Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 109
57. Let y =cos −1 x.Thencos y = x and 0 ≤ y ≤ π ⇒ −sin y dy
dx =1
dy
dx = − 1
sin y = − 1
1 − cos2 y = − 1
√
1 − x
2 .
⇒
[Notethatsin y ≥ 0 for 0 ≤ y ≤ π.]
59. x 2 + y 2 = r 2 is a circle with center O and ax + by =0is a line through O [assume a
and b are not both zero]. x 2 + y 2 = r 2 ⇒ 2x +2yy 0 =0 ⇒ y 0 = −x/y,sothe
slope of the tangent line at P 0 (x 0,y 0) is −x 0/y 0. The slope of the line OP 0 is y 0/x 0,
which is the negative reciprocal of −x 0/y 0. Hence, the curves are orthogonal, and the
families of curves are orthogonal trajectories of each other.
61. y = cx 2 ⇒ y 0 =2cx and x 2 +2y 2 = k [assume k>0] ⇒ 2x +4yy 0 =0 ⇒
2yy 0 = −x ⇒ y 0 = − x
2(y) = − x
2(cx 2 ) = − 1 , so the curves are orthogonal if
2cx
c 6= 0.Ifc =0, then the horizontal line y = cx 2 =0intersects x 2 +2y 2 = k orthogonally
at ± √
k, 0 , since the ellipse x 2 +2y 2 = k has vertical tangents at those two points.
63. To find the points at which the ellipse x 2 − xy + y 2 =3crosses the x-axis, let y =0and solve for x.
y =0 ⇒ x 2 − x(0) + 0 2 =3 ⇔ x = ± √ 3. So the graph of the ellipse crosses the x-axis at the points ± √ 3, 0 .
Using implicit differentiation to find y 0 ,weget2x − xy 0 − y +2yy 0 =0 ⇒ y 0 (2y − x) =y − 2x ⇔ y 0 = y − 2x
2y − x .
So y 0 at √ 3, 0 is 0 − 2 √ 3
2(0) − √ 3 =2and y0 at − √ 3, 0 is 0+2√ 3
2(0) + √ =2. Thus, the tangent lines at these points are parallel.
3
65. x 2 y 2 + xy =2 ⇒ x 2 · 2yy 0 + y 2 · 2x + x · y 0 + y · 1=0 ⇔ y 0 (2x 2 y + x) =−2xy 2 − y ⇔
y 0 = − 2xy2 + y
2x 2 y + x .So− 2xy2 + y
2x 2 y + x = −1 ⇔ 2xy2 + y =2x 2 y + x ⇔ y(2xy +1)=x(2xy +1) ⇔
y(2xy +1)− x(2xy +1)=0 ⇔ (2xy +1)(y − x) =0 ⇔ xy = − 1 2 or y = x.Butxy = − 1 2
⇒
x 2 y 2 + xy = 1 4 − 1 2 6=2,sowemusthavex = y. Then x2 y 2 + xy =2 ⇒ x 4 + x 2 =2 ⇔ x 4 + x 2 − 2=0 ⇔
(x 2 +2)(x 2 − 1) = 0. Sox 2 = −2, which is impossible, or x 2 =1 ⇔ x = ±1. Sincex = y, the points on the curve
where the tangent line has a slope of −1 are (−1, −1) and (1, 1).
67. (a) If y = f −1 (x),thenf(y) =x. Differentiating implicitly with respect to x and remembering that y is a function of x,
we get f 0 (y) dy dy
=1,so
dx dx = 1
f 0 (y)
⇒ f −10 (x) =
1
f 0 (f −1 (x)) .
(b) f(4) = 5 ⇒ f −1 (5) = 4. Bypart(a), f −10 (5) =
1
f 0 (f −1 (5)) = 1
f 0 (4) =1
2
3 =
3
. 2