Solutions to the quiz

Solutions to the quiz - we get u(x,y) = x^2 y^2/2 + f(y)...

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Question 1: The path x(t) = ( cos(2t), -sin(2t)) for time 0 <= t <= pi/4 traverses an anti-clockwise quarter-circle around the origin from (1,0) to (0,-1). Question 2: The path x (t) = (1-t, t) for time 0 <= t <= 1 will traverse a line segment from (1,0) to (0,1). Other parameterizations are possible, e.g. x(t) = (1-2t,2t) for time 0 <= t <= 1/2. Question 3: Each term has a magnitude which is 2|2x-1| times as large than the previous one. If 2|2x-1| < 1 , then the series converges by the ratio test; if 2|2x-1| > 1 , then the series diverges by the ratio test. When 2|2x-1| = 1 , then each term has the same magnitude, and one does divergence by the zero test. Thus one only has convergence when 2|2x-1| < 1 , i.e. when 0 < x < 1 . Question 4: Integrating (partial u(x,y))/(partial x) = x y^2
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Unformatted text preview: we get u(x,y) = x^2 y^2/2 + f(y) where f(y) is some undetermined function of y . Putting this back into the other equation we get x^2 y = x^2 y + f'(y) so f is actually a constant, i.e. u(x,y) = x^2 y^2/2 + C. Substituting x=0,y=0 we get C=1 , so u(1,1) = 3/2 . Question 5: Since ln|x| is an anti-derivative of 1/x for all x other than 0, we have from the Fundamental theorem of Calculus that the integral is ln|-2| - ln|-4| = ln 2 - ln 4 = ln (1/2) = - ln 2. Question 6: Since x ranges between -1 and 1, x^3 ranges between -1 and 1 and so 100 + x^3 ranges between 99 and 101. Thus 1/(100+x^3) <= 1/99. Since the integral of 1/99 from -1 to 1 is clearly equal to 2/99, we are done. Solutions to the quiz 1 of 1 9/19/2011 3:46 PM...
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This note was uploaded on 09/19/2011 for the course MATH 132 taught by Professor Grossman during the Spring '08 term at UCLA.

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