00:01
Hi there, so for this problem, we are given that the radius of this cylinder, let's label this as capital r, is equal to 3 feet, and its height is also given, let's label this as age, that is equal to 10 feet, and is standing on one end.
00:25
So if the 10 is initially full of water and the water leads from a circular hold with a radius, that we're going to label this, the radius lower, that is equal to 3 divided by 8 inches.
00:38
3 divided by 8 inches, okay? so with that set, first of all, we start by writing the volume of this cylinder.
00:56
We know that the volume of a cylinder is pi times its radius square times the height.
01:04
And then once we have this, we can solve for the height.
01:09
So the height is equal to, because remember that what we are asking here is to determine a differential equation for the height h of the water at a times t.
01:19
So h is going to be the volume divided by pi times the radius square.
01:28
Once we have that, we can take the rid of change of the, we can take the derivative in both sides of this expression with respect the time.
01:38
So for the left side, we obtain the rate of change of age with respect to time.
01:42
And for the right side, we obtain 1 divided by pi times the radius square, and this times the root of change of the volume with respect to time.
01:52
Now, we must first find the rate of change of the volume with respect to time, which means the rate at which the water flows out of the tank.
02:01
So first of all, we know that this is just simply equal to the product between the a, which is the area of the hold, and b, which is the speed of the water flowing out.
02:12
So we can write this as minus pi times the radius square, and then this times the speed.
02:23
Now, we must find an expression for the velocity.
02:26
So let us apply the conservation of energy for the potential energy of the water at a high age.
02:33
So we know that the kinetic energy, which is 1 divided by 2 times the mass times the speed square, equal to the potential energy, which is the mass times the acceleration due to gravity times the height.
02:46
So we can cancel the mass in here, it's solving for d sb, that will be the squared root of 2 times the acceleration due to gravity times the height.
02:55
Then we can substitute that into the rate of change of the volume with respect to time.
03:00
So the rate of change of the volume with respect to time now becomes minus pi times the radius squared, this times the rate of change of two times acceleration due to gravity times the height.
03:10
And then once we have this, we just substitute this expression into the rate of change of the height with respect to time.
03:19
So the rate of change of the height with respect to time now becomes 1 divided by pi times the radius squared, and this times minus pi times the radius of the whole to the square, this times the square root of two times the acceleration due to gravity times the height.
03:35
So simplifying things in here, as you can see we can cancel pie with this.
03:43
Then we will have the rate of change of the height with respect to time is equal to yes, so the root of change of this is the radius, well, minus the radius of the whole squared by the radius of the cylinder squared, this times the square root of two times the acceleration due to gravity times the height...