§ 9.1 (10.1) Parametric Curves ( 參數曲線)

§ 9.1 (10.1) Parametric Curves ( ^參數曲線 )

[Ex]Sketch and identify the curve defined by parametric equations [Ex]Sketch and identify the curve defined by parametric equations [Sol]:

6

, 2, 2 4 x = − t y = t − ≤ ≤ t

[ ]

(a) Sketch the curve by using the parametric equations to plot points:

(b) Eliminate the parameter to find a Cartesian equation ^{(笛卡兒方程式)} of the curve:

Substituting

t 2

^{= 6 4}

So the curve represented by the given parametric equations is .

2 t = y

2 2 2

6 6 (2 ) 6 4 x = − = − t y = − y

6 4

= −

The point (2,-1) is called the initial point and (-10,2) is called the terminal point.

起點 終點

2cos , 2sin ,

x

=

=

[Ex] What curve is represented by the parametric equations

0 ≤ ≤ 2 ?

[Sol]:

0 ≤ ≤

2 ? π

(a) Sketch the curve by using the parametric equations to plot points:

t x y

3

y

0 2 0

π/3 1

π/2 0 2

π -2 0

(b) Eliminate the parameter to find a Cartesian equation of the curve:

Observe that

^x

^{+ y}

⁼ ( ^{2cos t} ) (

^{+ 2sin t} )

^{= 4}

3π/2 0 -2

Observe that

So the curve represented by the given parametric equations is the unit

circle ^單位圓 .

( ^2cos ) ( ^2sin ) ⁴

x + y = t + t =

2 2

4

x + y = 4

x + y =

[Ex]What curve is represented by the parametric equations

1

2

1

2

(a) and (b) .

[Sol]:

1,

2

x

= −

=

−

= −

1,

=

− 2

[Sol]:

(a) (b)

( x + 1)

= + y 2 ( x + 1)

= + y 2, x ≥ − 1

[Ex] Find parametric equations for the portion of the parabola y = x² from (-1, 1) to (3, 9).

[Sol]:

Any equation of the form y = f (x) can be converted to parametric form by simply letting x equal t. Here, this gives us y = x² = t ² .

[Sol]:

Then is a parametric representation of the curve.

,

, for 1 3.

x = t y = t − ≤ ≤ t

[Ex5] [Ex5] Sketch the curve with parametric equations x

= sin ,

, y = sin

t .

[Sol]:

( ^sin )

y = t = x

∵

So the curve represented by the given parametric

equations is the parabola .

( ^sin )

y t x

∵

2

, 1 1 y = x − ≤ ≤ x

equations is the parabola .

Exercise

Sketch the curve with parametric Equations (a) x = 2cos t, y = 3sin t,

(b) x = 2+4cos t, y = 3 + 4sin t and (c) x = 3cos 2t, y = 3sin 2t all for

0 ≤ ≤

2 . π

(a) x = 2cos t y = 3sin t (b) x = 2+4cos t y = 3 + 4sin t (c) x = 3cos 2t y = 3sin 2t [Sol]:

(a) x = 2cos t, y = 3sin t (b) x = 2+4cos t, y = 3 + 4sin t (c) x = 3cos 2t, y = 3sin 2t

2 2

x y

⎛ ⎞ ⎛ ⎞ ₁ ( 2)

( 3)

16 ( ) x

+ ( ) y

9

2 3

x y

⎛ ⎞ + ⎛ ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2 2

( x − 2) + ( y − 3)

16 ( ) x + ( ) y

9

The Cycloid ( ^擺線 )

[Ex7][Ex7]

[Ex7][Ex7]

The curve traces out ^(描繪) by a point on the circumference ^(圓周) of a circle as the

circle rolls ^(滾動) along a straight line is called a cycloid, as shown below. If the circle circle rolls ^(滾動) along a straight line is called a cycloid, as shown below. If the circle has radius r and rolls along the x-axis and if one position of P is the origin, find

parametric equations for the cycloid.

