Multivariate Calculus - Bruce E. Shapiro

2 LECTURE 1. CARTESIAN COORDINATES
Example 1.1 Find the distance between P = (7, 5, −3) and Q = (12, 8, 6).
According to the distance formula,
|P Q| = √ (x 1 − x 2 ) 2 + (y 1 − y 2 ) 2 + (z 1 − z 2 ) 2
= √ (7 − 12) 2 + (5 − 8) 2 + (−3 − 6) 2
= √ (−5) 2 + (−3) 2 + (−9) 2
= √ 25 + 9 + 81 = √ 115 ≈ 10.72
Therefore the distance is approximately 10.72.
Example 1.2 Show that the triangle with vertices given by P = (2, 1, 6), Q =
(4, 7, 9) and R = (8, 5, −6) is a right triangle.
We use the fact that a right triangle must satisfy the Pythagorean theorem. The
lengths of the three sides are:
|P Q| = √ (4 − 2) 2 + (7 − 1) 2 + (9 − 6) 2 = √ 4 + 36 + 9 = √ 49 = 7
|P R| = √ (8 − 2) 2 + (5 − 1) 2 + (−6 − 6) 2 = √ 36 + 16 + 144 = √ 196 = 14
|QR| = √ (8 − 4) 2 + (5 − 7) 2 + (−6 − 9) 2 = √ 16 + 4 + 225 = √ 245
Since
|P Q| 2 + |P R| 2 = 49 + 196 = 245 = |QR| 2
we know that the triangle satisfies the Pythagorean theorem, and therefore it must
be a right triangle.
Definition 1.2 A sphere is locus of all points that are some distance r from some
fixed point C. The number r is called the radius of the sphere.
Suppose that C = (x 0 , y 0 , z 0 ). Then the distance between C and any other point
P = (x, y, z) is
If P is a distance r from C, then
|P C| = √ (x − x 0 ) 2 + (y − y 0 ) 2 + (z − z 0 ) 2 (1.2)
r = √ (x − x 0 ) 2 + (y − y 0 ) 2 + (z − z 0 ) 2 (1.3)
That is the condition that all points a distance r from C must satisfy. Equivalently,
the squares of both sides of (1.3) are equal to one another,
r 2 = (x − x 0 ) 2 + (y − y 0 ) 2 + (z − z 0 ) 2 (1.4)
Definition 1.3 Equation (1.4) is called the standard Equation of a Sphere of
radius r and center C = (x 0 , y 0 , z 0 ).
Revised December 6, 2006. Math 250, Fall 2006