More Applications of Derivatives 203[ [
022 π^32 π π
Direction Right
of Motion
Stopped StoppedLeft Righttv(t)0+++++–––––––0++++Figure 9.8-6
Step 4: Determine acceleration. (See
Figure 9.8-7.)[ [[ [[02 π02 π
[
0 2 π3 π
23 π
222π πtttSlowingdown Slowing
down
Speeding
up
Speeding
upStopped Stoppedv(t)a(t)Motion of
Particle
++++++ ––––– – – –– ++++––––– – – –– 0 +++++ + + +00ππFigure 9.8-7
Step 5: Draw the motion of the particle.
(See Figure 9.8-8.)t =^32 π
t = 2
z = 0t = πt = 2π–1 0s(t)
1πFigure 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 findthe 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
)
,orsimplifying,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 secondderivatives
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)