# Finding a functional connection.

1. Dec 6, 2006

### MathematicalPhysicist

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 dont know how to simplify it.

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

thanks.

2. Dec 8, 2006

### MathematicalPhysicist

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

3. Dec 8, 2006

### dextercioby

$$v=x\left(1-y^{2}\right)\frac{\partial u}{\partial y} +y\left(1-x^{2}\right)\frac{\partial u}{\partial x}$$

Is this what you were looking for?

Daniel.

4. Dec 8, 2006

### MathematicalPhysicist

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.

5. Dec 8, 2006

### HallsofIvy

Staff Emeritus
$cos(arcsin(x))= \sqrt{1- sin^2(arcsin(x)}= \sqrt{1- x^2}$

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))= $x\sqrt{1- y^2}+ y\sqrt{1-x^2}$.

In other words, v= sin(u)!

6. Dec 9, 2006

### MathematicalPhysicist

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