Question 1. A vessel contains gasoline that flows out from a...

Question 1. A vessel contains gasoline that flows out from a small hole at point O in the base of the vessel. At time $t$, in minutes, the height of the gasoline surface above O is $y$, and the rate at which $y$ is decreasing is inversely proportional to $y^{3/2}$. Write down the differential equation which expresses $\frac{dy}{dt}$ in terms of $y$. Given that $y = 25$ when $t = 0$ and that $y = 16$ when $t = 30$, find the value of $t$ when $y = 0$.

Transcript

0:00 Hi there.

00:01 So for this problem, we have a cylindrical tank that is initially filled with water to a death of 16 feet.

00:11 So we are given the initial condition that the death at zero at the time equals to zero is 16 feet.

00:20 And a bulb in the bottom is opened and the water runs out.

00:25 The death age of the water in the time decreases at a rate.

00:31 Proportional to the square root of the death.

00:33 So that is the following.

00:36 The rate change of the death is equal to minus a constant of proportionality k times the square root of h, where key is a constant between zero and one.

00:55 And t is the time expressed in hours.

00:59 So for part a of this problem, what we need to find is the solution of the differential equation for age in terms of the constant proportionality k and the time.

01:13 So what we need to do is to separate the variables in this differential equation.

01:20 So we pass the time to the right and the functions that depends on the death to the left.

01:29 So we obtain in here the integral of the different in the depth divided by the square root of h.

01:38 And this is equal to the integral of minus k times the differential in time.

01:48 Now, in the left side of this equation, we obtain that this is the integral of 1 over the square root of of h that we can also write as 1 over h elevated to 1 over 2.

02:10 So when we apply the integral of an exponent, we obtained that this will give us 2 times the square root of h, and this is equal to minus k times the time.

02:26 And because we are integrated in here, because the integral is indefinite, we need to add a constant of integration that we call c.

02:36 So solving for h, we obtain that the depth h is equal to 1 over 4 times minus k times the time plus c to the square.

02:51 Now we need to determine the value of the constant of integration c.

02:56 And for that, we use the initial value that at 0, this should be equal to 16.

03:02 Now if we substitute in here the time equals to 0, we obtain 1 over 4 times c square.

03:12 So from this we obtain that c square is equal to 64.

03:18 And if we take the square root of this, we obtain that c is equal to 8.

03:23 So with this, we can conclude for the first part of this problem, that the death in this cylindrical tank is 1 over 4 times minus k times the time plus 8 and that to the square.

03:45 So that's a solution for part a of this problem...

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