More Applications of Derivatives 191
Position, Speed, and Acceleration
When the motion of a particle is defined parametrically, its position is given by(x(t),y(t)).
The speed of the particle is
√(
dx
dt
) 2
+
(
dy
dt
) 2
〈 and its acceleration is given by the vector
d^2 x
dt^2
,
d^2 y
dt^2
〉
.
Example 1
Find the speed and acceleration of a particle whose motion is defined byx= 3 t and
y= 9 t− 3 t^2 whent=2.
Step 1: Differentiate
dx
dt
=3 and
dy
dt
= 9 − 6 t. Whent=2,
dx
dt
=3 and
dy
dt
=−3.
Step 2: Calculate the speed.
√
(3)^2 +(−3)^2 =
√
18 = 3
√
2
Step 3: Determine second derivatives.
d^2 x
dt^2
=0 and
d^2 y
dt^2
=−6. The acceleration vector is
〈
0,− 6
〉
.
Example 2
A particle moves along the curvey=
1
2
x^2 −
1
4
lnxso thatx=
1
2
t^2 andt>0. Find the speed
of the particle whent=1.
Step 1: Substitutex=
1
2
t^2 iny=
1
2
x^2 −
1
4
lnxto find
y(t)=
1
2
(
1
2
t^2
) 2
−
1
4
ln
(
1
2
t^2
)
=
1
8
t^4 −
1
4
ln
⎛
⎝^1
2
t^2
⎞
⎠
=
1
8
t^4 −
1
4
(−ln 2+2lnt)
=
t^4
8
+
ln 2
4
−
lnt
2
.
Step 2:
dx
dt
=tand
dy
dt
=
t^3
2
−
1
2 t
. Evaluated att=1,
dx
dt
=1 and
dy
dt
=0.
Step 3: The speed of the particle is
√
(1)^2 +(0)^2 =1.
Derivatives of Polar Equations
For polar representations, remember thatr = f(θ), so x=rcosθ = f(θ) cosθ and
y =rsinθ = f(θ) sinθ. Differentiating with respect toθ requires the product rule.
dx
dθ
=−rsinθ + cosθ
dr
dθ
and
dy
dθ
=rcosθ +sinθ
dr
dθ
. Dividing
dy
dθ
by
dx
dθ
gives
dy
dx
=
rcosθ+sinθdr/dθ
−rsinθ+cosθdr/dθ