polar coordinates - uplift education · 2015. 2. 23. · 1 polar coordinates polar coordinate...
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1
POLAR COORDINATES
Polar coordinate system: a pole (fixed point) and a polar axis (directed ray with endpoint at pole).
rectangular coordinates ⇒ polar coordinates polar coordinates ⇒ rectangular coordinates
𝑟 = √𝑥2 + 𝑦2, 𝜃 = 𝑎𝑟𝑐 𝑡𝑎𝑛𝑦
𝑥 𝑥 = 𝑟 𝑐𝑜𝑠 𝜃 𝑦 = 𝑟 𝑠𝑖𝑛 𝜃
The angle, θ, is measured from the polar axis to a line that passes through the point and the pole. If the angle is measured in a counterclockwise direction, the angle is positive. If the angle is measured in a clockwise direction, the angle is negative. The directed distance, r, is measured from the pole to point P. If point P is on the terminal side of angle θ, then the value of r is positive. If point P is on the opposite side of the pole, then the value of r is negative.
Problem : P (x, y) = (1, 3). Express it in polar coordinates (r, θ) two different ways such that 0≤ θ < 2
(r, θ) = (2, /3), (- 2, 4/3) .
Problem : P(x, y) = (-4, 0). Express it in polar coordinates (r, θ) two different ways such that 0≤ θ < 2.
(r, θ) = (4, ),(- 4, 0) .
Problem : P (x, y) = (-7, -7), express it in polar coordinates (r, θ) two different ways such that 0≤ θ < 2.
(r, θ) = (98, 5/4),(- 98, /4) .
Problem : Given a point in polar coordinates (r, θ) = (3, /4), express it in rectangular coordinates (x, y) . (x, y) = (3√2/2, 3√2/2) .
Problem : How many different ways can a point be expressed in polar coordinates such that r > 0 ?
An infinite number. (r, θ) = (r, θ +2n) , where n is an integer.
Problem : Transform the equation x2 + y2 + 5x = 0 to polar coordinate form.
x2 + y2 + 5x = 0 r2 + 5(r cos θ) = 0
r ( r + 5 cos θ) = 0 The equation r = 0 is the pole. Thus, keep only the other equation: r + 5 cos θ = 0
The location of a point can be named using many different pairs of polar coordinates. ← three different sets of polar coordinates for the point P (5, 60°).
The distance r and the angle 𝜃 are both directed--meaning that they represent the
distance and angle in a given direction. It is possible, therefore to have negative values for
both r and 𝜃. However, we typically avoid points with negative r , since they could just as
easily be specified by adding to 𝜃 .
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Problem : Transform the equation r = 4sin θ to Cartesian coordinate form. What is the graph? Describe it fully!!!
√𝑥2 + 𝑦2 = 4𝑦
√𝑥2+𝑦2
𝑥2 + 𝑦2 = 4 𝑦 𝑥2 + (𝑦 − 2)2 = 22 circle: r = 2 C(0, 2)
Problem : What is the maximum value of | r| for the following polar equations:
a) r = cos(2 θ) ;
b) r = 3 + sin(θ) ;
c) r = 2 cos(θ) - 1 .
a) The maximum value of | r| in r = cos(2 θ) occurs when θ = n/2 where n is an integer and | r| = 1 .
b) The maximum value of | r| in r = 3 + sin(θ) occurs when θ = /2+2n where n is an integer and | r| = 4 .
c) The maximum value of | r| in r = 2 cos(θ) - 1 occurs when θ = (2n + 1) where n is an integer and | r| = 3 .
Problem : Find the intercepts and zeros of the following polar equations: a) r = cos(θ) + 1 ; b) r = 4 sin(θ) .
a) Polar axis intercepts: (r, θ) = (2, 2n),(0, (2n + 1)) , where n is an integer.
Line θ = /2 intercepts: (r, θ) = (1, /2 + n) , where n is an integer. r = cos(θ) + 1 = 0 for θ = (2n + 1) , where n is an integer.
b) Polar axis intercepts: (r, θ) = (0, n) where n is an integer. Line θ = /2 intercepts: (r, θ) = (4, /2 +2n) where n is an integer.
r = 4 sin(θ) = 0 for θ = n, where n is an integer.
Problem : Sketch
Spiral of Archimedes: r = θ, θ ≥ 0
The curve is a nonending spiral.
