= = log 0.6732, and so on. r θ ≈ Observe that as increases its value, so does Therefore the matching graph should be C. r. θ 45. We can make a of 0 2π 3π θ π values of points satisfying 1 0 -1 0 r the equation to see which of the given graphs contains those points. So the matching graph should be D. 47. We can make a 0 2π θ π table of values of points satisfying the 1 1 + -1 1 + 1 1 - 3 1 - 1 r equation to see which of the given graphs contains those points. So the matching graph should be F. 49. = 3cos(θ) = 3rcos(θ) + = 3x in Cartesian coordinates. Using the method of r r2 x2 y2 completing square: , or . This is an equation of circle centered at with radius . Therefore the graph looks like: 51. We’ll start with a table of 0 2π θ π values, and plot them on the graph. 0 3 0 -3 0 3 0 -3 0 r So a graph of the equation is a 4-leaf rose.
θ = r = log ≈ 0.6732, and so on. Observe that as θ increases its value, so does r. Therefore the matching graph should be C. θ 0 π 2π 3π r 1 0 -1 0 45. We can make a of values of points satisfying the equation to see which of the given graphs contains those points. So the matching graph should be D. θ 0 π 2π r 1 1 + -1 1 + 1 1 - 3 1 - 1 47. We can make a table of values of points satisfying the equation to see which of the given graphs contains those points. So the matching graph should be F. 49. r = 3cos(θ) r2 = 3rcos(θ) x2 + y2 = 3x in Cartesian coordinates. Using the method of completing square: , or . This is an equation of circle centered at with radius . Therefore the graph looks like: θ 0 π 2π r 0 3 0 -3 0 3 0 -3 0 51. We’ll start with a table of values, and plot them on the graph. So a graph of the equation is a 4-leaf rose.
Chapter%208%20Solutions.docx