FUNCTIONS OF SEVERAL VARIABLES
365
41.20
41.21
41.22
41.23
41.24
41.25
41.26
41.27
Find the points at which the line
intersects the ellipsoid
The line can be written in parametric form as x = 6 + 3t, y = —2 — 6f, z = 2 + 4t. Hence, substituting in
the equation of the ellipsoid, we get
4(6 + 3f)
2 + 9(2 + 6f)
2 + 36(2 +
4f)
2 = 324, 26(36/
2 ) +26(360+ 4(81) = 324, t
2 + t = 0, t(t+l) = 0, t = 0 or <=-!. So, the points are
(6,-2,2) and (3,4,-2).
Show that the plane 2x - y - 2z = 10 intersects the paraboloid
at a single point, and find the
point.
Solve the equations simultaneously.
4*
2 +9y
2 -72x + 36y = -360, 4(x - 9)
2 +
9(y+ 2) =-360 + 324 + 36 = 0. Hence, the only solution is x = 9, y =-2. Then z = 5. Thus, the only
point of intersection is (9, -2,5), where the plane is tangent to the paraboloid.
Find the volume of the ellipsoid
The plane z = k cuts the ellipsoid in an ellipse
By Problem 20.72, the area of this ellipse is
Hence, by the
cross-section formula for volume.
Identify the graph of 9x
2 - y
2 + I6z
2 = 144.
This equation is equivalent to (x
2 /16) - (y
2 /144) + (z
2 /9) = 1. This is an elliptic hyperboloid of one sheet,
with axis the y-axis; see Problem 41.14. The cross sections y = k are ellipses (*
2
/16) + (z
2 /9) = 1 +
(A:
2
/144). The cross sections x = k are hyperbolas (z
2 /9) - (y
2 /144) = 1 - (fc
2 /16), and the cross sections z = k are hyperbolas (rVl6) - (y
2 /144) = 1 - (k
2 /9).
Identify the graph of 25x
2 - y
2 - z
2 = 25.
This is equivalent to x
2 — (y
2 /25) — (z
2 /25) = 1, which is a circular hyperboloid of two sheets; see Problem
41.15. Each cross section x = k is a circle y
2 + z
2 = 25(k
2 — 1). (We must have |fc|>l.) Each cross
section y = k is a hyperbola x
2 — (z
2 /25) = 1 + (k
2 /25), and each cross section z = k is a hyperbola x
2 -(y
2 /25) = 1 + (Jt
2 /25). The axis of the hyperboloid is the *-axis.
Identify the graph of x
2 + 4z
2 = 2y.
This is equivalent to (x
2 /4) + z
2 = y/2. This is an elliptic paraboloid, with the y-axis as its axis. For y =
k >0, the cross section is an ellipse (*
2
/4) + z
2 = k/2. (y = 0 is the origin, and the cross sections y = k <
0 are empty.) The cross section x = k is a parabola y/2 = z
2 + (k
2 /4). The cross section z = k is a
parabola y/2 = (*
2
/4) + k
2 .
Show that, if a curve "£ in the jry-plane has the equation f(x, y) = 0, then an equation of the cylinder generated
by a line intersecting <£ and moving parallel to the vector A = (a, b, 1) is f(x — az, y — bz) = 0.
The line !£ through a point (x 0 , y 0 ,0) on
if f(x - az, y- bz) = 0, then, setting x g = x - az, y 0 = y - bz, we see that (x a , y 0 ,0) is on <£ and
(x, y, z) is on the corresponding line X.
Find an equation of the cylinder generated by a line through the curve y
2 = x - y in the jcy-plane that moves
parallel to the vector A = (2,2,1).
By Problem 41.26, an equation is (y -2z)
2 = (x -2z) - (y -2z), (y - 2z)
2 =x-y.
365
41.20
41.21
41.22
41.23
41.24
41.25
41.26
41.27
Find the points at which the line
intersects the ellipsoid
The line can be written in parametric form as x = 6 + 3t, y = —2 — 6f, z = 2 + 4t. Hence, substituting in
the equation of the ellipsoid, we get
4(6 + 3f)
2 + 9(2 + 6f)
2 + 36(2 +
4f)
2 = 324, 26(36/
2 ) +26(360+ 4(81) = 324, t
2 + t = 0, t(t+l) = 0, t = 0 or <=-!. So, the points are
(6,-2,2) and (3,4,-2).
Show that the plane 2x - y - 2z = 10 intersects the paraboloid
at a single point, and find the
point.
Solve the equations simultaneously.
4*
2 +9y
2 -72x + 36y = -360, 4(x - 9)
2 +
9(y+ 2) =-360 + 324 + 36 = 0. Hence, the only solution is x = 9, y =-2. Then z = 5. Thus, the only
point of intersection is (9, -2,5), where the plane is tangent to the paraboloid.
Find the volume of the ellipsoid
The plane z = k cuts the ellipsoid in an ellipse
By Problem 20.72, the area of this ellipse is
Hence, by the
cross-section formula for volume.
Identify the graph of 9x
2 - y
2 + I6z
2 = 144.
This equation is equivalent to (x
2 /16) - (y
2 /144) + (z
2 /9) = 1. This is an elliptic hyperboloid of one sheet,
with axis the y-axis; see Problem 41.14. The cross sections y = k are ellipses (*
2
/16) + (z
2 /9) = 1 +
(A:
2
/144). The cross sections x = k are hyperbolas (z
2 /9) - (y
2 /144) = 1 - (fc
2 /16), and the cross sections z = k are hyperbolas (rVl6) - (y
2 /144) = 1 - (k
2 /9).
Identify the graph of 25x
2 - y
2 - z
2 = 25.
This is equivalent to x
2 — (y
2 /25) — (z
2 /25) = 1, which is a circular hyperboloid of two sheets; see Problem
41.15. Each cross section x = k is a circle y
2 + z
2 = 25(k
2 — 1). (We must have |fc|>l.) Each cross
section y = k is a hyperbola x
2 — (z
2 /25) = 1 + (k
2 /25), and each cross section z = k is a hyperbola x
2 -(y
2 /25) = 1 + (Jt
2 /25). The axis of the hyperboloid is the *-axis.
Identify the graph of x
2 + 4z
2 = 2y.
This is equivalent to (x
2 /4) + z
2 = y/2. This is an elliptic paraboloid, with the y-axis as its axis. For y =
k >0, the cross section is an ellipse (*
2
/4) + z
2 = k/2. (y = 0 is the origin, and the cross sections y = k <
0 are empty.) The cross section x = k is a parabola y/2 = z
2 + (k
2 /4). The cross section z = k is a
parabola y/2 = (*
2
/4) + k
2 .
Show that, if a curve "£ in the jry-plane has the equation f(x, y) = 0, then an equation of the cylinder generated
by a line intersecting <£ and moving parallel to the vector A = (a, b, 1) is f(x — az, y — bz) = 0.
The line !£ through a point (x 0 , y 0 ,0) on
if f(x - az, y- bz) = 0, then, setting x g = x - az, y 0 = y - bz, we see that (x a , y 0 ,0) is on <£ and
(x, y, z) is on the corresponding line X.
Find an equation of the cylinder generated by a line through the curve y
2 = x - y in the jcy-plane that moves
parallel to the vector A = (2,2,1).
By Problem 41.26, an equation is (y -2z)
2 = (x -2z) - (y -2z), (y - 2z)
2 =x-y.