( )

sin sin

x

=

θ −

θ =

θ − θ

( )

y = − r r cos θ = r 1 cos − θ , θ ∈R

§ 9.2 (10.2) Calculus with Parametric Curves ( ^{參數曲線的微積分} )

dy d ⎛ dy ⎞

Tangents ( ^切線 )

, if 0 dy

dy dt dx

dx dx dt

dt

= ≠ ²

2 , if 0

d dy

d y d dy dt dx dx

dx dx dx dx dt

⎛ ⎞

⎜ ⎟

⎛ ⎞ ⎝ ⎠

= ⎜⎝ ⎟⎠ = ≠

dt dt

[E 1][E 1]

[Ex1][Ex1]

A curve C is defined by the parametric equations . x

=

,

= −

3

(b) Find the points on C where the tangent is horizontal or vertical.

(c) Determine where the curve is concave upward or downward.

(d) Sketch the curve.

[Sol]:

(a)

{

^{=3 ⇒ = ±}

³

⁼

^{= 3}

^{− 3}

.

{

^{= −}

³

^{= 0 ⇒ = ±}

³

⁼

²

3 3

3

6 2

t

dy dx ^±

⇒

= ±

. (b) Horizontal tangent:

3

2

dx t^=±

( )

3

3

y = x − ^{y = −}

( ^{x − 3} )

0 0 1

dy dx

= ⇔

= ⇔ = ±

(b) Horizontal tangent:

So the corresponding points ^{(相對應的點)} on C are (1,-2) and (1,2).

Vertical tangent:

0 0 1

dy dx

⇔

⇔

±

0 0

2

dx dt

= = t ⇔ =

(c) (d)_{3 1} d ⎛⎜dy ⎞⎟ ⎛⎜1+ ⎞⎟

2 2

2

2 1

= 2

d y dt dx t

dx dx t

dt

⎜ ⎟ ⎜ + ⎟

⎝ ⎠ ⎝ ⎠

=

dt

2 2

d y dx

+

t < 0 t > 0

d y dx − ++

the curve C C.D C.U

[Ex2][Ex2]

(a) Find the tangent to the cycloid and at ^x

⁼

( ^{θ − sin θ} ) ^{y = r} ( ^{1 cos − θ} )

the point where .

(b) At what point is the tangent horizontal? When is it vertical?

[Sol]:

( ) ^y ( )

θ π= 3

[Sol]:

(a)

3

sin sin

= (1 cos ) (1 cos )

dy dy d r dy

dx dx d r dx ^{θ π}

θ θ θ

θ θ

θ

= =

− −

When , we have . So the equation of the tangent is

( 3 3 2 ) and 2

x = r π − y = r

θ π= 3

2 3( ( 3 3 2))

y −r = x−r π −

(b) horizontal tangent:

So the corresponding points are

0 and 1 cos 0 (2 1)

0 sin

dy dx

= ⇔

((2n+1)πr 2 )r So the corresponding points are . vertical tangent:

((2n+1) , 2 )πr r

lim dy ₌ lim

sin θ

cos θ

0

dx d

θ = ⇒

So the corresponding points are .

2 2 2 2

lim lim lim and lim

(1

cos

sin

n dx n n n dx

θ^→ π⁺ θ^→ π⁺

− θ

θ

(2n rπ ,0)

Exercise

Find the slope of the tangent line to the path of the Scrambler x = 2cos t + sin 2t Find the slope of the tangent line to the path of the Scrambler x 2cos t + sin 2t, y = 2sin t + cos 2t at (a) t = 0 ; (b) and (c) the point (0, -3) .t = π 4

Ans:

(a)

dy

(b)

dy

0 1

t

dy

dx

=

(b)

4

2 2 2

t

dy

dx

=

(c) d

(^{3 2})

lim dy

θ→ π dx = ∞

This says that the slope of the tangent line at (0,-3) is undefined.

The tangent line at the point (0, -3) is vertical.

Areas

The area under the curve ^y = ^F

( ) x

A = ∫

x dx )

The area under the curve from a to b is .

