Solução_Calculo_Stewart_6e

F.
154 ¤ CHAPTER 14 VECTOR FUNCTIONS ET CHAPTER 13
TX.10
15. a(t) =i +2j ⇒ v(t) = a(t) dt = (i +2j) dt = t i +2t j + C and k = v (0) = C,
so C = k and v(t) =t i +2t j + k. r(t) = v(t) dt = (t i +2t j + k) dt = 1 2 t2 i + t 2 j + t k + D.
But i = r (0) = D,soD = i and r(t) = 1
2 t2 +1 i + t 2 j + t k.
17. (a) a(t) =2t i +sint j +cos2t k ⇒
(b)
v(t) = (2t i +sint j +cos2t k) dt = t 2 i − cos t j + 1 2
sin 2t k + C
and i = v (0) = −j + C,soC = i + j
and v(t) = t 2 +1 i +(1− cos t) j + 1 sin 2t k.
2
r(t) = [ t 2 +1 i +(1− cos t) j + 1 sin 2t k]dt
2
= 1
3 t3 + t i +(t − sin t) j − 1 cos 2t k + D
4
But j = r (0) = − 1 4 k + D,soD = j + 1 4 k and r(t) = 1
3 t3 + t i +(t − sin t +1)j + 1
4 − 1 4 cos 2t k.
19. r(t) = t 2 , 5t, t 2 − 16t ⇒ v(t) =h2t, 5, 2t − 16i, |v(t)| = √ 4t 2 +25+4t 2 − 64t + 256 = √ 8t 2 − 64t + 281
and d dt |v(t)| = 1 2 (8t2 − 64t + 281) −1/2 (16t − 64). This is zero if and only if the numerator is zero, that is,
16t − 64 = 0 or t =4.Since d dt |v(t)| < 0 for t 0 for t>4, the minimum speed of √ 153 is attained
at t =4units of time.
21. |F(t)| =20N in the direction of the positive z-axis, so F(t) =20k. Alsom =4kg, r(0) = 0 and v(0) = i − j.
Since 20k = F(t) =4a(t), a(t) =5k. Thenv(t) =5t k + c 1 where c 1 = i − j so v(t) =i − j +5t k and the
speed is |v(t)| = √ 1+1+25t 2 = √ 25t 2 +2.Alsor(t) =t i − t j + 5 2 t2 k + c 2 and 0 = r(0), soc 2 = 0
and r(t) =t i − t j + 5 2 t2 k.
23. |v(0)| = 500 m/s and since the angle of elevation is 30 ◦ , the direction of the velocity is 1 2
√
3 i + j
.Thus
v(0) = 250 √ 3 i + j and if we set up the axes so the projectile starts at the origin, then r(0) = 0. Ignoring air resistance, the
only force is that due to gravity, so F(t) =−mg j where g ≈ 9.8 m/s 2 .Thusa(t) =−g j and v(t) =−gt j + c 1 .But
250 √ 3 i + j = v(0) = c 1 ,sov(t) =250 √ 3 i + (250 − gt) j and r(t) = 250 √ 3 t i + 250t − 1 2 gt2 j + c 2 where
0 = r(0) = c 2 .Thusr(t) =250 √ 3 t i + 250t − 1 2 gt2 j.
(a) Setting 250t − 1 2 gt2 =0gives t =0or t = 500
g
≈ 51.0 s. So the range is 250 √ 3 · 500
g
≈ 22 km.
(b) 0= d dt
250t −
1
2 gt2 = 250 − gt implies that the maximum height is attained when t = 250/g ≈ 25.5 s.
Thus, the maximum height is (250)(250/g) − g(250/g) 2 1 2 = (250)2 /(2g) ≈ 3.2 km.
(c) From part (a), impact occurs at t = 500/g ≈ 51.0. Thus, the velocity at impact is
v(500/g) =250 √ 3 i +[250− g(500/g)] j =250 √ 3 i − 250 j andthespeedis|v(500/g)| = 250 √ 3+1=500m/s.