Mathematics 205 HWK 3 Solutions Section 12.4 p588

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1 Mathematics 205 HWK 3 Solutions Section 12.4 p588 Problem 19, 12.4, p588. Find the 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 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 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 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. Page 1 of 5 A. Sontag September 13, 2003

2 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. Page 2 of 5 A. Sontag September 13, 2003

3 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. Page 3 of 5 A. Sontag September 13, 2003

4 Problem 15, 12.5, p594. x 2 y 2 + z 2 = 1. Solution. Rewrite the given as x 2 + 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 the cross sections perpendicular to the z-axis appear circular. The top sheet is the function z = x 2 + y 2 + 1, and the bottom sheet is the function z = x 2 + y Problem 16, 12.5, p594. x 2 + y 2 z 2 = 0. Solution. Rewrite as x 2 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. Page 4 of 5 A. Sontag September 13, 2003

5 Problem 17, 12.5, p594. x 2 + 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. x 2 + 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 f(x, y, z) = x2 4 + z2 = 1. In other words, for the function w = f(x, y, z) = x2 4 + z2, describe the level surface where w = 1. Solution. We are graphing the x2 4 + z2 = 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. Describe in words the level surfaces of the function g(x, y, z) = 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. Page 5 of 5 A. Sontag September 13, 2003

6 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. Page 6 of 5 A. Sontag September 13, 2003