More Applications of Derivatives 191Position, 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√
2Step 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 speedof the particle whent=1.
Step 1: Substitutex=
1
2
t^2 iny=1
2
x^2 −1
4
lnxto findy(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