Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 107
33. 9x 2 + y 2 =9 ⇒ 18x +2yy 0 =0 ⇒ 2yy 0 = −18x ⇒ y 0 = −9x/y ⇒
y · 1 − x · y
y 00 0 y − x(−9x/y)
= −9
= −9
= −9 · y2 +9x 2 9
= −9 · [since x and y must satisfy the original
y 2
y 2
y 3 y 3
equation, 9x 2 + y 2 =9].Thus,y 00 = −81/y 3 .
35. x 3 + y 3 =1 ⇒ 3x 2 +3y 2 y 0 =0 ⇒ y 0 = − x2
y 2
y 00 = − y2 (2x) − x 2 · 2yy 0
= − 2xy2 − 2x 2 y(−x 2 /y 2 )
= − 2xy4 +2x 4 y
= − 2xy(y3 + x 3 )
= − 2x
(y 2 ) 2 y 4
y 6 y 6 y , 5
since x and y must satisfy the original equation, x 3 + y 3 =1.
37. (a) There are eight points with horizontal tangents: four at x ≈ 1.57735 and
four at x ≈ 0.42265.
(b) y 0 =
3x 2 − 6x +2
2(2y 3 − 3y 2 − y +1)
⇒
⇒ y 0 = −1 at (0, 1) and y 0 = 1 at (0, 2).
3
Equations of the tangent lines are y = −x +1and y = 1 3 x +2.
(c) y 0 =0 ⇒ 3x 2 − 6x +2=0 ⇒ x =1± 1 3
√
3
(d) By multiplying the right side of the equation by x − 3, we obtain the first
graph. By modifying the equation in other ways, we can generate the other
graphs.
y(y 2 − 1)(y − 2)
= x(x − 1)(x − 2)(x − 3)
y(y 2 − 4)(y − 2)
= x(x − 1)(x − 2)
y(y +1)(y 2 − 1)(y − 2)
= x(x − 1)(x − 2)
(y +1)(y 2 − 1)(y − 2)
=(x − 1)(x − 2)
x(y +1)(y 2 − 1)(y − 2)
= y(x − 1)(x − 2)
y(y 2 +1)(y − 2)
= x(x 2 − 1)(x − 2)
y(y +1)(y 2 − 2)
= x(x − 1)(x 2 − 2)