5 Steps to a 5 AP Calculus BC 2019

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).

29. (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.