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Unformatted text preview: MA113 Calculus II The Cooper Union, Spring 2010 Prof. Alex Casti [email protected] Homework 3 Solutions Chapter 9 problems (Stein and Barcellos) Section 9.1 4ab Find the polar coordinates ( r , θ ) with ≤ θ < 2 π 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 θ , you must keep in mind the quadrant in which the point lies since the principal range of tan 1 z is between π 2 and π 2 ). r = p x 2 + y 2 = √ 2 + 2 = 2 θ = tan 1 y x = tan 1 ( 1 ) = π 4 → ( r , θ ) = 2 , π 4 . (b) ( x , y ) = ( 1 , √ 3 ) . This point is in quadrant II of the xyplane. r = p x 2 + y 2 = √ 1 + 3 = 2 θ = tan 1 y x = tan 1 ( √ 3 ) = π 3 → ( r , θ ) = 2 , 2 3 π . Notice that the angle θ = π 3 would put the point in quadrant IV, not quadrant II. To fix this, one must add π to the result, so that θ = π π 3 = 2 3 π . Observe that the angles θ = π 3 and θ = 2 3 π 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 θ . Since csc θ = 1 sin θ , we have r sin θ = 1. But y = r sin θ , 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 θ , r sin θ ) one obtains r 2 cos 2 θ + r 2 sin 2 θ = 4 r cos θ → r 2 = 4 r cos θ → r = 4cos θ . 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 θ and r = cos3 θ . 0.2 0.4 0.6 0.8 1 30 210 60 240 90 270 120 300 150 330 180 sin(3 θ ) cos(3 θ ) 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 , θ ) = ( , ) lies on the curve r = sin3 θ , and ( r , θ ) = ( , π 6 ) is on the curve r = cos3 θ (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 θ = cos3 θ → tan3 θ = 1 → 3 θ n = π 4 + π n θ n = π 12 + 1 3 π n → θ n = π ( 1 + 4 n ) 12 ( n = , 1 , 2 , 3 , 4 , 5 ) ....
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