5 Steps to a 5 AP Calculus BC 2019

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.