5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 15:14

More Applications of Derivatives 225

19. Step 1: s(t)=sint
    v(t)=cost
    a(t)=−sint
    Step 2: Setv(t)= 0 ⇒cost=0;
    t=
    π
    2
    and
    3 π
    2

.

Seta(t)= 0 ⇒−sint=0;

t=πand 2π.

Step 3: Determine the directions of

motion. (See Figure 10.7-6.)

[ [

02 π 2 32 π π

Direction Right

of motion

Stopped Stopped

Left Right

t

v(t)0+++++ ––––––– 0 ++++

Figure 10.7-6

Step 4: Determine acceleration. (See

Figure 10.7-7.)

[ [

[ [

[

02 π

02 π

[

(^02) π
3 π
2
3 π
2
π
2
ππ
π
2
t
t
t
Slowing
down
Slowing
down
Speeding
up
Speeding
up
Stopped Stopped
v(t)
a(t)
Motion of
particle
++++++ ––––––––– ++++
––––––––– 0 +++++++ +
00
Figure 10.7-7
Step 5: Draw the motion
of the particle. (See Figure 10.7-8.)
t =^32 π
t =π 2
z = 0
t = π
t = 2π

• 10

s(t)

1

Figure 10.7-8

The particle is initially at 0,s(0)=0. It

moves to the right but slows down to a

stop at 1 whent=

π

2

,s

(

π

2

)

=1. It then

turns and moves to the left speeding up

until it reaches 0, whent=π,s(π)=0 and

continues to the left, but slowing down to

a stop at−1 whent=

3 π

2

,s

(

3 π

2

)

=−1.

It then turns around again, moving to the

right, speeding up to 0 whent= 2 π,

s(2π)=0.

20. s(t)=− 16 t^2 +v 0 t+s 0
    s 0 =height of building andv 0 =0.
    Thus,s(t)=− 16 t^2 +s 0.
    When the coin hits the ground,s(t)=0,
    t= 10 .2. Thus, sets(t)= 0 ⇒
    − 16 t^2 +s 0 = 0 ⇒−16(10.2)^2 +s 0 = 0
    s 0 = 1664 .64 ft. The building is
    approximately 1665 ft tall.

10.8 Solutions to Cumulative Review Problems

21. Using product rule, letu=x;v=sin−^1 (2x).
    dy
    dx
    =(1) sin−^1 (2x)+

1

√

1 −(2x)^2

(2)(x)

=sin−^1 (2x)+

2 x

√

1 − 4 x^2

22. Lety= f(x)⇒y=x^3 − 3 x^2 + 3 x−1.
    To findf−^1 (x), switchxand
    y:x=y^3 − 3 y^2 + 3 y−1.
    dx
    dy
    = 3 y^2 − 6 y+ 3

dy

dx

=

1

dx/dy

=

1

3 y^2 − 6 y+ 3