Thread: 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...

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.

...

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...

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 centered at the origin ... points located on the surface of the sphere are not included.

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. is a circle with center at (0, 0) and radius 1. The point (2, 0), which is outside that circle, satisfies 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)
b)

As I said before, start by looking at the equality. . Simplify that by completing the square in x and y: => . 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 . 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...