第10章三角法のさらなる応用Further Applications of Trigonometry

Answer Key解答

Solution 1解答 1

$\begin{array}{l} α = 98^{\circ} a = 34.6 \\ β = 39^{\circ} b = 22 \\ γ = 43^{\circ} c = 23.8 \end{array}$

Solution 2解答 2

Solution 1

解1

\begin{array}{l} α = 80° & a = 120 \\ β \approx 83.2° & b = 121 \\ γ \approx 16.8° & c \approx 35.2 \end{array}

Solution 2

解2

\begin{array}{l} α^{'} = 80° a^{'} = 120 \\ β^{'} \approx 96.8° b^{'} = 121 \\ γ^{'} \approx 3.2° c^{'} \approx 6.8 \end{array}

Solution 3解答 3

$β \approx 5.7°, γ \approx 94.3°, c \approx 101.3$

Solution 4解答 4

two

二つ

Solution 5解答 5

about $8.2$ square feet

およそ $8.2$ 平方フィート

Solution 6解答 6

161.9 yd.

161.9ヤード

Solution 1解答 1

$a \approx 14.9,$ $β \approx 23.8°,$ $γ \approx 126.2° .$

Solution 2解答 2

$α \approx 27.7°,$ $β \approx 40.5°,$ $γ \approx 111.8°$

Solution 3解答 3

Area = 552 square feet

面積 = 552平方フィート

Solution 4解答 4

about 8.15 square feet

およそ8.15平方フィート

Solution 1解答 1

Polar grid with point (2, pi/3) plotted.

Solution 2解答 2

Points (2, 9pi/4) and (3, -pi/6) are plotted in the polar grid.

Solution 3解答 3

$(x, y) = (\frac{1}{2}, - \frac{\sqrt{3}}{2})$

Solution 4解答 4

$r = \sqrt{3}$

Solution 5解答 5

$x^{2} + y^{2} = 2 y$ or, in the standard form for a circle, $x^{2} + {(y - 1)}^{2} = 1$

英語のヒント

$x^{2} + y^{2} = 2 y$ 、または円の標準の形で $x^{2} + {(y - 1)}^{2} = 1$

Solution 1解答 1

The equation fails the symmetry test with respect to the line $θ = \frac{π}{2}$ and with respect to the pole. It passes the polar axis symmetry test.

英語のヒント

この式は、直線 $θ = \frac{π}{2}$ についての対称性の判定と、極についての対称性の判定に落ちる。極軸についての対称性の判定は通る。

Solution 2解答 2

Tests will reveal symmetry about the polar axis. The zero is $(0, \frac{π}{2}),$ and the maximum value is $(3, 0) .$

英語のヒント

判定により極軸についての対称性が分かる。零点は $(0, \frac{π}{2})$ 、最大値は $(3, 0)$ である。

Solution 3解答 3

Graph of the limaçon r=3-2cos(theta). Extending to the left.

Solution 4解答 4

The graph is a rose curve, $n$ even

英語のヒント

このグラフはバラ曲線で、 $n$ は偶数

Graph of rose curve r=4 sin(2 theta). Even - four petals equally spaced, each of length 4.

Solution 5解答 5

Graph of rose curve r=3cos(3theta). Three petals equally spaced from origin.

Rose curve, $n$ odd

バラ曲線、 $n$ は奇数

Solution 6解答 6

A dark blue spiral curve is plotted on a polar coordinate grid with concentric circles and radial lines. The curve starts near the origin and expands outwards, resembling an Archimedean spiral. The grid includes labeled axes from -15 to 15, with concentric circles indicating radii at intervals of 5, up to a maximum radius of 15.

Solution 1解答 1

Plot of 1+5i in the complex plane (1 along the real axis, 5 along the imaginary axis).

Solution 2解答 2

Solution 3解答 3

$| z | = \sqrt{50} = 5 \sqrt{2}$

Solution 4解答 4

$z = 3 (\cos (\frac{π}{2}) + i \sin (\frac{π}{2}))$

Solution 5解答 5

$z = 2 (\cos (\frac{π}{6}) + i \sin (\frac{π}{6}))$

Solution 6解答 6

$z = 2 \sqrt{3} - 2 i$

Solution 7解答 7

$z_{1} z_{2} = - 4 \sqrt{3}; \frac{z_{1}}{z_{2}} = - \frac{\sqrt{3}}{2} + \frac{3}{2} i$

Solution 8解答 8

$z_{0} = 2 (\cos (30°) + i \sin (30°))$

$z_{1} = 2 (\cos (120°) + i \sin (120°))$

$z_{2} = 2 (\cos (210°) + i \sin (210°))$

$z_{3} = 2 (\cos (300°) + i \sin (300°))$

Solution 1解答 1

$t$	$x (t)$	$y (t)$
$- 1$	$- 4$	$2$
$0$	$- 3$	$4$
$1$	$- 2$	$6$
$2$	$- 1$	$8$

A coordinate plane displays a blue line segment ascending from (-4, 2) to (0, 10). Arrows indicate its direction along grid lines, highlighting its positive slope.

Solution 2解答 2

$\begin{array}{l} x (t) = t^{3} - 2 t \\ y (t) = t \end{array}$

Solution 3解答 3

$y = 5 - \sqrt{\frac{1}{2} x - 3}$

Solution 4解答 4

$y = \ln \sqrt{x}$

Solution 5解答 5

$\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$

Solution 6解答 6

$y = x^{2}$

Solution 1解答 1

Graph of the given parametric equations with the restricted domain - it looks like the right half of an upward opening parabola.

Solution 2解答 2

Graph of the given equations - a horizontal ellipse.

