8/11/2011 FINAL EXAM PRACTICE III Maths 21a, O. Knill, Summer 2011Name:• Start by printing your name in the above box.• Try to answer each question on the same page as the question is asked. If needed, usethe back or the next empty page for work.• Do not detach pages from this exam packet or un s t ap l e the packet.• Please try to write nea t ly. Answers which are illegible for the grader ca n not be givencredit.• No not es, books, calculators, computers, or other electronic aids are allowed.• Problems 1-3 do not r equ i r e any justifications. For the rest of the p r ob l em s you have toshow your work. Even correct answers without derivation can not be gi ven credit.• You h ave 180 minutes time to complete your work.1 202 103 104 105 106 107 108 109 1010 1011 1012 1013 1014 10Total: 1501Problem 1) ( 20 points)1)T FThe qu ad r a ti c surface x2+ y − z2= −5 is a hyperbolic paraboloid.Solution:Write i t as y − 5 = z2− x2, to see it better.2)T FThere ar e vectors ~u and ~v such that |~u ×~v| > |~u||~v|.Solution:We have a general identity |~u ×~v| = |~u||~v|sin(alpha) which contradicts the claim.3)T FR2π0R50r dθ dr is the area of a disc of radius 5.Solution:There seems nothing wrong with that. But note that the 0 to 5 integration is paired withthe dθ.4)T FIf a vector field~F (x, y) satisfies curl(F )(x, y) = Qx− Py= 0 for all points(x, y) in the plane, then~F is a gradient field.Solution:True. We have derived this from Green’s theorem.5)T FThe jerk of a parameterized curve ~r(t) = hx(t), y(t), z(t)i is p ar al l el to theacceleration if the cu r ve ~r(t) is a line.Solution:The velo ci ty, the acceleration and the jerk are all parallel on a line.6)T FThe curvature of the curve ~r(t) = h 3 sin(t), 0, 3 cos(t)i is twice the curvatureof th e curve ~s ( t) = h6 + 6 sin(t), 6 cos(t), 0i.Solution:If we scale a curve by a factor 2, its curvature is divided by 2.27)T FThe curve ~r(t) = hsin(t), t2, cos(t)i for t ∈ [0, 10π] is located on a cylinder.Solution:Indeed, one can check tha t x(t)2+ z(t)2= 1.8)T FIf a function f(x, y) has t h e property that fx(x, y) is zero for all x, y, thenf is the constant function.Solution:No, for exampl e f(x, y) = y i s also a solution too and this solution is not constant.9)T FIf the unit tangent vector~T (t) of a curve ~r(t) is always parallel to a planeΣ, th en the curve is contained in a pla n e parallel to Σ.Solution:Indeed, we never leave the plane which goes through the i n i t i al point r(0) because alsor′(t) is always par al l el to Σ and after integration, the curve r(t) has to be in the plane.10)T FIf (x0, y0) is an extremum of f (x, y) under the constraint x2+ y2= 1, thenthe sam e point is an extremum of 10f(x, y) under the same constraint.Solution:The point is a solution to the same L a gr an g e equations.11)T FAt a critical point (x0, y0) of a function f(x, y) for which fxx(x0, y0) > 0,the cri t i cal point is always a minimum.Solution:No, we also need D > 0.12)T FIf a ve ct or field~F (x, y) is a gradient field, and C is a closed curve whichlooks like a figure 8, thenRC~F ·~dr is zer o .Solution:This fol l ows from the fundamental theorem of line integrals.313)T FIf C is part of a level curve of a function f(x, y) and~F = hfx, fyi is thegradient field of f, thenRC~F ·~dr = 0.Solution:The gra d i ent field is perpendicular to the level curves.14)T FThe divergence of the grad i ent vector field~F (x, y, z) = ∇f (x, y, z) is alwaysthe zero funct io n .Solution:The di vergence of the gradient of f i s the Laplacian of f15)T FThe line integral of the vector field~F (x, y, z) = hx, y, zi along a line segmentfrom ( 0, 0, 0) to (1, 1, 1) is 3/2.Solution:By the fundamental theorem of line integrals, we can take the difference of the potentialf(x, y, z) = x2/2 + y2/2 + z2/2, wh i ch is 1/2 + 1/2 + 1/2 = 3/2.16)T FThe area of a region G can be expressed as a line integral along its bound-ary.Solution:This i s a consequence of Green’s theorem and we have seen a few examples.17)T FThe flux of the vector field~F (x, y, z) = hx, y, −zi throu gh the boundary Sof a solid ellipsoid E is equal to the volume the ellipsoid.Solution:Indeed the di vergence of the field is 1 and we can apply the divergence theorem.18)T FIf~F is a vector field in space and S is a torus surface, then the flu x ofcurl(~F ) thr ough S is 0.Solution:This i s true by Stokes theorem.419)T FIf the divergence and the curl of a vector field~F are both zero, t h en it is aconstant field.Solution:Take~F (x, y, z) = hx, −y, 0i. It has zero cu r l and zero divergence but is not constant.20)T FFor any function f, the curl of~F = grad ( f ) is the zero field h0, 0, 0i.Solution:curl(grad(f) = h0, 0, 0i is an important identity.Problem 2) (1 0 points)a) (4 points) Match the regions with t he corresponding double integralsa0.00.20.40.60.81.00.20.40.60.81.0b0.00.20.40.60.81.00.20.40.60.81.0c0.00.20.40.60.81.00.20.40.60.81.0d0.00.20.40.60.81.00.20.40.60.81.0Enter a,b,c,d FunctionR10Rxx/2f(x, y) dydxR10Ry0f(x, y) dxd yR10Rx/20f(x, y) dydxR10R1y/2f(x, y) dxdyb) (6 points) Match the parametrized or implicit surfaces with their definitions5A B CD E FEnter A-F here Function or parametrizati on~r(u, v) = hcos(u), sin(u), vi~r(u, v) = hu − v, u + 2v, 2u + 3vix2+ y2/3 + z2/3 = 1~r(u, v) = h(sin(v) + 1) cos(u), (sin(v) + 1) sin(u), viz − x + sin(xy) = 0x2+ y2− z2= 0Solution:a) a c d bb) C E A D B FProblem 3) (1 0 points)a) (4 points) Match the vector fields and curves with the correspondin g line integralIII6III IVEnter I,II,III,IV Line integralR2π0hcos(t), sin(t)i · h−sin(t), cos(t)i dtR2π0h−t, t2i · h1, 1i dtR2π0ht2, ti · h1, 2ti dtR2π0h−3 sin(t), 3 cos(t)i·h−sin(t), cos (t ) i dtb) (6 points) Fill in from following choice: ”arc length”, ”surface area”, ”chain rule”,”volume of parallelepiped”, ”area of para ll el og r am ”, ”lin e integral”, ”flux integral”, ”cur-vature”.Formula Name of formula or rule or theoremR RR|~ru×~rv| d ud vddtf(~r(t)) = ∇f(~r(t)) ·~r′(t)Rba|~r′(t)| dt|~r′(t)×~r′′(t)||~r′(t)|3|~u ·(~v × ~w)|R10R10~F (~r(u, v)) · (~ru×~rv) d ud vSolution:a) IV, I, II, IIIb) surface area, chain rule, arc length, curvature, volume of parallelepiped, flux integral.Problem 4) (1 0 points)Given the line x − 1 = y − 2 = z − 3 and the point P = (8, 4, 5). Find the equationax + by + cz = d7of th e plane which contains the line and the point.Solution:From the symmetric equation, we
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