Trigonometry/ProOneSoftware-Trigonometry-Win31.iso/trig/chapter6.1r

Text File | 1995-04-09 | 8KB | 284 lines

 282
à 6.1ïPolar Coordinates

äïPlease change the following ordered pairs from rectangu-
êêlar form to polar form.
âê Change the ordered pair, (√3, 1), from rectangular form
êê to polar form.

#êë┌──────────
#ê r = á(√3)ì + 1ìï=ï2êïΘ¬ = Tanúî(1/√3)ï=ïπ/6

êThus, polar form of the ordered pair, (√3, 1), is (2, π/6).
éSêêè In the past, we have used a rectangular coordi-
@fig6101.bmp,15,35
êêêïnate system to locate points in the plane.ïIn
êêêïthe figure, if you move from the origin x units
êêêïto the right and y units up, you will arrive at
êêêïpoint, P.ïAnother way to arrive at P is to move
êêêïr units along the terminal side of angle Θ.
êêêïSometimes it is more convenient to use the second
method to locate points.ïThe first way to locate points is the rectan-
gular coordinate system, and the second method is called the polar co-
ordinate system.ïIn this system, the origin is called the pole, and the
positive half of the x-axis is called the polar axis.
è Using right triangle trigonometry, we see that x = r∙cos Θ, y =
#r∙sin Θ, r is the square root of xì plus yì, and Θ¬ = Tanúî(y/x), where
#Θ¬ is the reference angle for Θ.ïYou can see the similarities between
the polar coordinate system and the trigonometric form of a complex num-
ber.


êêêêë(more...)



è One very big difference between the rectangular coordinate system
and the polar coordinate system is that in the rectangular coordinate
system, the representation of a point is unique.ïIn the polar coordi-
nate system, each point has many representations.ïFor example, the or-
dered pair, (0, 1), has only one representation in the rectangular co-
ordinate system, but in the polar coordinate system it could be repre-
sented by (1, π/2), (-1, 3π/2), (-1, -π/2), (1, 5π/2), or infinitely
many other ways.ïThe first two ways are called the primary representa-
tions of (0, 1).ïIn general, the primary representations are (r, Θ),
where 0 ≤ Θ ≤ 2π and (-r, π + Θ), where 0 ≤ π + Θ ≤ 2π.
è In polar form, if Θ is positive, we measure the angle in the coun-
terclockwise direction.ïIf Θ is negative, the angle is measured clock-
wise.ïAlso, if r is positive, we measure r units along the terminal
#side of the angle, but if r is negative, we measure │r│ units along the
ray directed opposite to the terminal side of Θ.
@fig6101.bmp,200,335
 1
êêê Change (1, 1) to polar form.

êè A)ï(2, π/6)êêêB)ï(√2, π/3)

êè C)ï(√2, π/4)êêë D)ïå of ç
ü
êêê Change (1, 1) to polar form.
#êê ┌────────ë┌─
#êè r = á 1ì + 1ìï=ïá2ê Θ¬ = Tanúî(1/1) = π/4

#Since (1, 1) is in the first quadrant, Θ = Θ¬ = π/4.ïThus, the polar
form of (1, 1) is (√2, π/4).
Ç C
 2
êêê Change (1, -√3) to polar form.

êè A)ï(2, 300°)êêë B)ï(√2, 120°)

êè C)ï(√2, 60°)êêë D)ïå of ç
ü
êêê Change (1, -√3) to polar form.
#êê ┌────────────ë┌─
#êè r = á 1ì + (-√3)ìï=ïá4 = 2ë Θ¬ = Tanúî(-√3/1) = -60°

Since (1, -√3) is in the fourth quadrant, Θ = 300°.ïThus, the polar
form of (1, -√3) is (2, 300°).
Ç A
 3
êêê Change (-1, -1) to polar form.

êè A)ï(√2, 135°)êêëB)ï(√2, 225°)

êè C)ï(2, 45°)êêêD)ïå of ç
ü
êêê Change (-1, -1) to polar form.
#êê ┌──────────────ë┌─
#êè r = á (-1)ì + (-1)ìï=ïá2êΘ¬ = Tanúî(-1/-1) = 45°

Since (-1, -1) is in the third quadrant, Θ = 225°.ïThus, the polar
form of (-1, -1) is (√2, 225°).
Ç B
 4
êêê Change (2, -2√3) to polar form.

