You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**GurraTedden****Member**- Registered: 2005-07-20
- Posts: 19

Hi guys, I'm stuck again!

And it's mathematical, like always (regret those lazy days back in analytic math courses, haha).

I'm in this situation:

dx=asin(y)dt

(a is a constant value)

where y=y(t)

I want to solve this for x.

smthing like x = smthing + C (constant of integration, or what ever you want to call it)

I can recall it had something to do about, maybe substitution, but i can't see how thats going to happen...

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,935

GurraTedden wrote:

dx=asin(y)dt

(a is a constant value)

where y=y(t)

∫dx = ∫ asinyt dt

∫dx = a ∫sinyt dt

x = a(-cosyt/y) + Constant ???????????

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**GurraTedden****Member**- Registered: 2005-07-20
- Posts: 19

I'm not convinced... i dont get it really. Do you mead x = a(-cos(yt/y)) ?

How do i differentiate that one and get the same thing in the brickets sin(y).

Ganesh... maybe i didn't make my self clear that y is a function of t.?? Or is it

me that have misunderstood you? Coud you show me how you diferentiate the

expression?

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

dx=asiny(t)dt

∫dx= ∫asiny(t)dt

x=a ∫sin y(t)dt

Do we know what the function of t is? If so, the rest would be much easier...

Why did the vector cross the road?

It wanted to be normal.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,619

Forgive me if I am wrong, but it may simply be impossible without knowing the function y(t).

dx = a sin(y(t)) dt

Let us just call the funtion "" instead of y:

dx = a sin((t)) dt

If, *for example*, there were no sine or factor applied, we could have:

dx = (t) dt

Which is just a general statement.

So, yes, what is y(t) ?

Offline

**GurraTedden****Member**- Registered: 2005-07-20
- Posts: 19

So, can't we express it in the first derivates of y(t) ? Thats what i'm looking for.

some kind of chainrule or smthing?

Offline

**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

*** This is my interpretation of the problem:

x=x(t)

y=y(t)

dx=a.sin(y(t))dt

I want to solve this for x.

dx/dt = a.sin( y(t) )

∫ (dx/dt)dt = ∫a.sin( y(t) )dt

∫ dx = ∫a.sin( y(t) )dt

x(t)+C=∫a.sin( y(t) )dt

*** if..

y(t)=a.t

then

dy/dt=a

*** therefore...

∫a.sin( y(t) )dt = ∫a.sin( a.t) )dt

Let z=a.t => dz/dt = a

∫a.sin( a.t) )dt = ∫(dz/dt)sin(z)dt = ∫sin(z)dz = -cos(z)+C = -cos(a.t)+C

*** putting it all together...

x(t) = -cos(a.t)+C

y(t)= a.t

dx/dt= a.sin(a.t)

dx= a.sin(a.t) dt

dx=a sin(y(t)) dt <=> dx=a sin(y) dt

***I think this is a possible solution for your problem

Offline

Pages: **1**