306 STEP 4. Review the Knowledge You Need to Score High
Step 1. Distance Formula:
L=
√
(x− 4 )^2 +(y− 1 )^2
=
√
(x− 4 )^2 +
(
x^2
2
− 1
) 2
where the domain is all real
numbers.
Step 2. Enter√ y 1 =
((x−4)∧ 2 +(. 5 x∧ 2 −1)∧2).
Entery 2 =d(y1(x),x).
Step 3. Use the [Zero] function and
obtainx=2 fory 2.
Step 4. Use the First Derivative Test. (See
Figures 12.10-4 and 12.10-5.)
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 12.10-4
–0+
decr.^2 incr.
rel. min.
y 2 =(
dL (
dx
L
Figure 12.10-5
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.