8/11/2011 FINAL EXAM PRACTICE II 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. Even correct answers without derivation can not be gi ven 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 10Total: 1501Problem 1) ( 20 points)1)T FThe line ~r(t) = ht, t, ti is perpendicular to the plane x + y + z = 10.Solution:Yes, th e normal vector to the plane is h1, 1, 1i.2)T FThe qu ad r a ti c surface −x2+ y2+ z2= −1 is a one sheeted hyperboloid.Solution:It is a two sheeted hyperboloid.3)T FThe relation |~u ×~v| = |~u ·~v| is on l y possible if at least one of the vectors ~uand ~v i s the zero vector.Solution:It is also possible if they are nonzero but form a 45 degre e angle.4)T FRπ/20R10r3dθ dr =R10R10x2+ y2dxdy.Solution:The bounds are wrong. The second 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 plane, then~F is a constant field.Solution:There ar e more fields which satisfy this. An exa m p l e 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) for m three vectors which are mutually perpendicular.Solution:Already for a straight line, this is not the case.27)T FThe c u r vature of the curve ~r(t) = hsin(2t), 0, cos(2t)i is equal to the curva-ture o f the curve ~s(t) = h0, cos(3t), sin(3t) i .Solution:Both are circl es of radius 1.8)T FThe space curve ~r(t) = ht si n ( 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 i s located on a paraboloid.9)T FIf a smooth function f(x, y) has a global maximum, t h en this maximum isa crit ic al point.Solution:It is then also a local maximum.10)T FIf L(x, y) is the linearizatio n of f(x, y) and ~s(t) is the line tangent to th ecurve ~r(t) at t0. The n d/dtL(~s(t)) = d/dtf(~r(t)) at the time t = t0.Solution:This i s how the chain rule can be prove d .11)T FIf~F is a gradient field and ~r(t) is a flow line defined by ~r′(t) =~F (~r(t)),then the lin e integralR10~F ·~dr is eit h e r positive or zero.Solution:The power is positive.12)T FIf we extremize the function f(x, y) und er the constraint g(x, y) = 1, andthe functions are the same f = g, we have infinitely many extrema.Solution:Every point on the curve g(x, y) = 1 is a solution 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) without constraints.Solution:The gra d i ent of f does not have to be the zero vector.14)T FIf a vector field~F (x, y) is a gr adient field, the n any lin e integral along anyellipse is zero .Solution:This fol l ows from the fundamental theorem of line integrals.15)T FThe flux of an irrotational vector field is zero th r ou g h any surface S inspace.Solution:It is only zero through a closed surface.16)T FThe di vergence of a gradient field~F(x, y) = ∇f(x, y) i s zero.Solution:While the curl of a gradient is zero and the divergence of a curl is zero, the divergence ofa grad 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 zer o .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 d xd ydz.Solution:For smal l solids, the moment of inertia is small, for large s ol i d s, the moment of inertia islarge.419)T FThe cur vature of a circle is always larger than the acceleration .Solution:The accele ra t i on depends on the parametrization.20)T FThe di r ect i onal derivati ve of div(~F (x, y)) of the divergence of th e vectorfield~F = hP, Qi in the direction ~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) (1 0 points) Match objects wi t h defin i t i ons. No justifications necessary.Match the objects with their definitions1 23 45 678Enter 1-8 or 0 if no match Object defi n i t i on~r(t) = h(2 + cos(10t)) cos(t), (2 + cos(10t)) sin(t), sin ( 10 t )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) (1 0 points)a) (4 points) Check every box to the left, for which the missing part to the right is ∇f (1, 2).The function f(x, y) i s an arbitrary nice functio n like for example f(x, y) = x − yx + y2.The curve ~r(t), wherever it appears, parametrizes the level curve f (x, y) = f(1, 2) and hasthe p ro perty 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 .9 9) ∼ f(1, 2)+ ·h0.1, −0. 01 iDirectional 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 p a r am et r i zat i on or implicitly. Match them.Each surface matches one defi n i t i on .A B CD E F7Enter A-F here Function or parametrizati on~r(u, v) = hu2, v2, u2+ v2i~r(u, v) = h(1 + sin(u)) cos(v), (1 + sin(u)) sin(v), ui4x2+ y2− …
View Full Document
Unlocking...