# DiffEq prob 2

1. Sep 14, 2003

### 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.

2. Sep 17, 2003

### HallsofIvy

Staff Emeritus
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.