8/11/2011 FINAL EXAM PRACTICE V 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. Answers without derivation are not given 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 1015 10Total: 1601Problem 1) ( 20 points)1)T FIf ~u + ~v + ~w =~0 then ~u · (~v × ~w) = 0 .Solution:If ~u + ~v + ~w =~0, then ~u and ~v an d ~w are in the same plane and the tripl e scalar productis zero, because i t is the volume of the parallel epiped spanned by the t h r ee vectors.2)T FR50Rπ0r dθ dr is half the area of a disc radius 5 in th e plane.Solution:This i s indeed the formula of the area in polar coordinates.3)T FIf a vector field~F (x, y) satisfies curl(F )(x, y) = 0 for all points (x, y) in theplane, then~F is conservative.Solution:True. We have derived this from Greens theorem.4)T FIf the acceleration of a param et er i zed curve ~r(t) = (x(t), y(t), z(t)) is zerothen the cur ve ~r(t) is a line.Solution:If we integrate, then ~r′(t) is a constant vector ~v. Then ~r(t) = ~r(0) + t~v.5)T FA circle of radius 1/2 has a smaller curvature than a circle of radius 1.Solution:The cur vature of a circle of radius r is equal to 1/r.6)T FThe cur ve ~r(t) = (−sin(t), cos(t)) for t ∈ [0, π] is half a circle.Solution:True. Indeed, one can check that sin2(t) + cos2(t) = 1.27)T FThe function u (t , x) = sin(x + t ) is a solution of the partial differentialequation utx+ u = 0Solution:ux= −cos(x + t), ut= −cos(x + t) and uxt+ u = 0.8)T FThe length of a curve ~r(t) in space parameterized on a ≤ t ≤ b is the valueof th e integralR21|~T′(t)| dt, wh er e~T (t) is the unit tangent vector.Solution:The corr ec t soluti on isR21|~r′(t)| dt.9)T FLet (x0, y0) b e the maximum of f(x, y) under the constraint g( x, y) = 1.Then the gradient of g at (x0, y0) is parallel to the gradient of f at (x0, y0).Solution:This p ar a p h r as es indeed part of the Lagrange equations.10)T FAt a po int which is not a critical point, the directi onal derivativeD~vf(x0, y0, z0) can take both the negative and the positive sign.Solution:For ~v = ∇f, th e directiona l derivative is positive. For ~v = −∇f, the directional derivativeis nega t ive.11)T FIf a nonzero vector field~F (x, y) is a gradient field, we always can find acurve C for which the line integralRC~F ·~dr is positive.Solution:While there does not exist a closed curve with this property, there are many curves, if thefield~F is not zero. Just move a bit into a direction of~F at a point, wher e~F is not zero.12)T FIf C is a closed level curve of a function f(x, y) and~F = (fx, fy) is thegradient field of f, thenRC~F ·~dr = 0.Solution:The gra d i ent field is perpendicular to t he level curves.313)T FThe divergence of a grad i ent vector field~F (x, y, z) = ∇f (x, y, z) is alwayszero.Solution:Just take a simple example like f(x, y, z) = x2, where div(grad(f) = 2. Actually,div(grad( f ) = ∆f is the Laplacian of f.14)T FThe line integral of the vector field~F (x, y, z) = hx2, y2, z2i along a linesegment from (0, 0, 0) t o (1, 1, 1) is 1.Solution:By the fundamental theorem of line integrals, we can take the difference of the potentialf(x, y, z) = x3/3 + y3/3 + z3/3, wh i ch is 1/3 + 1/3 + 1/3.15)T FIf~F (x, y) = (x2− y, x) and C : ~r(t) = hqcos(t),qsin(t)i p a r am et er i zes theboundary of the region R : x4+ y4≤ 1, thenRCF · ds is twice the area ofR.Solution:This is a direct consequence of Green’s theorem and the fact that the two-dimensi o n alcurl Qx− Pyof~F = hP, Qi is equal to 2.16)T FThe flux of the vector field~F (x, y, z) = h0, y, 0i th r o u gh the bou n d a ry S ofa solid sphe r e E is equal to the volume the sphere.Solution:It is the volume of the sol i d toru s.17)T FThe qu ad r a ti c surface −x2+ y2+ z2= 5 is a one-sheeted hyperboloid.Solution:The tr aces are a circle and hyperbol oi d s.18)T FIf~F is a vector field in space and S is the bound ar y of a solid sphere thenthe fl u x of~curl(~F ) thro u gh S is 0.4Solution:This i s true by Stokes theorem.19)T FIf div(~F )(x, y, z) = 0 for all ( x, y, z) and S is a torus surface, then the fluxof~F throu g h S is zero.Solution:This i s a consequence of the divergence theorem.20)T FIn spherical coordinates, the equation ρ cos(φ) = ρ cos ( θ) sin(φ) defines aplane.Solution:True. It is the plane z = x.Problem 2) (1 0 points)I II III IVV VI VII VIII5Enter I,II,III,IV,V,VI,VII,VIII he r e Equ a t io nx2− y2+ z2= 1~r( t ) = hcos(3t), sin(2t)iz = f(x, y) = cos(3x) + sin(2y)~F (x, y) = h−y/√x2+ y2, x/√x2+ y2icos(3x) + sin(2y) = 1~F (x, y, z) = h−y, x, 1i~r( u, v) = hcos(3u), sin(2u), vi{(x, y) ∈ R2| | x2− y2| = 1 }Solution:Enter I,II,III,IV,V,VI,VII,VIII he r e Equ a t io nVIIIx2− y2+ z2= 1I~r( t ) = hcos(3t), sin(2t)iIVz = f(x, y) = cos(3x) + sin(2y)III~F (x, y) = h−y/√x2+ y2, x/√x2+ y2iVcos(3x) + sin(2y) = 1VI~F (x, y, z) = h−y, x, 1iII~r( u, v) = hcos(3u), sin(2u), viVII{(x, y) ∈ R2| | x2− y2| = 1 }Furthermore, fill in the peoples names, Green, Stokes, Gauss, Fubin i , Clairot. If there isno n am e associated t o the theorem, write the name of the theorem.Formula Name of the theoremRC~F ·~dr =R RScurl(~F ) · dSfxy(x, y) = fyx(x, y)RC~F · dr =R RRcurl(~F ) d x dyRba∇f(~r(t)) ·~r′(t)) dt = f (~r ( b) ) −f(~r(a))R RSF · dS =R R REdiv(F ) dVRbaRdcf(x, y) dxdy =RdcRbaf(x, y) dydx6Solution:Formula Name of the theoremRC~F ·~dr =R RScurl(~F ) · dSStokesfxy(x, y) = fyx(x, y) ClairotRC~F · dr =R RRcurl(~F ) d x dyGreenRba∇f(~r(t)) ·~r′(t)) dt = f (~r ( b) ) −f(~r(a))Fundamental theorem of line integralsR RSF · dS =R R REdiv(F …
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