# Particularly annoying

1. Aug 1, 2008

### latentcorpse

(dx/dt)^2 = (Mgh/A)*(2x-1)*(1-x)^2
where x=cos (theta)
and M,g,h and A are all constants

the solution is

sec(theta) = 1 + sech(sqrt(Mght/A))
if u can do it then could u please list your full answer cos i can get quite far into it and then the sech thing just confuses me

i think it involves using the weierstrass elliptic p function

HELP?!?!?!

2. Aug 1, 2008

### tiny-tim

Hi latentcorpse!

Since they give you the answer :

Hint: substitute y = 1/x - 1, and you get … ?

3. Aug 1, 2008

### latentcorpse

i forgot to mention that x = cos (theta)....my bad

4. Aug 1, 2008

### latentcorpse

also i can't get your substitution to work mainly because i dont see how that helps with the sech - i have little experience with sech so you might need to explain a bit more deeply how your getting that answer to work.

cheers guys

5. Aug 1, 2008

### latentcorpse

also i missed out the t.
ignor that first post.
here we go:

(dx/dt)^2 = (Mgh/A)*(2x-1)*(1-x)^2

the solution is

sec(theta) = 1 + sech((sqrt(Mght/A))t)

where x = cos theta

6. Aug 2, 2008

### tiny-tim

Substitute y = 1/x - 1, and show us what you get.

That'll give you something which you can conveniently take the square-root of.

Just try it!

7. Aug 2, 2008

### latentcorpse

ok, if i try y=1/x - 1, then x=1/(y+1) and dy = -x^(-2) dx so dx = a x^(2) dy

this means dx/dt = sqrt(Mgh/A) (2x-1)^(0.5) (1-x)

becomes

- [(y+1)^(-2) dy]/[(((2/y+1)-1)^(0.5))(1-(1/y+1))] = (mgh/A)^(0.5) t

which gives me the argument of the sech term on the right side but i cant get the rest out!!!

8. Aug 2, 2008

### tiny-tim

Hi latentcorpse!

(have a square-root: √ )
Yes, but why have you stopped there?

What is (2/y+1)-1?

And what is 1-(1/y+1)?

9. Aug 3, 2008

### latentcorpse

well if y=1/x - 1 then

x = 1/(y+1)

and so y+1 = 1/x

then

(2/(y+1))-1 = 2x-1 = 2*cos(theta) - 1

and 1 - 1/(y+1) = 1 - x = 1 - cos(theta)

and then cos(theta)=1/(y+1), we have -sin(theta) d(theta) = -(y+1)^(-2) dy, giving

-[sin(theta) d(theta)]/[((2cos(theta)-1)^(0.5))(1-cos(theta)] = (Mgh/A)^(0.5) t

but how does this help us to integrate? i still can't see how the LHS works?

10. Aug 3, 2008

### tiny-tim

Hi latentcorpse!

Forget theta … it isn't in the question, and it isn't really in the answer (sectheta is just 1/x).

And forget x … the whole point of substituting y for x is to get away from x!

In terms of y:

(2/y+1)-1 = … ?

And 1-(1/y+1) = … ?

11. Aug 3, 2008

### latentcorpse

ok well i can't see any way of rearranging them other than factorising y+1

so that

(2/(y+1)-1)^(0.5) = (y+1)^(0.5)[1-y]^(0.5)
and,
1-1/(y+1)=(y+1)[(y+1)y]

and then substituting in and collecting these(y+1) terms with that one already on the numerator we get

-dy/[(1-y)^(0.5) (y+1)^(0.5) y]

now what?

also i see that sec(theta) is just 1/x but i think the reason im finding this so hard is because im looking so hard for where the sech term is going to come out

12. Aug 3, 2008

### tiny-tim

Hi latentcorpse!
(hmm … what happened to that √ I gave you? )

Now write that as -dy/y√(1- y2).

Then make the obvious substitution.

13. Aug 3, 2008

### latentcorpse

ok
my bad about the root sign and damn i just realised i forgot to copy it again

so the obvious substitution would be some trigonometric function because of that
(1-y^2)^(0.5) term

now by y=1/x - 1 i get y=sec(theta) - 1 but that mucks things up

so i tried just y = cos(theta)

giving dy=-sin(theta) dtheta

then our LHS rearranges to

[sin(theta) dtheta]/[cos(theta) sin(theta)]

which is just the integral of sec(theta) dtheta

and the integral of sec is ln( sec(theta) + tan(theta) ) and surely i want something with a sech^(-1) in it so that i can get my sech(sqrt(Mgh/A)t) term on the RHS.

