How to Draw Intersection Curve With Surface and Plane Graph

The Intersection of Two Surfaces

Though the theme of this folio is the points that lie on both of two surfaces, permit the states begin with but one, the profile x two z - xy ii = four or substantially z = (xy 2 + 4)/x ii. Permit u.s. suppose that we want to find all the points on this surface at which a vector normal to the surface is parallel to the yz-plane. Information technology is not clear, at least to me, that there are whatsoever such points; equally I picture vectors perpendicular to the surface, they all seem to get forward or astern, at least slightly. But let u.s.a. come across what calculus tells united states of america.
All vectors normal to the surface at a point (ten,y,z) on information technology are multiples of the gradient (2xz-y 2)i - 2xy j + x 2 k at that point, so to say that the normal is parallel to the yz-plane is to say that twoxz - y 2 = 0. So the desired points are on both the original surface and the new surface adamant by this new equation (shown in bluish to the right).
Information technology appears from the diagram higher up that in that location are no points with positive x-values that are on both surfaces at the same fourth dimension; but in the region where x < 0 and z < 0, it appears that the two surfaces intersect, i.east., they have some points in common. Those points are bundled in two curves, drawn in green in this diagram.

Algebraically, we are looking for the points (x,y,z) that make both of the equations true:

ten ² z - xy ² = 4 2xz - y ² = 0 In some cases this may the simplest way to describe algebraically a bend in infinite: as the set of mutual solutions of two or more than equations. Simply in this case we can utilize another method, which may be more useful: giving the coordinates of the points on the curve by expressions in some common variable (called a "parameter") which may or may not be i of the coordinates. In this case we can limited y and z,and of course x itself, in terms of x on each of the two green curves, so nosotros tin "parametrize" the intersection curves by x: From the second equation we get y ² = 2xz, and substituting into the first equations gives x ⁱⁱ z - x(2xz) = 4, or z = -4/x ² -- from which we tin can run across immediately that the z-values will be negative. From the equation y ² = iixz we tin see that the x-values will besides be negative; and substituting into this equation our expression for z in terms of x shows that y = ±sqrt(-eight/x). So we accept the desired parametrizations of the intersection curves:
10 = x, y = ±sqrt(-viii/10), z = -4/x ²

Roughly, what we expect is that a single equation in 3 variables determines a surface in space; ii equations determine a bend or curves (in the sense that the common solutions (x,y) of both equations form 1 or more curves); and iii dermine a point or isolated points. Of class, exceptions abound: The solutions of the single equation x ⁱⁱ + y ² = 0 is the (one-dimensional) z-axis. And if we consider the infinitely many planes that all pass through the same line, then whatsoever ii or more of their corresponding (linear) equations volition even so determine that common line. Just "each new equation cuts downwardly the dimension by one" is a handy dominion of pollex.

Because I mentioned in a higher place that it is hard to see whether whatsoever points on the original surface x ² z - xy ^two = 4 have the property that the normals at that place are parallel to the yz-airplane, allow us take one of the points on our curve and expect at the slope vector there, to see whether information technology has the desired backdrop.

One betoken on the surface is (-two,2,-1), and the gradient there is 8j + 4thou, so it is clearly parallel to the yz-airplane. The real question is the more basic one: Is it true that the slope is always perpendicular to the contour at its base? Here is a slice of the original surface, with (-2,two,-1) at one corner and with the in a higher place slope vector drawn in nighttime green.

The reader is invited to download the respective Winplot file and rotate it to see that gradient is indeed perpendicular to the surface.

calvertracion.blogspot.com

Source: http://math.colgate.edu/math213/dlantz/examples/twosurf.html

Share this post