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Math  /  Trigonometry

QuestionFind the polar coordinates, 0≤θ<2π0 \leq \theta<2 \pi and r≥0r \geq 0, of the point given in Cartesian A) (42,3π4)\left(4 \sqrt{2}, \frac{3 \pi}{4}\right) B) (42,7π4)\left(4 \sqrt{2}, \frac{7 \pi}{4}\right) coordinates. (4,−4)(4,-4) C) (42,π4)\left(4 \sqrt{2}, \frac{\pi}{4}\right) D) (42,5π4)\left(4 \sqrt{2}, \frac{5 \pi}{4}\right)

Studdy Solution

STEP 1

1. Polar coordinates are given by (r,θ)(r, \theta), where rr is the radial distance from the origin and θ\theta is the angle measured from the positive x-axis.
2. Cartesian coordinates are given by (x,y)(x, y).
3. The conversion formulas from Cartesian to polar coordinates are: - r=x2+y2 r = \sqrt{x^2 + y^2} - θ=tan⁡−1(yx) \theta = \tan^{-1}\left(\frac{y}{x}\right)
4. Adjust θ\theta to ensure 0≤θ<2π0 \leq \theta < 2\pi.

STEP 2

1. Verify the Cartesian coordinates.
2. Convert Cartesian coordinates to polar coordinates.
3. Adjust θ\theta if necessary to ensure it lies within the specified range.

STEP 3

Verify the Cartesian coordinates provided in the options. Note that the given Cartesian coordinates for the point are (4,−4)(4, -4).

STEP 4

Calculate the radial distance rr using the formula r=x2+y2 r = \sqrt{x^2 + y^2} .
x=4,y=−4 x = 4, \quad y = -4 r=42+(−4)2 r = \sqrt{4^2 + (-4)^2} r=16+16 r = \sqrt{16 + 16} r=32 r = \sqrt{32} r=42 r = 4\sqrt{2}

STEP 5

Calculate the angle θ\theta using the formula θ=tan⁡−1(yx)\theta = \tan^{-1}\left(\frac{y}{x}\right).
θ=tan⁡−1(−44) \theta = \tan^{-1}\left(\frac{-4}{4}\right) θ=tan⁡−1(−1) \theta = \tan^{-1}(-1) θ=−π4 \theta = -\frac{\pi}{4}

STEP 6

Adjust θ\theta to ensure it lies within the range 0≤θ<2π0 \leq \theta < 2\pi.
Since θ=−π4\theta = -\frac{\pi}{4} is negative, add 2π2\pi to adjust:
θ=−π4+2π \theta = -\frac{\pi}{4} + 2\pi θ=7π4 \theta = \frac{7\pi}{4}
The polar coordinates of the point (4,−4)(4, -4) are:
(42,7π4) \boxed{\left(4\sqrt{2}, \frac{7\pi}{4}\right)}

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