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.)

Question

clalgee · Answer

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: $$x^2/A^2 + y^2/A^2 - (z - 500)^2 / C^2 = 1$$

clalgee · Answer

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

clalgee · Answer

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