1 1

Lesson １３

Application of Integrals (2)

１3A •  Arc length of a curve

Course II

2

Finding Arc Length by Integration (1)

Small segment

)(xfy =

Total length

→ { } dxxfdxdxdys

b

a

b

a ∫∫ ʹ+=⎟⎠

⎞⎜⎝

⎛+= 2

2

)(11

xxyyxs Δ⎟⎠

⎞⎜⎝

⎛Δ

Δ+≈Δ+Δ≈Δ

222 1

∑∑ Δ⎟⎠

⎞⎜⎝

⎛Δ

Δ+≈Δ≈ x

xyss2

1

Curve in a plane

3

Example Example 13-1 Find the length of circumference of a circle

Ans. Consider the length in quadrant I.

Graph

222 ryx =+

22 xry −=

Length

022 =+dxdyyx

dxyxdx

dxdys

rr

∫∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛−+=⎟

⎠

⎞⎜⎝

⎛+=

0

2

0

2

1 11

Polar coordinates θθ sin,cos ryrx ==

[ ] rrdrdrsr

2)sin(

sincos1 0

2

0

20

2

1π

θθθθθθ

ππ =−=−=−⎟⎠

⎞⎜⎝

⎛+= ∫∫

Length of circumference rss π24 1 ==

Finding Arc Length by Integration (2)

Small segment approximation

Total arc length

)(),( tgytfx ==

tty

txyxss Δ⎟

⎠

⎞⎜⎝

⎛Δ

Δ+⎟

⎠

⎞⎜⎝

⎛Δ

Δ=Δ+Δ=Δ≈ ∑∑ ∑

2222

{ } { } dttgtfdtdtdy

dtdxs

b

a

b

a ∫∫ ʹ+ʹ=⎟⎠

⎞⎜⎝

⎛+⎟

⎠

⎞⎜⎝

⎛= 22

22

)()(

Curve in a plane

Expression using a parameter

5

Example

Example 13-2 Find the length of the following cycloid line. [Note] A cycloid is the curve traced by a point on the rim of a circular wheel

as the wheel rolls along a straight line.

)20()cos1(2),sin(2 π≤≤−=−= ttyttx

x

y

6

Example

Example 13-2 Find the length of the following cycloid line. )20()cos1(2),sin(2 π≤≤−=−= ttyttx

Ans. t

dtdyt

dtdx sin2),cos1(2 =−=

Total arc length ( )

dtt

dttdttts

∫

∫∫=

−=+−=

π

ππ

2

0

2

2

0

2

0

22

2sin4

)cos1(22sin4cos14

02

sin ≥t π20 ≤≤ t

162

cos242

sin42

0

2

0=⎥⎦

⎤⎢⎣

⎡−== ∫

ππ tdtts

Since in

we have

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Exercise

Ans.

Ex.13-1 Answer the questions about the curve expressed by the polar coordinates. (1)  Represent this curve by the rectangular coordinate (x, y). (2)  Find the length of the curve.

θsin2=r ( )πθ ≤≤0

Pause the video and solve by yourself.

8

Answer to the Exercise

Ans.

Ex.13-1 Answer the questions about the curve expressed by the polar coordinates. (1)  Represent this curve by the rectangular coordinate (x, y)　. (2)  Find the length of the curve.

θsin2=r ( )πθ ≤≤0

θθθθ 2sincossin2cos === rxθθθ 2cos1sin2sin 2 −=== ry

12cos2sin)1( 2222 =+=−+ θθyxThis curve is a circle because

(1)

(2)

{ } { } ∫∫∫ ==+=⎟⎠

⎞⎜⎝

⎛+⎟

⎠

⎞⎜⎝

⎛=

ππππθθθ

θθ 00

22

0

22

222sin22cos2 ddtdtddy

ddxs

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Lesson １３

Application of Integrals

１３B •  Application to Physics

Course II

10

Point P moving on the x-axis

Position, Velocity and Acceleration

O x(t) x

Position

)(txx = )(tvv = )(taa =Velocity Acceleration dt

dxv =dtdva =

11

)( xdtvtxt

t+= ∫ 1

1

)( vdtatvt

t+= ∫

)(tv

Movement of an object P

Distance Traveled in a Straight Line

O x(t) x

A

B

C 0>v 0>v

0<v

Displacement from t=t1 to t=t2

Distance traveled in a straight line t=a to t=b

)()()( 122

1

txtxdttvst

t−== ∫

s

{ } dttvdttvdttvlt

t

t

t

t

t ∫∫∫ =−+=2

1

2

3

3

1

)()()(

2t

1t 3t

12

Distance Traveled in a Plane

Total arc length

)(),( tgytfx ==

Point P moves in curve C

⎟⎠

⎞⎜⎝

⎛=

dtdy

dtdxv ,

!Velocity

Distance traveled

dtdtdy

dtdxdtvl

t

t

t

t ∫∫ ⎟⎠

⎞⎜⎝

⎛+⎟

⎠

⎞⎜⎝

⎛==

2

1

2

1

22!

dtdtdy

dtdxl

t

t∫ ⎟⎠

⎞⎜⎝

⎛+⎟

⎠

⎞⎜⎝

⎛=

2

1

22

13

Example [Examples 13-3] Consider a ball being tossed upward from the ground with an initial velocity of 29.4m/s. The x-axis is taken vertically and its origin is located at the ground. The acceleration is -9.8m/s2. Answer the following questions. (1) Express the velocity as a function of time. (2) When the ball reached the highest point ? (3) When did the ball fall on the ground ? (4) Find the total distance traveled.

Ans. (1) Velocity ttv 8.94.29)( −=

(2) From sec 08.94.29 =−= tv 3=∴ t(3) The position is given by . Therefore, from 29.44.29 ttx −=

0)6(9.4 =−−= ttx 6=∴ t(4) The total distance traveled

sec

2.88)8.94.29()8.94.29(6

3

3

0

2

1

=+−+−== ∫∫∫ dttdttdtvlt

t

! m

x

O

)(tv

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Exercise

Ex.12-2 　The velocity of a point P moving on the x-axis is given

by . Find the distance traveled between t=0 and

t=6. Ans.

Pause the video and solve by yourself.

34)( 2 +−= tttv

Exercise

Ex.12-2 　The velocity of a point P moving on the x-axis is given

by . Find the distance traveled between t=0 and t=6.

Ans.

34)( 2 +−= tttv

0)3)(1()( =−−= tttv0)( ≥tv 1≤t 3≥t0)( ≤tv 31 ≤≤ t

We find when and when

From

Therefore, 34)( 2 +−= tttv

.

1≤t 3≥t31 ≤≤ t

when and when

The distance traveled

dtttdtttdtttdtvl ∫∫∫∫ +−++−−+−==6

3

23

1

21

0

26

0)34()34()34(

!

362

=∴ l