Section 10.1 Curves Deﬁned by Parametric Equations

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Section 10.1 Curves Defined by Parametric Equations 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

10.1 Curves Defined by Parametric Equations

1. = 1 − ²,  = 2 − ², −1 ≤  ≤ 2

 −1 0 1 2

 0 1 0 −3

 −3 0 1 0

2. = ³+ ,  = ²+ 2, −2 ≤  ≤ 2

 −2 −1 0 1 2

 −10 −2 0 2 10

 6 3 2 3 6

3. =  + sin ,  = cos , − ≤  ≤ 

 − −2 0 2 

 − −2 + 1 0 2 + 1 

 −1 0 1 0 −1

4. = 2 cos ,  =  − cos , 0 ≤  ≤ 2

 0 2  32 2

 2 0 −2 0 2

 −1 2

157

 + 1 414

32 471

2 − 1 528

5. = 2 − 1,  = ¹2 + 1 (a)

 −4 −2 0 2 4

 −9 −5 −1 3 7

 −1 0 1 2 3

(b)  = 2 − 1 ⇒ 2 =  + 1 ⇒  = ¹2 +¹₂, so

 = ¹₂ + 1 = ¹₂1 2 +¹₂

+ 1 = ¹₄ +¹₄+ 1 ⇒  = ¹4 +⁵₄

10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

10.1 Curves Defined by Parametric Equations

1. = 1 − ²,  = 2 − ², −1 ≤  ≤ 2

 −1 0 1 2

 0 1 0 −3

 −3 0 1 0

2. = ³+ ,  = ²+ 2, −2 ≤  ≤ 2

 −2 −1 0 1 2

 −10 −2 0 2 10

 6 3 2 3 6

3. =  + sin ,  = cos , − ≤  ≤ 

 − −2 0 2 

 − −2 + 1 0 2 + 1 

 −1 0 1 0 −1

4. = 2 cos ,  =  − cos , 0 ≤  ≤ 2

 0 2  32 2

 2 0 −2 0 2

 −1 2

157

 + 1 414

32 471

2 − 1 528

5. = 2 − 1,  = ¹2 + 1 (a)

 −4 −2 0 2 4

 −9 −5 −1 3 7

 −1 0 1 2 3

(b)  = 2 − 1 ⇒ 2 =  + 1 ⇒  = ¹2 +¹₂, so

 = ¹₂ + 1 = ¹₂1 2 +¹₂

+ 1 = ¹₄ +¹₄+ 1 ⇒  = ¹4 +⁵₄

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c 863 864 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

6.  = 3 + 2,  = 2 + 3 (a)

 −4 −2 0 2 4

 −10 −4 2 8 14

 −5 −1 3 7 11

(b)  = 3 + 2 ⇒ 3 =  − 2 ⇒  =¹3 −²3, so

 = 2 + 3 = 21 3 −²3

+ 3 = ²₃ −⁴3+ 3 ⇒  = ²3 +⁵₃

7.  = ²− 3,  =  + 2, −3 ≤  ≤ 3 (a)

 −3 −1 1 3

 6 −2 −2 6

 −1 1 3 5

(b)  =  + 2 ⇒  =  − 2, so

 = ²− 3 = ( − 2)²− 3 = ²− 4 + 4 − 3 ⇒

 = ²− 4 + 1, −1 ≤  ≤ 5

8.  = sin ,  = 1 − cos , 0 ≤  ≤ 2

(a)

 0 2  32 2

 0 1 0 −1 0

 0 1 2 1 0

(b)  = sin ,  = 1 − cos  [or  − 1 = − cos ] ⇒

²+ ( − 1)² = (sin )²+ (− cos )² ⇒ ²+ ( − 1)²= 1.

As  varies from 0 to 2, the circle with center (0 1) and radius 1 is traced out.

9.  = 1 − ²,  =  − 2, −2 ≤  ≤ 2 (a)

 −2 −1 0 1 2

 −3 0 1 0 −3

 −4 −3 −2 −1 0

(b)  =  − 2 ⇒  =  + 2, so  = 1 − ²= 1 − ( + 2)² ⇒

 = −( + 2)²+ 1, or  = −²− 4 − 3, with −4 ≤  ≤ 0

SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS ¤ 865

10.  =  − 1,  = ³+ 1, −2 ≤  ≤ 2 (a)

 −2 −1 0 1 2

 −3 −2 −1 0 1

 −7 0 1 2 9

(b)  =  − 1 ⇒  =  + 1, so  = ³+ 1 ⇒  = ( + 1)³+ 1, or  = ³+ 3²+ 3 + 2, with −3 ≤  ≤ 1

11. (a)  = sin¹₂,  = cos¹₂, − ≤  ≤ .