Solution 3解答 3

The graph of the parametric equations is in red and the graph of the rectangular equation is drawn in blue dots on top of the parametric equations.

英語のヒント

媒介変数表示のグラフは赤で、直交座標での方程式のグラフは媒介変数表示の上に青い点で描いてある。

Overlayed graph of the two versions of the ellipse, showing that they are the same whether they are given in parametric or rectangular coordinates.

Solution 1解答 1

A vector from the origin to (3,5) - a line with an arrow at the (3,5) endpoint.

Solution 2解答 2

$3 u = 〈 15, 12 〉$

Solution 3解答 3

$u = 8 i - 11 j$

Solution 4解答 4

$v = \sqrt{34} \cos (59°) i + \sqrt{34} \sin (59°) j$

Magnitude = $\sqrt{34}$

大きさ = $\sqrt{34}$

$θ = \tan^{- 1} (\frac{5}{3}) = 59.04°$

10.1 Section Exercises10.1 節末問題

Solution 1解答 1

The altitude extends from any vertex to the opposite side or to the line containing the opposite side at a 90° angle.

英語のヒント

高さは、どの頂点からでも、向かい合う辺、または向かい合う辺を含む直線へ90°の角で下ろす。

Solution 3解答 3

When the known values are the side opposite the missing angle and another side and its opposite angle.

分かっている値が、未知の角と向かい合う辺、もう一つの辺、そしてそれと向かい合う角であるとき。

Solution 5解答 5

A triangle with two given sides and a non-included angle.

二辺と、そのはさむ角でない角が与えられた三角形。

Solution 7解答 7

$β = 72°, a \approx 12.0, b \approx 19.9$

Solution 9解答 9

$γ = 20°, b \approx 4.5, c \approx 1.6$

Solution 11解答 11

$b \approx 3.78$

Solution 13解答 13

$c \approx 13.70$

Solution 15解答 15

one triangle, $α \approx 50.3°, β \approx 16.7°, a \approx 26.7$

三角形は一つ、 $α \approx 50.3°, β \approx 16.7°, a \approx 26.7$

Solution 17解答 17

two triangles, $γ \approx 54.3°, β \approx 90.7°, b \approx 20.9$ or $γ^{'} \approx 125.7°, β^{'} \approx 19.3°, b^{'} \approx 6.9$

三角形は二つ、 $γ \approx 54.3°, β \approx 90.7°, b \approx 20.9$ または $γ^{'} \approx 125.7°, β^{'} \approx 19.3°, b^{'} \approx 6.9$

Solution 19解答 19

two triangles, $β \approx 75.7°, γ \approx 61.3°, b \approx 9.9$ or $β^{'} \approx 18.3°, γ^{'} \approx 118.7°, b^{'} \approx 3.2$

三角形は二つ、 $β \approx 75.7°, γ \approx 61.3°, b \approx 9.9$ または $β^{'} \approx 18.3°, γ^{'} \approx 118.7°, b^{'} \approx 3.2$

Solution 21解答 21

two triangles, $α \approx 143.2°, β \approx 26.8°, a \approx 17.3$ or $α^{'} \approx 16.8°, β^{'} \approx 153.2°, a^{'} \approx 8.3$

三角形は二つ、 $α \approx 143.2°, β \approx 26.8°, a \approx 17.3$ または $α^{'} \approx 16.8°, β^{'} \approx 153.2°, a^{'} \approx 8.3$

Solution 23解答 23

no triangle possible

三角形はできない

Solution 25解答 25

$A \approx 47.8°$ or $A^{'} \approx 132.2°$

$A \approx 47.8°$ または $A^{'} \approx 132.2°$

Solution 27解答 27

$8.6$

Solution 29解答 29

$370.9$

Solution 31解答 31

$12.3$

Solution 33解答 33

$12.2$

Solution 35解答 35

$16.0$

Solution 37解答 37

$29.7°$

Solution 39解答 39

$x = 76.9° or x = 103.1°$

式中の英語：or：または

Solution 41解答 41

$110.6°$

Solution 43解答 43

$A \approx 39.4, C \approx 47.6, B C \approx 20.7$

Solution 45解答 45

$57.1$

Solution 47解答 47

$42.0$

Solution 49解答 49

$430.2$

Solution 51解答 51

$10.1$

Solution 53解答 53

$A D \approx 13.8$

Solution 55解答 55

$A B \approx 2.8$

Solution 57解答 57

$L \approx 49.7, N \approx 56.3, L N \approx 5.8$

Solution 59解答 59

51.4 feet

51.4フィート

Solution 61解答 61

The distance from the satellite to station $A$ is approximately 1716 miles. The satellite is approximately 1706 miles above the ground.

英語のヒント

人工衛星から局 $A$ までの距離はおよそ1716マイルである。人工衛星は地上およそ1706マイルにある。

Solution 63解答 63

2.6 ft

2.6フィート

Solution 65解答 65

5.6 km

Solution 67解答 67

371 ft

371フィート

Solution 69解答 69

5936 ft

5936フィート

Solution 71解答 71

24.1 ft

24.1フィート

Solution 73解答 73

19,056 ft²

19,056平方フィート

Solution 75解答 75

445,624 square miles

445,624平方マイル

Solution 77解答 77

8.65 ft²

8.65平方フィート

10.2 Section Exercises10.2 節末問題

Solution 1解答 1

two sides and the angle opposite the missing side.

英語のヒント

二辺と、未知の辺と向かい合う角。

Solution 3解答 3

$s$ is the semi-perimeter, which is half the perimeter of the triangle.

