SOLUTIONS TO PRACTICE EXAM FOR FINAL: 1/13/2002

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1 SOLUIONS O PRACICE EXAM OR INAL: 1/13/22 Math 21a Name: MW9 Sasha Braverman MW1 Ken Chung MW1 Jake Rasmussen MW1 WeiYang Qui MW1 Spiro Karigiannis MW11 Vivek Mohta MW11 Jake Rasmussen MW12 Ken Chung H1 Oliver Knill H11 Daniel Goroff Start by printing your name in the above box and check your section in the box to the left. ry to answer each question on the same page as the question is asked. If needed, use the back or next empty page for work. If you need additional paper, write your name on 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. Justify your answers. No notes, books, calculators, computers or other electronic aids are allowed. You have 18 minutes time to complete your work Which section-specific problem do you choose? Check exactly one problem. Only this problem can be graded. If you don t commit yourself here, the first attempted problem will be graded. 12a 12b 12c 12d 12 1 otal: 13

2 Problem 1) questions (2 points) Circle the correct letter. No justifications are needed. he length of the curve r(t) = (sin(t), t 4 + t, cos(t)) on t [, 1] is the same as the length of the curve r(t) = (sin(t 2 ), t 8 + t 2, cos(t 2 )) on [, 1]. rue. his is a consequence of the chain rule: b a r(s(t)) dt = b a r (s(t)) s (t) dt = s(b) s(a) r (s) ds. he parametric surface r(u, v) = (5u 3v, u v 1, 5u v 7) is a plane. rue. he coordinate functions r(u, v) = (x(u, v), y(u, v), z(u, v)) are all linear. Any function u(x, y) that obeys the differential equation u xx + u x u y = 1 has no local maxima. rue. If u = (u x, u y ) = (, ), then u xx = 1 which is incompatible with a local maximum, where u xx > by the second derivative test. he scalar projection of a vector a onto a vector b is the length of the vector projection of a onto b. rue. By definition. If f(x, y) is a function such that f x f y = then f is conservative. alse. he notion of conservative applies to vector fields and not to functions. (u v) w = (u w) v for all vectors u, v, w. alse. While (u w) v and (u w) v are both equal to the volume of the parallelepiped determined by u, v and w, the sign is different. An example: for u = 1,,, v =, 1,, w =,, 1, we have (u v) w = 1 and (u w) v = 1. he equation ρ = φ/4 in spherical coordinates is half a cone. alse. he equation ρ = φ/4 defines a heart shaped rotational symmetric surface. he surface φ = c = const would define half a cone for c [, π] { x if (x, y) (, ) he function f(x, y) = x 2 +y 2 if (x, y) = (, ) every point in the plane. is continuous at 2

3 alse. aking y =, we get f(x, ) = 1/x which is discontinuous. 1 x 1 dydx = 1/2. rue. his is the area of half of the unit square. Let a and b be two vectors which are perpendicular to a given plane Σ. hen a + b is also perpendicular to Σ. rue. If v is a vector in the plane, then a v = and b v = then also (a + b) v =. If g(x, t) = f(x vt) for some function f of one variable f(z) then g satisfies the differential equation g tt v 2 g xx =. rue. Actually one could show that g(x, t) = f(x vt) + h(x + vt) is the general solution of the wave equation g tt v 2 g xx =. If f(x, y) is a continuous function on R 2 such that D f da for any region D then f(x, y) for all (x, y). rue. Assume f(a, b) < at some point (a, b), then f(x, y) < in a small neighborhood D of (a, b) and also D f da < contradicting the assumption. Assume the two functions f(x, y) and g(x, y) have both the critical point (, ) which are saddle points, then f + g has a saddle point at (, ). alse. Example f(x, y) = x 2 y 2 /2, g(x, y) = x 2 /2 + y 2 have both a saddle point at (, ) but f + g = x 2 /2 + y 2 /2 has a minimum at (, ). If f(x, y) is a function of two variables and if h(x, y) = f(g(y), g(x)), then h x (x, y) = f y (g(y), g(x))g (y). alse. he correct identity would be h x (x, y) = f y (g(y), g(x))g (x) according to the chain rule. If we rotate a line around the z axes, we obtain a cylinder. alse. he surface could also be a one-sheeted hyperboloid or a cone. he line integral of (x, y) = (x, y) along an ellipse x 2 + 2y 2 = 1 is zero. rue. he curl Q x P y of the vector field (x, y) = (P, Q) is. By Green s theorem, the 3

4 line integral is zero. An other way to see this is that is a gradient field = f with f(x, y) = (x 2 + y 2 )/2. herefore is conservative: the line integral along any closed curve in the plane is zero. If u(x, y) satisfies the transport equation u x = u y, then the vector field (x, y) = u(x, y), u(x, y) is a gradient field. rue. = (P, Q) = (u, u). rom u x = u y we get Q x = P y which implies that is a gradient field. 3 grad(f) = d f(x + t, y + t, z + t). dt alse. he left hand side is a vector field, the right hand side a function. 1 2π/11 π ρ2 sin(φ) dφdθdρ = 4π/33. rue. he region is a lemon slice which is 1/11 th of a sphere. If is a vector field in space and f is equal to the line integral of along the straight line C from (,, ) to (x, y, z), then f =. alse. his would be true if were a conservative vector field. In that case, f would be a potential. In general this is false: for example if (x, y, z) = (, x, ), then C dr = x2 /2 and f(x, y, z) = (x,, ) which is different from. 4

