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Pre-Calc Exam Notes 150

Pre-Calc Exam Notes 150 - the circle so canceling r does...

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150 Chapter 6 Additional Topics §6.4 (b) Using formula (6.10) with x =− 5 and y =− 5, we get: tan θ = y x = 5 5 = 1 θ = 45 or θ = 225 Since θ = 225 is in the same quadrant (QIII) as the point ( x , y ) = ( 5 , 5), we can take r = radicalbig x 2 + y 2 = radicalbig ( 5) 2 + ( 5) 2 = 5 radicallow 2. Thus, ( r , θ ) = (5 radicallow 2 , 225 ) . Note that if we had used θ = 45 , then we would have ( r , θ ) = ( 5 radicallow 2 , 45 ). Example6.17 Write the equation x 2 + y 2 = 9 in polar coordinates. Solution: This is just the equation of a circle of radius 3 centered at the origin. Since r radicalbig x 2 + y 2 = ± radicallow 9, in polar coordinates the equation can be written as simply r = 3 . Example6.18 Write the equation x 2 + ( y 4) 2 = 16 in polar coordinates. Solution: This is the equation of a circle of radius 4 centered at the point (0 , 4). Expanding the equation, we get: x 2 + ( y 4) 2 = 16 x 2 + y 2 8 y + 16 = 16 x 2 + y 2 = 8 y r 2 = 8 r sin θ r = 8 sin θ Why could we cancel r from both sides in the last step? Because we know that the point (0
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Unformatted text preview: the circle, so canceling r does not eliminate r = 0 as a potential solution of the equation (since θ = ◦ would make r = 8 sin θ = 8 sin 0 ◦ = 0). Thus, the equation is r = 8 sin θ . Example 6.19 Write the equation y = x in polar coordinates. Solution: This is the equation of a line through the origin. So when x = 0, we know that y = 0. When x n= 0, we get: y = x y x = 1 tan θ = 1 θ = 45 ◦ Since there is no restriction on r , we could have r = 0 and θ = 45 ◦ , which would take care of the case x = 0 (since then ( x , y ) = (0,0), which is the same as ( r , θ ) = (0,45 ◦ )). Thus, the equation is θ = 45 ◦ ....
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