If , and the curve is traversed ^(橫越) once as t increases from α to β, then

( )

y = F

x

)

a F

A = ∫ x dx ( ),

x = f

t

if is the leftmost endpoint.( ( ), ( ))f α g α

b α

∫ ∫

( ) ( )

b

A a ydx ^β g t f t dt

α ′

=

∫

∫

if is the leftmost endpoint.

( ) ( )

b

A a ydx ^α g t f t dt

β ′

=

∫

∫

[Ex3][Ex3]

Find the area under one arch ^(拱形) of the cycloid ^x

⁼

( ^{θ − sin θ} ) ^,

⁼

( ^{1 c − os θ} ) ^.

[Sol]:

∫

∫

[Sol]:

0 y 0 (1 cos )r(1 cos ) A=

∫

∫

( )

2 2

2 2 2 2

0 (1 cos ) 0 1 2cos cos

r ^π θ dθ r ^π θ θ θd

=

∫

∫

3 1 2^π 3

⎡ ⎤ ⎛ ⎞

( )

0 0

∫ ∫

2 2 0

1

1 2 c os 2 (1 c os 2 )

r ^{π ⎛} θ θ ^⎞ dθ

=

∫

2 2 2

0

3 1 3

2 2sin 4sin 2 2 2 3

r ^⎡ θ θ θ^⎤ r ^⎛ π ^⎞ πr

= ⎢⎣ − + ⎥⎦ = ⎜⎝ ⋅ ⎟⎠ =

Arc length

Suppose that C is given by the parametric equations , and , then the arc length of C

( ), ( )

x = f t y = g t ( ) 0

f t

dx dt

α ≤ ≤t β

, where

2 2

dx dy

ds dt

L d

t dt

β α

⎛ ⎞ +⎛ ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

=

∫

∫

[Ex5][Ex5]

Find the length of arch of the cycloid ^{g y x}

⁼

( ( ^{θ − sin θ} ) ) ^{, ,}

⁼

( ( ^{1 cos − θ} ) ) ^.

[Sol]: _{2 2}

2 2 2 2

cos sin

dx y r

d r

d

L = ^{π ⎛}⎜⎝ ^⎞⎟⎠ +^⎛⎜⎝ d ^⎞⎟ θ = ^π − θ + θ θ

∫

∫

d d θ θ θ

θ

θ θ

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

∫ ∫

2 2 2 2 2

0^π r 1 2cosθ cos θ sin θ θd r 0^π 2 1 cosθ θd

=

∫

∫

2

2 2 2

0 4sin ( 2) 2 0 sin( 2)

r ^π θ dθ r ^π θ dθ

=

∫

∫

[

]

2r 2 cos(θ ) ^π 2 (2 2) 8r r

= − = + =

[Ex]Find the arc length of the plane curve x = cos 5t, y = sin 7t, for 0

≤ ≤

π

[Sol]:

Lissajous curve j

§ 9.3 (10.3) Polar Coordinates

2 2 2

r = x + y cos x = r θ

( )

( , ) ,

P r θ = P x y

i

sin y = r θ tanθ = y

θ

r

tanθ x

[Ex4][Ex4] What curve is represented by the polar equation ^極方程

r = 2

O x

[Ex4][Ex4] What curve is represented by the polar equation ^極方程 ? [Ex] Sketch the polar curve .

2 r =

[Sol]:

θ π = 3

[Sol]:

[Ex6][Ex6]

(a) Sketch the curve with polar equation r = 2cosθ (a) Sketch the curve with polar equation . (b) Find a Cartesian equation for this curve.

2cos

r = θ

[Sol]:

[Sol]:

(a) Sketch the curve by using the parametric equations to plot points:

(b)

Completing the square ^(配方), we have

2 2 2

2cos 2 cos 2

r = θ ⇒ r = r θ ⇒ x + y = x

2 2

(

− 1) +

= 1

which is an equation of a circle with center (1,0) and radius 1.