英語のヒント

$s$ は半周長、すなわち三角形の周の長さの半分である。

Solution 5解答 5

The Law of Cosines must be used for any oblique (non-right) triangle.

英語のヒント

斜(直角でない)三角形には余弦定理を使わなければならない。

Solution 7解答 7

11.3

Solution 9解答 9

34.7

Solution 11解答 11

26.7

Solution 13解答 13

$c = 257.3, 96.7$

Solution 15解答 15

not possible

不可能

Solution 17解答 17

95.5°

Solution 19解答 19

26.9°

Solution 21解答 21

$B \approx 45.9°, C \approx 99.1°, a \approx 6.4$

Solution 23解答 23

$A \approx 20.6°, B \approx 38.4°, c \approx 51.1$

Solution 25解答 25

$A \approx 37.8°, B \approx 43.8, C \approx 98.4°$

Solution 27解答 27

177.56 in²

177.56平方インチ

Solution 29解答 29

0.04 m²

Solution 31解答 31

0.91 yd²

0.91平方ヤード

Solution 33解答 33

3.0

Solution 35解答 35

29.1

Solution 37解答 37

0.5

Solution 39解答 39

70.7°

Solution 41解答 41

77.4°

Solution 43解答 43

25.0

Solution 45解答 45

9.3

Solution 47解答 47

43.52

Solution 49解答 49

1.41

Solution 51解答 51

0.14

Solution 53解答 53

18.3

Solution 55解答 55

48.98

Solution 57解答 57

A triangle. One angle is 52 degrees with opposite side = x. The other two sides are 5 and 6.

Solution 59解答 59

7.62

Solution 61解答 61

85.1

Solution 63解答 63

24.0 km

Solution 65解答 65

99.9 ft

99.9フィート

Solution 67解答 67

37.3 miles

37.3マイル

Solution 69解答 69

2371 miles

2371マイル

Solution 71解答 71

Angle BO is 9.1 degrees, angle PH is 150.2 degrees, and angle DC is 20.7 degrees.

Solution 73解答 73

292.4 miles

292.4マイル

Solution 75解答 75

65.4 cm²

65.4平方センチメートル

Solution 77解答 77

468 ft²

468平方フィート

10.3 Section Exercises10.3 節末問題

Solution 1解答 1

For polar coordinates, the point in the plane depends on the angle from the positive x-axis and distance from the origin, while in Cartesian coordinates, the point represents the horizontal and vertical distances from the origin. For each point in the coordinate plane, there is one representation, but for each point in the polar plane, there are infinite representations.

英語のヒント

極座標では、平面上の点は x軸の正の部分からの角と原点からの距離で決まるが、デカルト座標では、点は原点からの水平と鉛直の距離を表す。座標平面の各点には表しかたが一つしかないが、極平面の各点には表しかたが無限にある。

Solution 3解答 3

Determine $θ$ for the point, then move $r$ units from the pole to plot the point. If $r$ is negative, move $r$ units from the pole in the opposite direction but along the same angle. The point is a distance of $r$ away from the origin at an angle of $θ$ from the polar axis.

英語のヒント

その点の $θ$ を決め、そして極から $r$ だけ動いて点を打つ。 $r$ が負なら、同じ角に沿って極から逆の向きに $r$ だけ動く。この点は、極軸から $θ$ の角のところで原点から $r$ だけ離れている。

Solution 5解答 5

The point $(- 3, \frac{π}{2})$ has a positive angle but a negative radius and is plotted by moving to an angle of $\frac{π}{2}$ and then moving 3 units in the negative direction. This places the point 3 units down the negative y-axis. The point $(3, - \frac{π}{2})$ has a negative angle and a positive radius and is plotted by first moving to an angle of $- \frac{π}{2}$ and then moving 3 units down, which is the positive direction for a negative angle. The point is also 3 units down the negative y-axis.

英語のヒント

点 $(- 3, \frac{π}{2})$ は角が正で動径が負であり、角 $\frac{π}{2}$ まで動き、そして負の向きに3だけ動いて打つ。これでこの点は y軸の負の部分を3だけ下ったところに来る。点 $(3, - \frac{π}{2})$ は角が負で動径が正であり、まず角 $- \frac{π}{2}$ まで動き、そして3だけ下って打つ。負の角ではこれが正の向きである。この点も y軸の負の部分を3だけ下ったところにある。

Solution 7解答 7

$(- 5, 0)$

Solution 9解答 9

$(- \frac{3 \sqrt{3}}{2}, - \frac{3}{2})$

Solution 11解答 11

$(2 \sqrt{5}, 0.464)$

Solution 13解答 13

$(\sqrt{34}, 5.253)$

Solution 15解答 15

$(8 \sqrt{2}, \frac{π}{4})$

Solution 17解答 17

$r = 4 \csc θ$

Solution 19解答 19

$r = \sqrt[3]{\frac{s i n θ}{2 c o s^{4} θ}}$

Solution 21解答 21

$r = 3 \cos θ$

Solution 23解答 23

$r = \frac{3 \sin θ}{\cos (2 θ)}$

Solution 25解答 25

$r = \frac{9 \sin θ}{\cos^{2} θ}$

Solution 27解答 27

$r = \sqrt{\frac{1}{9 \cos θ \sin θ}}$

Solution 29解答 29

$x^{2} + y^{2} = 4 x$ or $\frac{{(x - 2)}^{2}}{4} + \frac{y^{2}}{4} = 1;$ circle

$x^{2} + y^{2} = 4 x$ または $\frac{{(x - 2)}^{2}}{4} + \frac{y^{2}}{4} = 1;$ 円

Solution 31解答 31

$3 y + x = 6;$ line

$3 y + x = 6;$ 直線

Solution 33解答 33

$y = 3;$ line

$y = 3;$ 直線

Solution 35解答 35

$x y = 4;$ hyperbola

$x y = 4;$ 双曲線

Solution 37解答 37

$x^{2} + y^{2} = 4;$ circle

$x^{2} + y^{2} = 4;$ 円

Solution 39解答 39

$x - 5 y = 3;$ line

$x - 5 y = 3;$ 直線

Solution 41解答 41

$(3, \frac{3 π}{4})$

Solution 43解答 43

$(5, π)$

Solution 45解答 45

Polar coordinate system with a point located on the second concentric circle and two-thirds of the way between pi and 3pi/2 (closer to 3pi/2).

