Well hello to all my new followers! Hope you like $\LaTeX{}$ as i'm particularly fond of it and so you will see it all over my blog. My exams are finally over for this semester, and as a treat, I am going to show you another question I had to answer! However its too late tonight and I have a 9am tomorrow..... So heres a question to keep you all intrigued...

I have two children, and at least one of them is a son. What is the probability that the other is also a son?

These are the whimsical thoughts that I shall be having during my times at university...

## Monday, 31 January 2011

## Sunday, 23 January 2011

### One side of hand-written notes only...

It's finally over! My week of dreaded exams... Not that they went to plan. Now with most exams, it is normally customary for the paper to be proof read, i.e before the actual exam is sat. Well that certainly does NOT seem to be the case with the folks here at the University of Portsmouth! Thats right, we were due to sit our exam at 11:15 on a Thursday... At 11:16 - I raised my hand with a puzzled look on my face.

'I'm afraid theres a mistake with the paper'

Looking through, every symbol had been replaced with a black dot. Every bracket, $\pi$, $\rho$, $\Omega$,.. well you get the idea. What with this being an exam on astrophysics, those symbols are pretty much used through out, and so the whole exam was a write off. after half an hour, it was finally decided that the exam would be cancelled, and so we all had to leave one at a time (as there were still exams happening) which took another 10 minutes. It was finally decided that another exam would be held tomorrow (the Friday).

After we vented our rage upon the Head of Department, and various other people, we eventually managed to sit our exam. As I went to the trouble of learning it so much, I'm actually going to include one of the questions I had to answer in this blog, and I'll also show the sheet of notes I took in... (the blog title is a clue).

1) Show that a sphere of uniform density will collapse to zero radius in a "free-fall" time of

\[t_{FF} = \left[\frac{3\pi}{32G\rho}\right ] ^{\frac{1}{2}}\]

We can start solving this by stating the following formula:

\[\rho\frac{d^{2}r}{dt^{2}}=-\frac{GM}{r^{2}}\rho-\frac{dP}{dr}\]

We are told in the question that we are working in uniform density. As of that, $\rho$ can be 'ignored' as it is a constant. WE also know that we are in free-fall, so $P=0$. That leaves us with the following:

\[\frac{d^{2}r}{dt^{2}}=-\frac{GM}{r^{2}}\]

We can start by multiplying through by $\frac{dr}{dt}$, to give:

\[\frac{dr}{dt}\left(\frac{d^{2}r}{dt^{2}\right)}=-\frac{GM}{r^{2}}\frac{dr}{dt}\]

We can integrate both sides of this equation with respect to $t$, however before we do that, it is possible to make some substitutions on the Left hand side.

\[x=\frac{dr}{dt}\]

\[\frac{dx}{dt}=\frac{d^{2}r}{dt^{2}}\]

We can now re-write as:

\[\int x\left(\frac{dx}{dt}\right)dt=- \int \frac{GM}{r^{2}}\left(\frac{dr}{dt}\right)dt\]

We can cancel the $dt's$ on both sides of the equation, which leaves us with:

\[\int x dx=- \int \frac{GM}{r^{2}}dr\]

\[\Rightarrow\frac{x^{2}}{2}=\frac{GM}{r}+c\]

We can state that when $t=0, r=r_{0}$, where $r_{0}$ is the starting radius. We stated above that $x=\frac{dr}{dt}$, and so now $t=0$, we can rewrite the above formula to show that:

\[c=-\frac{GM}{r_{0}}\]

\[\therefore \frac{x^{2}}{2}=\frac{GM}{r}-\frac{GM}{r_{0}}\]

\[\Rightarrow x^{2}=\frac{2GM}{r}-\frac{2GM}{r_{0}}\]

\[\Rightarrow \left(\frac{dr}{dt}\right)^{2}=\frac{2GM}{r}-\frac{2GM}{r_{0}}\]

Now we need to integrate, with $t=t_{FF}$, and $r=0$

\[t_{FF}\Rightarrow \int_0^{t_{FF}} dt= \int_{r_{0}}^0 \left(\frac{dt}{dr}\right)dr\]

By inversing the right hand side, we can show that

\[\int_{r_{0}}^0 \left(\frac{dr}{dt}\right)^{-1}dr\]

Now we are basically going to perform some shuffling about with the equations, which involves square-rooting $\frac{dr}{dt}$, and then inversing. This can be shown:

\[\left(\frac{dr}{dt}\right)^{2}=\frac{2GM}{r}-\frac{2GM}{r_{0}}\Rightarrow \frac{dr}{\sqrt{\frac{2GM}{r}-\frac{2GM}{r_{0}}}}=dt\]

We can no integrate this, though we can take out some constants to make life a bit easier.

