Convert (3, -1) from rectangular coordinates to polar coordinates

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

The Correct Answer and Explanation is :

To convert rectangular coordinates to polar coordinates, we use the following formulas:

1. Polar Radius (r):
   [
   r = \sqrt{x^2 + y^2}
   ]
2. Polar Angle (θ):
   [
   \theta = \tan^{-1}\left(\frac{y}{x}\right)
   ]

Part 1: Converting (3, -1) to Polar Coordinates

Given the point ((x, y) = (3, -1)):

1. Calculate (r):
   [
   r = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \approx 3.162
   ]
2. Calculate (θ):
   [
   \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}
   ]

Thus, the polar coordinates are approximately:
[
(r, \theta) \approx (3.162, -0.322)
]

Part 2: Convert (r = \tan(\theta) \sec(\theta)) to Rectangular Form

Using the relationships:

• (\tan(\theta) = \frac{y}{x})
• (\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{r}{x})

We rewrite the equation:
[
r = \tan(\theta) \sec(\theta) = \frac{y}{x} \cdot \frac{r}{x}
]

Multiplying both sides by (x^2) gives:
[
xr = y \implies x^2 = y
]

Part 3: Convert (2x^2 + 3y^2 = 6x) to Polar Form

In polar coordinates:

• (x = r \cos(\theta))
• (y = r \sin(\theta))

Substituting these into the equation:
[
2(r \cos(\theta))^2 + 3(r \sin(\theta))^2 = 6(r \cos(\theta))
]

This simplifies to:
[
2r^2 \cos^2(\theta) + 3r^2 \sin^2(\theta) = 6r \cos(\theta)
]

Factoring out (r^2):
[
r^2(2\cos^2(\theta) + 3\sin^2(\theta)) = 6r \cos(\theta)
]

Dividing both sides by (r) (assuming (r \neq 0)):
[
r(2\cos^2(\theta) + 3\sin^2(\theta)) = 6 \cos(\theta)
]

Thus, the polar form of the equation is:
[
r = \frac{6 \cos(\theta)}{2\cos^2(\theta) + 3\sin^2(\theta)}
]

Conclusion

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.