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anonymous

  • one year ago

A cooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet. The diameter at the base is 300 m and the minimum diameter, 500 m above the base, is 200 m. Find an equation for the tower. (Assume the center is at the origin with axis the z-axis and the minimum diameter is at the center.)

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  1. Clalgee
    • one year ago
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    Alright, Since minimum diameter is at z = 500, then centre of hyperboloid is (0, 0, 500). Right? Now here is the equation for hyperboloid: \[x^2/A^2 + y^2/B^2 - (z - 500)^2 / C^2 = 1\] Horizontal Cross-sections are circles in other words: A = B New equation: \[x^2/A^2 + y^2/A^2 - (z - 500)^2 / C^2 = 1\]

  2. Clalgee
    • one year ago
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    When z = 500, radius = 200/2 = 100 \[x^2 / A^2 + y^2 / B^2 - (500 - 500)^2 / C^2 = 1\] \[x^2 / A^2 + y^2 /A^2 = 1\] A= 100

  3. Clalgee
    • one year ago
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    When z = 0, radius = 300/2 = 150 (cross-section: x² + y² = 150²) \[x^2 / 100^2 + y^2 / 100^2 - (0 - 500)^2 / C^2 = 1\] \[x^2 / 100^2 + y^2 = 500^2 / C^2 + 1\] \[x^2 + y^2 = 100^2 (500^2 / C^2 + 1) = 150^2\] \[100^2 (500^2 / C^2 + 1) = 150^2 \] \[(500^2 / C^2 + 1) = 150^2 /100^2 = 9/4 \] \[500^2 / C^2 = 5/4\] \[C^2 / 500^2 = 4/5 \] ^- Multiply Above \[C^2 = 200,000 \] Lastly: x²/10,000 + y²/10,000 − (z−500)²/200,000 = 1

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