5 Problem 2) (1 points) Match the equations with the curves. No justifications are needed. I II III IV Enter I,II,III,IV here III I II IV Equation r(t) = (sin(t), t(2π t)) r(t) = (cos(5t), sin(7t)) r(t) = (t cos(t), sin(t)) r(t) = (cos(t), sin(6/t)) 5

6 Problem 3) (1 points) In this problem, vector fields are written as = (P, Q). We use abbreviations curl( ) = Q x P y and div( ) = P x +Q y. When stating curl( )(x, y) = we mean that curl( )(x, y) = vanishes for all (x, y). he statement curl( ) means that curl( )(x, y) does not vanish for at least one point (x, y). he same remark applies if curl is replaced by div. Check the box which match the formulas of the vectorfields with the corresponding picture I,II,III or IV. Mark also the places, indicating the vanishing or not vanishing of curl and div. In each of the four lines, you should finally have circled three boxes. No justifications are needed. Vectorfield I II III IV curl( ) = curl( ) div( ) = div( ) (x, y) = (, 5) X X X (x, y) = (y, x) X X X (x, y) = (x, y) X X X (x, y) = (2, x) X X X I II III IV 6

7 Problem 4) (1 points) a) ind the scalar projection of the vector v = (3, 4, 5) onto the vector w = (2, 2, 1). b) ind the equation of a plane which contains the vectors 1, 1, and, 1, 1 and contains the point (, 1, ). a) v w / w = ( )/3 = 19/3 b) (1, 1, ) (, 1, 1) = (1, 1, 1). he plane has the form x y + z = d and d = 1 is obtained by plugging in the point (, 1, ). he solution is x y + z = 1 Problem 5) (1 points) ind the surface area of the ellipse cut from the plane z = 2x+2y+1 by the cylinder x 2 +y 2 = 1. Parameterize the surface r(u, v) = (u, v, 2u + 2v + 1) on the disc R = {u 2 + y 2 1}. We get r u r v = 1,, 2, 1, 2 = 2, 2, 1 and r u r v = 3. he surface integral R r u r v dudv = R 3 dudv = 3 R dudv which is 3 times the area of the disc R: Solution: 3π. Problem 6) (1 points) Sketch the plane curve r(t) = (sin(t)e t, cos(t)e t ) for t [, 2π] and find its length. r (t) = cos(t)e t + sin(t)e t, sin(t)e t + cos(t)e t satisfies r (t) = 2e t so that 2π r (t) dt = 2(e 2π 1). Problem 7) (1 points) Let f(x, y, z) = 2x 2 +3xy +2y 2 +z 2 and let R denote the region in R 3, where 2x 2 +2y 2 +z 2 1. ind the maximum and minimum values of f on the region R and list all points, where said maximum and minimum values are achieved. Distinguish between local extrema in the interior and extrema on the boundary. a) Extrema in the interior of the ellipsoid 2x 2 + 2y 2 + z 2 < 1. f(x, y, z) = 4x + 3y, 3x + 4y, 2z =,, for (x, y, z) = (,, ). One has f(,, ) =. 7

8 he discriminant (also called Hessian determinant) D = f xx f yy f 2 xy at (,, ) is D = 7 > and f xx > so that (,, ) is a local minimum. b) o get the extrema on the boundary g(x, y, z) = 2x 2 +2y 2 +z 2 1 = we solve the Lagrange equations f = λ g, g =. hey are 4x + 3y = λ4x 3x + 4y = λ4y 2z = λ2z 2x 2 + 2y 2 + z 2 = 1 We obtain z =, x = ±y or z = ±1, x = y = giving 6 critical points (1/2, 1/2, ), ( 1/2, 1/2, ), (1/2, 1/2, ), ( 1/2, 1/2, ), (,, 1), (,, 1). c) Comparing the values f(,, ) =, f(1/2, 1/2, ) = f( 1/2, 1/2, ) = 7/4 and f(1/2, 1/2, ) = f( 1/2, 1/2, ) = 1/2 and f(,, ±1) = 1 shows that (1/2, 1/2, ) and ( 1/2, 1/2, ) are maxima and that (,, ) is the minimum. Problem 8) (1 points) Sketch the region of integration of the following iterated integral and then evaluate the integral: ( π π ( x ) ) sin(xy)dy dx dz. z he region is contained inside the cube [, π [, π] [, π]. It is bounded by the surfaces x = z, x = y, z =, y = π (see picture). he integral can not be solved in the given order. Using the picture as a guide, we write the integral as π x 2 x Solve the most inner integral: π x 2 sin(xy) dydzdx cos(xy) x π x 2 dzdx = (1 cos(x 2 ))/x dzdx x Now solve the z integral: 3 2 z y x = π to finally get x 2 (1 cos(x 2 ))/x dx = π x(1 cos(x 2 )) dx 1 he answer is π 2. = ( sin(x2 ) 2 + x2 2 ) π = π 2. 8

