Questions: Express the point given in Cartesian coordinates in cylindrical coordinates (r, θ, z). A) (9(1/2), 9(sqrt(3)/2),-6)=() () () B) (-9(1/2), 9(sqrt(3)/2),-6)=() () () C) (9(1/2),-9(sqrt(3)/2),-6)=() () () D) (-9(1/2),-9(sqrt(3)/2),-6)=() () ()

Express the point given in Cartesian coordinates in cylindrical coordinates (r, θ, z).
A) (9(1/2), 9(sqrt(3)/2),-6)=()
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B) (-9(1/2), 9(sqrt(3)/2),-6)=()
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C) (9(1/2),-9(sqrt(3)/2),-6)=()
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D) (-9(1/2),-9(sqrt(3)/2),-6)=()
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Transcript text: (1 point) Express the point given in Cartesian coordinates in cylindrical coordinates $(r, \theta, z)$. A) $\left(9\left(\frac{1}{2}\right), 9\left(\frac{\sqrt{3}}{2}\right),-6\right)=($ $\square$ $\square$ $\square$ ) B) $\left(-9\left(\frac{1}{2}\right), 9\left(\frac{\sqrt{3}}{2}\right),-6\right)=($ $\square$ $\square$ $\square$ ) C) $\left(9\left(\frac{1}{2}\right),-9\left(\frac{\sqrt{3}}{2}\right),-6\right)=($ $\square$ $\square$ $\square$ ) D) $\left(-9\left(\frac{1}{2}\right),-9\left(\frac{\sqrt{3}}{2}\right),-6\right)=($ $\square$ $\square$ $\square$ )
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Solution

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Solution Steps

To convert Cartesian coordinates \((x, y, z)\) to cylindrical coordinates \((r, \theta, z)\), we use the following formulas:

  1. \( r = \sqrt{x^2 + y^2} \)
  2. \( \theta = \text{atan2}(y, x) \)
  3. The \( z \)-coordinate remains the same.

We will apply these formulas to each of the given points.

Step 1: Convert Point A

For point A, given the Cartesian coordinates \( \left(9\left(\frac{1}{2}\right), 9\left(\frac{\sqrt{3}}{2}\right), -6\right) \):

  • Calculate \( r \): \[ r = \sqrt{\left(9\left(\frac{1}{2}\right)\right)^2 + \left(9\left(\frac{\sqrt{3}}{2}\right)\right)^2} = 9.0 \]
  • Calculate \( \theta \): \[ \theta = \text{atan2}\left(9\left(\frac{\sqrt{3}}{2}\right), 9\left(\frac{1}{2}\right)\right) = 1.0472 \text{ radians} \]
  • The \( z \)-coordinate remains the same: \( z = -6 \).

Thus, the cylindrical coordinates for point A are \( (9.0, 1.0472, -6) \).

Step 2: Convert Point B

For point B, given the Cartesian coordinates \( \left(-9\left(\frac{1}{2}\right), 9\left(\frac{\sqrt{3}}{2}\right), -6\right) \):

  • Calculate \( r \): \[ r = \sqrt{\left(-9\left(\frac{1}{2}\right)\right)^2 + \left(9\left(\frac{\sqrt{3}}{2}\right)\right)^2} = 9.0 \]
  • Calculate \( \theta \): \[ \theta = \text{atan2}\left(9\left(\frac{\sqrt{3}}{2}\right), -9\left(\frac{1}{2}\right)\right) = 2.0944 \text{ radians} \]
  • The \( z \)-coordinate remains the same: \( z = -6 \).

Thus, the cylindrical coordinates for point B are \( (9.0, 2.0944, -6) \).

Step 3: Convert Point C

For point C, given the Cartesian coordinates \( \left(9\left(\frac{1}{2}\right), -9\left(\frac{\sqrt{3}}{2}\right), -6\right) \):

  • Calculate \( r \): \[ r = \sqrt{\left(9\left(\frac{1}{2}\right)\right)^2 + \left(-9\left(\frac{\sqrt{3}}{2}\right)\right)^2} = 9.0 \]
  • Calculate \( \theta \): \[ \theta = \text{atan2}\left(-9\left(\frac{\sqrt{3}}{2}\right), 9\left(\frac{1}{2}\right)\right) = -1.0472 \text{ radians} \]
  • The \( z \)-coordinate remains the same: \( z = -6 \).

Thus, the cylindrical coordinates for point C are \( (9.0, -1.0472, -6) \).

Final Answer

  • Point A: \( \boxed{(9.0, 1.0472, -6)} \)
  • Point B: \( \boxed{(9.0, 2.0944, -6)} \)
  • Point C: \( \boxed{(9.0, -1.0472, -6)} \)
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