Here it is shown in detail from θ = 0 to θ = 2π
Problem : Sketch Lima¸cons (Snail): 𝑟 = 1 − cos 𝜃
θ 0 π/4 π/3 π/2 2 π/3 3 π/4 π 5 π/4 4 π/3 3 π/2 5 π/3 7 π/4 2 π
r – 1 –0.41 0 1 2 2.41 3 2.41 2 1 0 –0.41 –1
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Problem : Sketch Lima¸cons (Snail): 𝑟 = 𝑎 + 𝑏 cos 𝜃
The general shape of the curve depends on the relative magnitudes of |a| and |b|.
𝑟 = 3 + cos 𝜃 𝑟 =3
2+ cos 𝜃 𝑟 = 1 + cos 𝜃 𝑟 =
1
2+ cos 𝜃
convex limacon limacon with a dimple carotid limacon with an inner loop
Problem : Sketch Cardioids (Heart-Shaped): r = 1 ± cosθ , r = 1 ± sinθ
𝑟 = 1 + cos 𝜃 𝑟 = 1 + sin 𝜃 𝑟 = 1 − cos 𝜃 𝑟 = 1 − sin 𝜃
Flowers
Problem : Sketch Petal Curve: r = cos 2 θ
Problem : Sketch Petal Curves: r = a cos n θ, r = a sin n θ
r = sin 3θ r = cos 4 θ
• If n is odd, there are n petals. • If n is even, there are 2n petals.
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First and second derivative r = r():
𝑻𝒂𝒏𝒈𝒆𝒏𝒕 𝒍𝒊𝒏𝒆 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝜽 = 𝜽𝟎 𝑜𝑓 𝑎 𝑐𝑢𝑟𝑣𝑒 𝒓 = 𝒓(𝜽): 𝑦 = 𝑚 (𝑥 – 𝑥0) + 𝑦0
𝑟0 = 𝑟(𝜃0) ⟹𝑥0 = 𝑟0𝑐𝑜𝑠 𝜃0
𝑦0 = 𝑟0𝑠𝑖𝑛 𝜃0 𝑎𝑛𝑑 𝑚 =
𝑑𝑦
𝑑𝑥 𝑎𝑡 𝑟0 = 𝑟(𝜃0)
𝑵𝒐𝒓𝒎𝒂𝒍 𝒍𝒊𝒏𝒆: 𝑦 = − 1
𝑚 (𝑥 – 𝑥0) + 𝑦0
𝑯𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝒍𝒊𝒏𝒆: 𝑤𝑖𝑙𝑙 𝑜𝑐𝑐𝑢𝑟 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑖𝑠 𝑧𝑒𝑟𝑜 ⟹ 𝑑𝑦
𝑑𝑥= 0
𝑑𝑦
𝑑𝑥= 0 ⟹
𝑑𝑦
𝑑𝜃= 0 ⟹ 𝜃0 (𝑐ℎ𝑒𝑐𝑘
𝑑𝑥
𝑑𝑡|
𝜃0
≠ 0) ⟹ 𝑟0 = 𝑟(𝜃0) ⟹ 𝑦0
= 𝑟0 𝑠𝑖𝑛𝜃0 𝑒𝑞: 𝑦 = 𝑦0
𝑽𝒆𝒓𝒕𝒊𝒄𝒂𝒍 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝒍𝒊𝒏𝒆: 𝑤𝑖𝑙𝑙 𝑜𝑐𝑐𝑢𝑟 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑: 𝑑𝑦
𝑑𝑥= ∞
𝑑𝑦
𝑑𝑥= ∞ ⟹
𝑑𝑥
𝑑𝜃= 0 ⟹ 𝜃0 (𝑐ℎ𝑒𝑐𝑘
𝑑𝑦
𝑑𝑡|
𝜃0
≠ 0) ⟹ 𝑟0 = 𝑟(𝜃0) ⟹ 𝑥0 = 𝑟0 𝑐𝑜𝑠𝜃0 𝑒𝑞: 𝑥 = 𝑥0
𝑪𝒐𝒏𝒄𝒂𝒗𝒊𝒕𝒚 𝒂𝒕 𝒑𝒐𝒊𝒏𝒕 (𝒙𝟏, 𝒚𝟏) 𝑜𝑟 𝒕𝟏 𝑜𝑟 𝜽𝟏 ∶ 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑒 𝑑2𝑦
𝑑𝑥2 𝑎𝑡 𝑡ℎ𝑎𝑡 𝑝𝑜𝑖𝑛𝑡.
