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 position of the hyperboloid is such that the center is at the origin with its axis along the z-axis, and the minimum diameter at the center.)

Step-by-step explanation:

Equation for a hyperboloid of one sheet, with center at the origen and axis along z-axis is:

(x/a)²  +  (y/b)²   -  (z/c)²  =  1                         (1)

We have to find a , b, and c

We can express equation (1)

(x/a)²  +  (y/b)²    =  (z/c)² + 1                 (2)

Now if we cut the hyperboloid with planes parallel to xy plane we get for  z = k       ( K = 1 , 2 , 3  and so on ) circles of different radius

(x/a)²  +  (y/b)²    =  (k/c)² + 1

at z = k = 0 at the base of the hyperboloid  d = 300   or r = 150 m

we have

(x/a)²  +  (y/b)²   = 1      

x²  +  y²   =   a²                a² = (150)²       a = b = 150

and    x²  +  y²  = (150)²

Now the other condition is at 200 m above the base d = 500 m   r = 250 m  minimum diameter then in equation (2)  we have:

(x/a)²  +  (y/b)²    =  (z/c)² + 1        

(1/a)² [ x² + y² ]  = (z/c)² + 1  

but   x²  +  y²  = r²    and in this case   r  =  250 m  then

(250)²/(150)²   =  (z/c)² + 1    ⇒ (62500/ 22500)  =  (200/c)² + 1

2,78  =  40000/c² + 1

2.78c²  =  40000  + c²

1.78c² = 40000

c²  =  40000/1.78

c²  = 22471.91

c = 149,91

Then we finally have the equation:

x²/(150)²   + y² /(150)² - z²/149,91  = 1

10% of increase is the final answer! : )

phosphate, calcium ion, chloride ion, neon atom

step-by-step explanation: