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Solucionario Cálculo Multivariable - Dennis G. Zill. 4ta Edición.pdf, Ejercicios de Cálculo diferencial y integral

Universidad del CaucaCálculo diferencial y integral

Solucionario del libro Cálculo Multivariable - Dennis G. Zill. 4ta Edición.pdf

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Chapter 10
Conics and Polar Coordinates
10.1 Conic Sections
y
x
1. vertex: (0,0)
focus: (1,0)
directrix: x=−1
axis: y= 0
y
x
2. vertex: (0,0)
focus: (7/8,0)
directrix: x=−7
8
axis: y= 0
y
x
3. vertex: (0,0)
focus: (0,−4)
directrix: y= 4
axis: x= 0
668
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Chapter 10

Conics and Polar Coordinates

10.1 Conic Sections

y

x

  1. vertex: (0, 0) focus: (1, 0) directrix: x = − 1 axis: y = 0

y

x

  1. vertex: (0, 0) focus: (7/ 8 , 0) directrix: x = − (^78) axis: y = 0

y

x

  1. vertex: (0, 0) focus: (0, −4) directrix: y = 4 axis: x = 0

668

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10.1. CONIC SECTIONS 669

y

x

  1. vertex: (0, 0) focus:

directrix: y = − 401 axis: x = 0

y

x

  1. vertex: (0, 1) focus: (4, 1) directrix: x = − 4 axis: y = 1

y

x

  1. vertex: (− 2 , −3) focus: (− 4 , −3) directrix: x = 0 axis: y = − 3

y

x

  1. vertex: (− 5 , −1) focus: (− 5 , 2) directrix: y = 0 axis: x = − 5

10.1. CONIC SECTIONS 671

y

x

  1. (x − 1)^2 = 4(y − 4) vertex: (1, 4) focus: (1, 5) directrix: y = 3 axis: x = 1

y

x

  1. (y − 4)^2 = −2(x + 3) vertex: (3, 4) focus: (5/ 2 , 4) directrix: x = 7/ 2 axis: y = 4

y

x

  1. (y − 2)^2 = 4

x + (^14)

vertex: (− 1 / 4 , 2) focus: (3/ 4 , 2) directrix: x = − 5 / 4 axis: y = 2

  1. x^2 = 28
  2. y^2 = − 16 x
  3. y^2 = 10x
  4. x^2 = − 40 y
  5. The parabola is of the form (y − k)^2 = 4p(x − h) with (h, k) = (− 2 , −7) and p = 3. Thus the equation is (y + 7)^2 = 12(x + 2).
  6. The parabola is of the form (x−h)^2 = 4p(y −5) with (h, k) = (2, 0) and p = 3, so the equation of the parabola is (x − 2)^2 = 12y.

672 CHAPTER 10. CONICS AND POLAR COORDINATES

  1. The parabola is of the form x^2 = 4py with (−2)^2 = 4p(8). Thus p = 18 and the equation is x^2 = 12 y.
  2. The parabola is of the form y^2 = 4px with

4

= 4p(1) so p = 641. Thus the equation is y^2 = 161 x.

  1. To find the x-intercept set y = 0. Solving 4^2 = 4(x + 1) gives x = 3. The x-intercept is (3, 0). To find the y-intercept set x = 0. Solving (y + 4)^2 = 4 gives y = − 4 ± 2. The y-intercepts are (0, −2) and (0, −6).
  2. To find the x-intercept set y = 0. Solving (x − 1)^2 = 2 gives x = 1 ±
  1. The x-intercepts are (1 +