Take out:

\[-\frac{1}{\sqrt{2GM}}\]

To give

\[-\frac{1}{\sqrt{2GM}}\int_{r_{0}}^0 \left(\frac{r}{1-\frac{r}{r_{0}}}\right)^{\frac{1}{2}}dr\]

We can now introduce some more variables in which we can sub in:

\[r=r_{0}\sin^{2}\theta\]

\[dr=2r_{0}\cos\theta\sin\theta d\theta\]

If we substite $r$ into $\left(\frac{r}{1-\frac{r}{r_{0}}}\right)$, we get the following:

\[\left(\frac{r}{1-\frac{r}{r_{0}}}\right) \Rightarrow\left(\frac{r_{0}\sin^{2}\theta}{1-\sin^{2}\theta}\right)=\left(\frac{r_{0}\sin^{2}\theta}{\cos^{2}\theta}\right)=\sqrt{r_{0}}\tan\theta\]

Now that has been simplified we can substitute our new values into the intergral, which gives us the following:

\[t_{FF}=-\frac{1}{\sqrt{2GM}}\int_{r_{0}}^0\left(2r_{0}\cos\theta\sin\theta\sqrt{r_{0}}\tan\theta d\theta\]

this can be simplified further by saying:

\[2r_{0}\times r_{0}^{\frac{1}{2}}=2r_{0}^{\frac{3}{2}}\]

and

\[\tan\theta\times\cos\theta\times\sin\theta=\sin^{2}\theta\]

\[\sin^{2}\theta=\frac{1-\cos 2\theta}{2}\]

We can take out from the intergral that $2r_{0}^{\frac{3}{2}}$ is a constant, leaving us with:

\[-\frac{2r_{0}^{3}}{\sqrt{2GM}}\int_{r_{0}}^0\frac{1}{2}\left(1-\cos 2\theta\right)d\theta\]

so when $r=r_{0}$, $\sin^{2}\theta=1\Rightarrow\theta=\frac{\pi}{2}$

\[\therefore =-\frac{2r_{0}^{3}}{\sqrt{2GM}}\left[\frac{1}{2}\theta-\frac{1}{4}\sin2\theta\right]_{2\pi}^{0}\]

As we know both of the limits would cancel out $\sin$, we can cancel through on the second half of the intergral, which we then get left with:

\[\frac{\pi}{4}\left(\frac{-2r_{0}^{3}}{GM}\right)^\frac{1}{2}\]

Once again stating formula, we can say that:

\[M=\frac{4}{3}\pi r_{0}^{3}\rho\]

We can substitute that into above to give:

\[\frac{\pi}{4}\left( \frac{-2r_{0}^{3}}{{G\left(\frac{4}{3}\pi r_{0}^{3}\rho}}\right)}\right)^{\frac{1}{2}}\]

This is where the magic happens - you can see that the $r_{0}^{3}$'s can cancel out, we can take the $\frac{\pi}{4}$ into the square root section by squaring (and then we can cancel the bottom $\pi$, giving:

\[\left(\frac{-2\pi}{16G{\frac{4}{3}}\rho}\right)\]

taking the $3$ upto the top, and then dividing through by $2$ can give:

\[\left(\frac{3\pi}{32G\rho}\right)^{\frac{1}{2}}\]

Which is what we were after!

'I'm afraid theres a mistake with the paper'

Looking through, every symbol had been replaced with a black dot. Every bracket, $\pi$, $\rho$, $\Omega$,.. well you get the idea. What with this being an exam on astrophysics, those symbols are pretty much used through out, and so the whole exam was a write off. after half an hour, it was finally decided that the exam would be cancelled, and so we all had to leave one at a time (as there were still exams happening) which took another 10 minutes. It was finally decided that another exam would be held tomorrow (the Friday).

After we vented our rage upon the Head of Department, and various other people, we eventually managed to sit our exam. As I went to the trouble of learning it so much, I'm actually going to include one of the questions I had to answer in this blog, and I'll also show the sheet of notes I took in... (the blog title is a clue).

