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HARVARD MATH 21A - Final Exam Practice IV

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8/11/2011 FINAL EXAM PRACTICE IV 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.• Justify your answers. Answers without derivation can not be given credit.• Do not detach pages from this exam packet or un s t ap l e the packet.• Please wr i t e neatly. Answers which are il l egi b l e for the grader can not be given credit.• No not es, books, calcu l at or s , computers, or other electronic aids can be allowed.• You h ave 180 minutes time to complete your work.1 202 103 104 105 106 107 108 109 1010 1011 1012 1013 1014 1015 10Total: 1601Problem 1) ( 20 points)1)T FFor any two nonzero vectors ~v, ~w the vector ~v − ~w is perpendicular to ~v × ~w.Solution:Indeed, both ~v and ~w are perpendicular. So also their difference.2)T FThe cros s product satisfies the law (~u ×~v) × ~w = ~u × (~v × ~w).Solution:Take ~v = ~w, then t h e right hand side is the zero vector while the left hand side is notzero in genera l (for example if u = i, v = j).3)T FIf the curvature of a smooth curve ~r(t) in space is defined and zero for allt, th e n the curve is part of a line.Solution:One can see that with the formula κ(t ) = |r′(t) × r′′(t)|/|r′(t)|3which shows that theacceleration r′′(t) is in the velocity direction at all times. One can also see it intuitivelyor with the defin i t io n κ(t) = |T′(t)|/|r′(t)|. If curve is not part of a line, t h en T has tochange which means that κ is not zero somewhere.4)T FThe curve ~r(t) = (1 − t)A + tB, t ∈ [0, 1] connects the point A with thepoint B.Solution:The curve is a parameterization of a li n e an d for t = 0, one has ~r(0) = A and for t = 1one h as ~r(1) = B.5)T FFor every c, the function u(x, t) = (2 cos(ct) + 3 sin(ct)) sin(x) is a sol u t i onto th e wave equation utt= c2uxx.Solution:Just d i ffe re ntiate.6)T FThe arc length of ~r(t) = (t, sin(t) ) , t ∈ [0, 2π] isR2π0q1 + cos2(t)dt.2Solution:The speed at time t is |~r′(t)| =q1 + cos2(t).7)T FLet (x0, y0) b e the maximum of f(x, y) under the constraint g(x, y) = 1.Then fxx(x0, y0) < 0.Solution:While this would be true for g(x, y) = f(y), where the constraint is a straight line parallelto th e y axes, it is false in general.8)T FThe function f(x, y, z) = x2− y2− z2decreases in the direction(2, −2, −2)/√12 at the point (1, 1, 1).Solution:It increases in that direction.9)T F~F is a vector field for which |~F (x, y, z)| ≤ 1. For every curve C : ~r(t) witht ∈ [0, 1], the line integralRC~F ·~dr is ≤ the arc length of C.Solution:|~F ·~r′| ≤ |~F ||~r′| ≤ |~r′|.10)T FLet~Fbe a vector field and C is a curve which is a flow line, thenRC~F·~dr> 0.Solution:The vector field p o ints in the same direction than the velocity vector so that the dotproduct is positive at each point.11)T FThe di vergence of the gradient of any f(x, y, z) is always zero.Solution:div(grad( f )) = ∆f is the Laplacian of f.312)T FFor every function f, one has div(curl( gr a d ( f ))) = 0.Solution:Both because div(curl(F ) = 0 and curl(grad(f) = 0.13)T FIf for two vector fields~F and~G one has curl(~F ) = curl(~G), then~F =~G + (a, b, c), where a, b, c are constants.Solution:One can also have~F =~G + grad( f ) which are vectorfields with the same curl.14)T FFor every vector field~Fthe id entity grad(div(~F)) =~0holds.Solution:F = (x2, y2, z2) has div(F ) = (2x, 2y, 2z) which has a nonzero gradient.15)T FIf a nonempty quadric surface g(x, y, z) = ax2+ by2+ cz2= 5 can becontained inside a finite box, then a, b, c ≥ 0.Solution:If one or two of the constants a, b, c are negative, we have a hyperboloid which all can notbe contained into a fini t e space. If all three are negative, then the surface is empty.16)T FIf~F is a vector field in space then the flux of~F through any closed surfaceS is 0.Solution:While it is true that the flux of curl(F ) vanishes through every cl ose d surface, thi s is nottrue for~F itself. Take for example F = (x, y, z).17)T FIf div(~F )(x, y, z) = 0 for al l (x, y, z), then curl(~F ) = (0, 0, 0) for all (x, y, z).Solution:Take (−y, x, 0) for example.418)T FThe flux of the vector field~F (x, y, z) = (y + z, y, −z) through th e boundaryof a solid region E is equal to the volume of E.Solution:By the divergence theorem, the flux thr ough the boundary isR R REdiv(F ) dV butdiv(F ) = 0. So the flux is zero.19)T FIf in spherical coordinates the equation φ = α (with a constant α) definesa plane, then α = π/2.Solution:Otherwise, it is would be a cone (or for α = 0 or α = π a half line).20)T FFor every function f(x, y, z), there exists a vector field~F such that div(~F ) =f.Solution:In order to solve Px+ Qy+ Rz= f just take F = (0, 0,Rz0f(x, y, w) dw).5Problem 2) (1 0 points)-2 -1 1 2-2-112-1 -0.5 0.5 1-1-0.50.51I II-1 -0.5 0.5 1-1-0.50.51-1 -0.5 0.5 1-1-0.50.51III IVFor the sign of t h e curl or divergence, where either + (positive), − (negative) or 0 for zero.The vector fields are considered on the square [−1/2, 1/2]x[−1/2, 1/2] in this problem.6Enter I,II,III,IV here Vector fi el d curl sign divergence signF (x, y) = (x, y2)F (x, y) = (1 − y, x)F (x, y) = (y − x, −y)F (x, y) = (−x, y3)Solution:Enter I,II,III,IV here Vector fi e ld curl sign divergence signIIF (x, y) = (x, y2) 0 +IF (x, y) = (1 − y, x) + 0IVF (x, y) = (y − x, −y) - -IIIF (x, y) = (−x, y3) 0 -Problem 3) (1 0 points)Mark with a cross in th e column below ”conservative” if a vector fields is conservative(that is if curl(~F )(x, y, z) = (0, 0, 0) for all points (x, y, z)). Similarly, ma r k the fieldswhich are incompressib l e ( t h a t is if div(~F )(x, y, z) = 0 for all (x, y, z) ) . No justification sare nee d ed .Vectorfield conservative incompressiblecurl(~F ) =~0 div(~F ) = 0~F (x, y, z) = (−5, 5, 3)~F (x, y, z) = (x, y, z)~F (x, y, z) = (−y, x, z)~F (x, y, z) = (x2+ y2, xyz, x − y + z)~F (x, y, z) = (x − 2yz, y − 2zx, z − 2xy)7Solution:Vectorfield conservative incompressiblecurl(~F ) =~0 div(~F ) = 0~F (x, y, z) …


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HARVARD MATH 21A - Final Exam Practice IV

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