êè A)ï(-4, 60°)êêë B)ï(4, 60°)

êè C)ï(4, 300°)êêë D)ïå of ç
ü
êêê Change (2, -2√3) to polar form.
#êè ┌─────────────ë┌──
#êr = á 2ì + (-2√3)ìï=ïá16 = 4ë Θ¬ = Tanúî(-2√3/2) = -60°

Since (2, -2√3) is in the fourth quadrant, Θ = 300°.ïThus, the polar
form of (2, -2√3) is (4, 300°).
Ç C
 5
êêê Change (0, -1) to polar form.

êè A)ï(-1, 180°)êêëB)ï(1, 270°)

êè C)ï(-1, -90°)êêëD)ïå of ç
ü
êêê Change (0, -1) to polar form.
#êê ┌────────────è ┌─
#êè r = á 0ì + (-1)ìï=ïá1 = 1ë Θ¬ = Tanúî(-1/0), undefined
êêêêêêïby inspection Θ = 270°
Thus, the polar form of (0, -1) is (1, 270°).
Ç B
 6
êëChange (3, 7) to polar form with Θ in radian measure.

êè A)ï(2.38, 2.137)êêïB)ï(7.616, 1.1659)

êè C)ï(1.782, 1.782)êê D)ïå of ç
ü
êêê Change (3, 7) to polar form.
#êê ┌────────────ë┌──
#êè r = á (3)ì + (7)ìï=ïá58êΘ¬ = Tanúî(7/3) ≈ 1.1659
êë ≈ 7.616
Since (3, 7) is in the first quadrant, Θ = 1.1659.ïThus, the polar
form of (3, 7) is (7.616, 1.1659).
Ç B
 7
êêê Change (0, 0) to polar form.

êè A)ï(0, 0°)êêê B)ï(0, 180°)

êè C)ï(0, -90°)êêë D)ïAll of ç
ü
êêê Change (0, 0) to polar form.
#êê ┌──────────ë┌─
#êè r = á 0ì + (0)ìï=ïá0 = 0ë Θ¬ = Tanúî(0/0), indetermi-
êêêêêê nant form, but by inspection
êêêêêê Θ can be any angle.
Thus, the polar form of (0, 0) is all of the above.
Ç D
äïPlease change the following ordered pairs from polar form
êêto rectangular form.
â
êêèChange (2, 2π/3) to rectangular form.

êêïx = r∙cos Θ = 2∙cos 2π/3 = 2∙(-1/2) = -1
êêïy = r∙sin Θ = 2∙sin 2π/3 = 2∙(√3/2) = √3

êëThus, the rectangular form of (2, 2π/3) is (-1, √3).
éSïTo change the ordered pair, (2, 2π/3), from polar form to rec-
tangular form, you should use the formulas x = r∙cos Θ and y = r∙sin Θ.
êêë x = 2∙cos 2π/3 = 2∙(-1/2) = -1
êêë y = 2∙sin 2π/3 = 2∙(√3/2) = √3
Thus, the rectangular form of (2, 2π/3) is (-1, √3).
 8
êêè Change (5, 5π/4) to rectangular form.

êëA)ï(-5/√2, -5/√2)êë B)ï(-√2, -√2)

êëC)ï(√2, √2)êêëD)ïå of ç
ü
êêè Change (5, 5π/4) to rectangular form.

êêx = r∙cos Θ = 5∙cos 5π/4 = 5∙(-1/√2) = -5/√2
êêy = r∙sin Θ = 5∙sin 5π/4 = 5∙(-1/√2) = -5/√2

ë Thus, the rectangular form of (5, 5π/4) is (-5/√2, -5/√2).
Ç A
 9
êêè Change (3, -π/6) to rectangular form.

êëA)ï(3√3/2, -3/2)êêB)ï(2, -√3)

êëC)ï(-√3, 3)êêëD)ïå of ç
ü
êêè Change (3, -π/6) to rectangular form.

