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Math Help - Circles should be easy to understand...

  1. #1
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    Unhappy 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...
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  2. #2
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    Quote Originally Posted by tinyone View Post
    ...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.
    ...
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  3. #3
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    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...
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  4. #4
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    Quote Originally Posted by tinyone View Post
    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.
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  5. #5
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    Quote Originally Posted by tinyone View Post
    ...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...
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