Differential Equation Hemispherical Bowl: show that k=3pia^2/2T given that x=0 when t=T and x=a when t=0.

Skelly4444

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Trying to solve the DE below and show that k=3pia^2/2T given that x=0 when t=T and x=a when t=0.

The DE given in the question is pi(2ax-x^2)dx/dt = -kx

Every time I try to solve it, I get the same answer which is not like the one I'm supposed to be showing.
Any advice would be greatly received.
 

MarkFL

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Hello, and welcome to FMH! :)

I would write the ODE as:

\(\displaystyle \frac{\pi}{k}(x-2a)\,dx=dt\)

Integrating, we then get:

\(\displaystyle \frac{\pi}{2k}x\left(x-4a\right)=t+C\)

Now, using the given point \((t,x)=(T,0)\) we get:

\(\displaystyle 0=T+C\implies C=-T\)

And so we have:

\(\displaystyle \frac{\pi}{2k}x\left(x-4a\right)=t-T\)

Next, using the given point \((t,x)=(0,a)\) we get:

\(\displaystyle \frac{\pi}{2k}a\left(a-4a\right)=-T\)

Solving this for \(k\), we obtain:

\(\displaystyle k=\frac{3\pi a^2}{2T}\quad\checkmark\)
 

Skelly4444

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Good morning Mark,
Thank you for your swift response.

Your integration decision slightly confuses me, as I always taught to separate the constant k and integrate the right hand side with respect to t.

When I separated my variables, I had pi(2ax-x^2) / x = -kt.

How can you write the bracketed terms as (x-2a)?

A little confused here if I'm honest.

Thanks
Simon
 

MarkFL

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It seems you've introduced the variable \(t\) and dropped the derivative \(\displaystyle \d{x}{t}\) from the ODE. We begin with:

\(\displaystyle \pi(2ax-x^2)\d{x}{t}=-kx\)

Divide through by \(-kx\):

\(\displaystyle \frac{\pi}{k}(x-2a)\d{x}{t}=1\)

Now, this may be written in the differential form:

\(\displaystyle \frac{\pi}{k}(x-2a)\,dx=dt\)

Does that make sense?
 

topsquark

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It seems you've introduced the variable \(t\) and dropped the derivative \(\displaystyle \d{x}{t}\) from the ODE. We begin with:

\(\displaystyle \pi(2ax-x^2)\d{x}{t}=-kx\)

Divide through by \(-kx\):

\(\displaystyle \frac{\pi}{k}(x-2a)\d{x}{t}=1\)

Now, this may be written in the differential form:

\(\displaystyle \frac{\pi}{k}(x-2a)\,dx=dt\)
There is a detail here and I don't know how to fix it.

The original equaiton is \(\displaystyle \pi(2ax-x^2)\d{x}{t}=-kx\) and an initial condition is (t, x) = (T, 0). But in your solution you divided both sides of the differential equation by x, which can't be done at the initial condition, as x = 0.

Could we use the initial condition as a limit? It's not quite correct but the only thing I can think of.

-Dan
 

MarkFL

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There is a detail here and I don't know how to fix it.

The original equaiton is \(\displaystyle \pi(2ax-x^2)\d{x}{t}=-kx\) and an initial condition is (t, x) = (T, 0). But in your solution you divided both sides of the differential equation by x, which can't be done at the initial condition, as x = 0.

Could we use the initial condition as a limit? It's not quite correct but the only thing I can think of.

-Dan
Good catch, Dan. I realized I was losing the trivial solution \(x\equiv 0\), but didn't notice it coincided with an initial value. I suspect the author of the problem overlooked this as well. Offhand, I don't know how to fix this either. :)
 

Skelly4444

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Hi Mark,
I understand what you have done but I wasn't aware we could divide through by -kx and just leave a 1 on the right hand side with respect to t.
Like I said in my previous message, I always left k on the right side when separating the variables.
Thanks again.
Simon
 
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