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To convert the given Cartesian coordinates [tex]\((\sqrt{3}, -\sqrt{7})\)[/tex] into polar coordinates, we follow these steps:
### Step 1: Calculate the radius [tex]\( r \)[/tex]
The radius [tex]\( r \)[/tex] is the distance from the origin to the point [tex]\((\sqrt{3}, -\sqrt{7})\)[/tex], which can be found using the Pythagorean theorem:
[tex]\[ r = \sqrt{x^2 + y^2} \][/tex]
Substituting the given values [tex]\( x = \sqrt{3} \)[/tex] and [tex]\( y = -\sqrt{7} \)[/tex]:
[tex]\[ r = \sqrt{ (\sqrt{3})^2 + (-\sqrt{7})^2 } \][/tex]
[tex]\[ r = \sqrt{3 + 7} \][/tex]
[tex]\[ r = \sqrt{10} \][/tex]
However, after calculating this with more precision, the radius [tex]\( r \)[/tex] is approximately 3.1622776601683795.
### Step 2: Calculate the angle [tex]\( \theta \)[/tex]
The angle [tex]\( \theta \)[/tex] can be found using the arctangent function, but since the point is in the fourth quadrant, we use the `atan2(y, x)` function which considers the signs of both coordinates to determine the correct quadrant:
[tex]\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \][/tex]
Given [tex]\( x = \sqrt{3} \)[/tex] and [tex]\( y = -\sqrt{7} \)[/tex]:
[tex]\[ \theta = \tan^{-1}\left(\frac{-\sqrt{7}}{\sqrt{3}}\right) \][/tex]
Since [tex]\( y \)[/tex] is negative, we subtract the value from [tex]\( 2 \pi \)[/tex] to get a positive angle in the correct range [tex]\([0, 2\pi]\)[/tex]:
[tex]\[ \theta = 2\pi - \tan^{-1}\left(\sqrt{\frac{7}{3}}\right) \][/tex]
The precise calculation reveals that the angle [tex]\( \theta \)[/tex] is approximately 5.292028720748394.
### Conclusion
Thus, the polar coordinates [tex]\((r, \theta)\)[/tex] for the given Cartesian coordinates [tex]\((\sqrt{3}, -\sqrt{7})\)[/tex] are approximately [tex]\((3.1622776601683795, 5.292028720748394)\)[/tex]. Comparing this with the given options:
(a) [tex]\(\left(\sqrt{10}, 2 \pi + \tan^{-1}\left(\sqrt{\frac{7}{3}}\right)\right)\)[/tex]
(b) [tex]\(\left(\sqrt{10}, 2 \pi - \tan^{-1}\left(\sqrt{\frac{7}{3}}\right)\right)\)[/tex]
(c) [tex]\(\left(10, \pi - \tan^{-1}\left(\sqrt{\frac{7}{3}}\right)\right)\)[/tex]
(d) [tex]\(\left(10, 2 \pi - \tan^{-1}\left(\sqrt{\frac{7}{3}}\right)\right)\)[/tex]
After matching the radius and angle, the correct option is:
(b) [tex]\(\left(\sqrt{10}, 2 \pi - \tan^{-1}\left(\sqrt{\frac{7}{3}}\right)\right)\)[/tex].
### Step 1: Calculate the radius [tex]\( r \)[/tex]
The radius [tex]\( r \)[/tex] is the distance from the origin to the point [tex]\((\sqrt{3}, -\sqrt{7})\)[/tex], which can be found using the Pythagorean theorem:
[tex]\[ r = \sqrt{x^2 + y^2} \][/tex]
Substituting the given values [tex]\( x = \sqrt{3} \)[/tex] and [tex]\( y = -\sqrt{7} \)[/tex]:
[tex]\[ r = \sqrt{ (\sqrt{3})^2 + (-\sqrt{7})^2 } \][/tex]
[tex]\[ r = \sqrt{3 + 7} \][/tex]
[tex]\[ r = \sqrt{10} \][/tex]
However, after calculating this with more precision, the radius [tex]\( r \)[/tex] is approximately 3.1622776601683795.
### Step 2: Calculate the angle [tex]\( \theta \)[/tex]
The angle [tex]\( \theta \)[/tex] can be found using the arctangent function, but since the point is in the fourth quadrant, we use the `atan2(y, x)` function which considers the signs of both coordinates to determine the correct quadrant:
[tex]\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \][/tex]
Given [tex]\( x = \sqrt{3} \)[/tex] and [tex]\( y = -\sqrt{7} \)[/tex]:
[tex]\[ \theta = \tan^{-1}\left(\frac{-\sqrt{7}}{\sqrt{3}}\right) \][/tex]
Since [tex]\( y \)[/tex] is negative, we subtract the value from [tex]\( 2 \pi \)[/tex] to get a positive angle in the correct range [tex]\([0, 2\pi]\)[/tex]:
[tex]\[ \theta = 2\pi - \tan^{-1}\left(\sqrt{\frac{7}{3}}\right) \][/tex]
The precise calculation reveals that the angle [tex]\( \theta \)[/tex] is approximately 5.292028720748394.
### Conclusion
Thus, the polar coordinates [tex]\((r, \theta)\)[/tex] for the given Cartesian coordinates [tex]\((\sqrt{3}, -\sqrt{7})\)[/tex] are approximately [tex]\((3.1622776601683795, 5.292028720748394)\)[/tex]. Comparing this with the given options:
(a) [tex]\(\left(\sqrt{10}, 2 \pi + \tan^{-1}\left(\sqrt{\frac{7}{3}}\right)\right)\)[/tex]
(b) [tex]\(\left(\sqrt{10}, 2 \pi - \tan^{-1}\left(\sqrt{\frac{7}{3}}\right)\right)\)[/tex]
(c) [tex]\(\left(10, \pi - \tan^{-1}\left(\sqrt{\frac{7}{3}}\right)\right)\)[/tex]
(d) [tex]\(\left(10, 2 \pi - \tan^{-1}\left(\sqrt{\frac{7}{3}}\right)\right)\)[/tex]
After matching the radius and angle, the correct option is:
(b) [tex]\(\left(\sqrt{10}, 2 \pi - \tan^{-1}\left(\sqrt{\frac{7}{3}}\right)\right)\)[/tex].
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