Solução_Calculo_Stewart_6e

F.
142 ¤ CHAPTER 3 PROBLEMS PLUS
TX.10
tangent line is y − y 0 = − 4 x 0
(x − x 0) or 4x 0x +9y 0y =4x 2 0 +9y
9 y
0. 2 This can be written as x 0x
0 9 + y 0y
4 = x2 0
9 + y2 0
4 =1,
because (x 0 ,y 0 ) lies on the ellipse. So an equation of the tangent line is x 0x
9 + y 0y
4 =1.
Therefore, the x-intercept x T for the tangent line is given by x 0x T
9
=1 ⇔ x T = 9 x 0
,andthey-intercept y T is given
by y0yT
4
=1 ⇔ y T = 4 y 0
.
So as x 0 takes on all values in (0, 3), x T takes on all values in (3, ∞),andasy 0 takes on all values in (0, 2), y T takes on
all values in (2, ∞).
1
At the point (x 0,y 0) on the ellipse, the slope of the normal line is −
y 0 (x = 9 ,andits
0,y 0) 4 x 0
y 0
y 0
equation is y − y 0 = 9 (x − x 0). Sothex-intercept x N for the normal line is given by 0 − y 0 = 9 (x N − x 0)
4 x 0 4 x 0
y 0
⇒
x N = − 4x0
9 + x 0 = 5x0
9 ,andthey-intercept y N is given by y N − y 0 = 9 4
y 0
(0 − x 0 ) ⇒ y N = − 9y0
x 0 4 + y 0 = − 5y0
4 .
So as x 0 takes on all values in (0, 3), x N takes on all values in 0, 5 3
,andasy0 takes on all values in (0, 2), y N takes on
all values in − 5 2 , 0 .
17. (a) If the two lines L 1 and L 2 have slopes m 1 and m 2 and angles of
inclination φ 1 and φ 2 ,thenm 1 =tanφ 1 and m 2 =tanφ 2 . The triangle
in the figure shows that φ 1 + α +(180 ◦ − φ 2 ) = 180 ◦ and so
α = φ 2 − φ 1 . Therefore, using the identity for tan(x − y),wehave
tan α =tan(φ 2 − φ 1 )= tan φ 2 − tan φ 1
m2 − m1
and so tan α = .
1+tanφ 2 tan φ 1 1+m 1m 2
(b) (i) The parabolas intersect when x 2 =(x − 2) 2 ⇒ x =1.Ify = x 2 ,theny 0 =2x, so the slope of the tangent
to y = x 2 at (1, 1) is m 1 =2(1)=2.Ify =(x − 2) 2 ,theny 0 =2(x − 2), so the slope of the tangent to
y =(x − 2) 2 at (1, 1) is m 2 =2(1− 2) = −2. Therefore, tan α =
so α =tan −1 4
3
≈ 53 ◦ [or 127 ◦ ].
m2 − m1
= −2 − 2
1+m 1m 2 1+2(−2) = 4 3 and
(ii) x 2 − y 2 =3and x 2 − 4x + y 2 +3=0intersect when x 2 − 4x +(x 2 − 3) + 3 = 0 ⇔ 2x(x − 2) = 0 ⇒
x =0or 2,but0 is extraneous. If x =2,theny = ±1.Ifx 2 − y 2 =3then 2x − 2yy 0 =0 ⇒ y 0 = x/y and
x 2 − 4x + y 2 +3=0 ⇒ 2x − 4+2yy 0 =0 ⇒ y 0 = 2 − x .At(2, 1) the slopes are m 1 =2and
y
m 2 =0,so tan α = 0 − 2
1+2· 0 = −2 ⇒ α ≈ 117◦ .At(2, −1) the slopes are m 1 = −2 and m 2 =0,
so tan α =
0 − (−2)
1+(−2) (0) =2 ⇒ α ≈ 63◦ [or 117 ◦ ].