MA 3972-MA-Book April 11, 2018 16:1
Areas and Volumes 313
Step 3: Use the [Zero] function and
obtainx=2 fory 2.
Step 4: Use the First Derivative Test.
(See Figures 13.9-4 and 13.9-5.)
Atx=2,Lhas a relative
minimum. Since atx=2,Lhas
the only relative extremum, it is
an absolute minimum.
[–3, 3] by [– 15 , 15]
Figure 13.9-4
decr. 2 incr.
rel. min.
y 2 =(
dL (
dx
L
Figure 13.9-5
Step 5: Atx=2,y=
1
2
(
x^2
)
=
1
2
(
22
)
= 2 .Thus, the point on
y=
1
2
(
x^2
)
closest to the point
(4, 1) is the point (2, 2).
- (a)s(0)=0 and
s(t)=
∫
v(t)dt=
∫
tcos(t^2 +1)dt.
Enter
∫
(x∗cos(x∧ 2 +1), x)
and obtain
sin(x^2 +1)
2
.
Thus, s(t)=
sin(t^2 +1)
2
+C.
Sinces(0)= 0 ⇒
sin(0^2 +1)
2
+C= 0
⇒
. 841471
2
+C= 0
⇒C=− 0. 420735 =− 0. 421
s(t)=
sin(t^2 +1)
2
− 0. 420735
s(2)=
sin(2^2 +1)
2
− 0. 420735
=− 0. 900197 ≈− 0. 900.
(b)v(2)=2 cos(2^2 +1)=2 cos(5)=
0. 567324
Sincev(2)>0, the particle is moving
to the right att=2.
(c) a(t)=v′(t)
Enterd(x∗cos(x∧ 2 +1), x)|x=2 and
obtain 7.95506.
Thus, the velocity of the particle is
increasing att=2, sincea(2)>0.
- (See Figure 13.9-6.)
[–π, π] by [–1, 2]
Figure 13.9-6
