1. ## 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

2. 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:

$\displaystyle \displaystyle c={{(x^2)+(y^2)}\over{2x}}$

$\displaystyle \displaystyle 2c={{(x^2)+(y^2)}\over{x}}$

$\displaystyle \displaystyle 2c=(x)+{{y^2}\over{x}}$

What you need after your line, $\displaystyle c(2) x-x^2=y^2$ is:

$\displaystyle 0=x^2-2cx+y^2$, then add $\displaystyle \displaystyle c^2$ complete the square.

$\displaystyle c^2=x^2-2cx+c^2+y^2$

Finally you have $\displaystyle c^2=(x-c)^2+y^2$.

Do you recognize this?

3. 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

4. ## Calc II, contours

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, $\displaystyle c^2=(x-c)^2+y^2$ 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, $\displaystyle c^2=(x-c)^2+y^2$ is a family of such circles.

Therefore, the equation, $\displaystyle \displaystyle z={{(x^2)+(y^2)}\over{2x}}$, describes a pair of oblique circular cones, each with a vertex at
(0, 0, 0) and axis along the line $\displaystyle z=x$ 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 $\displaystyle z={4\over3}x$ in the x-z plane. (I haven't figured out the eccentricity of the ellipses.)

5. $\displaystyle c= \frac{x^2+ y^2}{2x}$ gives $\displaystyle 2cx= x^2+ y^2$, $\displaystyle x^2- 2cx+ y^2= 0$
Now "complete the square": $\displaystyle x^2- 2cx+ c^2+ y^2= c^2$, $\displaystyle (x- c)^2+ y^2= c^2$,
a circle with center at (c, 0) and radius c so it is tangent to the y-axis at (0, 0).