𝐼𝑓 𝑑2𝑦
𝑑𝑥2< 0 → 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑑𝑜𝑤𝑛. 𝐼𝑓
𝑑2𝑦
𝑑𝑥2> 0 → 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑢𝑝
𝑥 = 𝑟𝑐𝑜𝑠𝜃 ⟹ 𝑑𝑥
𝑑𝜃=
𝑑𝑟
𝑑𝜃 𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
𝑦 = 𝑟𝑠𝑖𝑛𝜃 ⟹ 𝑑𝑦
𝑑𝜃=
𝑑𝑟
𝑑𝜃 𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝒅𝒚
𝒅𝒙=
𝑑𝑦𝑑𝜃𝑑𝑥𝑑𝜃
=
𝑑𝑟𝑑𝜃
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝑑𝑟𝑑𝜃
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃
𝑑2𝑦
𝑑𝑥2=
𝑑
𝑑𝑥[𝑑𝑦
𝑑𝑥] =
𝑑𝑑𝑡
[𝑑𝑦𝑑𝑥
]
𝑑𝑥𝑑𝑡
𝒅𝟐𝒚
𝒅𝟐𝒙=
𝑑
𝑑𝑥 (
𝑑𝑦
𝑑𝑥) =
1
𝑑𝑥𝑑𝜃
𝑑
𝑑𝜃 (
𝑑𝑟𝑑𝜃
𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
𝑑𝑟𝑑𝜃
𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃 )
and now good luck
Note that rather than trying to remember this formula it would probably be easier to
remember how we derived it .
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Area enclosed by a polar curve r = r():
For a very small 𝜃 (𝑑𝜃), the curve could be approximated by a straight line and the area could be found using the triangle formula:
𝑑𝐴 =1
2(𝑟 𝑑𝜃)𝑟 =
1
2𝑟2𝑑𝜃
1 ≤ ≤ 2
𝐴 = ∫ 𝑑𝐴 =1
2 ∫ 𝑟2 𝑑𝜃
𝜃2
𝜃1
𝜃2
𝜃1
Example: Find the area enclosed by:
example: Find the area of the inner loop of r = 2 + 4 cos θ
𝑟 = 2(1 + 𝑐𝑜𝑠𝜃)
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Length of a Polar Curve:
𝑆 = ∫ 𝑑𝑠 = ∫ √(𝑑𝑥)2 + (𝑑𝑦)2 =𝑏
𝑎
𝑏
𝑎
∫ √ (𝑑𝑥
𝑑𝜃)
2
+ (𝑑𝑦
𝑑𝜃)
2
𝑑𝜃 𝜃2
𝜃1
𝜃1 𝑎𝑛𝑑 𝜃2 𝑎𝑟𝑒 𝑎𝑛𝑔𝑙𝑒𝑠 𝑐𝑜𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑠 𝑎 𝑎𝑛𝑑 𝑏
𝑥 = 𝑟𝑐𝑜𝑠𝜃 ⟹ 𝑑𝑥
𝑑𝜃=
𝑑𝑟
𝑑𝜃 𝑐𝑜𝑠 𝜃 − 𝑟 sin 𝜃 𝑎𝑛𝑑 𝑦 = 𝑟𝑠𝑖𝑛𝜃 ⟹
𝑑𝑦
𝑑𝜃=
𝑑𝑟
𝑑𝜃 𝑠𝑖𝑛 𝜃 + 𝑟𝑐𝑜𝑠 𝜃
(𝑑𝑥
𝑑𝜃)
2
+ (𝑑𝑦
𝑑𝜃)
2
= (𝑑𝑟
𝑑𝜃)
2
𝑐𝑜𝑠2𝜃 − 2𝑟𝑑𝑟
𝑑𝜃𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 + 𝑟2𝑠𝑖𝑛2𝜃 + (
𝑑𝑟
𝑑𝜃)
2
𝑠𝑖𝑛2𝜃 + 2𝑟𝑑𝑟
𝑑𝜃𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 + 𝑟2𝑐𝑜𝑠2𝜃 = (
𝑑𝑟
𝑑𝜃)
2
+ 𝑟2
𝑆 = ∫ 𝑑𝑠 = ∫ √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫ √ 𝑟2 + (𝑑𝑟
𝑑𝜃)
2
𝑑𝜃 𝜃2
𝜃1
𝑏
𝑎
𝑏
𝑎
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EXAMPLE: limaçon: r = 0.5 + cos θ
1. Find the area of the inner circle. 2. Find all vertical and horizontal tangents. 3. Find the points with two tangent lines. Find tangents.
