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#1 2005-07-26 23:27:13

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

Integration...

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...

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#2 2005-07-27 00:01:50

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,522

Re: Integration...

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 ??????????? roll


Character is who you are when no one is looking.

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#3 2005-07-27 05:22:03

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

Re: Integration...

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?

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#4 2005-07-27 08:52:31

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

Re: Integration...

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.

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#5 2005-07-27 09:34:29

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,535

Re: Integration...

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 "ƒ" instead of y:

dx = a sin(ƒ(t)) dt

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

dx = ƒ(t) dt

Which is just a general statement.

So, yes, what is y(t) ?


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#6 2005-07-27 21:39:01

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

Re: Integration...

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?

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#7 2005-07-28 02:31:37

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

Re: Integration...

*** 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

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