Level curves in contour map f(x,y)=y/(x^2+y^2), f(x,y)=y sec

M

MarkSA

Junior Member
Joined
Sep 8, 2007
Messages
243
Hello,

Suppose I have the question:
Draw the contour map of the function showing several level curves.
f(x,y) = y/(x^2 + y^2)
f(x,y) = y*sec(x)

What is the best approach to take with these problems? How would I draw a contour map for these strange functions. Is there an easy way of doing it? I'm able to do it with easier functions (ie: f(x,y) = y - lnx) where I recognize the function, but for these i'm not sure.
 
Re: Level curves in contour maps

Hello, MarkSA!

Draw the contour map of the function showing several level curves.
. . f(x,y) = yx2+y2\displaystyle f(x,y) \:= \:\frac{y}{x^2 + y^2}
Click to expand...

Let z = f(x,y) take on various constant values.\displaystyle \text{Let }z\:=\:f(x,y)\text{ take on various constant values.}
. . We can do this in  general.\displaystyle \text{We can do this }in\;general.

Let:   z  =  yx2+y2  =  a     for any real number a.\displaystyle \text{Let: }\;z\;=\; \frac{y}{x^2+y^2} \;=\;a\;\;\text{ for any real number }a.

Then we have:  y = ax2+ay2⇒ax2+ay2−y  =  0⇒x2+y2−1ay  =  0\displaystyle \text{Then we have: }\:y \:=\:ax^2 + ay^2 \quad\Rightarrow\quad ax^2 + ay^2 - y \;=\;0 \quad\Rightarrow\quad x^2 + y^2 - \frac{1}{a}y \;=\;0

Complete the square:   x2+y2−1ay+14a2  =  14a2\displaystyle \text{Complete the square: }\;x^2 + y^2 - \frac{1}{a}y + \frac{1}{4a^2} \;=\;\frac{1}{4a^2}

. . and we have:  x2+(y−12a)2  =  (12a)2\displaystyle \text{and we have: }\:x^2 + \left(y - \frac{1}{2a}\right)^2 \;=\;\left(\frac{1}{2a}\right)^2


For z=a, we have a circle . . . center (0, 12a),   radius ∣12a∣\displaystyle \text{For }z = a\text{, we have a circle . . . center }\left(0,\:\frac{1}{2a}\right),\;\text{ radius }\left|\frac{1}{2a}\right|

Its center is in the y-direction and the circle passes through the z-xis.\displaystyle \text{Its center is in the y-direction and the circle passes through the z-xis.}


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ - -

As z gets larger, the center approaches (0,0) and the radius approaches 0.\displaystyle \text{As }z\text{ gets larger, the center approaches }(0,0)\text{ and the radius approaches 0.}

That is, as we go "higher", we have a "cone" that leans towards the z-axis.\displaystyle \text{That is, as we go "higher", we have a "cone" that leans towards the z-axis.}


As z→0, the center moves out in the y-direction\displaystyle \text{As }z \to0\text{, the center moves out in the y-direction}
. . and the radius gets larger (of course).\displaystyle \text{and the radius gets larger (of course).}


I believe a side view would look like this.
Code: 
    z |
      |*
      |
      | *
      |  *
      |    *
      |       *
      |             *
      |                     * 
  - - + - - - - - - - - - - - - - y
      |

where all horizontal cross-sections are circles.\displaystyle \text{where all horizontal cross-sections are circles.}


I’ll let you describe what happens below the xy-plane.\displaystyle \text{I'll let }you\text{ describe what happens }below\text{ the }xy\text{-plane.}

 
Re: Level curves in contour maps

thanks for the reply. below the x-y plane, would it be the same shape (cone leaning toward z axis), but with the radius being really large for small negative values of z and shrinking for large negative values?
 
You must log in or register to reply here.
Top