1) Show that a sphere of uniform density will collapse to zero radius in a "free-fall" time of

\[t_{FF} = \left[\frac{3\pi}{32G\rho}\right ] ^{\frac{1}{2}}\]

We can start solving this by stating the following formula:

\[\rho\frac{d^{2}r}{dt^{2}}=-\frac{GM}{r^{2}}\rho-\frac{dP}{dr}\]

We are told in the question that we are working in uniform density. As of that, $\rho$ can be 'ignored' as it is a constant. WE also know that we are in free-fall, so $P=0$. That leaves us with the following:

\[\frac{d^{2}r}{dt^{2}}=-\frac{GM}{r^{2}}\]

We can start by multiplying through by $\frac{dr}{dt}$, to give:

\[\frac{dr}{dt}\left(\frac{d^{2}r}{dt^{2}\right)}=-\frac{GM}{r^{2}}\frac{dr}{dt}\]

We can integrate both sides of this equation with respect to $t$, however before we do that, it is possible to make some substitutions on the Left hand side.

\[x=\frac{dr}{dt}\]

\[\frac{dx}{dt}=\frac{d^{2}r}{dt^{2}}\]

We can now re-write as:

\[\int x\left(\frac{dx}{dt}\right)dt=- \int \frac{GM}{r^{2}}\left(\frac{dr}{dt}\right)dt\]

We can cancel the $dt's$ on both sides of the equation, which leaves us with:

\[\int x dx=- \int \frac{GM}{r^{2}}dr\]

\[\Rightarrow\frac{x^{2}}{2}=\frac{GM}{r}+c\]

We can state that when $t=0, r=r_{0}$, where $r_{0}$ is the starting radius. We stated above that $x=\frac{dr}{dt}$, and so now $t=0$, we can rewrite the above formula to show that:

\[c=-\frac{GM}{r_{0}}\]

\[\therefore \frac{x^{2}}{2}=\frac{GM}{r}-\frac{GM}{r_{0}}\]

\[\Rightarrow x^{2}=\frac{2GM}{r}-\frac{2GM}{r_{0}}\]

\[\Rightarrow \left(\frac{dr}{dt}\right)^{2}=\frac{2GM}{r}-\frac{2GM}{r_{0}}\]

Now we need to integrate, with $t=t_{FF}$, and $r=0$

\[t_{FF}\Rightarrow \int_0^{t_{FF}} dt= \int_{r_{0}}^0 \left(\frac{dt}{dr}\right)dr\]

By inversing the right hand side, we can show that

\[\int_{r_{0}}^0 \left(\frac{dr}{dt}\right)^{-1}dr\]

Now we are basically going to perform some shuffling about with the equations, which involves square-rooting $\frac{dr}{dt}$, and then inversing. This can be shown:

\[\left(\frac{dr}{dt}\right)^{2}=\frac{2GM}{r}-\frac{2GM}{r_{0}}\Rightarrow \frac{dr}{\sqrt{\frac{2GM}{r}-\frac{2GM}{r_{0}}}}=dt\]

We can no integrate this, though we can take out some constants to make life a bit easier.

Take out:

\[-\frac{1}{\sqrt{2GM}}\]

To give

\[-\frac{1}{\sqrt{2GM}}\int_{r_{0}}^0 \left(\frac{r}{1-\frac{r}{r_{0}}}\right)^{\frac{1}{2}}dr\]

We can now introduce some more variables in which we can sub in:

\[r=r_{0}\sin^{2}\theta\]

\[dr=2r_{0}\cos\theta\sin\theta d\theta\]

If we substite $r$ into $\left(\frac{r}{1-\frac{r}{r_{0}}}\right)$, we get the following:

\[\left(\frac{r}{1-\frac{r}{r_{0}}}\right) \Rightarrow\left(\frac{r_{0}\sin^{2}\theta}{1-\sin^{2}\theta}\right)=\left(\frac{r_{0}\sin^{2}\theta}{\cos^{2}\theta}\right)=\sqrt{r_{0}}\tan\theta\]

Now that has been simplified we can substitute our new values into the intergral, which gives us the following:

\[t_{FF}=-\frac{1}{\sqrt{2GM}}\int_{r_{0}}^0\left(2r_{0}\cos\theta\sin\theta\sqrt{r_{0}}\tan\theta d\theta\]

this can be simplified further by saying:

\[2r_{0}\times r_{0}^{\frac{1}{2}}=2r_{0}^{\frac{3}{2}}\]

and

\[\tan\theta\times\cos\theta\times\sin\theta=\sin^{2}\theta\]

\[\sin^{2}\theta=\frac{1-\cos 2\theta}{2}\]

We can take out from the intergral that $2r_{0}^{\frac{3}{2}}$ is a constant, leaving us with:

\[-\frac{2r_{0}^{3}}{\sqrt{2GM}}\int_{r_{0}}^0\frac{1}{2}\left(1-\cos 2\theta\right)d\theta\]

so when $r=r_{0}$, $\sin^{2}\theta=1\Rightarrow\theta=\frac{\pi}{2}$

\[\therefore =-\frac{2r_{0}^{3}}{\sqrt{2GM}}\left[\frac{1}{2}\theta-\frac{1}{4}\sin2\theta\right]_{2\pi}^{0}\]

