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HARVARD MATH 21A - Final Exam Practice II

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8/5/2010 FINAL EXAM PRACTICE II Maths 21a, O. Knill, Summer 2010Name:• 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 unstaple the packet.• Please try to write neatly. Answers which are illegible for the grader can not be givencredit.• No n ot 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 given credit.• You have 180 m i nutes 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 q u adratic surface x2+ y − z2= −5 is a hyperbolic paraboloid.Solution:Write it as y − 5 = z2− x2, t o see it better.2)T FThere are vectors ~u and ~v such that |~u ×~v| > |~u||~v|.Solution:We have a genera l 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 n ot h i n g 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 t h e 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 curve ~r(t) is a line.Solution:The velocity, the acceler at i on and the jerk are all parallel on a line.6)T FThe curvature of the curve ~r(t) = h3 sin(t), 0, 3 cos(t)i is twice the curvatureof t h 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 t hat x(t)2+ z(t)2= 1.8)T FIf a function f(x, y) has the property that fx(x, y) is zero for all x, y, thenf is the constant function.Solution:No, for example f(x, y) = y is 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, t h en the curve is contained in a plane parallel to Σ.Solution:Indeed, we never leave the plane which goes through the initial point r(0) because alsor′(t) is always parallel 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 same point is an extremum of 10f(x, y) under the same con st r a int.Solution:The point is a solution to the same Lagrange equations.11)T FAt a critical point (x0, y0) of a function f (x, y) for which fxx(x0, y0) > 0,the critical point is always a minimum.Solution:No, we also need D > 0.12)T FIf a vector field~F (x, y) is a gradient field , and C is a closed curve whichlooks like a figure 8, thenRC~F ·~dr is zero.Solution:This follows 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 g r ad 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 fun ct i on .Solution:The divergence of the gradient of f is 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, which 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 is a consequenc e of Green’s theorem and we have seen a few exam p l es.17)T FThe flux of the vector field~F (x, y, z) = hx, y, −zi through the bounda r y Sof a solid ellipso id E is equal to the volume the ellipsoid.Solution:Indeed the divergence of the field is 1 and we can a p p l y the divergence theorem.18)T FIf~F is a vector field in s p ace and S is a torus surface, then the flux ofcurl(~F ) th r o u gh S is 0.Solution:This is true by Stokes theorem.419)T FIf the divergence and the curl of a vector field~F are both zero, then it is aconstant field.Solution:Take~F (x, y, z) = hx, −y, 0i. It has zero curl 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) (10 points)a) ( 4 points) Match the regions with the 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) dxdyR10Rx/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 parametriza t io n~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) (10 points)a) ( 4 points) Match the vector fields and curves with the corresponding 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 par al le lo gr am ” , ”line 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) (10 points)Given the line x − 1 = y − 2 = z − 3 and the point P = (8, 4, 5). Find the equationax + by + cz = d7of t h e plane which contains the line and the point.Solution:From the symmetric equation, we see that the line


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HARVARD MATH 21A - Final Exam Practice II

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