# Thread: Circles should be easy to understand...

1. ## Circles should be easy to understand...

...But I'm struggling!

A circle in R^3, ie 3D space is written as: f(x,y)= x^2+ y^2
If we want to draw level curves we give the function different values of C, for example:
C= 1 --> x^2+ y^2 = 1, This will give us a circle with the center in (0,0) and the radius = sqrt(1)=1

BUT what changes when we have x^2+ 4y^2
What does the 4 change, does it change the centre of the circle?

Also, how should one sketch 1< x^2+y^2 (less than or equal to 4) in the XY-plane?

Also 2: The temperature in a point (x,y,z) in a body is described by the function T(x,y,z)=x^2+y^2+z^2+2x-2y (degrees celsius). Describe the areas of the body where the temperature is :
a) larger than 2 degrees celsius
b) smaller than 3 degrees celsius

The last example is like Spanish for me, and I'm not a Spanish speaking person...

An answer to any of the questions is highly appreciated...

2. Originally Posted by tinyone
...But I'm struggling!

A circle in R^3, ie 3D space is written as: f(x,y)= x^2+ y^2
If we want to draw level curves we give the function different values of C, for example:
C= 1 --> x^2+ y^2 = 1, This will give us a circle with the center in (0,0) and the radius = sqrt(1)=1

BUT what changes when we have x^2+ 4y^2
What does the 4 change, does it change the centre of the circle?

... the circle becomes an ellipse

Also, how should one sketch 1< x^2+y^2 (less than or equal to 4) in the XY-plane?

1 < x^2+y^2 < 4
all points in the plane between the two circles centered at the origin of radius 1 and radius 2 ... includes points on the circle of radius 2 , does not include all points on the circle of radius 1.

Also 2: The temperature in a point (x,y,z) in a body is described by the function T(x,y,z)=x^2+y^2+z^2+2x-2y (degrees celsius). Describe the areas of the body where the temperature is :
a) larger than 2 degrees celsius
b) smaller than 3 degrees celsius

2 < x^2+y^2+z^2+2x-2y < 3

completing the square in x and y ...

2+1+1 < x^2+2x+1 + y^2-2y+1 + z^2 < 3+1+1

4 < (x+1)^2 + (y-1)^2 + (z-0)^2 < 5

all points in the sphere centered at (-1,1,0) that are greater than 2 units from the center but less than sqrt(5) units from the center.
...

3. Thank you so much! I think I understand the XY-plane well but how would one draw x^2+y^2+z^2 < 5...

Im not good with 3D...

4. Originally Posted by tinyone
Thank you so much! I think I understand the XY-plane well but how would one draw x^2+y^2+z^2 < 5...

Im not good with 3D...
all the points inside the 3D space of a sphere of radius $\sqrt{5}$ centered at the origin ... points located on the surface of the sphere are not included.

5. Originally Posted by tinyone
...But I'm struggling!

A circle in R^3, ie 3D space is written as: f(x,y)= x^2+ y^2
This is not true. z= f(x,y)= x^2+ y^2 is a cone, not a circle. A single equation in three dimentions is always a surface not a curve.

If we want to draw level curves we give the function different values of C, for example:
C= 1 --> x^2+ y^2 = 1, This will give us a circle with the center in (0,0) and the radius = sqrt(1)=1

BUT what changes when we have x^2+ 4y^2
What does the 4 change, does it change the centre of the circle?
That would be an elliptic cone. The center (actually the axis of the cone) remains the z-axis the the x direction is "squashed".

Also, how should one sketch 1< x^2+y^2 (less than or equal to 4) in the XY-plane?
The simplest way to solve or draw complicated inequalities is to start with the associated equation. $x^2+ y^2= 1$ is a circle with center at (0, 0) and radius 1. The point (2, 0), which is outside that circle, satisfies $x^2+ y^2> 1$ while the point (.5, 0), which is inside the circle does not. The solution set for this inequality is the set of all points outside the unit circle.

Also 2: The temperature in a point (x,y,z) in a body is described by the function T(x,y,z)=x^2+y^2+z^2+2x-2y (degrees celsius). Describe the areas of the body where the temperature is :
a) larger than 2 degrees celsius
b) smaller than 3 degrees celsius
a) $x^2+ y^2+ z^2+ 2x- 2y> 2$
b) $x^2+ y^2+ z^2+ 2x- 2y< 3$

As I said before, start by looking at the equality. $x^2+ y^2+ z^2+ 2x- 2y= 2$. Simplify that by completing the square in x and y: $x^2+ 2x+ 1+ y^2- 2y+ 1+ z^2= 2+ 1+ 1$=> $(x+ 1)^2+ (y- 1)^2+ z^2= 4$. That is the equation of a sphere with center at (1, -1, 0) and radius 2. The solution set for the first inequality, (a), is the set of all points outside that sphere.

Do the same thing with $x^2+ y^2+ z^2+ 2x- 2y= 3$. You will again get a sphere and the solution set for the second inequality, (b), is the set of all point inside that sphere.

The last example is like Spanish for me, and I'm not a Spanish speaking person...

An answer to any of the questions is highly appreciated...