More Applications of Derivatives 203
[ [
022 π^32 π π
Direction Right
of Motion
Stopped Stopped
Left Right
t
v(t)0+++++–––––––0++++
Figure 9.8-6
Step 4: Determine acceleration. (See
Figure 9.8-7.)
[ [
[ [
[
02 π
02 π
[
0 2 π
3 π
2
3 π
2
2
2
π π
t
t
t
Slowingdown Slowing
down
Speeding
up
Speeding
up
Stopped Stopped
v(t)
a(t)
Motion of
Particle
++++++ ––––– – – –– ++++
––––– – – –– 0 +++++ + + +
00
π
π
Figure 9.8-7
Step 5: Draw the motion of the particle.
(See Figure 9.8-8.)
t =^32 π
t = 2
z = 0
t = π
t = 2π
–1 0
s(t)
1
π
Figure 9.8-8
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.
- 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. - Whenx=
− 1
2
=cost−1, cost=
1
2
,
andt=
π
3
, and soy=sin
(
π
3
)
+
π
3
=
√
3
2
+
π
3
. Find
dx
dt
=−sintand
dy
dt
=cost+1, and divide to find
dy
dx
=
cost+ 1
−sint
. Evaluate att=
π
3
to find
the slopem=
cos(π/3)+ 1
−sin(π/3)
=
3 / 2
−
√
3 / 2
=−
√
- Therefore, the
equation of the tangent line is
y −
(√
3
2
+
π
3
)
=−
√
3
(
x+
1
2
)
,or
simplifying,y=−
√
3 x+
π
3
.
Differentiate to find
dx
dt
= 2 e^2 t+1 and
dy
dt
=et. The speed of the object is
√
(2e^2 t+1)^2 +(et)^2
√
√^4 e^4 t+^5 e^2 t+1. Whent=2,
4 e^4 t+ 5 e^2 t+ 1 ≈ 110 .444. Find second
derivatives
d^2 x
dt^2
= 4 e^2 tand
d^2 y
dt^2
=etand
evaluate att=2 to find the acceleration
vector〈 4 e^2 t,et〉≈〈 218 .393, 7. 389 〉.
- Sincex=rcosθandy=sinθ,
dy
dx
=
dy
dθ
÷
dx
dθ
=
rcosθ+sinθ(dr/dθ)
−rsinθ+cosθ(dr/dθ)