Differential Equation of a hemispherical bowl

[jamesbob]

A hemispherical bowl has a radious of R (metres) and at a time t = 0 is full of water. At that moment a circular hole of radius a (centimetres) is opened in the bottom of the bowl. Let y be the vertical height of the water above the hole at time t. Then y is governed by the differential equation

\pi(Ry-y^2)\frac{dy}{dt} = -\pi(a10^{-2})^2\sqrt{2gy},​

where g = 9.8m/s^2 is gravity.

Solve this differential equation to find y as an implicit function of t.


I need help with this. I am unsure of how to start. Do i try gather all the y terms on one side and everything else on the other?

 

[daveb]

Do you notice the equation is separable, so that f(y)dy = Cdt for some constant C?

 

[jamesbob]

Yes, can i say:

\frac{(Ry-y^2)dy}{\sqrt{2gy}} = \frac{-\pi(a10^{-2})^2dt}{\pi} ?

 

[daveb]

Now you should be able to integrate both sides (don't forget the constant of integration afterwards).

 

[Hurkyl]

I've merged the old thread into this one.

 

[jamesbob]

im having difficulty integrating this. how do i get it into a form that is managable or atleast more recognisable??

thanks

 

[daveb]

You have
(Ry-y^2)dy/sqrt(2gy) = Rydy/sqrt(2gy) - y^2dy/sqrt(2gy). Reduce the fractions, and you should be able to find the antiderivative of each term. Does that help?

 

[jamesbob]

um kinda. I am not 100% sure on integrating them.

would i get:

\frac{2Ry^2}{\sqrt{2gh}} = \frac{3y^2}{\sqrt{2gh}} ??

 

[HallsofIvy]

um kinda. I am not 100% sure on integrating them.

would i get:

\frac{2Ry^2}{\sqrt{2gh}} = \frac{3y^2}{\sqrt{2gh}} ??

Where did "h" come from?? There was no h in anything you wrote before.

Before, you had
\frac{(Ry-y^2)dy}{\sqrt{2gy}}
= \frac{1}{\sqrt{2g}}\left(Ry^{\frac{1}{2}}- y^{\frac{3}{2}}\right)dy
That should be easy to integrate.