Mathematics 205
HWK 3 Solutions
Section 12.4 p588
Problem 19, §12.4, p588. Find the equation of the linear function z = c + mx + ny whose
graph intersects the xz-plane in the line z = 3x +4 and intersects the yz-plane in the line z = y + 4.
Solution. The xz-plane is the same as the plane y = 0. Thus z = c + mx + ny meets the plane
y = 0 in the line z = 3x +4. Substituting y = 0 into the equation z = c+mx + ny gives z = c+mx,
so the line z = c + mx must be the same as the line z = 3x + 4. Consequently m = 3 and c = 4.
Similarly, using x = 0 in the equation z = c + mx + ny gives us the line z = c + ny, which has to
be the same as the line z = y + 4. This gives us c = 4 (which is consistent with what we already
found) and n = 1. Thus the linear function we want is z = 4 + 3x + y.
Problem 25, §12.4, p588. Sketch the graph for the function z = 2 − x − 2y by locating the x-,
y-, and z-intercepts and joining them in a triangle. Adjust the perspective of the three coordinate
axes, if necessary.
Solution. We know that the graph will be a plane. Rewriting the given formula gives us the
equation x + 2y + z = 2. Putting x = y = 0 gives us the z-intercept z = 2. Put x = z = 0 to
find the y-intercept: y = 1. Similarly, find the x-intercept, which is x = 2. Plot the corresponding
points on a set of coordinate axes. Join them to show a triangular portion of the given plane.
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Math 205 HWK 3 Solns continued
§12.4 p588
Problem 26, §12.4, p588. Sketch the graph for the function z = 6 − 2x − 3y by locating the x-,
y-, and z-intercepts and joining them in a triangle. Adjust the perspective of the three coordinate
axes, if necessary.
Solution. We are graphing the plane 2x + 3y + z = 6. The intercepts are at x = 3, y = 2, and
z = 6. The portion of the triangle in the first octant is shown below.
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Math 205 HWK 3 Solns continued
§12.4 p588
Problem 27, §12.5, p594. Sketch the graph for the function z = 4 + x − 2y by locating the x-,
y-, and z-intercepts and joining them in a triangle. Adjust the perspective of the three coordinate
axes, if necessary.
Solution. The plane is given by −x + 2y + z = 4. The three intercepts are x = −4, y = 2, and
z = 4. The graph is shown below.
Since the x-intercept is negative this time, it might have been better to adjust the perspective a
little.
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§12.4 p588
Problem 15, §12.5, p594.
equation
Use the catalog on page 593 to identify the surface given by the
−x2 − y 2 + z 2 = 1.
Solution. Rewrite the given equation as x2 + y 2 − z 2 = −1. This is a hyperboloid of two sheets.
See Figure 12.74, and note that we have a = b = c = 1. Here’s a computer-generated sketch. Note
how p
the cross sections perpendicular to the z-axis appear circular.
p The top sheet is the function
2
2
z = x + y + 1, and the bottom sheet is the function z = − x2 + y 2 + 1.
Problem 16, §12.5, p594.
equation
Use the catalog on page 593 to identify the surface given by the
−x2 + y 2 − z 2 = 0.
Solution. Rewrite as x2 − y 2 + z 2 = 0. This is a cone. See Figure 12.75, and note that the roles
of y and z have been reversed here, so the cone will have the y-axis as its central axis, rather than
the z-axis. Moreover, we have a = b = c = 1.
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§12.4 p588
Problem 17, §12.5, p594.
equation
Use the catalog on page 593 to identify the surface given by the
x2 + y 2 − z = 0.
Solution. This is an elliptical paraboloid. See Figure 12.70, and note that we have a = b = 1.
Problem 18, §12.5, p594.
equation
Use the catalog on page 593 to identify the surface given by the
x2 + z 2 = 1.
Solution. Since one variable is missing, this is a cylinder. In fact it is a right circular cylinder.
See Figure 12.77, but note that the roles of y and z have been interchanged. The cylinder in this
problem would be obtained by taking a circle in the xz-plane, with radius 1 and center at the
origin, and then moving that circle along the y-axis. So the central axis of the cylinder will be the
y-axis, not the z-axis.
Problem 21, §12.5, p594. Describe, in words, the level surface
x2
+ z 2 = 1.
4
f (x, y, z) =
In other words, for the function w = f (x, y, z) =
x2
4
+ z 2 , describe the level surface where w = 1.
2
Solution.
We are graphing the equation x4 + z 2 = 1. Since one variable is missing, this
is a cylinder. The section in the plane y = 0 (the xz-plane) is an ellipse, centered at the origin,
intercepting the x-axis at ±2, and intercepting the z-axis at ±1. Take this ellipse and move it along
the y-axis, generating what we might call a right elliptic cylinder. This cylinder is the required
graph.
Problem 23, §12.5, p594.
x + y + z.
Describe in words the level surfaces of the function g(x, y, z) =
Solution. Each of the level surfaces is the graph of x + y + z = c for some fixed value of c. These
surfaces are planes. Each plane has its three intercepts all equal to c. The planes are parallel to
each other. The higher the value of c, the higher the plane is in the z-direction.
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Math 205 HWK 3 Solns continued
§12.4 p588
Problem 25, §12.5, p594. Describe in words the level survaces of f (x, y, z) = sin(x + y + z).
Solution. Each level surface is the graph of sin(x + y + z) = c for some fixed value of c. If
c > 1 or if c < −1, then the surface will not have any points at all, since the sine function can
only yield values between −1 and 1. The level surfaces will be the same planes as they were for
Problem 23, but the corresponding c-values (or w-values) will be different. Moreover, each fixed
c-value will correspond to infinitely many parallel planes. For instance, with c = 0, the graph for
sin(x + y + z) = c will be the infinitely many planes x + y + z = 0, ±π, ±2π, and so forth.
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