Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 417
11. x =4+t 2 , y = t 2 + t 3 ⇒ dy
dx = dy/dt
dx/dt
d 2 y
dx 2 = d
dx
=
2t +3t2
2t
=1+ 3 2 t ⇒
dy
= d(dy/dx)/dt = (d/dt) 1+ 3 t 2
= 3/2
dx dx/dt
2t 2t = 3 4t .
The curve is CU when d2 y
dx 2 > 0,thatis,whent>0.
13. x = t − e t , y = t + e −t ⇒
dy
dx = dy/dt
dx/dt = 1 − 1 − 1 e t − 1
d dy
e−t
= e t
1 − e t 1 − e = e t
t 1 − e = t −e−t ⇒ d2 y
dx = dt dx
=
2 dx/dt
The curve is CU when e t < 1 [since e −t > 0] ⇒ t