2 , 0) and (1 −

2 , 0). To find the y-intercept set x = 0. Solving 1 = −2(y − 1) gives y = 12. The y-intercept is (0, 1 /2). y

x

  1. center: (0, 0) foci: (0, ±

vertices: (0, ±4) endpoints of the minor axis: (± 1 , 0)

eccentricity:

y

x

  1. center: (0, 0) foci: (± 4 , 0) vertices: (± 5 , 0) endpoints of the minor axis: (0, ±3) eccentricity:

y

x

x^2 16

y^2 9 = 1 center: (0, 0) foci: (±

vertices: (± 4 , 0) endpoints of the minor axis: (0, ±3)

eccentricity:

www.elsolucionario.org

674 CHAPTER 10. CONICS AND POLAR COORDINATES

y

x

  1. center: (− 5 , −2) foci: (− 5 , − 2 ±

vertices: (− 5 , −6), (− 5 , 2) endpoints of the minor axis: (− 6 , −2), (− 4 , −2)

eccentricity:

y

x

  1. center: (3, −4) foci: (3, − 4 ±

vertices: (3, −13), (3, 5) endpoints of the minor axis: (− 5 , −4), (11, −4)

eccentricity:

y

x

  1. x^2 +

y + (^12)

center: (0, − 1 /2) foci: (0, − 1 / 2 ±

vertices: (0, − 5 /2), (0, 3 /2) endpoints of the minor axis: (− 1 , − 1 /2), (1, − 1 /2)

eccentricity:

www.elsolucionario.org

10.1. CONIC SECTIONS 675

y

x

(x + 2)^2 2

(y − 4)^2 72

center: (− 2 , 4) foci: (− 2 , 4 ±

vertices: (− 2 , 4 ± 6

endpoints of the minor axis: (− 2 ±

eccentricity:

y

x

(x − 7)^2 9

(y + 1)^2 25

center: (2, −1) foci: (2, −5), (2, 3) vertices: (2, −6), (2, 4) endpoints of the minor axis: (− 1 , −1), (5, −1) eccentricity:

y

x

(x + 1)^2 5

(y − 1)^2 9

center: (− 1 , 1) foci: (− 1 , −1), (− 1 , 3) vertices: (− 1 , −2), (− 1 , 4) endpoints of the minor axis: (− 1 ±

eccentricity:

10.1. CONIC SECTIONS 677

  1. The center is (0, 0) with the y-axis as the major axis. c =

5 and a = 8, so b =

  1. Thus the equation is x^2 59

y^2 64

  1. The center is (0, 0) with the y-axis as the major axis. c = 3 thus 9 = a^2 − b^2 and a =

9 + b^2.

Thus the equation is of the form

x^2 b^2

y^2 9 + b^2 = 1. The ellipse passes through the point

(− 1 , 2

2), thus

(−1)^2

b^2

2)^2

9 + b^2

= 1. Solving this for b, we obtain b =

  1. Thus a^2 = 12 and

the equation is x^2 3

y^2 12

  1. The center is (0, 0) with the x-axis as the major axis and a = 5. The equation is of the form x^2 25

y^2 b^2

= 1. The ellipse passes through the point (

5 , 4) so

b^2

= 1. Solving for b^2 , we

obtain b^2 = 20. Thus the equation of the ellipse is

x^2 25

y^2 20

  1. The y-axis as the major axis with c = 3 and a = 4. Thus b =

7 and the equation of the ellipse is

(x − 1)^2 7

(y − 3)^2 16

  1. The center is (15/ 2 , 4) with the x-axis as the major axis. a = 1/2 and c = 7/ 2 , thus b =

Thus the equation is (x − 15 /2)^2 (11/2)^2

(y − 4)^2 18

y

x

  1. center: (0, 0) foci: (±

vertices: (± 4 , 0) asymptotes: y = ± 54 x

eccentricity:

y

x

  1. center: (0, 0) foci: (±

vertices: (± 2 , 0) asymptotes: y = ±x eccentricity:

678 CHAPTER 10. CONICS AND POLAR COORDINATES

y

x

y^2 20

x^2 4

= 1 center: (0, 0) foci: (0, ± 2

vertices: (0, ± 2

asymptotes: y = ±

5 x eccentricity:

y

x

y^2 9

x^2 16

= 1 center: (0, 0) foci: (0, ±5) vertices: (0, ±3) asymptotes: y = ± 34 x eccentricity:

y

x

  1. center: (5, −1) foci: (5 ±

vertices: (3, −1), (7, −1) asymptotes: y = − 1 ± 72 (x − 5)

eccentricity:

