Areas, Volumes, and Arc Lengths 291
Example 2
Find the length of the spiralr=eθfromθ=0toθ=π.
Step 1. Differentiate
dr
dθ
=eθ.
Step 2. Squarer^2 =e^2 θ.
Step 3. L=
∫π
0
√
e^2 θ+e^2 θdθ=
∫π
0
√
2 e^2 θdθ=
√
2
∫π
0
eθdθ=
√
2 eθ
∣
∣π 0
=
√
2 eπ−
√
2
Integration of a Vector-Valued Function
Integrating a Vector Function
For a vector-valued functionr(t)=
〈
x(t),y(t)
〉
,
∫
r(t)dt=
∫
x(t)dt·i +
∫
y(t)
dt·j.
Example 1
The acceleration vector of a particle at any timet≥0isa(t)=
〈
et,e^2 t
〉
. If at timet=0, its
velocity isi+jand its displacement is 0, find the functions for the position and velocity at
any timet.
Step 1. a(t)=
〈
et,e^2 t
〉
,sov(t)=
∫
a(t)dt=
∫
x(t)dt·i+
∫
y(t)dt·j
v(t)=
∫
etdt·i+
∫
e^2 tdt·j=et·i+
e^2 t
2
·j+C. Since velocity att = 0 is known
to bei + j,i+
1
2
·j+C=i+j, andC=
1
2
·j; therefore,v(t)=et·i+
e^2 t
2
·j+
1
2
·j=
〈
et,
e^2 t+ 1
2
〉
.
Step 2. The position functions(t)=
∫
v(t)dt=
∫
etdt·i+
∫
e^2 t+ 1
2
dt·j.
s(t)=
∫
v(t)dt=
∫
etdt·i+
∫
e^2 t+ 1
2
dt· j=et·i+
(
e^2 t+ 2 t
4
)
·j+C.
Displacement is 0 at t =0,soi +
1
4
·j +C =0 and C =−i −
1
4
·j.
The position functions(t)=et·i+
(
e^2 t+ 2 t
4
)
·j −i−
1
4
·j =(et− 1 )i+
(
e^2 t+ 2 t− 1
4
)
j=
〈
et−1,
e^2 t+ 2 t− 1
4
〉
.
Length of a Vector Curve
The length of a curve defined by the vector-valued functionr(t)=
〈
x(t),y(t)
〉
traced from
t=atot=biss=
∫b
a
∥∥
r′(t)
∥∥
dt.