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test2_sol

# test2_sol - Mathematics 20E Spring 2009 Name ID Test 2...

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Unformatted text preview: Mathematics 20E Spring 2009 Name: ID: Test 2 Section: Be sure to show all work. Answers without full justification are worth no credit. Don't forget that you may be able to simplify some problems by being clever. You may use a single page of notes, but no other aids are allowed. This test consists of 4 questions. Each question is worth a total of 5 points. Answers should all be in closed form, but need not be fully simplified. 1 2 3 4 /5 /5 /5 /5 /20 1) Let C be the oriented curve given by the graph of y = 1 x2 , 0 x 2 2 going from left to right. (a) Give a parameterization of C. Be sure to state the domain of your parameterization. Because C is the graph of a function, we can choose the input, x as the parameter. Thus, we use the function c(t) = as a parameterization. 1 t, t2 , 2 0t2 (b) Compute g ds C where g(x, y) = 2x(1 + 2y). Using the above parameterization, this becomes 2 g ds = C 0 2 g(c(t)) c(t) dt 2t(1 + t2 ) 12 + t2 dt 2t(1 + t2 ) 2 dt 0 2 0 3 = 0 2 = 5 2 = (1 + t2 ) 2 5 2 5 (5 2 - 1). = 5 2) Let T : 2 2 be given by T(u, v) = (u(1 + v), (1 + u)v). Take as fact that T maps [0, 1] [0, 1] onto 1 R = {(x, y) : x 0, y 0, 2x - 2 y x + 1}. 2 (a) Compute DT, the total derivative of T. Here, we have DT = 1+v u v 1+u . (b) Using T, compute (x - y) dAx,y . R In order to compute the given integral using T, we have (x - y) dAx,y = R 1 1 0 1 1 0 1 (u(1 + v) - (1 + u)v) |det(DT)| dAu,v [0,1][0,1] = 0 (u - v)(1 + u + v) du dv u - v + u2 - v 2 du dv u=1 = 0 1 1 2 u - uv + u3 - uv 2 3 0 2 1 5 = - v - v 2 dv 0 6 v=1 1 1 5 v - v2 - v3 = 6 2 3 v=0 = 0. = dv u=0 3) Let F(x, y) = xyi+(x+y)j. Find the work done by F around the boundary of the region D = [0, 1] [0, 1] oriented counter clockwise. Here, in a formula, we're being asked to compute F ds. D Using Green's theorem, this becomes F ds = D D xy dx + (x + y) dy ( D = = D 1 (x + y) - (xy)) dAx,y x y 1 - x dAx,y 1 0 1 = 0 1 - x dx dy x=1 = 0 1 = 0 1 x - x2 2 1 dy 2 y=1 y=0 dy x=0 1 = y 2 1 . = 2 4) Let S be the triangle in 3 with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1) oriented with normal pointing upwards. A parameterization of S is given by defined as Compute T : {(u, v) : u 0, v 0, u + v 1} 3 T(u, v) = (u, v, 1 - u - v). F dS S where F(x, y, z) = (x - z)i + yj + k. Let us call the domain of T the set D. First, notice that and and so Then, we compute F dS = S D Tu (u, v) = (1, 0, -1) Tv (u, v) = (0, 1, -1) Tu (u, v) Tv (u, v) = (1, 1, 1). F(T(u, v)) (Tu (u, v) Tv (u, v)) dAu,v (u - (1 - u - v), v, 1) (1, 1, 1) dAu,v D 1 1-u = = 0 1 2u + 2v dv du 0 = 0 1 2uv + v 2 1 - u2 du u=1 u=0 v=1-u v=0 du = 0 1 = u - u3 3 2 = . 3 ...
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