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Convert the point \( (x, y, z)=(2,3,4) \) to cylindrical coordinates. Give answers either as expressions, or decimals to two decimal places, with positive values for \( \theta \) and \( r \). \( (r, \theta, z)= \)

Ask by Hill Gonzalez. in the United States
Jan 27,2025

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Answer

\( (r, \theta, z) = (3.61, 0.98, 4) \)

Solution

To convert the Cartesian coordinates \((x, y, z) = (2, 3, 4)\) to cylindrical coordinates \((r, \theta, z)\), follow these steps: 1. **Calculate \( r \):** \[ r = \sqrt{x^2 + y^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61 \] 2. **Calculate \( \theta \):** \[ \theta = \arctan\left(\frac{y}{x}\right) = \arctan\left(\frac{3}{2}\right) \approx 0.98 \text{ radians} \] (Alternatively, in degrees: \( \theta \approx 56.31^\circ \)) 3. **Retain \( z \):** \[ z = 4 \] **Cylindrical Coordinates:** \[ (r, \theta, z) = \left(3.61, 0.98, 4\right) \]

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To convert the Cartesian coordinates \( (x, y, z) = (2, 3, 4) \) to cylindrical coordinates, we first calculate \( r \) and \( \theta \). The radius \( r \) is calculated as: \[ r = \sqrt{x^2 + y^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61 \] Next, we find the angle \( \theta \) using the arctangent function: \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{3}{2}\right) \approx 0.982 \text{ radians} \] Remember, since both \( x \) and \( y \) are positive, \( \theta \) remains in the first quadrant. The height \( z \) remains the same. Putting all together, the cylindrical coordinates are: \[ (r, \theta, z) = \left(\sqrt{13}, \tan^{-1}\left(\frac{3}{2}\right), 4\right) \approx (3.61, 0.98, 4) \]

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