Solução_Calculo_Stewart_6e

F.
110 ¤ CHAPTER 13 VECTORSANDTHEGEOMETRYOFSPACE ET TX.10 CHAPTER 12
|F| ≈ (−200) 2 + 100 √ 3 2 √ √
= 70,000 = 100 7 ≈ 264.6 N. Let θ be the angle F makes with the positive x-axis.
Then tan θ = 100 √ √
3 3
−200 = − 2
θ =tan −1 −
√
3
2
and the terminal point of F lies in the second quadrant, so
+180 ◦ ≈−40.9 ◦ + 180 ◦ =139.1 ◦ .
31. With respect to the water’s surface, the woman’s velocity is the vector sum of the velocity of the ship with respect to the water,
and the woman’s velocity with respect to the ship. If we let north be the positive y-direction, then
v = h0, 22i + h−3, 0i = h−3, 22i. The woman’s speed is |v| = √ 9 + 484 ≈ 22.2mi/h. Thevectorv makes an angle θ
with the east, where θ =tan −1 22
≈ 98 ◦ . Therefore, the woman’s direction is about N(98 − 90) ◦ W=N8 ◦ W.
33. Let T 1 and T 2 represent the tension vectors in each side of the
clothesline as shown in the figure. T 1 and T 2 have equal vertical
−3
components and opposite horizontal components, so we can write
T 1 = −a i + b j and T 2 = a i + b j [a, b > 0]. By similar triangles, b a = 0.08
4
⇒
a =50b. The force due to gravity
acting on the shirt has magnitude 0.8g ≈ (0.8)(9.8) = 7.84 N, hence we have w = −7.84 j. The resultant T 1 + T 2
of the tensile forces counterbalances w, soT 1 + T 2 = −w ⇒ (−a i + b j)+(a i + b j) =7.84 j ⇒
(−50b i + b j)+(50b i + b j) =2b j =7.84 j ⇒ b = 7.84
2
=3.92 and a =50b = 196. Thus the tensions are
T 1 = −a i + b j = −196 i +3.92 j and T 2 = a i + b j = 196 i +3.92 j.
Alternatively, we can find the value of θ and proceed as in Example 7.
35. The slope of the tangent line to the graph of y = x 2 at the point (2, 4) is
dy
dx =2x
=4
x=2 x=2
and a parallel vector is i +4j which has length |i +4j| = √ 1 2 +4 2 = √ 17, so unit vectors parallel to the tangent line are
± 1 √
17
(i +4j).
37. BytheTriangleLaw, −→
AB + −→ −→ −→ −→ −→ −→ −→ −→ −→ −→
BC = AC. ThenAB + BC + CA = AC + CA,butAC + CA = AC + − −→
AC
So −→
AB + −→
BC +
−→
CA = 0.
39. (a), (b) (c) From the sketch, we estimate that s ≈ 1.3 and t ≈ 1.6.
(d) c = s a + t b ⇔ 7=3s +2t and 1=2s − t.
Solving these equations gives s = 9 7 and t = 11 7 .
= 0.
41. |r − r 0 | is the distance between the points (x, y, z) and (x 0 ,y 0 ,z 0 ), so the set of points is a sphere with radius 1 and
center (x 0 ,y 0 ,z 0 ).
Alternate method: |r − r 0| =1 ⇔ (x − x 0) 2 +(y − y 0) 2 +(z − z 0) 2 =1 ⇔
(x − x 0 ) 2 +(y − y 0 ) 2 +(z − z 0 ) 2 =1, which is the equation of a sphere with radius 1 and center (x 0 ,y 0 ,z 0 ).