Solução_Calculo_Stewart_6e

F.
SECTION 14.4 TX.10 MOTION IN SPACE: VELOCITY AND ACCELERATION ET SECTION 13.4 ¤ 153
(b) We can estimate the velocity at t =1by averaging the average velocities over the time intervals [0.5, 1] and [1, 1.5]:
v(1) ≈ 1 [(2 i − 2.4 j − 0.6 k)+(2.8 i +0.8 j − 0.4 k)] = 2.4 i − 0.8 j − 0.5 k. Then the speed is
2
|v(1)| ≈ (2.4) 2 +(−0.8) 2 +(−0.5) 2 ≈ 2.58.
3. r(t) = − 1 2 t2 ,t ⇒ At t =2:
v(t) =r 0 (t) =h−t, 1i
a(t) =r 00 (t) =h−1, 0i
v(2) = h−2, 1i
a(2) = h−1, 0i
|v(t)| = √ t 2 +1
5. (t) =3cost i +2sint j ⇒ At t = π/3:
v(t) =−3sint i +2cost j
a(t) =−3cost i − 2sint j
v
π
3 = −
3 √ 3
a π
3
i + j
2
= −
3
i − √ 3 j
2
|v(t)| = 9sin 2 t +4cos 2 t = 4+5sin 2 t
Notice that x 2 /9+y 2 /4=sin 2 t +cos 2 t =1, so the path is an ellipse.
7. r(t) =t i + t 2 j +2k ⇒ At t =1:
v(t) =i +2t j v(1) = i +2j
a(t) =2j
a(1) = 2 j
|v(t)| = √ 1+4t 2
Here x = t, y = t 2 ⇒ y = x 2 and z =2, so the path of the particle is a
parabola in the plane z =2.
9. r(t) = t 2 +1,t 3 ,t 2 − 1 ⇒ v(t) =r 0 (t) = 2t, 3t 2 , 2t , a(t) =v 0 (t) =h2, 6t, 2i,
|v(t)| = (2t) 2 +(3t 2 ) 2 +(2t) 2 = √ 9t 4 +8t 2 = |t| √ 9t 2 +8.
11. r(t) = √ 2 t i + e t j + e −t k ⇒ v(t) =r 0 (t) = √ 2 i + e t j − e −t k, a(t) =v 0 (t) =e t j + e −t k,
|v(t)| = √ 2+e 2t + e −2t = (e t + e −t ) 2 = e t + e −t .
13. r(t) =e t hcos t, sin t, ti ⇒
v(t) =r 0 (t) =e t hcos t, sin t, ti + e t h− sin t, cos t, 1i = e t hcos t − sin t, sin t +cost, t +1i
a(t) =v 0 (t) =e t hcos t − sin t − sin t − cos t, sin t +cost +cost − sin t, t +1+1i
= e t h−2sint, 2cost, t +2i
|v(t)| = e t cos 2 t +sin 2 t − 2cost sin t +sin 2 t +cos 2 t +2sint cos t + t 2 +2t +1
= e t√ t 2 +2t +3