Thread: consider

f(x,y)=(2/sqrt(x^2 + y^2 -4)) - 1

1. describe the domain & range of f, and provide graph of the domain. be sure to use shading and dashed or solid curves as approriate.

2. draw a level curves for the values c=0 & c= 1. Be sure lable your axes to indicate the scale, and graph the equation f(x,y) = c, you will want to rearrange the equation to put it in recognizable form. (hint: get variables out of the denominator and elminitae the radical.

3. determined any y-intercepts of f. write answers as ordered triples and not round any of the coordinates.

Originally Posted by ggw

f(x,y)=(2/sqrt(x^2 + y^2 -4)) - 1

1. describe the domain & range of f, and provide graph of the domain. be sure to use shading and dashed or solid curves as approriate.

The Domain
The the given fraction cannot have the denominator zero.
Nor can it have the expression inside the square root to be negative.

Thus, we conclude that,
x^2+y^2-4>0
x^2+y^4>4
Which is a circle of radius 2, but all the points are strictly OUTSIDE the circle.

The Range
This is tricker.
Let use write,

z=2/sqrt(x^2+y^2-4) -1

We need to find all possible z, such that the equation above has at least one solution (x,y) in the domain.

Add 1 to both sides,

z+1=2/sqrt(x^2+y^2-4).

If z=-1 then we have zero on left and and expression which cannot be zero on right. Thus, z not = -1.

Since z not = -1 that means z+1 is non-zero. And we can take reciprocals of both sides.

Thus,

1/(z+1)=sqrt(x^2+y^2-4)/2

Multiply both sides by 2,

2/(z+1)=sqrt(x^2+y^2-4)

To have a solution we need that the expression on the left to be positive (because square roots are always positive).

Thus we require that,
2/(z+1)>0 solve inequality z>-1.

Now square both sides,

4/(z+1)=x^2+y^2-4

Add 4 to both sides,

4/(z+1)+4 = x^2+y^2

Now here is the important part.
So far we have shown that z>-1 is the necessary condition for solutions to exists. But is that enought.
Meaning does the equation above solution in x,y?
Well, you might say, just make x=0 and solve for y.
But it is not so simple, because the domain is x^2+y^2>4.
Thus, we are asking can we solve for x,y such that x^2+y^2>4.
Yes, we can!
Because the left hand term is,
4/(z+1)+4>4 because 4/(z+1)>0.

Thus, z>-1 is the range.

i got the curve level, which is cordinates are (sqrt(8),-sqrt(8)) & (sqrt(5),-sqrt(5)). I am not sure if it is right.

determined the y-intercepts, is that set x=o to find the positions for y ?