Solution 47解答 47

Polar coordinate system with a point located midway between the third and fourth concentric circles and midway between 3pi/2 and 2pi.

Solution 49解答 49

Polar coordinate system with a point located on the fifth concentric circle and pi/2.

Solution 51解答 51

Polar coordinate system with a point located on the third concentric circle and 2/3 of the way between pi/2 and pi (closer to pi).

Solution 53解答 53

Polar coordinate system with a point located on the second concentric circle and midway between pi and 3pi/2.

Solution 55解答 55

$r = \frac{6}{5 \cos θ - \sin θ}$

Plot of given line in the polar coordinate grid

Solution 57解答 57

$r = 2 \sin θ$

Plot of given circle in the polar coordinate grid

Solution 59解答 59

$r = \frac{2}{\cos θ}$

Solution 61解答 61

$r = 3 \cos θ$

Solution 63解答 63

$x^{2} + y^{2} = 16$

Plot of circle with radius 4 centered at the origin in the rectangular coordinates grid.

Solution 65解答 65

$y = x$

Plot of line y=x in the rectangular coordinates grid.

Solution 67解答 67

$x^{2} + {(y + 5)}^{2} = 25$

Plot of circle with radius 5 centered at (0,-5).

Solution 69解答 69

$(1.618, - 1.176)$

Solution 71解答 71

$(10.630, 131.186°)$

Solution 73解答 73

$(2, 3.14) o r (2, π)$

Solution 75解答 75

A vertical line with $a$ units left of the y-axis.

英語のヒント

y軸の左 $a$ のところの鉛直な直線。

Solution 77解答 77

A horizontal line with $a$ units below the x-axis.

x軸の下 $a$ のところの水平な直線。

Solution 79解答 79

Graph of shaded circle of radius 4 with the edge not included (dotted line) - polar coordinate grid.

Solution 81解答 81

Graph of ray starting at (2, pi/4) and extending in a positive direction along pi/4 - polar coordinate grid.

Solution 83解答 83

Graph of the shaded region 0 to pi/3 from r=0 to 2 with the edge not included (dotted line) - polar coordinate grid

10.4 Section Exercises10.4 節末問題

Solution 1解答 1

Symmetry with respect to the polar axis is similar to symmetry about the $x$ -axis, symmetry with respect to the pole is similar to symmetry about the origin, and symmetric with respect to the line $θ = \frac{π}{2}$ is similar to symmetry about the $y$ -axis.

英語のヒント

極軸についての対称性は $x$ 軸についての対称性に似ており、極についての対称性は原点についての対称性に似ており、直線 $θ = \frac{π}{2}$ についての対称性は $y$ 軸についての対称性に似ている。

Solution 3解答 3

Test for symmetry; find zeros, intercepts, and maxima; make a table of values. Decide the general type of graph, cardioid, limaçon, lemniscate, etc., then plot points at $θ = 0, \frac{π}{2},$ $π$ and $\frac{3 π}{2},$ and sketch the graph.

英語のヒント

対称性を調べる。零点、座標軸との交点、最大値を求める。値の表を作る。心臓形、リマソン、レムニスケートなど、グラフのおおまかな種類を決め、そして $θ = 0, \frac{π}{2}$ $π$ $\frac{3 π}{2}$ で点を打ち、グラフの概形を描く。

Solution 5解答 5

The shape of the polar graph is determined by whether or not it includes a sine, a cosine, and constants in the equation.

英語のヒント

極座標のグラフの形は、その式に正弦、余弦、定数が入っているかどうかで決まる。

Solution 7解答 7

symmetric with respect to the polar axis

英語のヒント

極軸について対称

Solution 9解答 9

symmetric with respect to the polar axis, symmetric with respect to the line $θ = \frac{π}{2},$ symmetric with respect to the pole

極軸について対称、直線 $θ = \frac{π}{2}$ について対称、極について対称

Solution 11解答 11

symmetric with respect to the line $θ = \frac{π}{2}$

直線 $θ = \frac{π}{2}$ について対称

Solution 13解答 13

Symmetric with respect to line $θ = \frac{π}{2}$ (y-axis)

英語のヒント

直線 $θ = \frac{π}{2}$ (y軸)について対称

Solution 15解答 15

symmetric with respect to the pole

極について対称

Solution 17解答 17

circle

円

Solution 19解答 19

cardioid

心臓形

Solution 21解答 21

cardioid

心臓形

Solution 23解答 23

one-loop/dimpled limaçon

英語のヒント

一重の輪／くぼんだリマソン

Solution 25解答 25

one-loop/dimpled limaçon

一重の輪／くぼんだリマソン

Solution 27解答 27

inner loop/two-loop limaçon

内側に輪を持つ／二重の輪のリマソン

Solution 29解答 29

inner loop/two-loop limaçon

内側に輪を持つ／二重の輪のリマソン

Solution 31解答 31

inner loop/two-loop limaçon

内側に輪を持つ／二重の輪のリマソン

Solution 33解答 33

lemniscate

レムニスケート

Solution 35解答 35

lemniscate

レムニスケート

Solution 37解答 37

rose curve

バラ曲線

Solution 39解答 39

rose curve

バラ曲線

Solution 41解答 41

Archimedes’ spiral

アルキメデスの渦巻き

Solution 43解答 43

Archimedes’ spiral

アルキメデスの渦巻き

Solution 45解答 45

Solution 47解答 47

Graph of given hippopede (two circles that are centered along the x-axis and meet at the origin)

Solution 49解答 49

Solution 51解答 51

Graph of given equation. Similar to original Archimedes' spiral.

