Thread: calcii: contours

Hi, I am having trouble with contours, is there a procedure to do these?

the question I am working on is when z=c

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

what i tried to do:

c2x=(x^2)+(y^2)
c2x-x^2=y^2
x(2c-x)=y^2
2c-x=(y^2)/x
2c=((y^2)/x)+x

and that is where I got stuck. I cant visualize what is going on here

thanks for any replys

Originally Posted by jameskw

Hi, I am having trouble with contours, is there a procedure to do these?

the question I am working on is when z=c

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

what i tried to do:

c2x=(x^2)+(y^2)
c2x-x^2=y^2
x(2c-x)=y^2
2c-x=(y^2)/x
2c=((y^2)/x)+x

and that is where I got stuck. I cant visualize what is going on here

thanks for any replies

What you have done is equivalent to:







What you need after your line, is:

, then add complete the square.



Finally you have .

Do you recognize this?

Well, it looks like its close to a circle because of its form. As c increases radius increases and x gets displaced.
sounds like 2 cones with a slope, how would i find a slope?

thanks

Originally Posted by jameskw

Well, it looks like it's close to a circle because of its form. As c increases radius increases and x gets displaced.
sounds like 2 cones with a slope, how would i find a slope?

thanks

Yes! For any fixed value of c, is the equation of a circle with radius, c, centered on the x-axis at x=c. This circle passes through the origin.

When looked at in a more general way, is a family of such circles.

Therefore, the equation, , describes a pair of oblique circular cones, each with a vertex at
(0, 0, 0) and axis along the line in the x-z plane. One opens upward, the other downward.

You can also look at this as a pair of right elliptical cones, each with a vertex at (0, 0, 0) and axis along the line in the x-z plane. (I haven't figured out the eccentricity of the ellipses.)

gives ,
Now "complete the square": , ,
a circle with center at (c, 0) and radius c so it is tangent to the y-axis at (0, 0).