I am learning the math. But how do I sketch a three dimensional/multivariable shape, plane, ect? I am talking about physically sketching on a piece of scrap paper. Thank you for your help.
Examples:
Sketch the parallelogram spanned by (2,1,0) and (-3,1,0) in the xy-plane, and
compute its area (got the area and everything...the sketching is the problem?)
If you "got the area and everything", you should understand that this is NOT a three dimensional problem- every thing is in the xy-plane.
Mark the points (0, 0), (2, 1), (-3, 1), and (-1, 2) as the vertices. Do you see where I got (0, 0) and (-1, 2)? Do you understand that this is the same as the four points (0, 0, 0), (2, 1, 0), (-3, 1, 0), and (-1, 2, 0) in three dimensions?
and/or
Sketch the surface z = x^2 + 2y^2
If x= 0 this is the same as z= y^2, the equation of a parabola. If y= 0, this is z= x^2, the equation of a parabola. Draw a yz-coordinate system on one piece of paper and graph z= y^2. Draw a xz-coordinate system on another piece of paper and graph z= x^2. Cut
down the z-axis to (0, 0) on one and
up the z-axis to (0, 0) on the other. That will allow you to put the two graphs together at right angles. You should be able to see that the actual graph, a paraboloid, rotates smoothly through those.
Or, change to "cylindrical coordinates"( polar coordinates in the xy-plane) so the equation is z= r^2 for any [itex]\theta[/itex]. Again, you can graph z= r^2, a parabola and then imagine it rotating around the z-axis.
Or, graph different values of z. If z= 0, we have x^2+y^2= 0, satisfied only by the single point (0, 0). If z= 1, we have x^2+ y^2= 1, a circle with center at (0, 0, 1) and radius 1 If z= 2 we have x^2+ y^2= 2, a circle with center at (0, 0, 2) and radius sqrt(2). If z= 3, we have x^2+ y^2= 3, a circle with center at (0, 0, 3) and radius sqrt(3). "Stacking" those we have a stack of circles with increasing radius but more slowly than z so we have the circles
inside the cone z= r.
and/or
Sketch the surface z = 2y^2 4x^2
Sorry but this makes no sense. Do you mean z= 2y^2+ 4x^2 or z= 2y^2- 4x^2?
If the first, it is the same as the previous example, a paraboloid. z= 2y^2 is a parabola, z= 4x^2 is a "thinner" parabola. The three dimensional figure is a paraboloid a little "fatter" in the x direction than in the y direction. z= 2y^2- 4x^2 is a hyperboloid. z= 2y^2 is a parabola, z= -4x^2 is a parabola opening downward. Again, you can imagine those two parabolas at right angles, with the surface connecting them going smoothly from one to the other. Also, z will be 0 when 2x^2- 4y^2= (sqrt(2)x- 2y)(sqrt(2)x+ 2y)= 0 which means that either sqrt(2)x- 2y= 0 so that y= x/\sqrt{2}, a line in the xy plane through the origin or sqrt(2)x+ 2y= 0 so that y= -x/\sqrt{2}, another line.
