Thomas Calculus 13th [Solutions]

Section 3.7 Implicit Differentiation 167
2 2
2
41. Solving x xy y 7 and y 0 x 7 x 7 7,0 and 7,0 are the points where the curve
2 2
2x y
crosses the x -axis. Now x xy y 7 2x y xy 2yy 0 ( x 2 y)
y 2x y y
x 2 y
m 2x y
x 2y
the slope at
2 7
2 7
7,0 is m 2 and the slope at 7, 0 is m 2. Since the
7
7
slope is 2 in each case, the corresponding tangents must be parallel.
42.
43.
2 0 dy 2 dy 0 dy y 2
xy x y x y
dx dx dx 1 x
; the slope of the line 2x y 0 is 2. In order to be
parallel, the normal lines must also have slope of 2. Since a normal is perpendicular to a tangent, the slope
of the tangent is 1 2 . Therefore, y 2 1 2y 4 1 x x 3 2 y . Substituting in the original equation,
1 x 2
2
y( 3 2 y) 2( 3 2 y)
y 0 y 4y 3 0 y 3 or y 1. If y 3, then x 3 and
y 3 2( x 3) y 2x 3. If y 1, then x 1 and y 1 2( x 1) y 2x
3.
4 2 2 3
3
y y x 4y y 2yy 2x 2(2 y y) y 2 x y x ; the slope of the tangent line at
3
y 2y
3 3
, is
4 2
x
y 2y
x
y 2y
3
4
3 1 2
3 1
, 2 8
4 2
3 1
4 4
3 3 6 3 1 3
3 3
, 2 8 2 4
4 2
2 3
4 2
3
1
2 3
1; the slope of the tangent line at
3
, 1 is
4 2
44.
45.
46.
2 3
y (2 x) x 2 yy (2 x)
4
2
2 2
2 2
y 2 2
3
( 1) 3 x y y x
;
2 y(2 x)
the slope of the tangent line is y 3x
m
2 y(2 x) (1,1)
y x the normal line is y 1 1 ( x 1) y 1 x 3
2
2 2
2 the tangent line is y 1 2( x 1) 2 1;
3 3
3 3
4 2 4 2 3
3 3
3
y 4y
x 9x 4y y 8yy
4x 18 x y (4y 8 y)
4x 18x y 4x 18x 2x 9x
4y 8y 2y 4 y
2
x(2x
9)
( 3)(18 9)
m; ( 3, 2): m 27 ;( 3, 2): m 27 ;(3, 2): m 27 ;(3, 2): m 27
2
y(2 y 4)
2(8 4) 8 8
8 8
3 3
x y 9xy
2 2
0 3x 3y y 9xy 9y
(a) y 5 and 4
(4, 2) 4
y (2, 4) 5 ;
(b)
(c)
y
0
2
3y x
2
y 3x
2
0 3y
x 0
3 3
x ( x 54) 0 x 0 or
y
2
0 y (3y 9 x)
2 2 3
x 3
x x
3 3
2
9y 3x y
2 2
9 y 3x 3y x
2 2
3y 9x y 3x
2
6 3
9x x 0 x 54x
0
3
3 3
x 54 3 2 there is a horizontal tangent at
3
x 3 2. To find the
corresponding y -value, we will use part (c).
2
dx y 3x
2
3
3
3 3/2
0 0 y 3x 0 y 3 x; y 3x
x 3x 9x 3x
0 x 6 3x
0
dy
2
3y x
3/2 3/2
3/2
x x 6 3 0 or x 0 or 3/2 3 3
x 6 3 x 0 or x 108 3 4. Since the equation
3 3
x y 9xy 0 is symmetric in x and y , the graph is symmetric about the line y x . That is, if ( a, b ) is
a point on the folium, then so is ( b, a ). Moreover, if y m, then y 1 . Thus, if the folium has
( a, b) ( a, b)
m
a horizontal tangent at ( a, b ), it has a vertical tangent at ( b, a ) so one might expect that with a horizontal
tangent at x 3 54 and a vertical tangent at 33
3
x 4, the points of tangency are
3 54, 3 4 and
3
3
3 3
3 4, 54 , respectively. One can check that these points do satisfy the equation x y 9xy
0.
Copyright
2014 Pearson Education, Inc.