Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION 0 169
9. r (t) = (t2 + t, e - t , t 3 ) => v(t) = r' (t) = (2t + 1, 2t - 1, 3t 2 ), a(t) = v' (t) = (2, 2, 6t),
lv (t)i = .j(2t + 1)2 + (2t- 1)2 + (3t 2 )2 = v'9t4 + 8t 2 + 2.
11. r (t) = v'2t i + e' j + e-t k => v (t) = r'(t) = v'2i + e' j - e-' k, a(t) = v'(t) = e' j + e-' k,
lv(t)l = v'2 + e 2 ' + e zt = J(e' + e-') 2 = e' + e-' .
13. r (t) = e'(cost,sint, t) =>
v (t) = r'(t) = e'(cost,.sin t, t~ + et (-sint,cost, 1) = et (cost - sint,sint +cost, t + 1)
a(t) = v'-(t) = 'e'(cost- sint - sin t - cost, sin t +cost+ cost - sin t , t + 1 + 1)
= et(- 2sin t , 2 cost, t + 2)
lv(t)i = etJcos 2 t + sin 2 t - 2costsint + sin 2 t + cos 2 t + 2sin tcost + t 2 + 2t + 1
= e' .Jt 2 + 2t + 3
15. a(t) = i + 2j => v(t) = J a(t) dt = J(i + 2j) dt = t i + 2tj + C and k = v {0) = C,
.soC = k and v(t) = t i + 2tj + k. , r{t) = J v(t)"dt = J (t i + 2tj + k) dt = ~t 2 i + t2 j + t k + D .
But i = r (0) = D, soD= i and r(t) = (tt 2 + 1) i + t 2 j + tk.
17 .. (a) a(t) = 2t i + sintj +cos 2t k => ·
v( t) = J (2t i + sin t j + cos 2t k) dt = t2 i - cost j + t sin 2t k + C
and i = v (O) = -j + C, so C = i + j
and v(t) = (t 2 + 1) i + (1- cost)j + t sin2tk.
r (t) = J[(t 2 + 1) i + (1- cost)j + t sin2tk)dt
= (tt 3 + t) i + (t- sin t)j- ~cos 2t k + D
(b)
0.6
0.4
% 0.2
0
Butj = r(O) = -i k+ D , soD = j + {kand r (t) = (tt 3 +t) i+ (t -sint + 1)j + (!- { cos2t) k.
19. r (t) = (t 2 ' 5t, e- 16t) =} v(t) = (2t, 5, 2t - ,16), lv(t)i = v'4t 2 + 25 + 4t 2 - 64t + 256 = v'8t2 - 64t + 281
and dd lv(t)i = t(se - 64t + 281)- 1 1 2 (16t - 64) . This is zero if and only if the numerator is zero, that is,
t .
16t - 64 = 0 or t = 4. Since :t lv(t)l < 0 fort < 4 a~d :t lv(t)l > 0 fort > 4, the minimum speed of -/153 is atta.ined
at t = 4 units of time.
21 . IF (t)i = 20 N in the direction of the positive z-axis, so F (t) = 20 k. Also m = 4 kg, r{O) = 0 and v {O) = i - j .
Since 20k = F(t) = 4 a(t), a (t) = 5k. Then v (t) = 5t k + c1 where c1 = i-j so v(t) = i - j + 5tk and the
speed is lv (t)l = v 1 + 1 + 25t 2 = v'25t 2 + 2. Also r (t) = t i - t j + ~t 2 k + cz and 0 = r (O), so C 2 = 0
and r (t) = t i - t j + ~ t 2 k.
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