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Unformatted text preview: MA113 Calculus II The Cooper Union, Spring 2010 Prof. Alex Casti casti@cooper.edu Homework 3 Solutions Chapter 9 problems (Stein and Barcellos) Section 9.1 4ab Find the polar coordinates ( r , ) with < 2 and r > , for the point whose rectangular coordinates are (a) ( x , y ) = 2 , 2 . This point is in quadrant I of the xyplane (when using the inverse tangent function to calculate , you must keep in mind the quadrant in which the point lies since the principal range of tan 1 z is between 2 and 2 ). r = p x 2 + y 2 = 2 + 2 = 2 = tan 1 y x = tan 1 ( 1 ) = 4 ( r , ) = 2 , 4 . (b) ( x , y ) = ( 1 , 3 ) . This point is in quadrant II of the xyplane. r = p x 2 + y 2 = 1 + 3 = 2 = tan 1 y x = tan 1 ( 3 ) = 3 ( r , ) = 2 , 2 3 . Notice that the angle = 3 would put the point in quadrant IV, not quadrant II. To fix this, one must add to the result, so that =  3 = 2 3 . Observe that the angles = 3 and = 2 3 lie along the same ray and are both equal to tan 1 ( 3 ) . This underscores the care one must take in using the inverse tangent function. 6 Transform the following polar equation into one in rectangular coordinates. r = csc . Since csc = 1 sin , we have r sin = 1. But y = r sin , so the same equation in rectangular coordinates is y = 1 . 12 Transform the following equation in rectangular coordinates into polar coordinates: x 2 + y 2 = 4 x . Upon applying the polar coordinate transform ( x , y ) = ( r cos , r sin ) one obtains r 2 cos 2 + r 2 sin 2 = 4 r cos r 2 = 4 r cos r = 4cos . The only danger here is in canceling a factor of r in the final answer, which risks losing a description of the point at r = 0. However, notice that the rectangular version of the equation is equivalent to ( x 2 ) 2 + y 2 = 4, which corresponds to a circle of radius 2 centered at ( 2 , ) . The origin r = 0 is not on the given curve, so the cancellation of a factor of r is a legitimate maneuver. 1 27 Find the intersection of the polar curves r = sin3 and r = cos3 . 0.2 0.4 0.6 0.8 1 30 210 60 240 90 270 120 300 150 330 180 sin(3 ) cos(3 ) Figure 1: Polar plot for the two curves of problem 27, section 9 . 1. Figure 1 shows that these two curves intersect at 4 points, including the origin r = 0. To see that the origin is a point of intersection, observe that ( r , ) = ( , ) lies on the curve r = sin3 , and ( r , ) = ( , 6 ) is on the curve r = cos3 (in fact, each of these curves hits the origin r = a total of 6 times). Now look for the intersections with r 6 = 0. Setting the two functions equal to one another gives sin3 = cos3 tan3 = 1 3 n = 4 + n n = 12 + 1 3 n n = ( 1 + 4 n ) 12 ( n = , 1 , 2 , 3 , 4 , 5 ) ....
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