( )

[Ex7][Ex7]Sketch the curve .r = +1 sinθ

[Ex8][Ex8]Sketch the curve .r = cos 2θ

[ ][ ]

[Ex]

Sketch the curve r =θ Sketch the curve .r =θ

[Ex]

Sketch the curve r = +3 2cosθ.

[Ex]

Sketch the curve r = −1 2sin .θ

1 sin r = + c θ

The graph of

Tangents to Polar Curves ( ^{極座標曲線的切線} )

sin cos

= , if 0

i

dy dr

dy d d r dx

dx dr

dx d

θ θ

θ θ

θ θ θ

= + ≠

cos sin

dx dr

dx d

d d r

θ θ θ

θ θ ⁻

[Ex9][Ex9]

(a) For the cardioid ^(心臟線) of Example 7, find the slope of the tangent line hen

1 sin

r

= + θ

θ 3

tangent line when .

(b) Find the points on the cardioid where the tangent is horizontal or vertical.

θ π= 3 [Sol]:

[Sol]:

Since , we have and , so

1 sin

= + θ

sin (1 sin )sin y = r θ = + θ θ

cos (1 sin ) cos

x = r θ = + θ θ

sin (1 sin )sin , so

y r θ + θ θ

((1 sin )sin ) cos sin (1 sin ) cos (1 2sin ) cos

=

dy d

dy d d

d d

θ θ θ θ θ θ θ θ

θ θ ^{+ + + +}

= = =

cos cos (1 sin )sin (1 sin )(1 2sin ) ((1 sin ) cos )

dx d

dx

d d

θ θ θ θ θ θ

θ θ

θ θ ^{+ − + + −}

(a) The slope of the tangent at the point where isθ π= ₃

3 3

cos sin (1 sin ) cos cos cos (1 sin )sin 1 dy

dx ^{θ π θ π}

θ θ θ θ

θ θ θ θ

= =

= + + = −

− +

(b) Observe that

3 7 11

(1 2 i ) 0

dy θ θ θ π π,³ ,⁷π ,¹¹π 2 2 6 6

(1 2sin )cos 0 dy

d

π π π π

θ θ θ

θ ^{= + = ⇒ =}

3 5

2 6 6, ,

(1 sin )(1 2sin ) 0 dx

d

π π π

θ θ θ

θ ^{= + − = ⇒ =}

So, there are horizontal tangents at and

vertical tangents at and . Besides,

2 6 6

dθ

(

) (

) (

)

(3 2 π 6)

(

)

vertical tangents at and . Besides,(3 2,π 6)

(

)

(3 2) (3 2) (3 2) (3 2)

(1 2sin ) cos 1 sin

lim lim lim lim

(1 2sin ) (1 sin ) 3 cos

dy

θ π dx θ π θ π θ π

θ θ θ

θ θ θ

− − − −

→ → → →

+ −

= = − = ∞

− +

Similarly,

( ) ( )

(3 2) (3 2)

1 sin

lim lim

3 cos

dy

θ π dx θ π

θ θ

+ +

→ →

= − − = −∞

Thus, there is a vertical tangent at which is the pole.

( ⁰

)

Exercise

Find the slope of the tangent line to the three-leaf rose ^{(三瓣玫瑰線)} at and .

sin 3

r

= θ

θ π = 4 θ = 0

Ans:

(1)

1 d

0 0

dy

dx ^θ= =

(2)

4

1 2 dy

dx ^θ=π =

§ 9.4 (10.4) Areas and Lengths in Polar Coordinates ( 極座標曲線的面積與弧長 )

( 極座標曲線的面積與弧長 )

* 2 1

1 2

lim ( ( )) ( ( ))

2

1 2

n n i

i

b

f a f

A θ θ θ dθ

→∞ =

=

∑

∫

The area of the region bounded by the polar curve and the raysr = f

( ) θ

i=1

( )

f

( )

and

,

0,

0 2 ,

a b b a

θ = θ = θ ≥ < − ≤ π

1

2

A

= ∫

θ

( ( ))2

1 o

2 r

b f

A

= ∫ ∫

₂

θ ∫

₂

cos 2

r

=

[Ex1][Ex1]

Find the area enclosed by one loop of the four leaved rose r

= cos 2

Find the area enclosed by one loop of the four-leaved rose .

