A= 1, B = 1, C = 2, D = 1, E = 3, F = 1
Write the equation for the object and identify it.
Solution:
Ax 2 + B xy + C y 2 + Dx + Ey + F = 0
So: x 2 + xy + 2 y 2 + x + 3y + 1 = 0
When graphed using MathCad:
Discloses a parabolic cylinder.
2) Describe and sketch each of the following:
a) x 2 + y 2 = a 2
b) x 2 + y 2 + z 2 + 4x - 6y = 3
Solutions:
a) The graph is for the cross sectional area of a circular cylinder parallel to the z-axis, and sketched below (from the graphic shown in the blog post, changed coordinates)
b) We let z = 0 in the equation then graph the remaining equation to obtain the circular cross section for what will be a sphere (e.g. in 3D):
From this MathCad plotted graph we infer the sphere will have a center at: h = -2, k = 3, m = 0
And radius a = 4
To satisfy the eqn: (x - h) 2 + (y - k ) 2 + (z - m ) 2 = a 2
We check, by substituting the values, h = -2, k = 3 in the sphere equation (with z = 0, m = 0), and radius a = 4, to get:
(x - (-2)) 2 + (y - 3) 2 = 16
= (x + 2) 2 + (y - 3) 2 = 16
So that: (x 2 + 4x + 4) + (y 2 - 6y + 9) = 16
Then: x 2 + 4x + y 2 - 6y = 16 - 9 - 4
x 2 + 4x + y 2 - 6y = 3
Which is the original equation.
For those lacking a program like MathCad a simpler graphing program (such as Graphmat) can be used to obtain the sphere center coordinates. This can be done by graphing a form for the original equation (again z = 0) based on completing the square, e.g. for the x 2 + 4x part and the y 2 - 6y part. Then:
( x 2 + 4x + 4) + ( y 2 - 6y + 9) = 3 + 9 + 4 = 16
(Since 9 and 4 are added to each side in the process). Then on graphing the circle in the x-y plane, e.g. (x + 2) 2 + (y - 3) 2 = 16, we obtain:
Where the reader can see by inspection the center is at (-2, 3) This then forms the circular cross section in xy-plane for the sphere sketched below in a (somewhat) 3 D aspect:
Where x = -6 and x = 2 form the left & right limits along the x-dimension, and y = -1 to y = 7 form the limits along the y-axis. In each case confirming the radius r = 4.
3) Sketch (and identify) the surface given by:
f(x,y) = 4 - x 2 + y 2
This can also be written (since we are discussing a 3D surface):
z = 4 - x 2 + y 2
When y = 0, the surface's trace on the x-z plane is y = 4 - x 2 and is a parabola that opens up in the negative z direction. In the xy plane we see:
When x =0, the trace on the yz plane is: z = 4 + y 2 and opens up in the positive z direction. The result is a hyperbolic paraboloid.
When one cuts through the plane such that z1 > o a hyperbola is obtained, e.g.
The MathCad graph of the surface is:
No comments:
Post a Comment