= 0.375 (4/3 − 2/3) + 0.5(𝑠𝑖𝑛 4/3 – 𝑠𝑖𝑛 2/3) + 0.125 (𝑠𝑖𝑛 8/3 – 𝑠𝑖𝑛 4/3)
= 0.25 − 0.5 3 + 0.125 3 𝑨 = 𝟎. 𝟐𝟓 − 𝟎. 𝟑𝟕𝟓 √𝟑
2. 𝑑𝑦
𝑑𝑥=
𝑑𝑦𝑑𝜃𝑑𝑥𝑑𝜃
𝑥 = 𝑟 𝑐𝑜𝑠 = 0.5 𝑐𝑜𝑠 + 𝑐𝑜𝑠2 𝑑𝑥
𝑑𝜃 = – 0.5 𝑠𝑖𝑛 – 2 𝑐𝑜𝑠 𝑠𝑖𝑛
𝑦 = 𝑟 𝑠𝑖𝑛 = 0.5 𝑠𝑖𝑛 + 𝑐𝑜𝑠 𝑠𝑖𝑛 𝑑𝑦
𝑑𝜃 = 0.5 𝑐𝑜𝑠 – 𝑠𝑖𝑛2 + 𝑐𝑜𝑠2𝜃 = 0.5 𝑐𝑜𝑠 + 2𝑐𝑜𝑠2𝜃 – 1
ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠: 𝑑𝑦𝑑𝑥
= 0 → 𝑑𝑦𝑑𝜃
= 0 & 𝑑𝑥𝑑𝜃
≠ 0
2cos2 + 0.5 cos – 1 = 0 cos = (– 0.5 ± √8.25)/4
cos = 0.593 1 = 0.936 rad 2 = 5.347 rad
cos = – 0.843 3 = 2.572 rad 4 = 3.710 rad
each equation is: y = r sin
𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑠: 𝑑𝑦
𝑑𝑥= ∞ →
𝑑𝑥
𝑑𝜃= 0 &
𝑑𝑦
𝑑𝜃 ≠ 0
𝑠𝑖𝑛 + 4 𝑐𝑜𝑠 𝑠𝑖𝑛 = 0 𝑠𝑖𝑛(1 + 4 𝑐𝑜𝑠) = 0
sin = 0 = 0 tangent: x = 1.5 the same for cos = - ¼
3. r = 0.5 + cos θ will have two tangents at point r = 0 = 2/3 and = 4/3
slope = . 𝑑𝑦
𝑑𝑥=
𝑑𝑦
𝑑𝜃𝑑𝑥
𝑑𝜃
= 0.5 cos + 2cos2 – 1
– 0.5 sin – 2 cos sin calculate that for both = 2/3 and = 4/3
first tangent line at r = 0 = 2/3 (y1 = 0, x1 = 0) y = y’ (x – x1) + y1 second tangent line at r = 0 = 4/3 (y1 = 0, x1 = 0) y = y’ (x – x1) + y1
table:
r
0 1.5
/6 1.37
/3 1
2/3 0
5/6 -0.367
-0.5
7/6 -0.367
4/3 0
5/3 1
11/6 1.37
2 1.5
𝐴 =1
2∫ 𝑟2𝑑𝜃
4𝜋/3
2𝜋/3
= 1
2∫ (0.5 + 𝑐𝑜𝑠 𝜃)2𝑑𝜃
4𝜋/3
2𝜋/3
=1
2∫ (0.25 + cos 𝜃 + 𝑐𝑜𝑠2𝜃) 𝑑𝜃
4𝜋/3
2𝜋/3
=1
2∫ (0.25 + cos 𝜃 + 0.5 𝑐𝑜𝑠 2𝜃 + 0.5) 𝑑𝜃
4𝜋/3
2𝜋/3