Finding a functional connection.

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Discussion Overview

The discussion revolves around finding a functional connection between two variables, u and v, defined in terms of arcsine and square root functions. Participants explore the relationships and derivatives involved, aiming to express one variable as a function of the other.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant presents the equations u=arcsin(x)+arccos(y) and v=xsqrt(1-y^2)+ysqrt(1-x^2), seeking to find a functional connection between u and v.
  • Another participant suggests a derivative relationship, stating v'_x/u'_x=dv/du, and provides a form for dv/du as sqrt(1-x^2)sqrt(1-y^2)-xy.
  • A different participant proposes a formula involving partial derivatives: v=x(1-y^2)(∂u/∂y) + y(1-x^2)(∂u/∂x), questioning if this aligns with the original inquiry.
  • One participant clarifies their initial mistake regarding the definition of u, correcting it to u=arcsin(x)+arcsin(y) and reiterating their request for assistance.
  • Another participant derives a relationship, stating that v can be expressed as sin(u), based on trigonometric identities involving arcsine functions.
  • One participant confirms the derivative approach, relating the expressions for v'_x and u'_x, and suggests that the squared form of the derivative relationship leads to a connection with v.

Areas of Agreement / Disagreement

Participants express various approaches to finding the connection between u and v, with no consensus on a single method or solution. Multiple competing views and interpretations of the relationships remain present throughout the discussion.

Contextual Notes

Participants rely on specific mathematical identities and derivatives, but the discussion does not resolve the complexities or assumptions underlying these relationships. The exploration of functional connections remains open-ended.

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im given:
u=arcsin(x)+arccos(y)
v=xsqrt(1-y^2)+ysqrt(1-x^2)
and i need to find the functional connection between u and v.

i know that:
v'_x/u'_x=dv/du
and i have got:
dv/du=sqrt(1-x^2)sqrt(1-y^2)-xy
now i need to show the rhs as a function of v or u, obviously of v should be much easier, the problem is i don't know how to simplify it.

i mean i tried adding and substracting terms, but with no success can someone help me on this?

thanks.
 
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loop quantum gravity said:
im given:
u=arcsin(x)+arcsin(y)
v=xsqrt(1-y^2)+ysqrt(1-x^2)
and i need to find the functional connection between u and v.

i know that:
v'_x/u'_x=dv/du
and i have got:
dv/du=sqrt(1-x^2)sqrt(1-y^2)-xy
now i need to show the rhs as a function of v or u, obviously of v should be much easier, the problem is i don't know how to simplify it.

i mean i tried adding and substracting terms, but with no success can someone help me on this?

thanks.
i fixed my mistake, the first equation is u=arcsin(x)+arcsin(y)
can now someone help me on this?
thanks.
 
How about

[tex]v=x\left(1-y^{2}\right)\frac{\partial u}{\partial y} +y\left(1-x^{2}\right)\frac{\partial u}{\partial x}[/tex]

Is this what you were looking for?

Daniel.
 
i need to find a functional connection like u=u(v) or v=v(u), or in simple terms v is a function of u or otherwise.
 
[itex]cos(arcsin(x))= \sqrt{1- sin^2(arcsin(x)}= \sqrt{1- x^2}[/itex]

Since sin(a+ b)= sin(a)cos(b)+ cos(a)cos(b), sin(arcsin(x)+ arcsin(y))= sin(arcsin(x))cos(arcsin(y))+ cos(arcsin(x))sin(arcsin(y)).

That is, sin(arcsin(x)+ arcsin(y))= [itex]x\sqrt{1- y^2}+ y\sqrt{1-x^2}[/itex].

In other words, v= sin(u)!
 
yes i also found it by using derivatives we know that:
v'_x=sqrt(1-y^2)-xy/sqrt(1-x^2)
and u'_x=1/sqrt(1-x^2)
so dv/du=sqrt(1-x^2)sqrt(1-y^2)-xy
(sqrt(1-x^2)sqrt(1-y^2)-xy)^2=1-v^2
 

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