???

14. Aug 3, 2008

### tiny-tim

Yes, when you have a √(1 - y2), the usual substitution is either sin or cos, because sin2 + cos2 = 1.

But you can also use either sech or tanh, because sech2 + tanh2 = 1 also.

As you say, using cos gives you ln (sec + tan), which is really awkward.

So try sech intead of cos, and see what you get!

15. Aug 3, 2008

### latentcorpse

ok
well if y = sech(theta), dy = -sech(theta) tanh(theta) dtheta

and 1-y^2 = [tanh(theta)]^2 and so the denominator is sech(theta) tanh(theta)

then -dy/[y(1-y^2)^(0.5)] becomes

[sech(theta) tanh(theta) dtheta]/[sech(theta) tanh(theta)] but then everything cancels and we just have the integral of dtheta do we not?

16. Aug 3, 2008

### tiny-tim

Yes!!

And the integral of dtheta is … ?

17. Aug 3, 2008

### latentcorpse

yeah but the integral of dtheta is theta
and that's equal to my right hand side of sqrt(Mgh/A) t
but if u look back to my first post that's not the solution i was aiming for

we're trying to get

sec(theta) = 1 + sech(sqrt(Mgh/A)t)

18. Aug 3, 2008

### latentcorpse

actually might be able to do a part without you!! yay!

is this because if we take sech of both sides

but then sech(theta) = y = sec(theta) -1 and so it all rearranges

fantastic!
cheers for all your help m8!

legend!

19. Aug 3, 2008

### tiny-tim

D'oh!

That's a different theta!!

Try it with phi instead of theta …

EDIT: ah … you beat me to it … well done!!

20. Aug 4, 2008

### Marin

Hi there!

Here's my solution:

$$(\frac{dx}{dt})^2=\frac{mgh}{A}(2x-1)(1-x)^2$$
$$\frac{dx}{dt}=\sqrt{\frac{mgh}{A}(2x-1)(1-x)^2}$$
$$\frac{dx}{dt}=(1-x)\sqrt{\frac{mgh}{A}(2x-1)}$$
$$\frac{mgh}{A}=K$$
$$\frac{dx}{(1-x)\sqrt{K(2x-1)}}=dt$$
$$\displaystyle{\frac{1}{\sqrt{K}}\int\frac{dx}{(1-x)\sqrt{2x-1}}}=\displaystyle{\int}dt$$

$$2x-1=z^2 <=> x=\frac{z^2+1}{2}$$
$$2dx=2zdz <=> dx=zdz$$

$$\displaystyle{\int}\frac{1}{(1-x)\sqrt{2x-1}} dx}=\displaystyle{\int}\frac{z}{(1-\frac{z^2+1}{2})z} dz=\displaystyle{2\int}\frac{1}{1-z^2} dz=\displaystyle{2\int}(\frac{1}{2(1-z)}-\frac{1}{2(1+z)}) dz=\displaystyle{\int}(\frac{1}{(1-z)}-\frac{1}{(1+z)}) dz=\ln{\frac{1-z}{1+z}$$

$$z=\sqrt{2x-1}$$

$$\ln{\frac{1-z}{1+z}=\ln{\frac{\sqrt{2x-2}}{\sqrt{2x}}=\ln\sqrt{\frac{2-2x}{2x}}=\displaystyle{\frac{1}{2}}\ln{2\frac{1-x}{x}}=\displaystyle{\frac{1}{2}}\ln{\frac{1-x}{x}}+c$$

$$\displaystyle{\frac{1}{\sqrt{K}}\int\frac{dx}{(1-x)\sqrt{2x-1}}}=\displaystyle{\int}dt$$
$$\frac{1}{2\sqrt{K}}\ln{\frac{1-x}{x}}+c=t$$
$$\ln{\frac{1-x}{x}}=2\sqrt{K}(t-c)$$
$$\displaystyle{\frac{1-x}{x}}=e^{2\sqrt{K}(t-c)}$$
$$1-x=xe^{2\sqrt{K}(t-c)}$$
$$x(1+e^{2\sqrt{K}(t-c)})=1$$
$$x=\displaystyle{\frac{1}{1+e^{2\sqrt{K}(t-c)}}$$
$$\cos{\theta}=\displaystyle{\frac{1}{1+e^{2\sqrt{K}(t-c)}}$$
$$\sec{\theta}=1+e^{2\sqrt{K}(t-c)}$$
$$\sec{\theta}=1+e^{2\sqrt{\frac{mgh}{A}}(t-c)}$$

which is very similar to the solution given :) - s.o. has to prove the last term