As we know both of the limits would cancel out $\sin$, we can cancel through on the second half of the intergral, which we then get left with:

\[\frac{\pi}{4}\left(\frac{-2r_{0}^{3}}{GM}\right)^\frac{1}{2}\]

Once again stating formula, we can say that:

\[M=\frac{4}{3}\pi r_{0}^{3}\rho\]

We can substitute that into above to give:

\[\frac{\pi}{4}\left( \frac{-2r_{0}^{3}}{{G\left(\frac{4}{3}\pi r_{0}^{3}\rho}}\right)}\right)^{\frac{1}{2}}\]

This is where the magic happens - you can see that the $r_{0}^{3}$'s can cancel out, we can take the $\frac{\pi}{4}$ into the square root section by squaring (and then we can cancel the bottom $\pi$, giving:

\[\left(\frac{-2\pi}{16G{\frac{4}{3}}\rho}\right)\]

taking the $3$ upto the top, and then dividing through by $2$ can give:

\[\left(\frac{3\pi}{32G\rho}\right)^{\frac{1}{2}}\]

Which is what we were after!

## Friday, 14 January 2011

### A lovely little Question I Like

Well its been a quite time on here, that can only mean it's exam time. As i'm in final year I've been taking it full force - university at 10 am every morning, decent breakfast, the whole 9 yards. the best part is that I feel great!. As I have been revising, that unfortunatly means nothing of interest will occur. Although Next weekend (22$^{nd}$ of January) should be interesting...

In the meantime, here is one of my favourite simple maths questions, that I find really enjoyable.

A Curious Exchange.

Mrs. Smith: The product of their ages is 36, and the sum of their ages is the address on our door here.

So analysing the question, we are told "The product of their ages is 36"

So we can say that we need 3 numbers that have the product of 36.

1 - 1 - 36

1 - 2 - 18

1 - 3 - 12

1 - 4 - 9

1 - 6 - 6

2 - 2 - 9

2 - 3 - 6

Now we have this list, we then analyse the second part of what we are told. " and the sum of their ages is the address on our door here"

Now we take the sum of all these:

We can see that 2 of the groups of numbers sum to 13. We now no that the ages of the 3 daughters are either (1, 6, 6) or (2, 2, 9).

Like the census taker in the question, we cannot yet work out who the ages, however what we are told next makes that possible. "My eldest daughter has red hair"

As a 'Red' herring, we can finally say that the age of the daughters are 2, 2, and 9. This is as we know that there is an eldest daughter, so we can eliminate the other set of numbers.

Isnt it nice?

In the meantime, here is one of my favourite simple maths questions, that I find really enjoyable.

A Curious Exchange.

Mrs. Smith: The product of their ages is 36, and the sum of their ages is the address on our door here.

Census Taker: (after some figuring) I’m afraid I can’t determine their ages from that …

Mrs. Smith: My eldest daughter has red hair.

Census Taker: Oh, thanks, now I know.

How old are the three girls?

If this were a blog that people actually read and commented on, then I would have probably delayed the answer for a while, however as I am probably the only one to read this, I shall give the answer below.

I would like to hope that people would at least attempt this though.

Its easier than you think.

So analysing the question, we are told "The product of their ages is 36"

So we can say that we need 3 numbers that have the product of 36.

1 - 1 - 36

1 - 2 - 18

1 - 3 - 12

1 - 4 - 9

1 - 6 - 6

2 - 2 - 9

2 - 3 - 6

Now we have this list, we then analyse the second part of what we are told. " and the sum of their ages is the address on our door here"

Now we take the sum of all these:

1 - 1 - 36 = 38

1 - 2 - 18 = 21

1 - 3 - 12 = 16

1 - 4 - 9 = 14

1 - 6 - 6 = 13

2 - 2 - 9 = 13

2 - 3 - 6 = 11

We can see that 2 of the groups of numbers sum to 13. We now no that the ages of the 3 daughters are either (1, 6, 6) or (2, 2, 9).

Like the census taker in the question, we cannot yet work out who the ages, however what we are told next makes that possible. "My eldest daughter has red hair"

As a 'Red' herring, we can finally say that the age of the daughters are 2, 2, and 9. This is as we know that there is an eldest daughter, so we can eliminate the other set of numbers.

Isnt it nice?

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