Use E q .(4) and Torricelli's Law [E q .(5)]. Atankhas the...

Use $E q .(4)$ and Torricelli's Law $[E q .(5)]$. Atankhas the shape of the parabola $y=a x^{2}$ (where $a$ is a constant) revolved around the $y$-axis. Water drains from a hole of area $B \mathrm{m}^{2}$ at the bottom of the tank. (a) Show that the water level at time $t$ is $$y(t)=\left(y_{0}^{3 / 2}-\frac{3 a B \sqrt{2 g}}{2 \pi} t\right)^{2 / 3}$$ where $y_{0}$ is the water level at time $t=0$. (b) Show that if the total volume of water in the tank has volume $V$ at time $t=0,$ then $y_{0}=\sqrt{2 a V / \pi} .$ Hint: Compute the volume of the tank as a volume of rotation. (c) Show that the tank is empty at time $$t_{e}=\left(\frac{2}{3 B \sqrt{g}}\right)\left(\frac{2 \pi V^{3}}{a}\right)^{1 / 4}$$ We see that for fixed initial water volume $V,$ the time $t_{e}$ is proportional to $a^{-1 / 4} .$ A large value of $a$ corresponds to a tall thin tank. Such a tank drains more quickly than a short wide tank of the same initial volume.

Use $E q .(4)$ and Torricelli's Law $[E q .(5)]$.

Atankhas the shape of the parabola $y=a x^{2}$ (where $a$ is a constant) revolved around the $y$-axis. Water drains from a hole of area $B \mathrm{m}^{2}$ at the bottom of the tank.
(a) Show that the water level at time $t$ is
$$y(t)=\left(y_{0}^{3 / 2}-\frac{3 a B \sqrt{2 g}}{2 \pi} t\right)^{2 / 3}$$
where $y_{0}$ is the water level at time $t=0$.
(b) Show that if the total volume of water in the tank has volume $V$ at time $t=0,$ then $y_{0}=\sqrt{2 a V / \pi} .$ Hint: Compute the volume of the tank as a volume of rotation.
(c) Show that the tank is empty at time
$$t_{e}=\left(\frac{2}{3 B \sqrt{g}}\right)\left(\frac{2 \pi V^{3}}{a}\right)^{1 / 4}$$
We see that for fixed initial water volume $V,$ the time $t_{e}$ is proportional to $a^{-1 / 4} .$ A large value of $a$ corresponds to a tall thin tank. Such a tank drains more quickly than a short wide tank of the same initial volume.

Key Concepts

Torricelli's Law

This is a principle in fluid dynamics that relates the speed of fluid flowing out of an orifice to the height of the fluid above it. It is typically derived from Bernoulli’s equation and states that the velocity of the fluid exiting a hole is proportional to the square root of the gravitational acceleration multiplied by the height of the fluid. This law provides the fundamental relationship between the water level in the tank and the speed at which water drains, which is essential when setting up the differential equation for the rate of change of the water level.

Volume of Revolution

This concept is used in calculus to calculate the volume of a solid formed by revolving a curve around an axis. When a tank is described by the rotation of a curve, such as a parabola, about an axis, the volume of the tank can be computed by integrating the area of circular cross-sections along the axis of revolution. The technique is particularly useful for determining relationships between water volume and height, which is necessary for setting the initial conditions in the draining problem.

Differential Equations in Fluid Dynamics

In the context of a draining tank, differential equations are used to model the change in water level over time. These equations incorporate physical principles such as Torricelli’s Law and the relationship between the volume of fluid and its height in the tank. By setting up a balance between the rate of change of volume and the outflow through the orifice, one can derive and solve a differential equation that describes how the water level decreases over time.

Geometry-Dependent Flow Characteristics

The shape of the tank, which is defined by its geometric parameters, directly affects how the water drains. In problems involving tanks with shapes generated by rotating curves (such as a parabola), the parameter defining the curve’s steepness influences the cross-sectional area at different heights. This geometric relationship plays a crucial role in determining the draining time, with different shapes (tall thin vs. short wide) resulting in different rates of decrease in the water level even for the same initial water volume.

Transcript

00:01 That's in the shape of the parabola y equals a x squared rotated around the y axis has a hole in it area b and we know at time zero the height is y zero and the volume is v zero and we'll use that in a little bit to find out how long it takes to empty the tank with respect to the initial volume okay so first we've got to fix our formula up we got to find out what the area of a slice is.

00:33 Well, a slice here is a circle, so its area is pi r squared.

00:41 R is x, so pi x squared, which will be pi y over a.

00:51 So, d, y, dt equals minus b, squared to 2g, y to the one -half, pi y over a, which will put the a on the top.

01:05 So, d, y, dt minus a, b, square root of 2g, pi over y to the one -half, because these two cancel.

01:18 Separate the variables, and you get y to the one -half, d -y equals minus a, b, square to 2g, pi, d -t.

01:32 Integrate, two -thirds, y to the three - halves, equals minus a, b, square, to 2g pi t plus c all right now initially at times zero y is y zero so two -thirds y -0 to the three -haves equals oh wait before i do that let's multiply those three halves over there or multiply both sides by three halves so we have just white to the three halves here so minus three a b square root of 2g over 2 pi plus c all right now let's plug in um that's a t there plus c let's plug in t equals 0 and get y sub 0 so y sub 0 to the 3 halves equals 0 plus c so then y to the 3 halves equal y 0 to the 3 halves minus 3ab okay now i'm going to simplify this square root of two and this square root of two or this two and leave a square root of two pi i still got a square to g and a t there okay y zero okay minus three a b square to two g by t all right and then let's just raise both sides to the two -thirds power all right so that's my formula for how much water is in there at any time t.

03:33 And that's with respect to y zero.

03:36 So let's fix it.

03:37 So instead of the initial height, let's figure with the initial volume.

03:43 Okay, so the volume of that is, we can find it by integrating.

03:50 Let's use the disk method.

03:52 Okay, here's a disk, volume of a disk, pi r squared h, r is x, h is d .y.

04:06 So the volume is pi integral 0 to y not x squared, which is y over a, d .y.

04:18 And that's equal to v .0 because i put y sub zero there.

04:26 Okay, so v sub zero equals pi a, y squared over 2, from zero to y sub zero.

04:35 So v -0 equals pi over 2a y -subs -0 squared.

04:44 So y -sub -0 equals 2a v -0 over pi to the one -half power...