What is the radius of the circular hole in the bottom of the water tank?

[mailman85]

A water tank has the shape obtained by revolving the parabola x^2=by around the y-axis. The water depth is 4ft at noon when a circular plug at the bottom of the tank is removed. At 1pm the depth of the water is 1ft.
a)find the depth y(t) of water remaining after t hours
b)when will the tank be empty?
c)if the initial radius of the top surface of teh water is 2ft, what is the radius of the circuluar hole in the bottom

by using dV/dt=-a(2y)^(1/2) where a is the area of the bottom of the hole of the tank, I solved parts a and b. I found a) y=(8-7t)^(2/3) and b) 1.143 hrs. I am having trouble with part c. Can you help? Thanks.

 

[HallsofIvy]

Since you were able to do parts (a) and (b), I presume that you arrived at the general result that y(t)= (C- ([sqrt](2)a/([pi]b))t)². Then, since y(0)= 4 and y(1)= 1, that C= 1 and
[sqrt](2)a/([pi]b)= 3 so that y(t)= (4- 3t)² and the tank will be empty at 1:20 p.m.

Knowing that, at 12:00, when the height of the water was 4 feet, the top had radius 2 ft. Allows you to find b: the tank was formed by rotating the graph of x²= by around the y axis: when x (the radius) is 2, y (the height) is 4: 2²= b(4) so b=1.

You also know that [sqrt](2)a/([pi]b)= 3. Since b=1, you can easily solve for a.

 

[tacman]

To find the radius of the circular hole in the bottom of the water tank, we can use the formula for the volume of a paraboloid, V=(1/2)πr^2h, where r is the radius of the top surface of the water and h is the depth of the water.

Since we know the depth of the water at 1pm is 1ft, we can substitute this into the equation and solve for r:

1 = (1/2)πr^2(1)
1 = (1/2)πr^2
2 = πr^2
r^2 = 2/π
r = √(2/π) = 0.7979 ft

Therefore, the radius of the circular hole in the bottom of the water tank is approximately 0.7979 ft.