Convert (3, -1) from rectangular coordinates to polar coordinates. Use radian measure and round your answers to three decimal places. 2 Sin II) =zx\/zv=D 17. (2pts) Convert the equation from polar form to rectangular form. r = tan 8. sec 18. (2pts) Convert the equation from rectangular form to polar form and solve for r. 2×2 + 3y2 = 6x<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To convert rectangular coordinates to polar coordinates, we use the following formulas:<\/p>\n\n\n\n Given the point ((x, y) = (3, -1)):<\/p>\n\n\n\n Thus, the polar coordinates are approximately: Using the relationships:<\/p>\n\n\n\n We rewrite the equation: Multiplying both sides by (x^2) gives: In polar coordinates:<\/p>\n\n\n\n Substituting these into the equation: This simplifies to: Factoring out (r^2): Dividing both sides by (r) (assuming (r \\neq 0)): Thus, the polar form of the equation is: We successfully converted the rectangular coordinates (3, -1) into polar coordinates ((3.162, -0.322)). We then transformed the polar equation (r = \\tan(\\theta) \\sec(\\theta)) into rectangular form, yielding (x^2 = y). Finally, the equation (2x^2 + 3y^2 = 6x) was converted into polar coordinates as (r = \\frac{6 \\cos(\\theta)}{2\\cos^2(\\theta) + 3\\sin^2(\\theta)}). These conversions illustrate the relationship between the rectangular and polar systems, making it easier to analyze geometric properties.<\/p>\n","protected":false},"excerpt":{"rendered":" Convert (3, -1) from rectangular coordinates to polar coordinates. Use radian measure and round your answers to three decimal places. 2 Sin II) =zx\/zv=D 17. (2pts) Convert the equation from polar form to rectangular form. r = tan 8. sec 18. (2pts) Convert the equation from rectangular form to polar form and solve for r. 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[
r = \\sqrt{x^2 + y^2}
]<\/li>\n\n\n\n
[
\\theta = \\tan^{-1}\\left(\\frac{y}{x}\\right)
]<\/li>\n<\/ol>\n\n\n\nPart 1: Converting (3, -1) to Polar Coordinates<\/h3>\n\n\n\n
\n
[
r = \\sqrt{3^2 + (-1)^2} = \\sqrt{9 + 1} = \\sqrt{10} \\approx 3.162
]<\/li>\n\n\n\n
[
\\theta = \\tan^{-1}\\left(\\frac{-1}{3}\\right) \\approx -0.3217 \\text{ radians}
]
Since this angle is in the fourth quadrant, we can keep it as is, or we can convert it to a positive angle by adding (2\\pi):
[
\\theta = -0.3217 + 2\\pi \\approx 5.9615 \\text{ radians}
]<\/li>\n<\/ol>\n\n\n\n
[
(r, \\theta) \\approx (3.162, -0.322)
]<\/p>\n\n\n\nPart 2: Convert (r = \\tan(\\theta) \\sec(\\theta)) to Rectangular Form<\/h3>\n\n\n\n
\n
[
r = \\tan(\\theta) \\sec(\\theta) = \\frac{y}{x} \\cdot \\frac{r}{x}
]<\/p>\n\n\n\n
[
xr = y \\implies x^2 = y
]<\/p>\n\n\n\nPart 3: Convert (2x^2 + 3y^2 = 6x) to Polar Form<\/h3>\n\n\n\n
\n
[
2(r \\cos(\\theta))^2 + 3(r \\sin(\\theta))^2 = 6(r \\cos(\\theta))
]<\/p>\n\n\n\n
[
2r^2 \\cos^2(\\theta) + 3r^2 \\sin^2(\\theta) = 6r \\cos(\\theta)
]<\/p>\n\n\n\n
[
r^2(2\\cos^2(\\theta) + 3\\sin^2(\\theta)) = 6r \\cos(\\theta)
]<\/p>\n\n\n\n
[
r(2\\cos^2(\\theta) + 3\\sin^2(\\theta)) = 6 \\cos(\\theta)
]<\/p>\n\n\n\n
[
r = \\frac{6 \\cos(\\theta)}{2\\cos^2(\\theta) + 3\\sin^2(\\theta)}
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n