, z + t).False. The left hand side is a vector field, the right hand side a function.T FR10R2π/110Rπ0ρ2sin(φ) dφdθdρ = 4π/33.True. The region is a ”lemon slice” which is 1/11’th of a sphere.T FIf F is a vector field in space and f is equal to the line integral ofF along the straight line C from (0, 0, 0) to (x, y, z), then ∇f = F.False. This would be true if F were a conservative vector field. In that case, f would be apotential. In general this is false: for example if F(x, y, z) = (0, x, 0), thenRCF ·dr = x2/2 and∇f(x, y, z) = (x, 0, 0) which is different from F.4Problem 2) (10 points)Match the equations with the curves. No justifications are needed.I IIIII IVEnter I,II,III,IV here EquationIIIr(t) = (sin(t), t(2π − t))Ir(t) = (cos(5t), sin(7t))IIr(t) = (t cos(t), sin(t))IVr(t) = (cos(t), sin(6/t))5Problem 3) (10 points)In this problem, vector fields F are written as F = (P, Q). We use abbreviations curl(F ) =Qx−Pyand div(F ) = Px+Qy. When stating curl(F )(x, y) = 0 we mean that curl(F )(x, y) = 0vanishes for all (x, y). The statement curl(F ) 6= 0 means that curl(F )(x, y) does not vanish forat least one point (x, y). The same remark applies if curl is replaced by div.Check the box which match the formulas of the vectorfields with the corresponding pictureI,II,III or IV. Mark also the places, indicating the vanishing or not vanishing of curl and div. Ineach of the four lines, you should finally have circled three boxes. No justifications are needed.Vectorfield I II III IV curl(F ) = 0 curl(F ) 6= 0 div(F ) = 0 div(F ) 6= 0F(x, y) = (0, 5) X X …
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