Question about water container with hole at the bottom

[songoku]

Homework Statement

Let the function A = f(x) , 0 ≤ x ≤ X, gives the area (in meter square) of the top surface of water in a container to a depth of x meters. Initially, the container filled with water, and there is a small
hole with area p meter square at the bottom of the container.

(a) Find a function V = g(x), the volume of water in the tank (in meter cube) in terms of its depth.

(b) Use a suitable fundamental theorem to express the rate of change of volume with respect to time.

(c) Find the function t = y(x) for the time for the water to flows out from the container until the container is empty (assume initially it is filled to depth x), by taking the speed of the water to be $v=\sqrt{2gx}$

(d) Find the time function for container in the shape of a paraboloid of revolution with height X meters and radius R meters. Assume the tank opens up, like a typical bowl.

(e) The container in part (d) is filled to its maximum depth. What fraction of the total emptying time does it take for half the water to drain from the container?

Relevant Equations

Differentiation

Integration

(a)
$$V=\int_{0}^{x} A~dx$$
$$=\int_{0}^{x} f(u) dx , \text{u is dummy variable}$$

Is this the answer? Or there is something else I can do to continue the working?

(b)
$$\frac{dV}{dt}=\frac{d}{dt} \int_{0}^{x} f(u) dx=f(x).\frac{dx}{dt}$$

Is this correct?

(c)
$$\text{time}=\frac{\text{rate of change of volume}}{\text{area of hole} \times \text{speed}}$$
$$=\frac{\int_{0}^{x} f(u) dx}{p \sqrt{2gx}}$$

Can this be simplified further?

(d) I googled paraboloid of revolution to know the shape

So I need to find the equation of this shape to be able to find the time?

Thanks

 

[Orodruin]

a) You cannot have $x$ as your integration variable at the same time as it is one of the boundaries of the integral. You will need to use a different variable for the integration variable.

b) It is difficult to judge whether your reasoning here is correct due to the issues with (a). The end result is correct.

c) This is not correct. You need to solve the differential equation in (b).

d) A paraboloid by definition has the form $z = k r^2$.

 

[songoku]

a) You cannot have $x$ as your integration variable at the same time as it is one of the boundaries of the integral. You will need to use a different variable for the integration variable.

Ah yes, I suppose you mean $dx$ should be $du$?

My teacher also said the same thing but without explanation. Can you tell me why integration variable can not be the same as boundary of the integral?

b) It is difficult to judge whether your reasoning here is correct due to the issues with (a). The end result is correct.

$$\frac{dV}{dt}=\frac{d}{dt} \int_{0}^{x}f(u)~du=f(x).\frac{dx}{dt}$$

c) This is not correct. You need to solve the differential equation in (b).

$$\frac{dV}{dt}=\frac{d}{dt} \int_{0}^{x}f(u)~du=f(x).\frac{dx}{dt}$$
$$\int_{0}^{x}\frac{1}{f(b).\sqrt{2gb}}dV=\int_{0}^{t}dw$$
$$\text{where b and w are dummy variables to represent depth and time}$$

Is this what you mean? If yes, how to solve the integration in left hand side? The variable is $b$ but with respect to $dV$

Thanks

 

[Orodruin]

My teacher also said the same thing but without explanation. Can you tell me why integration variable can not be the same as boundary of the integral?

You are integrating with respect to the integration variable, which is therefore a dummy variable that will disappear upon the integration. The $x$ is something that the entire expression depends upon.

$$\frac{dV}{dt}=\frac{d}{dt} \int_{0}^{x}f(u)~du=f(x).\frac{dx}{dt}$$
$$\int_{0}^{x}\frac{1}{f(b).\sqrt{2gb}}dV=\int_{0}^{t}dw$$
$$\text{where b and w are dummy variables to represent depth and time}$$

Is this what you mean? If yes, how to solve the integration in left hand side? The variable is $b$ but with respect to $dV$

I do not understand what you did for the second line, nor what $b$ is supposed to represent. You have an expression for $dV/dt$ as a function of $x$. What you need to do is to insert that into the differential equation in the first line and solve the resulting separable differential equation.

 

[songoku]

I do not understand what you did for the second line, nor what $b$ is supposed to represent. You have an expression for $dV/dt$ as a function of $x$. What you need to do is to insert that into the differential equation in the first line and solve the resulting separable differential equation.

$$\frac{dV}{dt}=f(x).\frac{dx}{dt}$$
$$\frac{dV}{f(x).\frac{dx}{dt}}=dt$$
$$\frac{dV}{f(x).v}=dt$$
$$\frac{dV}{f(x)\sqrt{2gx}}=dt$$

then integrate both sides.

This is what I did, although I am not sure that this is what you meant.

Do I need to use the area of hole at the bottom of the container (which is $p ~ m^2$)? I thought I need to use that to find the time for the water to flow out from that hole so I substituted $f(x)$ to $p$

Thanks