# calcii: contours

• December 7th 2010, 04:53 PM
jameskw
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 :)
• December 7th 2010, 09:27 PM
SammyS
Quote:

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:

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

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

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

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

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

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

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

Do you recognize this?
• December 7th 2010, 10:08 PM
jameskw
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 :)
• December 8th 2010, 07:05 PM
SammyS
Calc II, contours
Quote:

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

Therefore, the equation, $\displaystyle z={{(x^2)+(y^2)}\over{2x}}$, describes a pair of oblique circular cones, each with a vertex at
(0, 0, 0) and axis along the line $z=x$ 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 $z={4\over3}x$ in the x-z plane. (I haven't figured out the eccentricity of the ellipses.)
• December 9th 2010, 03:46 AM
HallsofIvy
$c= \frac{x^2+ y^2}{2x}$ gives $2cx= x^2+ y^2$, $x^2- 2cx+ y^2= 0$
Now "complete the square": $x^2- 2cx+ c^2+ y^2= c^2$, $(x- c)^2+ y^2= c^2$,
a circle with center at (c, 0) and radius c so it is tangent to the y-axis at (0, 0).