Thomas Calculus 13th [Solutions]

Section 3.7 Implicit Differentiation 169
52.
53.
54.
1 2
y x b,
y
3
x
4
4
3
x
3 dy 1 dy 2 dy 2
x
and 2y
3x
3x
dx 3 dx dx 2y
4 3
x 4x
0
x
3 ( x 4) 0 x 0 or x 4. If 0
2
2 2
1 3x
1 x
3 2 y
2
2
(0)
x y
2
0 and
y
2
x
2 3
x
2
2
1 3x
3 2 y
(4)
indeterminate at (0, 0). If x 4 y 8. At (4, 8), y 1 x b 8 1 (4) b b 28 .
2
3 3 3
3 2
xy x y
6
3 2 2
also, xy x y 6 x (3 y )
2 dy 3 2 dy
dy 2 2 3 dy y 2xy
y 2xy
x 3y y x 2xy
0 3xy x y 2xy
;
dx
dx
dx
dx 2 2
2 2
3xy
x 3xy
x
y 3 dx 2
dy
x y (2 x dx
dy
) 0 dx 3
( 2 )
dy y xy 2 2
2 2
3xy x dx 3xy
x
;
dy 3
y 2xy
The two different treatments view the graphs as functions symmetric across the
thus dx
dy appears to equal 1
dy
dx
line y x , so their slopes are reciprocals of one another at the corresponding points ( , )
3
a b and ( b, a).
3 2 2 2
x y sin 3 2 dy
dy dy
y x y (2sin )(cos )
dx
dx dx
y 2sin y cos y ) 2 dy
2
3x
3x
dx 2 y 2sin y cos y
2
3x
2sin y cos y 2 y ; x 3 y 2 sin
2 2
2 sin y cos y 2 y
y 3x dx 2y 2 sin y cos y dx
; thus dx appears to
dy dy 2
3x
dy
equal 1
dy
The two different treatments view the graphs as functions symmetric across the line y x so their
dx
slopes are reciprocals of one another at the corresponding points ( a, b ) and ( b, a).
3
1is
55.
d d
dy dy
dx dx dx dx
dy d 1 1
dx dx
x
1 x2
y sin x x sin y ( x) sin y 1 cos y . Since
1 1
2 2
cos y 1 sin y 1 x . Therefore,
sin .
cos y
2 2
sin y cos y 1,
56. (a)
1 2 1 1 2sin 1
dx d (sin x) 2sin x
dx d sin x
x
1 x2
(b)
d 1 1 d 1 1 1 1 1
dx dx 1 x x 1 x x 1
1
sin x
2 x 1 2 2
or
1 2
1 1
x
x2 x2
57-64. Example CAS commands:
Maple:
q1: x^3-x*y y^3 7;
pt : [x 2, y 1];
p1: implicitplot( q1, x -3..3, y -3..3 ):
p1;
eval( q1, pt );
q2 : implicitdiff( q1, y, x );
m : eval( q2, pt );
tan_
line : y 1 m*(x-2);
p2 : implicitplot( tan_
line, x -5..5, y -5..5, color green ):
p3 : pointplot( eval([x, y],pt), color blue):
display( [p1,p2,p3], "Section 3.7 #57(c)" );
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