8/17/2004 FINAL EXAM PRACTICE Maths 21a, O. Knill, Summer 2004Name:• 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. If you need additional paper, write your nameon it.• Do not detach pages from this exam packet or unstaple the packet.• Please write neatly. Answers which are illegible for the grader can not be given credit.• No notes, books, calculators, computers, or other electronic aids can be allowed.• You have 180 minutes time to complete your work.1 202 103 104 105 106 107 108 109 1010 1011 1012 1013 1014 10Total: 150Problem 1) (20 points)T FFor any two nonzero vectors ~v, ~w the vector ((~v × ~w) ×~v) ×~v) is parallel to~w.Solution:Take v = (1, 0, 0), w = (0, 1, 0) so that ~v × ~w = (0, 0, 1) and (~v × ~w) × ~v) = (0, 1, 0) and((~v × ~w) ×~v) ×~v) = (0, 0, 1).T FThe cross product satisfies the law (~u ×~v) × ~w = ~u × (~v × ~w).Solution:Take ~v = ~w, then the right hand side is the zero vector while the left hand side is notzero in general (for example if u = i, v = j).T FIf the curvature of a smooth curve ~r(t) in space is defined and zero for allt, then the curve is part of a line.Solution:One can see that with the formula κ(t) = |r0(t) × r00(t)|/|r0(t)|3which shows that theacceleration r00(t) is in the velocity direction at all times. One can also see it intuitivelyor with the definition κ(t) = |T0(t)|/|r0(t)|. If curve is not part of a line, then T has tochange which means that κ is not zero somewhere.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 line and for t = 0, one has ~r(0) = A and for t = 1one has ~r(1) = B.T FFor every c, the function u(x, t) = (2 cos(ct) + 3 sin(ct)) sin(x) is a solutionto the wave equation utt= c2uxx.Solution:Just differentiate.T FThe length of the curve ~r(t) = (t, sin(t)), where t ∈ [0, 2π] isR2π0q1 + cos2(t) dt.Solution:The speed at time t is |~r0(t)| =q1 + cos2(t).T FLet (x0, y0) be 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 the y axes, it is false in general.T FThe function f(x, y, z) = x2− y2− z2decreases in the direction(2, −2, −2)/√8 at the point (1, 1, 1).Solution:It increases in that direction.T FAssume~F is a vector field satisfying |~F (x, y, z)| ≤ 1 everywhere. For everycurve C : ~r(t) with t ∈ [0, 1], the line integralRC~F ·~dr is less or equal thanthe arc length of C.Solution:|~F ·~r0| ≤ |~F ||~r0| ≤ |~r0|.T FLet~F be a vector field and C is a curve which is a flow line, thenRC~F ·~dr > 0.Solution:The vector field points in the same direction than the velocity vector so that the dotproduct is positive at each point.T FThe divergence of the gradient of any f (x, y, z) is always zero.Solution:div(grad(f)) = ∆f is the Laplacian of f .T FFor every function f, one has div(curl(grad(f))) = 0.Solution:Both because div(curl(F ) = 0 and curl(grad(f) = 0.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.T FFor every vector field~Fthe identity grad(div(~F)) =~0holds.Solution:F = (x2, y2, z2) has div(F ) = (2x, 2y, 2z) which has a nonzero gradient.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 finite space. If all three are negative, then the surface is empty.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 closed surface, this is nottrue for~F itself. Take for example F = (x, y, z).T FIf div(~F)(x, y, z) = 0 for all (x, y, z), then curl(~F) = (0, 0, 0) for all (x, y, z).Solution:Take (−y, x, 0) for example.T FThe flux of the vector field~F (x, y, z) = (y + z, y, −z) through the boundaryof a solid region E is equal to the volume of E.Solution:By the divergence theorem, the flux through the boundary isR R REdiv(F ) dV butdiv(F ) = 0. So the flux is zero.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).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).Problem 2) (10 points)Match the equations with the objects. No justifications are needed.I II III IVV VI VII VIIIEnter I,II,III,IV,V,VI,VII,VIII here Equationg(x, y, z) = cos(x) + sin(y) = 1y = cos(x) − sin(x)~r(t) = (cos(t), sin(t))~r(u, v) = (cos(u), sin(v), cos(u) sin(v))~F (x, y, z) = (cos(x), sin(x), 1)z = f(x, y) = cos(x) + sin(y)g(x, y) = cos(x) − sin(y) = 1~F (x, y) = (cos(x), sin(x))Solution:Enter I,II,III,IV,V,VI,VII,VIII here EquationVg(x, y, z) = cos(x) + sin(y) = 1VIIIy = cos(x) − sin(x)I~r(t) = (cos(t), sin(t))II~r(u, v) = (cos(u), sin(v), cos(u) sin(v))VI~F (x, y, z) = (cos(x), sin(x), 1)IVz = f(x, y) = cos(x) + sin(y)VIIg(x, y) = cos(x) − sin(y) = 1III~F (x, y) = (cos(x), sin(x))Problem 3) (10 points)Mark with a cross in the column below ”conservative” if a vector fields is conservative (thatis if curl(~F )(x, y, z) = (0, 0, 0) for all points (x, y, z)). Similarly, mark the fields which areincompressible (that is if div(~F )(x, y, z) = 0 for all (x, y, z)). No justifications are needed.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)Solution:Vectorfield conservative incompressiblecurl(~F ) =~0 div(~F ) = 0~F (x, y, z) = (−5, 5, 3)X X~F (x, y, z) = (x, y, z)X~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)XProblem 4) (10 points)Let E be a parallelogram in three dimensional space defined by two vectors ~u and ~v.a) (3 points) Express …
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