For the rectangular coordinates ((5sqrt2)/(2),-(5sqrt2)/(2)), (a) plot the point and (b) give two pairs of polar coordinates for the point, where 0^(@) <= theta < 360^(@). (a) Choose the correct plot of ((5sqrt2)/(2),-(5sqrt2)/(2)) below. A. B. c. D. (b) Give two pairs of polar coordinates for the point, where 0^(@) <= theta < 360^(@). (Type an ordered pair. Simplify your answer, including any radicals. Use integers or fractions for any numbers in the text.
The Correct Answer and Explanation is:
Let’s solve this step by step:
(a) Plotting the point (522,−522)(\frac{5\sqrt{2}}{2}, -\frac{5\sqrt{2}}{2})(252,−252) in rectangular coordinates:
The given point is in rectangular coordinates (x,y)=(522,−522)(x, y) = \left(\frac{5\sqrt{2}}{2}, -\frac{5\sqrt{2}}{2}\right)(x,y)=(252,−252). Here’s how we can plot it:
- The xxx-coordinate is positive, and the yyy-coordinate is negative, meaning the point lies in the fourth quadrant of the coordinate plane.
- The magnitude of both coordinates is the same (52/25\sqrt{2}/252/2), so the point is equidistant from the xxx- and yyy-axes.
(b) Converting to polar coordinates:
Polar coordinates (r,θ)(r, \theta)(r,θ) are given by:
- r=x2+y2r = \sqrt{x^2 + y^2}r=x2+y2, the radial distance from the origin.
- θ=tan−1(yx)\theta = \tan^{-1}\left(\frac{y}{x}\right)θ=tan−1(xy), the angle with respect to the positive xxx-axis.
- Finding rrr:
r=(522)2+(−522)2r = \sqrt{\left(\frac{5\sqrt{2}}{2}\right)^2 + \left(-\frac{5\sqrt{2}}{2}\right)^2}r=(252)2+(−252)2 r=(52)24+(52)24=504+504=1004=25=5r = \sqrt{\frac{(5\sqrt{2})^2}{4} + \frac{(5\sqrt{2})^2}{4}} = \sqrt{\frac{50}{4} + \frac{50}{4}} = \sqrt{\frac{100}{4}} = \sqrt{25} = 5r=4(52)2+4(52)2=450+450=4100=25=5
So, r=5r = 5r=5.
- Finding θ\thetaθ:
θ=tan−1(yx)=tan−1(−522522)=tan−1(−1)\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-\frac{5\sqrt{2}}{2}}{\frac{5\sqrt{2}}{2}}\right) = \tan^{-1}(-1)θ=tan−1(xy)=tan−1(252−252)=tan−1(−1)
Since the point is in the fourth quadrant, the angle corresponding to tan−1(−1)\tan^{-1}(-1)tan−1(−1) is θ=−45∘\theta = -45^\circθ=−45∘, or equivalently, θ=315∘\theta = 315^\circθ=315∘ in the range 0∘≤θ<360∘0^\circ \leq \theta < 360^\circ0∘≤θ<360∘.
Thus, the first pair of polar coordinates is (r,θ)=(5,315∘)(r, \theta) = (5, 315^\circ)(r,θ)=(5,315∘).
- Second pair of polar coordinates:
In polar coordinates, we can also represent the point by adding 180∘180^\circ180∘ to the angle (this is the same point, but with the opposite direction for the angle):
θ′=315∘+180∘=495∘\theta’ = 315^\circ + 180^\circ = 495^\circθ′=315∘+180∘=495∘
However, we want θ\thetaθ to be in the range 0∘≤θ<360∘0^\circ \leq \theta < 360^\circ0∘≤θ<360∘, so we subtract 360∘360^\circ360∘ from 495∘495^\circ495∘: 495∘−360∘=135∘495^\circ – 360^\circ = 135^\circ495∘−360∘=135∘
Thus, the second pair of polar coordinates is (r,θ)=(5,135∘)(r, \theta) = (5, 135^\circ)(r,θ)=(5,135∘).
Final Answer:
- (a) The point (522,−522)\left(\frac{5\sqrt{2}}{2}, -\frac{5\sqrt{2}}{2}\right)(252,−252) lies in the fourth quadrant.
- (b) Two pairs of polar coordinates for the point are: (5,315∘)and(5,135∘)(5, 315^\circ) \quad \text{and} \quad (5, 135^\circ)(5,315∘)and(5,135∘)
