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Unformatted text preview: SECTION 13.6 CYLINDERS AND QUADRIC SURFACES ET SECTION 12.6 263 (b) Since the equation y : er doesn’t (c) The equation 2 : 6" doesn’t involve ac,
involve 2, horizontal traces are so vertical traces in a: = k (parallel to the copies of the curve y = 61. The yzplane) are copies of the curve z : ey. rulings are parallel to the zaxis. The rulings are parallel to the m—axis. E.3. kince a: is missing from the equation, the vertical traces 4. Since y is missing from the equation, each vertical trace ,
y2 + 422 = 4, :L' 2 k, are copies ofthe same ellipse in the z = 4 — 332, y : k, is a copy ofthe same parabola in the plane a: = k. Thus, the surface 3;2 + 422 = 4 is an elliptic plane 3/ = k. Thus, the surface 2 = 4 — 1’2 is a parabolic ;
cylinder with rulings parallel to the xaxis. cylinder with rulings parallel to the yaxis. . l /{Rr‘ )
lnce z is missing, each horizontal trace :5 = y2, z 2 k, 6. Since a: is missing, each vertical trace yz = 4, x z k is a is a copy of the same parabola in the plane 2 = 19. Thus, copy of the same hyperbola in the plane a: : k. Thus, the '
the surface ac — 3/2 = 0 is a parabolic cylinder with rulings surface 1/2 : 4 is a hyperbOliC Cylinder With ﬁllings parallel to the zaxis. parallel to the maXis. __‘, zzk. 1. ISince y is missing, each vertical trace 2 = cos 2:, y = k is 8. Since 2 is missing, each horizontal trace :1; — y y : k. Thus, the 4“ a copy of a cosine curve in the plane 2 = k is a copy of the same hyperbola in the plane ce z = cos a: is a cylindrical surface with rulings : 1 is a hyperbolic cylinder Thus, the surface 11:2 — y2 surfa
with rulings parallel to the z—axis. , V '
‘Tuv 11» , b parallel to the yaxis. 9. (a) The traces of m2 + y2 — Z2 : 1 in {I} = k are y?‘ — 2:2 : 1 — 192, a family of hyperbolas. (Note that the hyperbolas are
oriented differently for —1 < k < 1 than for k < —1 or k > 1.) The traces in y = k are x2 — z2 = 1 — 102, a similar
= 1 + k2, a family of circles. For k = 0, the trace in the family of hyperbolas. The traces in z = k are m2 + 3/2 s 1. As [kl increases, so does the radius of the circles This behavior, combined with the 1 nay—plane, the circle is of radiu it hyperbolic vertical traces, gives the graph of the hyperboloid of on e sheet in Table l. (b) The shape of the surface is unchanged, but the hyperboloid is rotated so that its axis is the y—axis. Traces in y : k are circles, 5 while traces in m : k and z = k are hyperbolas. (c) Completing the square in y gives 1:2 + (y + 1)2 — 2:2 = 1. The surface is a hyperboloid identical to the one in part (a) but shifted one unit in the negative ydirection. 10. (a) The traces of —:c2 — y2 + 22 = 1 in m : k are ~y2 + 22 = 1 + W, a family of hyperbolas, as are the traces in y = k,
‘332 + Z2 = 1 + ’92 The traces in Z : k are 0:2 + y2 = k2 — 1, a family of circles for > 1. As 1k! increases, the ladiir , [kl < 1. This behavior, combined with the vertical traces, gives the graph ° 7 of the circles increase; the traces are empty for the hyperboloid of two sheets in Table 1. SECTION 13.6 CYLINDERS AND QUADRIC SURFACES ET SECTION 12.6 265 (b) The graph has the same shape as the hyperboloid in part (a) but is rotated so Z y axis is the asaxis. Traces in as = k, [k\ > 1, are circles, while traces that its
in y = k and z z k are hyperbolas.