²+ ²= sin^{2 1}₂ + cos^{2 1}₂ = 1. For − ≤  ≤ 0, we have

−1 ≤  ≤ 0 and 0 ≤  ≤ 1. For 0   ≤ , we have 0   ≤ 1 and 1   ≥ 0. The graph is a semicircle.

(b)

12. (a)  = ¹₂cos ,  = 2 sin , 0 ≤  ≤ .

(2)²+₁

22

= cos² + sin² = 1 ⇒ 4²+¹₄²= 1 ⇒

² (12)² +²

2² = 1, which is an equation of an ellipse with

-intercepts ±¹2 and -intercepts ±2. For 0 ≤  ≤ 2, we have

1

2 ≥  ≥ 0 and 0 ≤  ≤ 2. For 2   ≤ , we have 0   ≥ −¹2

and 2   ≥ 0. So the graph is the top half of the ellipse.

(b)

13. (a)  = sin   = csc , 0    ^₂.  = csc  = 1 sin  =1

. For 0   ^₂, we have 0    1 and   1. Thus, the curve is the portion of the hyperbola  = 1 with   1.

(b)

14. (a)  = ^−2= (^)⁻²= ⁻²= 1²for   0 since  = ^. (b)

SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS ¤ 865

10.  =  − 1,  = ³+ 1, −2 ≤  ≤ 2 (a)

 −2 −1 0 1 2

 −3 −2 −1 0 1

 −7 0 1 2 9

(b)  =  − 1 ⇒  =  + 1, so  = ³+ 1 ⇒  = ( + 1)³+ 1, or  = ³+ 3²+ 3 + 2, with −3 ≤  ≤ 1

11. (a)  = sin¹₂,  = cos¹₂, − ≤  ≤ .

²+ ²= sin^{2 1}₂ + cos^{2 1}₂ = 1. For − ≤  ≤ 0, we have

−1 ≤  ≤ 0 and 0 ≤  ≤ 1. For 0   ≤ , we have 0   ≤ 1 and 1   ≥ 0. The graph is a semicircle.

(b)

12. (a)  = ¹₂cos ,  = 2 sin , 0 ≤  ≤ .

(2)²+1 22

= cos² + sin² = 1 ⇒ 4²+¹₄²= 1 ⇒

² (12)² +²

2² = 1, which is an equation of an ellipse with

-intercepts ±¹2 and -intercepts ±2. For 0 ≤  ≤ 2, we have

1

2 ≥  ≥ 0 and 0 ≤  ≤ 2. For 2   ≤ , we have 0   ≥ −¹2

and 2   ≥ 0. So the graph is the top half of the ellipse.

(b)

13. (a)  = sin   = csc , 0    ^₂.  = csc  = 1 sin  =1

. For 0   ^₂, we have 0    1 and   1. Thus, the curve is the portion of the hyperbola  = 1 with   1.

(b)

14. (a)  = ^−2= (^)⁻²= ⁻²= 1²for   0 since  = ^. (b)

866 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

15. (a)  = ln  ⇒  = ^, so  = ²= (^)²= ^2. (b)

16. (a)  =√

 + 1 ⇒ ²=  + 1 ⇒  = ²− 1.

 =√

 − 1 =

(²− 1) − 1 =√

²− 2. The curve is the part of the hyperbola ²− ²= 2with  ≥√

2and  ≥ 0.

(b)

17. (a)  = ^2 ⇒ 2 = ln  ⇒  =¹2ln .

 =  + 1 = ¹₂ln  + 1.

(b)

18. (a)  = tan²,  = sec , −2    2.

1 + tan² = sec² ⇒ 1 +  = ² ⇒  = ²− 1. For

−2   ≤ 0, we have  ≥ 0 and  ≥ 1. For 0    2, we have 0  and 1  . Thus, the curve is the portion of the parabola  = ²− 1 in the ﬁrst quadrant. As  increases from −2 to 0, the point ( ) approaches (0 1) along the parabola. As  increases from 0 to 2, the point ( ) retreats from (0 1) along the parabola.

(b)

19.  = 5 + 2 cos ,  = 3 + 2 sin  ⇒ cos  =  − 5

2 , sin  =  − 3

2 . cos²() + sin²() = 1 ⇒

 − 5 2

2

+

 − 3 2

2

= 1. The motion of the particle takes place on a circle centered at (5 3) with a radius 2. As  goes

from 1 to 2, the particle starts at the point (3 3) and moves counterclockwise along the circle

 − 5 2

2

+

 − 3 2

2

= 1to (7 3)[one-half of a circle].

20.  = 2 + sin ,  = 1 + 3 cos  ⇒ sin  =  − 2, cos  = − 1

3 . sin² + cos² = 1 ⇒ ( − 2)²+

 − 1 3

2

= 1.