www.elsolucionario.org

680 CHAPTER 10. CONICS AND POLAR COORDINATES

y

x

(x − 3)^2 5

(y − 1)^2 25

= 1 center: (3, 1) foci: (3 ±

vertices: (3 ±

asymptotes: y = 1 ±

5(x − 3) eccentricity:

y

x

(x + 1)^2 10

(y − 1 /2)^2 50

= 1 center: (− 1 , 1 /2) foci: (− 1 ±

vertices: (− 1 ±

asymptotes: y = 1/ 2 ±

5(x + 1) eccentricity:

y

x

(x − 2)^2 6

(y − 1)^2 5

= 1 center: (2, 1) foci: (2 ±

vertices: (2 ±

asymptotes: y = 1 ±

(x − 2)

eccentricity:

www.elsolucionario.org

10.1. CONIC SECTIONS 681

y

x

(x − 8)^2 25

(y − 3)^2 16

= 1 center: (8, 3) foci: (8 ±

vertices: (13, 3), (3, 3) asymptotes: y = 3 ± 4 /5(x − 8)

eccentricity:

y

x

  1. (y − 3)^2 −

(x − 1)^2 1 / 4

= 1 center: (1, 3)

foci: (1, 3 ±

vertices: (1, 2), (1, 4) asymptotes: y = 3 ± 2(x − 1)

eccentricity:

y

x

(y + 5)^2 18

(x + 1)^2 4

= 1 center: (− 1 , −5) foci: (− 1 , − 5 ±

vertices: (− 1 , − 5 ±

asymptotes: y = − 5 ±

√ 18 2 (x^ + 1) eccentricity:

  1. The center is (0, 0) with the y-axis as the transverse axis. c = 4 and a = 2, thus b =

The equation is

y^2 4

x^2 12

  1. The center is (0, 0) with the y-axis as the transverse axis. c = 3 and a = 3/2, thus b =

10.1. CONIC SECTIONS 683

is found by solving (175)^2 = 400y. Solving this equation yields y = 76.5625. Therefore the towers are 76.5625 ft above the road.

  1. We place the coordinate axes so that the origin is at the vertex of the parabola. The parabola is of the form x^2 = 4py and contains the point (125, 75). Thus the equation of the parabola is x^2 = 62512. We need to find the y-value of the point on the parabola when we are 50 ft from the tower or when x = 75f t. Hence this y-value is found by solving the equation 75^2 = 6253 y which yield the solution y = 27 ft. The height of the cable above the roadway at a point 50 ft from one of the towers is 27 ft.
  2. We place the coordinate axes so that the origin is at end of the pipe with the parabola in Quadrants 3 and 4. The equation is of the form x^2 = 4py and the point (4, −2) lies on the parabola. Therefore the equation is x^2 = − 8 y. The water hits the ground at y = − 20. The point on the parabola with y-value -20 is found by solving x^2 = −8(−20). This point is x = 12. 65. Thus the water hits the ground 12.65 m from the point on the ground directly beneath the end of the pipe.
  3. We place the coordinate axes with the x-axis along the ground and the y-axis to be through the dart thrower. Thus the dart was released at the point (0, 5) and hits the ground at the point (

10 , 0). The parabola is of the form x^2 = 4p(y−5) and contains the point (

Therefore the equation of the parabola is x^2 = − 200 y. To find the height of the dart 10 ft from the thrower, we need to find the y-value of the point on the parabola corresponding to the x-value of 10. Hence, we need to solve the equation 10^2 = − 200 y which yields y = − 1 / 2 f t. So 10 feet from the thrower the dart will be 0.5 ft below the thrower or it will be 4.5 ft from the ground.

y

x

3.6x10 7 a

b

c a-c

3.52x10 7

  1. Taking the center of the ellipse to be at the origin, we have a = 3. 6 × 107 and b = 3. 52 × 107. Since c^2 = a^2 − b^2 , c^2 = 12. 96 × 1014 − 12. 3904 × 1014 = 0. 5696 × 1014 and c ≈ 0. 75 × 107. The perihelion or least distance is a − c ≈
    1. 85 × 107 miles or 28.5 million miles. And the aphelion or greatest distance is a + c ≈ 4. 35 × 107 miles or 43.5 million miles.
  2. Using a = 3. 6 × 107 and c = 0. 75 × 107 , we compute the eccentricity e =

0. 75 × 107

3. 6 × 107

  1. From a = 1. 67 × 109 and 4. 25 × 108 we obtain

c^2 = a^2 − b^2 = 2. 7889 × 1018 − 18. 0625 × 106 = 2. 78. 89 × 1016 − 18. 0625 × 1016 = 260. 8275 × 1016 = 2. 608275 × 1018.

