Page 200 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 191

Denoting the expressions in braces by A; B; C respectively so that rFj P ¼ Ai þ Bj þ Ck,we see
that the required equation is Aðx x 0 Þþ Bð y y 0 Þþ Cðz z 0 Þ¼ 0. This can be written in spherical
coordinates by using the transformation equations for x, y; and z in these coordinates.
(b)We have F ¼ r 4cos ¼ 0. Then @F=@r ¼ 1, @F=@ ¼ 4 sin , @F=@ ¼ 0.
ffiffiffi
p
Since r 0 ¼ 2 2; 0 ¼ =4; 0 ¼ 3 =4, we have from part (a), rFj P ¼ Ai þ Bj þ Ck ¼ i þ j.
ffiffiffi p
p
From the transformation equations the given point has rectangular coordinates ð 2; ffiffiffi 2; 2Þ, and
p ffiffiffi p ffiffiffi
2Þj þðz 2Þk.
so r r 0 ¼ðx þ 2Þi þð y
ffiffiffi ffiffiffi ffiffiffi
p p p
2Þ¼ 0or y x ¼ 2 2.In sphe-
The required equation of the plane is thus ðx þ 2Þþð y
ffiffiffi
p
rical coordinates this becomes r sin sin r sin cos ¼ 2 2.
2
2
2
In rectangular coordinates the equation r ¼ 4cos becomes x þ y þðz 2Þ ¼ 4 and the tangent
plane can be determined from this as in Problem 8.1. In other cases, however, it may not be so easy to
obtain the equation in rectangular form, and in such cases the method of part (a)is simpler to use.
(c) The equations of the normal line can be represented by
ffiffiffi ffiffiffi
p p
2 2 z 2
x þ y
1 ¼ 1 ¼ 0
the significance of the right-hand member being that the line lies in the plane z ¼ 2. Thus, the required
line is given by
p ffiffiffi p ffiffiffi
2 2
; z ¼ 0 or x þ y ¼ 0; z ¼ 0
x þ y
1 ¼ 1
TANGENT LINE AND NORMAL PLANE TO A CURVE
8.5. Find equations for the (a) tangent line and (b) normal plane to the curve x ¼ t cos t,
1
y ¼ 3 þ sin 2t, z ¼ 1 þ cos 3t at the point where t ¼ .
2
(a) The vector from origin O (see Fig. 8-2, Page 183) to any point of curve C is R ¼ðt cos tÞiþ
1
ð3 þ sin 2tÞj þð1 þ cos 3tÞk. Then a vector tangent to C at the point where t ¼ is
2

dR
¼ð1 þ sin tÞi þ 2cos 2t j 3 sin 3t kj t¼1=2 ¼ 2i 2j þ 3k

T 0 ¼
dt t¼1=2
1
1
The vector from O to the point where t ¼ is r 0 ¼ i þ 3j þ k.
2
2
The vector from O to any point ðx; y; zÞ on the tangent line is r ¼ xi þ yj þ zk.
1
Then r r 0 ¼ðx Þi þ y 3Þj þðz 1Þk is collinear with T 0 ,sothat the required equation is
2

i j
k
1

ðr r 0 Þ T 0 ¼ 0; i:e:; x y 3 z 1 ¼ 0

2
2 2 3
x y 3 z 1
1
1
and the required equations are 2 ¼ ¼ or in parametric form x ¼ 2t þ , y ¼ 3 2t,
2
z ¼ 3t þ 1: 2 2 3
(b)Let r ¼ xi þ yj þ zk be the vector from O to any point ðx; y; zÞ of the normal plane. The vector from O
1
1
1
to the point where t ¼ is r 0 ¼ i þ 3j þ k. The vector r r 0 ¼ðx Þi þð y 3Þj þðz 1Þk lies
2 2 2
in the normal plane and hence is perpendicular to T 0 . Then the required equation is ðr r 0 Þ T 0 ¼ 0or
1
2ðx Þ 2ð y 3Þþ 3ðz 1Þ¼ 0.
2
2
2
8.6. Find equations for the (a) tangent line and (b) normal plane to the curve 3x y þ y z ¼ 2,
2
2xz x y ¼ 3at the point ð1; 1; 1Þ.
(a) The equations of the surfaces intersecting in the curve are
2
2
2
F ¼ 3x y þ y z þ 2 ¼ 0; G ¼ 2xz x y 3 ¼ 0