8/5/2010 FINAL EXAM PRACTICE I 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 l i n e ~r(t) = ht, t, ti is perpendicular to the plane x + y + z = 10.Solution:Yes, t h e normal vector to the plane is h1, 1, 1i.2)T FThe q u adratic surface −x2+ y2+ z2= −1 is a one sheeted hyperboloid.Solution:It i s a two sheeted hyperboloid.3)T FThe relation |~u ×~v| = |~u ·~v| is only possible if at least one of the vectors ~uand ~v is the zero vector.Solution:It i s also possible if they are nonzero but form a 45 degree angle.4)T FRπ/20R10r3dθ dr =R10R10x2+ y2dxdy.Solution:The bounds are wrong. The s econ d integral integrates over a square.5)T FIf a vector field~F (x, y) satisfies curl(~F )(x, y) = Qx(x, y) −Py(x, y) = 0 anddiv(~F )(x, y) = Px(x, y) +Qy(x, y) = 0 for all points (x, y) in the plan e, then~F is a constant field.Solution:There are mo re fields which satisfy this. An example is~F (x, y) = hx + y, x − yi.6)T FThe acceleration vector ~r′′(t) = hx(t), y(t)i, the velocity vector ~r′(t) and~r′(t) ×~r′′(t) form three vectors which are mutually perpendicular.Solution:Already for a strai ght line, this is not the case.27)T FThe curvature of the curve ~r(t) = hsin ( 2t ) , 0, cos(2t)i is equal to the curva-ture of the curve ~s(t) = h0, cos(3t), sin ( 3t ) i .Solution:Both are ci r cle s of radius 1.8)T FThe space curve ~r(t) = ht sin(t), t cos(t), t2i for t ∈ [0, 10π] is located on acone.Solution:It is not x( t )2+ z( t)2= z2but x(t)2+ y(t)2= z(t) so that it is located on a paraboloid.9)T FIf a smooth funct i on f(x, y) has a global maximum, then this maximum isa cri t i cal point.Solution:It i s then also a local maximum.10)T FIf L(x, y) is the linearization of f(x, y) and ~s(t) is the line tangent to thecurve ~r(t) at t0. The n d/dtL(~s(t)) = d/dt f (~r(t)) at the time t = t0.Solution:This is how the chain rule can be proved.11)T FIf~F is a gradient field and ~r(t) is a flow line defined by ~r′(t) =~F (~r(t)),then the l ine integralR10~F ·~dr is either positive or zero.Solution:The power is positive.12)T FIf we extremize the function f(x, y) un d er the constraint g(x, y) = 1, andthe functions are t h e same f = g, we have infinitely many extrema.Solution:Every point on the curve g(x, y) = 1 is a s ol u t i on to the Lagrange equations.313)T FIf a point (x0, y0) is a critical point of f(x, y) under the constraint g(x, y) =1, then it is also a critical point of the function f(x, y) withou t constraints.Solution:The g r ad i ent of f does not have to be the zero vector.14)T FIf a vector field~F (x, y) is a gradient field, then any line integral along anyellipse is z er o.Solution:This follows from the fundamental theorem of line integrals.15)T FThe flux of an irrotational vector field is zero through any surface S inspace.Solution:It i s only zero through a closed surface.16)T FThe divergence of a gradient field~F(x, y) = ∇f(x, y) is zero.Solution:While the curl of a gradient is zero and the divergence of a curl is zero, the d i vergence ofa gr ad i ent is the Laplacian of f and not necessarily zero.17)T FThe line integral of a vector field~F (x, y, z) = hx, y, zi along a circle in thexy− plane is zero.Solution:Yes, by the fundamental theorem of line integrals.18)T FFor any solid E, the moment of inertiaRRREx2+ y2dxdydz is always largerthan the volumeRRRE1 dxdydz.Solution:For sm al l solid s, the moment of inertia is small, for large solids, the moment of inertia islarge.419)T FThe c u r vature of a circle is always larger than the acceleration.Solution:The a ccel er at i on depends on t he par am et r i za ti o n .20)T FThe di r ect i onal der i vative of div(~F (x, y)) of the divergence of the vectorfield~F = hP, Qi in th e directi on ~v = h1, 0i is Pxx+ Qxy.Solution:Yes by definition div(~F (x, y)) = Px(x, y) + Qy(x, y). The directional d er ivative in theh1, 0i direction is the partial derivative.5Problem 2) (10 points) Match objects with definitions. No justifications necessary.Match the objects with their definitions1 23 45 678Enter 1-8 or 0 i f no match Object definition~r(t) = h(2 + cos(10t)) c os ( t ) , (2 + cos(10t)) sin(t), sin(1 0t ) i~F (x, y, z) = h−y, x, 2i~r(t, s) = h(2 + cos(s)) cos(t), (2 + cos(s)) sin(t), sin(s)i{(x, y, z) | sin(x2) − cos(y2) = 1 }~F (x, y) = hx − y2, y − x2ixyz = 0x2+ y2− z2= 1{(x, y) | sin ( x2sin(x))y + sin(y − x ) = c }~r(t) = hsin(t) + cos(5t), cos(t) + cos(6t)i6Solution:8,2,6,0,1,3,0,4,5Problem 3) (10 points)a) (4 points) Check every box to the left, for which t h e missing part to the right is ∇f(1, 2).The function f(x, y) i s an arbitrary nice function like for exampl e f(x, y) = x − yx + y2.The curve ~r(t), wherever it appear s, parametrizes the level curve f(x, y) = f(1, 2) and hasthe property that ~r(0) = h1, 2i.Check Topic StatementLinearization L(x, y) = f(1, 2)+ ·hx − 1, y − 2iChain ruleddtf(~r(t))|t=0=·~r′(0)Steepest descent f decrea ses at (1, 2) most in the direction ofEstimation f(1 + 0. 1, 1.99) ∼ f(1, 2)+ ·h0.1, −0.01iDirectional derivative D~vf(1, 2) = ·~vLevel curve of f through (1, 2) has the form ·hx − 1, y − 2i = 0Vector projection of ∇f (1, 2) onto ~v is ~v(~v· )/|~v|2Tangent line of ~r(t) at (1, 2) is parametrized by~R(s) = h1, 2i + sb) (3 points) The surfaces are given either as a parametrization or implicit l y. Match th em.Each surface matches one definition.A B CD E F7Enter A-F here Function or parametrizatio n~r(u, v) = hu2, v2, u2+ v2i~r(u, v) = h(1 + sin(u)) cos(v), (1 + sin(u)) sin(v), ui4x2+ y2− 9z2= 1x − 9y2+ 4z2= 1~r(u, v) = hu, v, sin(u2+ v2)i4x2+ 9y2= 1c) ( 3 points) Match the solids with …
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