The motion of the particle takes place on an ellipse centered at (2 1). As  goes from 2 to 2, the particle starts at the point (3 1)and moves counterclockwise three-fourths of the way around the ellipse to (2 4).

1

(2)

15. (a)  = ln  ⇒  = ^, so  = ²= (^)²= ^2. (b)

16. (a)  =√

 + 1 ⇒ ²=  + 1 ⇒  = ²− 1.

 =√

 − 1 =

(²− 1) − 1 =√

²− 2. The curve is the part of the hyperbola ²− ²= 2with  ≥√

2and  ≥ 0.

(b)

17. (a)  = ^2 ⇒ 2 = ln  ⇒  =¹2ln .

 =  + 1 = ¹₂ln  + 1.

(b)

18. (a)  = tan²,  = sec , −2    2.

1 + tan² = sec² ⇒ 1 +  = ² ⇒  = ²− 1. For

−2   ≤ 0, we have  ≥ 0 and  ≥ 1. For 0    2, we have 0  and 1  . Thus, the curve is the portion of the parabola  = ²− 1 in the ﬁrst quadrant. As  increases from −2 to 0, the point ( ) approaches (0 1) along the parabola. As  increases from 0 to 2, the point ( ) retreats from (0 1) along the parabola.

(b)

19.  = 5 + 2 cos ,  = 3 + 2 sin  ⇒ cos  =  − 5

2 , sin  =  − 3

2 . cos²() + sin²() = 1 ⇒

 − 5 2

2

+

 − 3 2

2

= 1. The motion of the particle takes place on a circle centered at (5 3) with a radius 2. As  goes

from 1 to 2, the particle starts at the point (3 3) and moves counterclockwise along the circle

 − 5 2

2

+

 − 3 2

2

= 1to (7 3)[one-half of a circle].

20.  = 2 + sin ,  = 1 + 3 cos  ⇒ sin  =  − 2, cos  = − 1

3 . sin² + cos² = 1 ⇒ ( − 2)²+

 − 1 3

2

= 1.

The motion of the particle takes place on an ellipse centered at (2 1). As  goes from 2 to 2, the particle starts at the point (3 1)and moves counterclockwise three-fourths of the way around the ellipse to (2 4).

28. (a)  = ⁴−  + 1 = (⁴+ 1) −   0 [think of the graphs of  = ⁴+ 1and  = ] and  = ² ≥ 0, so these equations are matched with graph V.

(b)  =√

 ≥ 0.  = ²− 2 = ( − 2) is negative for 0    2, so these equations are matched with graph I.

(c)  = sin 2 has period 22 = . Note that

( + 2) = sin[ + 2 + sin 2( + 2)] = sin( + 2 + sin 2) = sin( + sin 2) = (), so  has period 2.

These equations match graph II since  cycles through the values −1 to 1 twice as  cycles through those values once.

(d)  = cos 5 has period 25 and  = sin 2 has period , so  will take on the values −1 to 1, and then 1 to −1, before  takes on the values −1 to 1. Note that when  = 0, ( ) = (1 0). These equations are matched with graph VI

(e)  =  + sin 4,  = ²+ cos 3. As  becomes large,  and ²become the dominant terms in the expressions for  and

, so the graph will look like the graph of  = ², but with oscillations. These equations are matched with graph IV.

(f )  = sin 2

4 + ²,  = cos 2

4 + ². As  → ∞,  and  both approach 0. These equations are matched with graph III.

29. Use  =  and  =  − 2 sin  with a -interval of [− ].

30. Use 1= , 1= ³− 4 and ²= ³− 4, ²= with a -interval of [−3 3]. There are 9 points of intersection; (0 0) is fairly obvious. The point in quadrant I is approximately (22 22), and by symmetry, the point in quadrant III is approximately (−22 −22). The other six points are approximately (∓19 ±05), (∓17 ±17), and (∓05 ±19).

31. (a)  = 1+ (2− ¹),  = 1+ (2− ¹), 0 ≤  ≤ 1. Clearly the curve passes through ¹(1 1)when  = 0 and through 2(2 2)when  = 1. For 0    1,  is strictly between 1and 2and  is strictly between 1and 2. For every value of ,  and  satisfy the relation  − ¹ = 2− ¹

2− ¹( − ¹), which is the equation of the line through

1(1 1)and 2(2 2).

Finally, any point ( ) on that line satisﬁes  − ¹

2− ¹ =  − ¹

2− ¹; if we call that common value , then the given parametric equations yield the point ( ); and any ( ) on the line between 1(1 1)and 2(2 2)yields a value of

in [0 1]. So the given parametric equations exactly specify the line segment from 1(1 1)to 2(2 2).

(b)  = −2 + [3 − (−2)] = −2 + 5 and  = 7 + (−1 − 7) = 7 − 8 for 0 ≤  ≤ 1.