9 Problem 9) (1 points) Evaluate the line integral C dr, where = (x + e x sin(y), x + e x cos(y)) and C is the right handed loop of the lemniscate described in polar coordinates as r 2 = cos(2θ). By Green s theorem, the integral is R curl( ) da, where R is the region enclosed by C. rom = (P, Q) = (x + e x sin(y), x + e x cos(y)) we calculate curl() = Q x P y = 1 + e x cos(y) cos(y)e x = 1 so that the result is the area of R which is π/4 cos(2θ)/2 dθ = sin(2θ) 4 π/4 π/4 = 1/2. π/4 Problem 1) (1 points) Evaluate the line integral C dr, where C is the planar curve r(t) = (t 2, t/ t + 2), t [, 2] and is the vector field (x, y) = (2xy, x 2 + y). Do this in two different ways: a) by verifying that is conservative and replacing the path with a different path connecting (, ) with (4, 1), b) by finding a potential U satisfying U =. a) o verify that = (P, Q) is conservative, it is enough to verify that curl() = Q x P y =. his is actually the case. o calculate the line integral, we therefore can replace the path with a straight line γ : r(t) = (4t, t) and calculate γ dr = 1 (8t2, 16t 2 +t) (4, 1) dt = (48t 2 +t) 1 = /2. b) A potential is f(x, y) = x 2 y + y 2 /2. he value of f at (4, 1) is /2. he value of f at (, ) is. he difference between the potential values is /2 again. Problem 11) (1 points) a) ind the line integral C dr of the vector field (x, y) = (xy, x) along the unit circle C : t r(t) = (cos(t), sin(t)), t [, 2π] by doing the actual line integral. b) ind the value of the line integral obtained in a) by evaluating a double integral. 9

10 a) 2π (cos(t) sin(t), cos(t)) ( sin(t), cos(t)) dt = 2π cos 2 (t) sin 2 (t) cos(t) dt = π+sin 3 (t)/3 2π = π b) curl( ) = Q x P y = 1 x. By Green s theorem, the line integral is a double integral which we evaluate using Polar coordinates D (1 x) da = 1 (1 cos(θ)r) dθdr = 2π/2 = π. 2π SECION SPECIIC PROBLEMS Math 21a Please choose one of the following problems and register your decision on the first page of this exam. Problem 12a) (1 points) Consider the surface given by the graph of the function z = f(x, y) = 1 sin ( π 1+x 2 +y 2 8 (x2 + y 2 ) ) in the region x 2 + y he surface is pictured to the right. A magnetic field B is given by the curl of a vector potential A. hat is, B = A = curl(a) and A is a vector field too. Suppose A = ( z sin(x 3 ), x ( 1 z 2), log(1 + e x+y+z ) ). Compute the flux of the magnetic field through this surface. (he surface has an upward pointing normal vector.) he surface S is bounded by the curve γ : r(t) = (4 cos(t), 4 sin(t), ). By Stokes theorem, the flux of the curl of A through the surface S is the line integral of A along γ: γ 16π. A dr = 2π (, 4 cos(t), log(1 + e x+y )) ( 4 sin(t), 4 cos(t), ) dt = 2π 16 cos 2 (t) dt = Problem 12b) (1 points) he annual rainfall in inches in a certain region is normally distributed with µ = 4 and σ = 4. What is the probability that starting with this year, it will take over 1 years before a year 1

11 occurs having a rainfall of over 5 inches. he probability p that no rainfall over 5 inches happens in one year is p = µ,σ (5) = 5 e (x 4)2 /16 / π16 he probability that it will take more than 1 years until a rainfall over 5 inches occurs is p 1. Problem 12c) (1 points) Let S be the surface given by the equations z = x 2 y 2, x 2 + y 2 4, with the upward pointing normal. If the vector field is given by the formula (x, y, z) = x, y, x 2 + y 2, find the flux of through S. Parameterize the surface by r(u, v) = u, v, u 2 v 2. hen r u r v = 1,, 2u, 1, 2v = 2u, 2v, 1. he flux integral is ds = u, v, u 2 + v 2 2u, 2v, 1 dxdy D D = 2(u 2 + v 2 ) + u 2 + v 2 da he answer is 64π/3. = D 2π 2 (2r 2 + r)r drdθ = 2π(2 2 4 / /3) = 64π/3. Problem 12d) (1 points) he random variable X has a variance 2 and that the random variable Y has a variance 1. Both random variables have zero expectation. ind the correlation between X + Y and X Y. so that Cov[X + Y, X Y ] E[(X + Y ) E[X + Y ], (X Y ) E[X Y ] ] E[(X + Y ), (X Y ) ] = E[X 2 Y 2 ] = E[X 2 ] E[Y 2 ] E[(X E[X]) 2 ] E[(Y E[Y ]) 2 ] = Var[X] Var[Y ] = 2 1 = 1, Corr[X + Y, X Y ] Cov[X + Y, X Y ]/(σ[x]σ[y ]) 1/( 2) he solution is 1/ 2. 11