êêx = r∙cos Θ = 3∙cos -π/6 = 3∙(√3/2) = 3√3/2
êêy = r∙sin Θ = 3∙sin -π/6 = 3∙(-1/2) = -3/2

ë Thus, the rectangular form of (3, -π/6) is (3√3/2, -3/2).
Ç A
 10
êêè Change (4.3, 29°) to rectangular form.

êëA)ï(6.12, -3.41)êêB)ï(4.31, 7.83)

êëC)ï(3.76, 2.08)êê D)ïå of ç
ü
êêè Change (4.3, 29°) to rectangular form.

êêëx = r∙cos Θ = 4.3∙cos 29° ≈ 3.76
êêëy = r∙sin Θ = 4.3∙sin 29° ≈ 2.08

ë Thus, the rectangular form of (4.3, 29°) is (3.76, 2.08).
Ç C
 11
êêè Change (-2, -π/4) to rectangular form.

êëA)ï(-2/√2, 2/√2)êêB)ï(√3, 1)

êëC)ï(-1, √3)êêëD)ïå of ç
ü
êêè Change (-2, -π/4) to rectangular form.

êêx = r∙cos Θ = -2∙cos -π/4 = -2∙(1/√2) = -2/√2
êêy = r∙sin Θ = -2∙sin -π/4 = -2∙(-1/√2) = 2/√2

ë Thus, the rectangular form of (-2, -π/4) is (-2/√2, 2/√2).
Ç A
äïPlease determine if the given point satisfies the given
êêequation.
â
êè Determine if (4, π/2) satisfies r = 2∙(1 + sin Θ)
êêêë r = 2∙(1 + sin Θ)
êêêë 4 = 2∙(1 + sin π/2)
êêêë 4 = 2∙(1 + 1)
êêêë 4 = 4
ê Thus, (4, π/2) satisfies the equation, r = 2∙(1 + sin Θ).
éSïA point in polar form satisfies an equation involving a func-
tion of nΘ if and only if at least one of the primary representations
satisfies the equation.
è In the example, the point (4, π/2) satisfies the equation r =
2∙(1 + sin Θ) since 4 = 2∙(1 + sin π/2).ïNotice that the other primary
solution, (-4, 3π/2), does not satisfy the equation.
è In another problem, the ordered pair (-1, 2π) does not satisfy the
equation, r = 2/(1 - cos Θ).ïHowever, the other primary solution,
(1, π), does satisfy the equation.ïTherefore, it is necessary to check
both of the primary representations before saying a point does not sat-
isfy an equation.
 12
êè Determine if (1/√3, π/6) satisfies r = (2/√3 - tan Θ).

êê A)ïyesêêêïB) no

ü
êêêë r = (2/√3 - tan Θ)

êêêë 1/√3 = (2/√3 - tan π/6)

êêêë 1/√3 = (2/√3 - 1/√3)
êêêë 1/√3 = 1/√3
Ç A
 13
êè Determine if (-3/2, 330°) satisfies r = 2 - sin Θ.

êê A)ïyesêêêïB) no

ü
êïr = 2 - sin Θêêè r = 2 - sin Θ

ë -3/2 = 2 - (-1/2)êêè3/2 = 2 - sin 150°

#ë -3/2 ƒ 5/2êêêè3/2 = 2 - 1/2
êêêêêë3/2 = 3/2
The primary representation, (-3/2, 330°), does not satisfy the equation,
but the other primary representation, (3/2, 150°), does satisfy the
equation.ïTherefore, the answer is yes.
Ç A
 14
êè Determine if (6, π/3) satisfies r = 4∙cos 3Θ.

êê A)ïyesêêêïB) no

üë(6, π/3)êêê (-6, 4π/3)
êïr = 4∙cos 3Θêêër = 4∙cos 3Θ

êï6 = 4∙cos πêêë -6 = 4∙cos 4π

#êï6 ƒ -4êêêè -6 ƒ 4

Neither of the primary representations satisfy the equation, therefore
the answer is no.
Ç B