Solução_Calculo_Stewart_6e

F.
68 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
TX.10
(b) The velocity of the particle is equal to the slope of the tangent line of the
graph. Note that there is no slope at the corner points on the graph. On the
interval (0, 1), the slope is 3 − 0 =3.Ontheinterval(2, 3), the slope is
1 − 0
1 − 3
3 − 1
= −2. On the interval (4, 6), the slope is
3 − 2 6 − 4 =1.
13. Let s(t) =40t − 16t 2 .
s(t) − s(2)
40t − 16t
2
− 16 −16t 2 +40t − 16 −8 2t 2 − 5t +2
v(2) = lim
=lim
=lim
=lim
t→2 t − 2 t→2 t − 2
t→2 t − 2
t→2 t − 2
−8(t − 2)(2t − 1)
=lim
= −8lim(2t − 1) = −8(3) = −24
t→2 t − 2
t→2
Thus, the instantaneous velocity when t =2is −24 ft/s.
1
s(a + h) − s(a) (a + h) − 1 15. v(a)= lim
= lim
2 a 2
h→0 h
h→0 h
a 2 − (a + h) 2
a
=lim
2 (a + h) 2
h→0 h
= lim
h→0
a 2 − (a 2 +2ah + h 2 )
ha 2 (a + h) 2
−(2ah + h 2 )
=lim
h→0 ha 2 (a + h) =lim −h(2a + h)
2 h→0 ha 2 (a + h) = lim −(2a + h)
2 h→0 a 2 (a + h) = −2a
2 a 2 · a = −2
2 a m/s 3
So v (1) = −2
−2
= −2 m/s, v(2) =
13 2 = − 1 −2
m/s, and v(3) = 3 4 3 = − 2 3 27 m/s.
17. g 0 (0) istheonlynegativevalue.Theslopeatx =4is smaller than the slope at x =2and both are smaller than the slope at
x = −2. Thus, g 0 (0) < 0