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(MATH100)worksheet-ch1-sol.pdf
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MATH 100 (L1) Introduction to Multivariable Calculus Spring 2006-07
. Class Worksheet Chapter 1 Vectors and Geometry of Space
1. (Graphs in three dimensions) Sketch the graphs of the equations in three-dimensional space.
(a)
z = 0. (c) z = 3. (e) z =1+ x.

(b)
z . 0. (d) x = 3. (f) z =1+ x . y.


Solution Identify the graphs as follows.
(a)
The equation z = 0 represents all points with coordinates (x, y, 0), that is, the xy-plane.

(b)
The inequality z . 0 represents all points on and above the xy-plane (i.e., half-space).

(c)
The equation z = 3 represents all points with coordinates (x, y, 3), that is, the horizontal plane passing through the point (0, 0, 3) on the z-axis.

(d)
The equation x = 3 represents the vertical plane perpendicular to the x-axis at x = 3. This plane lies parallel to the yz-plane and 3 units in front of it. The graph for this equation can be found in lecture notes (page 2).

(e)
The equation z = 1+ x represents the plane parallel to the y-axis and passing through the points (.1, 0, 0) and (0, 0, 1).

(f)
The equation z = 1+ x . y represents the plane passing through the three points (.1, 0, 0), (0, 1, 0) and (0, 0, 1).


2. (Graphs in three dimensions) Sketch the surface z = cos x in three-dimensional space.
Solution Using the cross-section method, we sketch the graph of z = cos x in R2 .rst and then the
graph of that in R3 .
z
z z = cos x in R3

z = cos x xy
x
R2
Cross section in
3. (Distance between two points) Find the distance between the points P1(1, 2, 5) and P2(3, 2, .5). Solution The distance between P1(1, 2, 5) and P2(3, 2, .5) is
..
.P1P2. = p(3 . 1)2 + (2 . 2)2 +(.5 . 5)2

= 4+0+100

4.
5.

6.

7.


(Distance between two points) Show that the triangle with vertices A = (1, .1, 2), B = (3, 3, 8) and C = (2, 0, 1) has a right angle.
Solution We calculate the lengths of the three sides of the triangle:
.
a = .BC. = p(2 . 3)2 + (0 . 3)2 + (1 . 8)2 = 59, .
b = .AC. = p(2 . 1)2 + (0 + 1)2 + (1 . 2)2 =3,
.
c = .AB. = p(3 . 1)2 + (3 + 1)2 + (8 . 2)2 = 56.
Observe that a 2 = b2 + c 2 (sketch the graph of the triangle). Then by the Pythagorean theorem, we know that CAB must be a right angle.
(Standard equation for sphere) Find the equation for the sphere (in space) of radius 4 with the center at the point (.3, 5, .4).
Solution Assume that P (x, y, z) can be any point on the sphere. Then, by the distance formula, we have
p(x + 3)2 +(y . 5)2 +(z + 4)2 =4.
The standard equation for the sphere is (x + 3)2 +(y . 5)2 +(z + 4)2 = 16.
(Standard equation for sphere) Identify the graphs of the equations: (a) y 2 +(z . 1)2 = 4;
2 222 222
(b) y +(z . 1)2 = 0; (c) x + y + z = 0 and (d) x + y + z = .1.
Solution
(a)
Since x is absent, the equation y 2 +(z . 1)2 = 4 represents an object parallel to the x-axis. In the yz-plane the equation represents a circle of radius