" ,orx : y2 + 422,the traces in m = k: are y2 + 422 : k. When k > 0 we
,When Is : 0 we have just a point at the origin, and have a family of ellipses. the trace is empty for k < O. The traces in y : k are x : 422 + W, a the positive w—direction. Similarly, the traces family of parabolas opening in in z = k are m = y2 + 4162, a family of parabolas opening in the positive
:c—direction. We recognize the graph as an elliptic paraboloid with axis the
x—axis and vertex the origin. 12. EM? — y2 + z2 : O. The traces in x = k are 3;2 — 22 = 913, a family of hyperbolas if k ;E 0 and two intersecting lines if k : 0. The traces in y : k are 9302 + 22 2 W, k 2 0, a family of ellipses; the traces in z : k are y2 — 95c2 : k2, a family of hyperbolas for k # O and two intersecting lines for k : 0. We recognize the graph as an elliptic cone with axis the y—axis and vertex the origin. 13. 1:2 : y? + 4z2. The traces in a: = k are the ellipses y2 + 4z2 : W. The
traces in y = k are :3 ~ 4Z2 = W, hyperbolas for k gé 0 and two
intersecting lines if k : 0. Similarly, the traces in z : k are
m2 e y2 = 4k2, hyperbolas for k # 0 and two intersecting lines if k : 0. We recognize the graph as an elliptic cone with axis the m—axis and vertex the origin. @251? + 4312 + 22 z 100. The traces in m : k are 4sz2 + Z2 : 100  25k2, a family of ellipses for [kl < 2. (The traces are a single point for [kl : 2
and are empty for > 2.) Similarly, the traces in y : k are the ellipses 25s2 + 22 : 100 — 418, \k} < 5, and the traces in z : k are the ellipses 25.752 + 4312 = 100 — k2, < 10. The graph is an ellipsoid centered at the origin with intercepts m : :2, y = i5, 2 : :10. u :w.m.u.uww,,w , SECTION 13.6 CYLINDERS AND QUADRIC SURFACES ET SECTION 12.6 y The traces in m = k: are 742 — 22 : k, two intersecting lines ’ whcnt‘k E 0 and a family of hyperbolas for k 7e 0 (oriented differently for > 0 than for k < O). The traces in y = k are the parabolas
I = 22 + k2, Opening in the negative xdirection, and the traces in z : k: are me parabolas m = y2 — k2 which open in the positive m—direction. The h is a hyperbolic paraboloid with saddle point (0, 0, 0). Slap y2 Z2
“' n is is the equation of an ellipsoid: 1:2 + 43/2 + Qz2 : 3:2 + (1/2)2 + (1/3)2 = 1, with mintercepts :1, y—intercepts ii
and zintercepts :té. So the major axis is the maxis and the only possible graph is VII.
,0, m2 y2 J’I‘his is the equation of an ellipsoid: 9:1:2 + 4y2 + 22 2 (1/302 + (1/2)2 + 22 : 1, with mintercepts ::%, y—intercepts 3% \M/ and zintercepts 551. So the major axis is the z—axis and the only possible graph is IV. 5' This is the e uation of a hyperboloid of one sheet, with a = b : c = 1. Since the coefﬁcient of y2 is negative, the axis of the
q hyperboloid is the yaxis, hence the correct graph is H. 24. This is a hyperboloid of two sheets, with a : b : c 2 1. This surface does not intersect the :cz—plane at all, so the axis of the hyperboloid is the yaxis and the graph is 111. (a.
@T here are no real values of x and 2 that satisfy this equation for y < 0, so this surface does not extend to the left of the :cz—plane. The surface intersects the plane y =2 k > 0 in an ellipse. Notice that 3/ occurs to the ﬁrst power whereas an and 2 occur to the second power. So the surface is an elliptic paraboloid with axis the yaxis. Its graph is VI. 26. This is the equation of a cone with axis the yaxis, so the graph is l. “\
{"27.\,This surface is a cylinder because the variable y is missing from the equation. The intersection of the surface and the mz—plane \JV
IS an ellipse. So the graph is VIII. 28. This is the equation of a hyperbolic paraboloid. The trace in the my—plane is the parabola y = $2. So the correct graph is V. 2
Z $2 2 2 2 29. Z2 :43? +9212 +360r—4932 —9y2 +22 = 36or 30.1? :21;2 +322 0er : ——1y/2 + 7/3 or? = 113— + 52—
2:2 112 22 —— — — + —— = 1 represents a hyperboloid of two 9 4 36 . the m—ax1s. sheets with axis the z—axis. represents an elliptic cone with vertex (0,0,0) and axis ...
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 Spring '09
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