Solution 53解答 53

Solution 55解答 55

They are both spirals, but not quite the same.

どちらも渦巻きだが、まったく同じではない。

Solution 57解答 57

Both graphs are curves with 2 loops. The equation with a coefficient of $θ$ has two loops on the left, the equation with a coefficient of 2 has two loops side by side. Graph these from 0 to $4 π$ to get a better picture.

英語のヒント

どちらのグラフも輪を二つ持つ曲線である。係数が $θ$ の式は左に輪が二つ、係数が2の式は輪が二つ並ぶ。0から $4 π$ まで描くと、より分かりやすい図になる。

Solution 59解答 59

When the width of the domain is increased, more petals of the flower are visible.

英語のヒント

定義域の幅を広げると、花びらがより多く見えるようになる。

Solution 61解答 61

The graphs are three-petal, rose curves. The larger the coefficient, the greater the curve’s distance from the pole.

どのグラフも三枚花びらのバラ曲線である。係数が大きいほど、曲線は極から遠くなる。

Solution 63解答 63

The graphs are spirals. The smaller the coefficient, the tighter the spiral.

どのグラフも渦巻きである。係数が小さいほど、渦は密になる。

Solution 65解答 65

$(4, \frac{π}{3}), (4, \frac{5 π}{3})$

Solution 67解答 67

$(\frac{3}{2}, \frac{π}{3}), (\frac{3}{2}, \frac{5 π}{3})$

Solution 69解答 69

$(0, \frac{π}{2}), (0, π), (0, \frac{3 π}{2}), (0, 2 π)$

Solution 71解答 71

There are five intersection points: the pole, and four points whose distance from the pole is $2^{- \frac{1}{4}}$ and whose angles are $\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ . Negative values of r must be allowed when tracing these two curves.

英語のヒント

交点は極（原点）と、極からの距離が $2^{- \frac{1}{4}}$ 、偏角が $\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ である4点の、合わせて5点である。二つの曲線をたどる際には、rの負の値も考える。極では、それぞれの曲線を表す偏角が異なっていても同じ点になる。

10.5 Section Exercises10.5 節末問題

Solution 1解答 1

a is the real part, b is the imaginary part, and $i = \sqrt{- 1}$

a が実部、b が虚部で、 $i = \sqrt{- 1}$ である

Solution 3解答 3

Polar form converts the real and imaginary part of the complex number in polar form using $x = r \cos θ$ and $y = r \sin θ .$

英語のヒント

極形式は、 $x = r \cos θ$ と $y = r \sin θ$ を使って複素数の実部と虚部を極形式に換える。

Solution 5解答 5

$z^{n} = r^{n} (\cos (n θ) + i \sin (n θ))$ It is used to simplify polar form when a number has been raised to a power.

英語のヒント

$z^{n} = r^{n} (\cos (n θ) + i \sin (n θ))$ 数が累乗されたときに極形式を簡単にするのに使う。

Solution 7解答 7

$5 \sqrt{2}$

Solution 9解答 9

$\sqrt{38}$

Solution 11解答 11

$\sqrt{14.45}$

Solution 13解答 13

$4 \sqrt{5} cis (333.4°)$

Solution 15解答 15

$2 cis (\frac{π}{6})$

Solution 17解答 17

$\frac{7 \sqrt{3}}{2} + i \frac{7}{2}$

Solution 19解答 19

$- 2 \sqrt{3} - 2 i$

Solution 21解答 21

$- 1.5 - i \frac{3 \sqrt{3}}{2}$

Solution 23解答 23

$4 \sqrt{3} cis (198°)$

Solution 25解答 25

$\frac{3}{4} cis (180°)$

Solution 27解答 27

$5 \sqrt{3} cis (\frac{17 π}{24})$

Solution 29解答 29

$7 cis (70°)$

Solution 31解答 31

$5 cis (80°)$

Solution 33解答 33

$5 cis (\frac{π}{3})$

Solution 35解答 35

$125 cis (135°)$

Solution 37解答 37

$9 cis (240°)$

Solution 39解答 39

$cis (\frac{3 π}{4})$

Solution 41解答 41

$3 cis (80°), 3 cis (200°), 3 cis (320°)$

Solution 43解答 43

$2 \sqrt[3]{4} cis (\frac{2 π}{9}), 2 \sqrt[3]{4} cis (\frac{8 π}{9}), 2 \sqrt[3]{4} cis (\frac{14 π}{9})$

Solution 45解答 45

$2 \sqrt{2} cis (\frac{7 π}{8}), 2 \sqrt{2} cis (\frac{15 π}{8})$

Solution 47解答 47

Plot of -3 -3i in the complex plane (-3 along real axis, -3 along imaginary axis).

Solution 49解答 49

Plot of -1 -5i in the complex plane (-1 along real axis, -5 along imaginary axis).

Solution 51解答 51

Plot of 2i in the complex plane (0 along the real axis, 2 along the imaginary axis).

