Page 201 - Schaum's Outline of Theory and Problems of Advanced Calculus

192 APPLICATIONS OF PARTIAL DERIVATIVES [CHAP. 8

The normals to each surface at the point Pð1; 1; 1Þ are, respectively,
2 2
N 1 ¼rFj P ¼ 6xyi þð3x þ 2yzÞj þ y k ¼ 6 þ j þ k
2
N 2 ¼rGj P ¼ð2z 2xyÞi x j þ 2xk ¼ 4i j þ 2k
Then a tangent vector to the curve at P is
T 0 ¼ N 1 N 2 ¼ð 6i þ j þ kÞ ð4 j þ 2kÞ¼ 3i þ 16j þ 2k
Thus, as in Problem 8.5(a), the tangent line is given by
ðr r 0 Þ T 0 ¼ 0 or fðx 1Þi þð y þ 1Þj þðz 1Þkg f3i þ 16j þ 2kg¼ 0
x 1 y þ 1 z 1
i.e., ¼ ¼ or x ¼ 1 þ 3t; y ¼ 16t 1; z ¼ 2t þ 1
3 16 2
(b)As in Problem 8.5(b)the normal plane is given by
ðr r 0 Þ T 0 ¼ 0 or fðx 1Þi þð y þ 1Þj þðz 1Þkg f3i þ 16j þ 2kg¼ 0

i.e., 3ðx 1Þþ 16ð y þ 1Þþ 2ðz 1Þ¼ 0 or 3x þ 16y þ 2z ¼ 11
The results in (a) and (b)can also be obtained by using equations (7) and (10), respectively, on Page
185.

8.7. Establish equation (10), Page 185.
Suppose the curve is defined by the intersection of two surfaces whose equations are Fðx; y; zÞ¼ 0,
Gðx; y; zÞ¼ 0, where we assume F and G continuously differentiable.
The normals to each surface at point P are given respectively by N 1 ¼rFj P and N 2 ¼rGj P . Then a
tangent vector to the curve at P is T 0 ¼ N 1 N 2 ¼rFj P rGj P . Thus, the equation of the normal plane is
ðr r 0 Þ T 0 ¼ 0. Now

T 0 ¼rFj P rGj P ¼fðF x i þ F y j þ F z kÞ ðG x i þ G y j þ G z kÞgj P

i j
k

F y F x F x F x
F z F y
¼ F x F y k
F z ¼ i þ j þ

G y G z G x G x P G x G y
P P
G x G y G z P

and so the required equation is

F y F z F x
ðr r 0 Þ rFj P ¼ 0 or F z F x F y ðz z 0 Þ¼ 0
ðx x 0 Þþ ð y y 0 Þþ
G y G z P G z G x P G x G y P
ENVELOPES
8.8. Prove that the envelope of the family ðx; y; Þ¼ 0, if it exists, can be obtained by solving
simultaneously the equations ¼ 0and ¼ 0.
Assume parametric equations of the envelope to be x ¼ f ð Þ; y ¼ gð Þ. Then ð f ð Þ; gð Þ; Þ¼ 0
identically, and so upon differentiating with respect to [assuming that , f and g have continuous deriva-
tives], we have
x f ð Þþ y g ð Þþ ¼ 0 ð1Þ
0
0
dy
The slope of any member of the family ðx; y; Þ¼ 0at ðx; yÞ is given by x dx þ y dy ¼ 0or ¼
dy dy=d 0 dx
x g ð Þ
. The slope of the envelope at ðx; yÞ is . Then at any point where the envelope and
¼ ¼
0
y dx dx=d f ð Þ
amember of the family are tangent, we must have
0
x g ð Þ
or x f ð Þþ y g ð Þ¼ 0
0
0
¼ ð2Þ
0
y f ð Þ
Comparing (2)with (1)we see that ¼ 0 and the required result follows.