Solução_Calculo_Stewart_6e

F.
108 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
39. From Exercise 29, a tangent to the lemniscate will be horizontal if y 0 =0 ⇒ 25x − 4x(x 2 + y 2 )=0 ⇒
x[25 − 4(x 2 + y 2 )] = 0 ⇒ x 2 + y 2 = 25 4
(1). (Note that when x is 0, y is also 0, and there is no horizontal tangent
at the origin.) Substituting 25 for 4
x2 + y 2 in the equation of the lemniscate, 2(x 2 + y 2 ) 2 =25(x 2 − y 2 ),weget
x 2 − y 2 = 25 (2). Solving (1)and(2), we have x 2 = 75 and 8 16 y2 = 25 , so the four points are ± 5√ 3
, ± 5 .
16 4 4
41.
x 2
a − y2
2x
=1 ⇒ 2 b2 a − 2yy0 =0 ⇒ y 0 = b2 x
2 b 2 a 2 y
⇒
an equation of the tangent line at (x 0 ,y 0 ) is
y − y 0 = b2 x 0
(x − x
a 2 0 ). Multiplying both sides by y0 y0y
gives
y 0 b2 b − y2 0
2 b 2
= x0x
a 2 − x2 0
a 2 .Since(x 0,y 0 ) lies on the hyperbola,
we have x0x
a 2
− y0y
b 2 = x2 0
a 2 − y2 0
b 2 =1.
43. If the circle has radius r, its equation is x 2 + y 2 = r 2 ⇒ 2x +2yy 0 =0 ⇒ y 0 = − x , so the slope of the tangent line
y
at P (x 0 ,y 0 ) is − x 0
−1
. The negative reciprocal of that slope is = y 0
, which is the slope of OP, so the tangent line at
y 0 −x 0 /y 0 x 0
P is perpendicular to the radius OP.
45. y =tan √ −1 x ⇒ y 0 1
= √ 2 ·
1+ x
d
√
x = 1
dx 1+x
1
2 x−1/2
=
1
2 √ x (1 + x)
47. y =sin −1 (2x +1) ⇒
y 0 =
1
· d
1 − (2x +1)
2 dx (2x +1)= 1
1 − (4x2 +4x +1) · 2= 2
√
−4x2 − 4x = 1
√
−x2 − x
49. G(x) = √ 1 − x 2 arccos x ⇒ G 0 (x) = √ 1 − x 2 ·
51. h(t) =cot −1 (t) +cot −1 (1/t) ⇒
h 0 (t) =− 1
1+t − 1
2 1+(1/t) · d 1
2 dt t = − 1
1+t − t2
2 t 2 +1 ·
−1
√ +arccosx · 1
1 − x
2 2 (1 − x2 ) −1/2 (−2x) =−1 − x √ arccos x
1 − x
2
− 1 t 2
= − 1
1+t 2 + 1
t 2 +1 =0.
Note that this makes sense because h(t) = π 2 for t>0 and h(t) =3π 2
for t