Solution 53解答 53

Plot of 6-2i in the complex plane (6 along the real axis, -2 along the imaginary axis).

Solution 55解答 55

Plot of 1-4i in the complex plane (1 along the real axis, -4 along the imaginary axis).

Solution 57解答 57

$3.61 e^{- 0.59 i}$

Solution 59解答 59

$- 2 + 3.46 i$

Solution 61解答 61

$- 4.33 - 2.50 i$

10.6 Section Exercises10.6 節末問題

Solution 1解答 1

A pair of functions that is dependent on an external factor. The two functions are written in terms of the same parameter. For example, $x = f (t)$ and $y = f (t) .$

英語のヒント

外部の要因に依存する関数の対。二つの関数は同じ媒介変数で書かれる。たとえば $x = f (t)$ と $y = f (t)$ である。

Solution 3解答 3

Choose one equation to solve for $t,$ substitute into the other equation and simplify.

英語のヒント

一方の式を選んで $t$ について解き、もう一方の式に代入して簡単にする。

Solution 5解答 5

Some equations cannot be written as functions, like a circle. However, when written as two parametric equations, separately the equations are functions.

英語のヒント

円のように、関数として書けない式もある。だが二つの媒介変数表示として書けば、それぞれの式は関数である。

Solution 7解答 7

$y = - 2 + 2 x$

Solution 9解答 9

$y = 3 \sqrt{\frac{x - 1}{2}}$

Solution 11解答 11

$x = 2 e^{\frac{1 - y}{5}}$ or $y = 1 - 5 l n (\frac{x}{2})$

$x = 2 e^{\frac{1 - y}{5}}$ または $y = 1 - 5 l n (\frac{x}{2})$

Solution 13解答 13

$x = 4 \log (\frac{y - 3}{2})$

Solution 15解答 15

$x = {(\frac{y}{2})}^{3} - \frac{y}{2}$

Solution 17解答 17

$y = x^{3}$

Solution 19解答 19

${(\frac{x}{4})}^{2} + {(\frac{y}{5})}^{2} = 1$

Solution 21解答 21

$y^{2} = 1 - \frac{1}{2} x$

Solution 23解答 23

$y = x^{2} + 2 x + 1$

Solution 25解答 25

$y = {(\frac{x + 1}{2})}^{3} - 2$

Solution 27解答 27

$y = - 3 x + 14$

Solution 29解答 29

$y = x + 3$

Solution 31解答 31

${\begin{array}{l} x (t) = t \\ y (t) = 2 \sin t + 1 \end{array}$

Solution 33解答 33

${\begin{array}{l} x (t) = \sqrt{t} + 2 t \\ y (t) = t \end{array}$

Solution 35解答 35

${\begin{array}{l} x (t) = 4 \cos t \\ y (t) = 6 \sin t \end{array};$ Ellipse

${\begin{array}{l} x (t) = 4 \cos t \\ y (t) = 6 \sin t \end{array};$ 楕円

Solution 37解答 37

${\begin{array}{l} x (t) = \sqrt{10} \cos t \\ y (t) = \sqrt{10} \sin t \end{array};$ Circle

${\begin{array}{l} x (t) = \sqrt{10} \cos t \\ y (t) = \sqrt{10} \sin t \end{array};$ 円

Solution 39解答 39

${\begin{array}{l} x (t) = - 1 + 4 t \\ y (t) = - 2 t \end{array}$

Solution 41解答 41

${\begin{array}{l} x (t) = 4 + 2 t \\ y (t) = 1 - 3 t \end{array}$

Solution 43解答 43

yes, at $t = 2$

ある、 $t = 2$ で

Solution 45解答 45

$t$	$x$	$y$
1	-3	1
2	0	7
3	5	17

Solution 47解答 47

answers may vary: ${\begin{array}{l} x (t) = t - 1 \\ y (t) = t^{2} \end{array} and {\begin{array}{l} x (t) = t + 1 \\ y (t) = {(t + 2)}^{2} \end{array}$

答えはいろいろありうる: ${\begin{array}{l} x (t) = t - 1 \\ y (t) = t^{2} \end{array} および {\begin{array}{l} x (t) = t + 1 \\ y (t) = {(t + 2)}^{2} \end{array}$

Solution 49解答 49

answers may vary: , ${\begin{array}{l} x (t) = t \\ y (t) = t^{2} - 4 t + 4 \end{array} and {\begin{array}{l} x (t) = t + 2 \\ y (t) = t^{2} \end{array}$

答えはいろいろありうる: 、 ${\begin{array}{l} x (t) = t \\ y (t) = t^{2} - 4 t + 4 \end{array} および {\begin{array}{l} x (t) = t + 2 \\ y (t) = t^{2} \end{array}$

10.7 Section Exercises10.7 節末問題

Solution 1解答 1

plotting points with the orientation arrow and a graphing calculator

英語のヒント

向きの矢印を付けて点を打つことと、グラフ電卓

Solution 3解答 3

The arrows show the orientation, the direction of motion according to increasing values of $t .$

英語のヒント

矢印は向き、すなわち $t$ の値が増えるのに従う運動の向きを示す。

Solution 5解答 5

The parametric equations show the different vertical and horizontal motions over time.

英語のヒント

媒介変数表示は、時間とともに変わる鉛直と水平の別々の運動を示す。

Solution 7解答 7

Graph of the given equations - looks like an upward opening parabola.

Solution 9解答 9

Graph of the given equations - a line, negative slope.

Solution 11解答 11

Graph of the given equations - looks like a sideways parabola, opening to the right.

Solution 13解答 13

Graph of the given equations - looks like the left half of an upward opening parabola.

