Assistance needed with a Torricelli's Law differential equation

[bitrex]

I would be grateful for any assistance with the following Torricelli's law problem :yuck:

Homework Statement

I have a cylindrical tank that is oriented horizontally, with an end radius of 3 feet and a length of 5 feet. (It's a math text, so no SI units for me.) The tank is half-full of liquid. A circular hole with a radius of 1 inch is opened in the bottom of the tank. How long will it take for the tank to drain?

Homework Equations

Torricelli's Law:

$$A(y)\frac{dy}{dt} = -a\sqrt{2gy}$$

Where $$A(y)$$ is the surface area of the liquid in the container as a function of height y, $$a$$ is the area of the hole at the bottom of the tank, and g = $$32 ft/sec^2$$

Length of a chord of a circle: $$2\sqrt{r^2-d^2}$$

The Attempt at a Solution

The surface area of the liquid in the tank is a series of rectangles going downwards following the contour of the circular ends, so $$A(y)$$ with y going downward is $$[5*6 - 5(2\sqrt{3^2-y^2})]$$. The area of the hole in the bottom of the tank is $$\pi(1/12)^2$$. When y = 3 the tank is half full, when y = 0 the tank is empty.

So the differential equation is $$[30-5(2\sqrt{9-y^2})]\frac{dy}/{dt} = -\pi(1/2)^2\sqrt{64y}$$

$$30-10\sqrt{9-y^2} \frac{dy}{dt} = -\frac{\pi}{18}$$

$$30-(90-10y^2)^{1/2} \frac{dy}{dt} = -\frac{\pi}{18}$$

$$\frac{30}{y^{1/2}} - \frac{(90-10y^2)^{1/2}}{y^{1/2}} \frac{dy}{dt} = -\frac{\pi}{18}$$

$$30y^{-1/2} - (90-10y^2)y^{-1} \frac{dy}{dt} = -\frac{\pi}{18}$$

$$30y^{-1/2} - 90y^{-1} - 10y dy = -\frac{\pi}{18} dt$$

I can do this integration, but unfortunately after finding the constant of integration I need to set y = 0 to find the time t for the tank to drain. This is a problem because after integrating the left hand side I have a term that involves ln(y). I've gone wrong somewhere either in the setup or execution of this problem but I can't see it. Thanks in advance for any help!

Edit: I see I made a mistake on the last step where I didn't multiply the first term by dt...perhaps I should have gotten rid of that term earlier? Still doesn't help the Ln(0) issue though...

 

[jbsteele]

SolutionThe surface area of the liquid in the tank is a series of rectangles going downwards following the contour of the circular ends, so A(y) with y going downward is [5*6 - 5(2\sqrt{3^2-y^2})] . The area of the hole in the bottom of the tank is \pi(1/12)^2 . When y = 3 the tank is half full, when y = 0 the tank is empty.So the differential equation is [30-5(2\sqrt{9-y^2})]\frac{dy}{dt} = -\pi(1/2)^2\sqrt{64y} 30-10\sqrt{9-y^2} \frac{dy}{dt} = -\frac{\pi}{18} dt 30-(90-10y^2)^{1/2} \frac{dy}{dt} = -\frac{\pi}{18} dt \frac{30}{y^{1/2}} - \frac{(90-10y^2)^{1/2}}{y^{1/2}} \frac{dy}{dt} = -\frac{\pi}{18} dt Integrating both sides we get \int_3^0{30y^{-1/2} - (90-10y^2)y^{-1} \frac{dy}{dt}} = -\frac{\pi}{18}\int_3^0{dt} 30\sqrt{y} - 30\ln(y) - 5(3y - y^3) = -\frac{\pi}{18}t 30\sqrt{0} - 30\ln(0) - 5(3\cdot 0 - 0^3) = -\frac{\pi}{18}t 30\cdot 0 - 0 - 0 = -\frac{\pi}{18}t t = \frac{540}{\pi} seconds

 

[thomate1]

Dear student,

Thank you for reaching out for assistance with your Torricelli's Law problem. It seems like you have made some progress in setting up the differential equation, but there are a few errors that need to be addressed.

Firstly, the surface area of the liquid in the tank should be a function of y, not of time (t). So the expression for A(y) should be [5(2√3^2-y^2)].

Secondly, when solving the differential equation, you have to use the chain rule to find dy/dt. So the correct equation should be:

30y^-1/2 - (90-10y^2)y^-1(dy/dt) = -π/18

To solve this, you can use the substitution u = √y, which will give you a simpler equation to integrate.

Finally, as you have mentioned, the constant of integration should be found by setting y = 0 (since the tank will be empty at that point). However, this will lead to ln(0) as you have pointed out. To avoid this, you can use the limit as y approaches 0 to find the constant of integration.

I hope this helps you in solving your problem. If you need further assistance, do not hesitate to reach out for help. Good luck!