Then c ≈ 1. 615 × 109 and the eccentricity is

c a

1. 615 × 109

1. 67 × 109

684 CHAPTER 10. CONICS AND POLAR COORDINATES

P

center of earth

center of elliptical orbit

satellite

200 mi 1000 mi

  1. We place the coordinate axes so that the origin is at the center of the ellipse. The length of the ma- jor axis is 200+2(4000)+1000 = 9200 so that a =
    1. Therefore c = 4600 − (200 + 4000) = 400. Then b^2 = a^2 − c^2 = 4600^2 − 4002 = (

21)^2 ,

and the equation is

x^2 46002

y^2 (

12)^2

  1. We place the coordinate axes so that the origin is at the point midway across the base. Thus

a = 5 and b = 15 so the equation of the doorway is y =

x^2 25

. The height of the

doorway at a point on the base 4 ft from the center is y =

= 12 ft.

  1. The base of the room is an ellipse with a = 20 and b = 16. We place the coordinate axes so that the origin is at the point in the center of the room and let the major axes be in the x-direction. c^2 = a^2 − b^2 , so c^2 = 20^2 − 162 , which gives c = 12. Thus the foci occur at the points (− 12 , 0) and (12, 0). Therefore the listening and whispering posts occur along the center line on the longer part of the base 12 ft in each direction from the center.

ellipse shifted ellipse center (0, 1) (4, 1) vertices (− 2 , 1), (2, 1), (0, −2), (0, 4) (2.1), (6.1), (4. − 2), (4.4) foci (0, 1 −

ellipse shifted ellipse center (1, 4) (− 4 , 7) vertices (− 2 , 4), (4, 4), (1, 3), (1, 5) (− 7 , 7), (− 1 , 7), (− 4 , 6), (− 4 .8) foci (1 − 2

  1. (a)

y^2 144

x^2 25

(b) Conjugate hyperbolas have the same asymptotes and do not intersect.

  1. (a) The equation of the hyperbola can be written as

(y + 5/2)^2 5

(x − 3 /2)^2 5

= 1, The equations of the asymptotes are y = 7/ 2 −x and x = 1+x. These lines are perpendicular, thus the hyperbola is a rectangular hyperbola. (b) The hyperbola in Problem 50 is a rectangular hyperbola.

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686 CHAPTER 10. CONICS AND POLAR COORDINATES

y

x

y

x

  1. y = (t^2 )^2 + 3t^2 − 1 = x^2 + 3x − 1; y = x^2 + 3x − 1 , x ≥ 0
  2. − 12 y = t^3 + t; x = − 12 y + 4; 2 x + y = 4
  3. x = cos 2t = cos^2 t − sin^2 t = 1 − 2 sin^2 t = 1 − 2 y^2 ; y = 1 − 2 y^2 , − 1 ≤ y ≤ 1
  4. ln x = t; y = ln(ln x), x > 1. Alternatively, ey^ = t; x = ee y , x > 1
  5. t = x^1 /^3 ; y = 3 ln x^1 /^3 ; y = ln x, x > 0
  6. x^2 tan^2 x, y^2 = sec^2 t; x^2 + 1 = tan^2 t + 1 = sec^2 t = y^2 ; y^2 − x^2 = 1. y ≥ 1

y

x

y

x

y = x x = sin t y = sin t

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10.2. PARAMETRIC EQUATIONS 687

y

x

y

x

y = x^2 x = −

t y = t t ≥ 0

y

x

y

x

y =

x^2 4

− 1 x = 2t y = t^2 − 1 − 1 ≤ t ≤ 2

y

x

y

x

y = −x^2 x = at y = −e^2 t t ≥ 0