Solution 15解答 15

Graph of the given equations - looks like a downward opening absolute value function.

Solution 17解答 17

Graph of the given equations - a vertical ellipse.

Solution 19解答 19

Graph of the given equations- line from (0, -3) to (3,0). It is traversed in both directions, positive and negative slope.

Solution 21解答 21

Solution 23解答 23

Graph of the given equations- looks like a downward opening parabola.

Solution 25解答 25

Graph of the given equations- horizontal ellipse.

Solution 27解答 27

Graph of the given equations- looks like the lower half of a sideways parabola opening to the right

Solution 29解答 29

Graph of the given equations- looks like an upwards opening parabola

Solution 31解答 31

Graph of the given equations- looks like the upper half of a sideways parabola opening to the left

Solution 33解答 33

Graph of the given equations- the left half of a hyperbola with diagonal asymptotes

Solution 35解答 35

Graph of the given equations - vertical periodic trajectory

Solution 37解答 37

Solution 39解答 39

There will be 100 back-and-forth motions.

行ったり来たりの運動が100回ある。

Solution 41解答 41

Take the opposite of the $x (t)$ equation.

英語のヒント

$x (t)$ の式の符号を変える。

Solution 43解答 43

The parabola opens up.

放物線は上に開く。

Solution 45解答 45

${\begin{cases} x (t) = 5 \cos t \\ y (t) = 5 \sin t \end{cases}$

Solution 47解答 47

Solution 49解答 49

Graph of the given equations - lines extending into Q1 and Q3 (in both directions) from the origin to 1 unit.

Solution 51解答 51

Graph of the given equations - lines extending into Q1 and Q3 (in both directions) from the origin to 3 units.

Solution 53解答 53

$a = 4,$ $b = 3,$ $c = 6,$ $d = 1$

Solution 55解答 55

$a = 4,$ $b = 2,$ $c = 3,$ $d = 3$

Solution 57解答 57

Solution 59解答 59

Solution 61解答 61

The $y$ -intercept changes.

$y$ 軸との交点が変わる。

Solution 63解答 63

$y (x) = - 16 {(\frac{x}{15})}^{2} + 20 (\frac{x}{15})$

Solution 65解答 65

${\begin{cases} x (t) = 64 t \cos (52 °) \\ y (t) = - 16 t^{2} + 64 t \sin (52 °) \end{cases}$

Solution 67解答 67

approximately 3.2 seconds

およそ3.2秒

Solution 69解答 69

1.6 seconds

1.6秒

Solution 71解答 71

Graph of the given equations - a hypocycloid

Solution 73解答 73

Graph of the given equations - a four petal rose

10.8 Section Exercises10.8 節末問題

Solution 1解答 1

lowercase, bold letter, usually $u, v, w$

小文字の太字、ふつうは $u, v, w$

Solution 3解答 3

They are unit vectors. They are used to represent the horizontal and vertical components of a vector. They each have a magnitude of 1.

英語のヒント

これらは単位ベクトルである。ベクトルの水平成分と鉛直成分を表すのに使う。どちらも大きさが1である。

Solution 5解答 5

The first number always represents the coefficient of the $i,$ and the second represents the $j .$

一つ目の数はつねに $i$ の係数、二つ目は $j$ の係数を表す。

Solution 7解答 7

$⟨ 7, - 5 ⟩$

Solution 9解答 9

not equal

等しくない

Solution 11解答 11

equal

等しい

Solution 13解答 13

equal

等しい

Solution 15解答 15

$- 7 i - 3 j$

Solution 17解答 17

$- 6 i - 2 j$

Solution 19解答 19

$u + v = ⟨ - 5, 5 ⟩, u - v = ⟨ - 1, 3 ⟩, 2 u - 3 v = ⟨ 0, 5 ⟩$

Solution 21解答 21

$- 10 i - 4 j$

Solution 23解答 23

$- \frac{2 \sqrt{29}}{29} i + \frac{5 \sqrt{29}}{29} j$

Solution 25解答 25

$- \frac{2 \sqrt{229}}{229} i + \frac{15 \sqrt{229}}{229} j$

Solution 27解答 27

$- \frac{7 \sqrt{2}}{10} i + \frac{\sqrt{2}}{10} j$

Solution 29解答 29

$| v | = 7.810, θ = 39.806 °$

Solution 31解答 31

$| v | = 7.211, θ = 236.310°$

Solution 33解答 33

$- 6$

Solution 35解答 35

$- 12$

Solution 37解答 37

Three graphs illustrate scalar multiplication of a vector. The first graph shows vector v. The second shows vector 3v, which is three times longer than v. The third shows vector 1/2v, half the length of v.

Solution 39解答 39

Plot of u+v, u-v, and 2u based on the above vectors. In relation to the same origin point, u+v goes to (0,3), u-v goes to (2,-1), and 2u goes to (2,2).

Solution 41解答 41

Plot of vectors u+v, u-v, and 2u based on the above vectors.Given that u's start point was the origin, u+v starts at the origin and goes to (2,-3); u-v starts at the origin and goes to (4,-1); 2u goes from the origin to (6,-4).

Solution 43解答 43

Plot of a single vector. Taking the start point of the vector as (0,0) from the above set up, the vector goes from the origin to (-1,-6).

Solution 45解答 45

Vector extending from the origin to (7,5), taking the base as the origin.

