3/14/2006 FIRST HOURLY FIRST PRACTICE Math 21a, Spring 2006Name:MWF 10 Samik BasuMWF 10 Joachim KriegerMWF 11 Matt LeingangMWF 11 Veronique GodinTTH 10 Oliver KnillTTH 115 Thomas Lam• Start by printing your name in the above box and checkyour section in the box to the left.• Do not detach pages from this exam packet or unstaplethe packet.• Please write neatly. Answers which are illegible f or thegrader can not be given credit.• No notes, books, calculators, computers, or other elec-tronic aids can be allowed.• You have 90 minutes time to complete your work.• The hourly exam itself will have space for work on eachpage. This space is excluded here in order to save print-ing resources.1 202 103 104 105 106 107 108 109 109 10Total: 110Problem 1) TF questions (20 points) No justifications needed1)T FThe length of the sum of two vectors is always the sum of the length of thevectors.2)T FFor any three vectors, ~v × (~w + ~u) = ~w ×~v + ~u ×~v.3)T FThe set of points which satisfy x2+ 2x + y2− z2= 0 is a cone.4)T FThe functions√x + y − 1 and log(x + y − 1) have the same domain ofdefinition.5)T FIf P, Q, R are 3 different points in space that don’t lie in a line, then~P Q×~RQis a vector orthogona l to the plane containing P, Q, R.6)T FThe line ~r(t) = h1 + 2t, 1 + 3t, 1 + 4ti hits the plane 2x + 3y + 4z = 9 at aright angle.7)T FThe graph of f(x, y) = cos(xy) is a level surface of a function g(x, y, z).8)T FFor any two vectors, ~v × ~w = ~w ×~v.9)T FIf |~v × ~w| = 0 for all vectors ~w, then ~v =~0.10)T FIf ~u and ~v are orthogonal vectors, then (~u ×~v) ×~u is par allel to ~v.11)T FEvery vector contained in the line ~r(t) = h1 + 2t, 1 + 3t, 1 + 4ti is parallelto the vector h1, 1, 1i.12)T FThe curvature of the curve 2~r(4t) at t = 0 is twice the curvature of thecurve ~r(t) at t = 0.13)T FThe set of points which satisfy x2− 2y2− 3z2= 0 f orm an ellipsoid.14)T FIf ~v × ~w = (0, 0, 0), then ~v = ~w.15)T FEvery vector contained in the line ~r(t) = h1 + 2t, 1 + 3t, 1 + 4ti is parallelto the vector h1, 1, 1i.16)T FTwo nonzero vectors are parallel if and only if their cross product is~0.17)T FThe function u(x, t) = x2/2 + t satisfies the heat equation ut= uxx.18)T FAny function of three variables f(x, y, z) satisfies the partial differentialequation fxyz+ fyzx= 2fzxy.19)T FIf fx(x, y) = fy(x, y) for all x, y, then f(x, y) is a constant.20)T FThe value of the function f (x, y) = sin(−x + 2y) a t (0.001, −0.002) can bylinear approximation be estimated as −0.003.Problem 2a) (5 points)Match the contour maps with the corresponding functions f(x, y) of two variables. No justifi-cations are needed.I II IIIIV V VIEnter I,II,III,IV,V or VI here Function f(x, y)f(x, y) = sin(x)f(x, y) = x2+ 2y2f(x, y) = |x| + |y|f(x, y) = sin(x) cos(y)f(x, y) = xe−x2−y2f(x, y) = x2/(x2+ y2)Problem 2b) ( 5 points)Match the parametric surfaces with their parameterization. No justification is needed.I IIIII IVEnter I,II,III,IV here Parameterization(u, v) 7→ (u, v, u + v)(u, v) 7→ (u, v, sin(uv))(u, v) 7→ (0.2 + u(1 − u2)) cos(v), (0.2 + u(1 − u2)) sin(v), u)(u, v) 7→ (u3, (u − v)2, v)Problem 3) (10 points)Use the technique of linear approximation to estimate f (log(2) + 0.001, 0.006) for f(x, y) =e2x−y. (Here, log means the natural logarithm).Problem 4) (10 points)Consider the equationf(x, y) = 2y3+ x2y2− 4xy + x4= 0It defines a curve, which you can see in the picture. Nearthe point x = 1, y = 1, the function can be written asa graph y = y(x). Find the slope of that g r aph at thepoint (1, 1).-2 -1.5 -1 -0.5 0.5 1x-2-1.5-1-0.50.51yProblem 5) (10 points)a) (6 points) Find a parameterization of the line of intersection of the planes 3x − 2y + z = 7and x + 2y + 3z = −3.b) (4 points) Find the symmetric equationsx − x0a=y − y0b=z − z0crepresenting that line.Problem 6) (10 points)a) (4 points) Find the area of the parallelogram with vertices P = (1, 0, 0) Q = (0, 2, 0),R = (0 , 0, 3) and S = (−1, 2, 3).b) (3 points) Verify that the triple scalar product has the property [~u+~v, ~v+ ~w, ~w+~u] = 2[~u,~v, ~w].c) (3 points) Verify that the triple scalar product [~u,~v, ~w] = ~u · (~v × ~w) has the property|[~u, ~v, ~w]| ≤ ||~u|| · ||~v|| · ||~w||Problem 7) (10 points)Find the distance between the two lines~r1(t) = ht, 2t, −tiand~r2(t) = h 1 + t, t, ti .Problem 8) (10 points)Find an equation for the plane that passes through the origin and whose normal vector isparallel to the line of intersection of the planes 2x + y + z = 4 and x + 3y + z = 2.Problem 9) (10 points)The intersection of the two surfaces x2+y22= 1 and z2+y22= 1 consists of two curves.a) (4 points) Parameterize each curve in the form ~r(t) = (x(t), y(t), z(t)).b) (3 points) Set up t he integral for the arc length of one of the curves.c) (3 points) What is the arc length of this curve?Problem 10) (10 points)a) (6 points) Find the curvature κ(t) of the space curve ~r(t) = h−cos(t), sin(t), −2ti at thepoint ~r(0).b) (4 points) Find the curvature κ(t) of the space curve ~r(t) = h−cos(5t), sin(5t), −10ti at thepoint ~r(0).Hint. Use one of the two formulas for the curvatureκ(t) =|~T′(t)||~r′(t)|=|~r′(t) ×~r′′(t)||~r′(t)|3,where~T (t) = ~r′(t)/|~r′(t)|. The curvatures in b) can be derived from the curvature in a).There is no need to redo the calculation, but we need a
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