Let

π π

and .

0 i.e. cos 2 0 2

,

r

π π 2 2

θ θ

= = ⇒ = −

θ

π 4

⇒ =

−

π 4

4 2 4 2

4 4

1 ( )

2

1 cos 2

A ^π

2

π

θ

θ θ

− = −

= ∫ ∫

4 1 4

π π

∫ ∫

1 1 4

sin 4 ^π π

θ θ

⎡ + ⎤

⎢ ⎥

4 2 4

0 0

( ) 1 ( )

cos 2

1 cos 4

π π

θ θ θ θ

=

∫

∫ +

sin 4 0

2 θ 4 θ 8

= ⎢⎣ + ⎥⎦ = Exercise

A

Find the area enclosed by one loop of the three-leaved rose .

π

sin 3

r

=

Ans:

12

π

[Ex2][Ex2]

Find the area of the region that lies inside the circle and outside

r = 3sin θ

[Sol]:

g the cardioid .

3sin 1 5

r ⁼ θ π π

⎧ ⎨

1 sin

r

= +

or 5

6 6

sin 1

1 sin 2

r

π π

θ θ

θ ^⇒ θ ^{= ⇒ = =}

= +

⎧ ⎨

⎩

5 6

1

1

3sin 1 sin

A

= ∫

3sin θ

1 sin + θ

θ

2 2

A d

π

θ + θ θ

∫

5 6 2

6

1 ( )

2 ^π

8sin 1 2sin d

π

θ θ θ

= ∫ − −

1

[ ]

1 3θ 2sin 2θ 2cosθ ^π π

= + =

5 6 6

1 ( )

2 ^π

3 4cos 2 2sin d

π

θ θ θ

= ∫ − −

[ ]

3 2sin 2 2cos 6

2 θ θ θ ^π π

= − + =

Exercise

Find the area of the region that lies inside the circle and outside the cardioid . r

= + 3 2cos

2

Ans:

. 11 7

2 3

3 −

[Ex3][Ex3]

Find all points of intersection of the curves and .r

= cos 2

1

= 2

d a po s o e sec o o e cu ves r

cos 2

r

2

(1) ^{cos 2} , ^{5 7} ,¹¹

6 6 6 6

cos 2 1 ,

1 2 2

r

r

θ π π π π

θ θ

= ⇒ = ⇒ =

=

⎧ ⎨

⎩

So there are four points of intersection:

⎩

1 1 5 1 7 1 11

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

(2)

1 1 5 1 7 1 11

, , , , , ,

2 6^π 2 6^π

,

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 4 5

cos 2 1

r = θ

(2)

⎧

S th th f i t f i t ti

2 4 5

, ,

3 3 3 3

cos 2

1 ,

cos 2

1 2 2

r

r

π π π π

θ θ θ

= ⇒ = − ⇒ =

= −

⎧ ⎨

⎩

So there are another four points of intersection:

1 1 2 1 4 1 5

,

^π

^π ,

^π

^π

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝

⎠ ⎝

⎠ ⎝ ,

⎠ ⎝

⎠

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Arc length

The length of a polar curve , isr = f ( ), θ a ≤ ≤θ b

2 2 2

b ⎛ dx ⎞ ⎛ dy ⎞ b ⎛ dr ⎞

∫

∫

L d r d

d d θ d θ

θ θ θ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

=

∫

∫

f ′

[Ex4][Ex4]

Find the length of the cardioid .r

= + 1 sin θ

[Sol]: ₂ ²

2

0 ( ) dr

d d

r

L ^π

θ θ

⎛ ⎞

=

∫

2^π 2 2sinθ θd 8

=

∫

2 2 2

0^π (1 si+ n

θ

θ

θ

=

∫

0 2 2sinθ θd 8

=

∫

Exercise

Set up the integral for the arc length of the curve .r

= sin 3 θ