Solution 47解答 47

$⟨ 4, 1 ⟩$

Solution 49解答 49

$v = - 7 i + 3 j$

Solution 51解答 51

$3 \sqrt{2} i + 3 \sqrt{2} j$

Solution 53解答 53

$i - \sqrt{3} j$

Solution 55解答 55

ⓐ58.7ⓑ12.5

Solution 57解答 57

$x = 7.13$ pounds, $y = 3.63$ pounds

$x = 7.13$ ポンド、 $y = 3.63$ ポンド

Solution 59解答 59

$x = 2.87$ pounds, $y = 4.10$ pounds

$x = 2.87$ ポンド、 $y = 4.10$ ポンド

Solution 61解答 61

4.635 miles, 17.764° N of E

4.635マイル、東から北へ17.764°

Solution 63解答 63

17 miles. 10.318 miles

17マイル。10.318マイル

Solution 65解答 65

Distance: 2.868. Direction: 86.474° North of West, or 3.526° West of North

英語のヒント

距離: 2.868。向き: 西から北へ86.474°、すなわち北から西へ3.526°

Solution 67解答 67

4.924°. 659 km/hr

4.924°。毎時659km

Solution 69解答 69

4.424°

Solution 71解答 71

$(0.081, 8.602)$

Solution 73解答 73

21.801°, relative to the car’s forward direction

英語のヒント

車の進む向きに対して21.801°

Solution 75解答 75

parallel: 16.28, perpendicular: 47.28 pounds

平行: 16.28ポンド、垂直: 47.28ポンド

Solution 77解答 77

19.35 pounds, 231.54° from the horizontal

19.35ポンド、水平から231.54°

Solution 79解答 79

5.1583 pounds, 75.8° from the horizontal

5.1583ポンド、水平から75.8°

Review Exercises復習問題

Solution 1解答 1

Not possible

不可能

Solution 3解答 3

$C = 120°, a = 23.1, c = 34.1$

Solution 5解答 5

distance of the plane from point $A :$ 2.2 km, elevation of the plane: 1.6 km

英語のヒント

点 $A :$ からこの飛行機までの距離: 2.2km、この飛行機の高度: 1.6km

Solution 7解答 7

$b = 71.0°, C = 55.0°, a = 12.8$

Solution 9解答 9

40.6 km

Solution 11解答 11

Polar coordinate grid with a point plotted on the fifth concentric circle 2/3 the way between pi and 3pi/2 (closer to 3pi/2).

Solution 13解答 13

$(0, 2)$

Solution 15解答 15

$(9.8489, 203.96°)$

Solution 17解答 17

$r = 8$

Solution 19解答 19

$x^{2} + y^{2} = 7 x$

Solution 21解答 21

$y = - x$

Plot of the function y=-x in rectangular coordinates.

Solution 23解答 23

symmetric with respect to the line $θ = \frac{π}{2}$

英語のヒント

直線 $θ = \frac{π}{2}$ について対称

Solution 25解答 25

Graph of the given polar equation - an inner loop limaçon.

Solution 27解答 27

Graph of the given polar equation - a cardioid.

Solution 29解答 29

Solution 31解答 31

$cis (- \frac{π}{3})$

Solution 33解答 33

$2.3 + 1.9 i$

Solution 35解答 35

$60 cis (\frac{π}{2})$

Solution 37解答 37

$3 cis (\frac{4 π}{3})$

Solution 39解答 39

$25 cis (\frac{3 π}{2})$

Solution 41解答 41

$5 cis (\frac{3 π}{4}), 5 cis (\frac{7 π}{4})$

Solution 43解答 43

Plot of -1 + 3i in the complex plane (-1 along the real axis, 3 along the imaginary).

Solution 45解答 45

$x^{2} + \frac{1}{2} y = 1$

Solution 47解答 47

${\begin{array}{l} x (t) = - 2 + 6 t \\ y (t) = 3 + 4 t \end{array}$

Solution 49解答 49

$y = - 2 x^{5}$

Solution 51解答 51

a. ${\begin{cases} x (t) = (80 \cos (40°)) t \\ y (t) = - 16 t^{2} + (80 \sin (40°)) t + 4 \end{cases}$ b. The ball is 14 feet high and 184 feet from where it was launched.c. 3.3 seconds

英語のヒント

b. 球は高さ14フィート、打ち出された場所から184フィートのところにある。c. 3.3秒

Solution 53解答 53

not equal

等しくない

Solution 55解答 55

Solution 57解答 57

$- \frac{3 \sqrt{10}}{10}$ i $- \frac{\sqrt{10}}{10}$ j

Solution 59解答 59

Magnitude: $3 \sqrt{2},$ Direction: $225°$

大きさ: $3 \sqrt{2}$ 向き: $225°$

Solution 61解答 61

$16$

Solution 63解答 63

Diagram of vectors u and v. Taking u's starting point as the origin, u goes from the origin to (4,1), and v goes from (4,1) to (6,0).

Practice Test実力テスト

Solution 1解答 1

$α = 67.1°, γ = 44.9°, a = 20.9$

Solution 3解答 3

$1712 miles$

Solution 5解答 5

$(1, \sqrt{3})$

Solution 7解答 7

$y = - 3$

Plot of the given equation in rectangular form - line y=-3.

Solution 9解答 9

Graph of the given equations - a cardioid.

Solution 11解答 11

$\sqrt{106}$

Solution 13解答 13

$\frac{- 5}{2} + i \frac{5 \sqrt{3}}{2}$

Solution 15解答 15

$4 cis (21°)$

Solution 17解答 17

$2 \sqrt{2} cis (18°), 2 \sqrt{2} cis (198°)$

Solution 19解答 19

$y = 2 {(x - 1)}^{2}$

Solution 21解答 21

Solution 23解答 23

−4i − 15j

Solution 25解答 25

$\frac{2 \sqrt{13}}{13} i + \frac{3 \sqrt{13}}{13} j$