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POLAR COORDINATES. There are many curves for which cartesian or parametric equations are unsuitable. For polar equations, the position of a point is defined.

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Presentation on theme: "POLAR COORDINATES. There are many curves for which cartesian or parametric equations are unsuitable. For polar equations, the position of a point is defined."— Presentation transcript:

1 POLAR COORDINATES

2 There are many curves for which cartesian or parametric equations are unsuitable. For polar equations, the position of a point is defined by its distance from a point O, called the pole, and the angular displacement of the line joining the point to O, from a fixed line, called the initial line. A θ O If the distance OA is 4, and the angle θ is then the point A has polar coordinates π3π3 ( 4, ). π3π3 Note that we normally take 0 ≤ θ < 2π. ( Or – π < θ ≤ π ).

3 Relationship between Polar and Cartesian Equations x y r θ A O ( r, θ ) y = r sin θ x = r cos θ x 2 + y 2 = r 2 tan θ = yxyx

4 Example 1: Find the cartesian equations for the following polar equations a) r = 3 b) θ = c) r = 4 cos θ d) r = 3 sin 2θ. π3π3 a) r = 3r 2 = 9x 2 + y 2 = 9 b) π3π3 θ = tan θ = tan π3π3 tan θ = yxyx = y = x c) r = 4 cos θ r = 4 xrxr r 2 = 4xx 2 + y 2 = 4x d) r = 3 sin 2θr = 6 sin θ cos θ r = 6 yryr xrxr r 3 = 6xy ( x 2 + y 2 ) = 6xy 3 2 x 2 + y 2 = r 2 y = r sin θx = r cos θ tan θ = yxyx

5 Sketching polar curves Given a polar equation r = f(θ), the shape can be found by plotting a few key points, and by the following observations. i)If r is a function of cos θ, since cos (– θ ) = cos θ, points can be plotted for 0 ≤ θ ≤ π, and completed by reflecting in the initial line. ii)If r is a function of sin θ, since sin (π – θ ) = sin θ, points can be plotted for – ≤ θ ≤, and completed by reflecting in the line θ =. π2π2 π2π2 π2π2 iii)If the curve passes through O and r = 0 when θ = α, then the line θ = α is a tangent to the curve at O.

6 r 012 we reflect in the initial line Example 1: Sketch the curve with polar equation r = 1 – cos θ, 0 ≤ θ ≤ 2π. Since r is a function of cos θ, Note that θ = 0 is a tangent to the curve at O. π2π2 θ0 π Plot some key points:

7 Example 2: Sketch the curve with polar equation r = 3 ( 1 + cos 2θ ), 0 ≤ θ ≤ 2π. r So, r can be written as a function of cos θ, Note that θ = is a tangent to the curve at O. π2π2 Now, cos 2θ = 2 cos 2 θ – 1 we reflect in the initial line. Also, cos 2θ = 1 – 2 sin 2 θ So, r can be written as a function of sin θ, we reflect in the line θ =. π2π2 π2π2 π6π6 0 θ Plot some key points:

8 Tangents parallel to the initial line: A tangent to a polar curve will be parallel to the initial line at points where Tangents perpendicular to the initial line: A tangent to a polar curve will be perpendicular to the initial line at points where d d θ (r cos θ) = 0 d d θ (r sin θ) = 0

9 cos θ + cos θ cos θ + sin θ (–sin θ ) = 0 cos θ + cos 2 θ – sin 2 θ = 0 cos θ + cos 2 θ – ( 1 – cos 2 θ ) = 0 2 cos 2 θ + cos θ – 1 = 0 ( 2 cos θ – 1 )( cos θ + 1 ) = 0 cos θ = – 1 cos θ = 1 2 θ = ± π3π3 giving: 3 2 π3π3 (, ) and Example 1: Find the points on the Polar curve shown, with equation r = 1 + cos θ where the tangents are parallel to the initial line. For tangents parallel to the initial line: π3π3 ( 3 2, ) – ( sin θ + cos θ sin θ) = 0 d d θ d d θ (r sin θ) = 0 r = 0 giving the pole.

10 Example 2: Find the points on the Polar curve shown, with equation r = 3 sin 2θ ; 0 ≤ θ ≤ where the tangents are perpendicular to the initial line. π2π2 For tangents perpendicular to the initial line: d d θ ( 3 sin 2θ cos θ ) = 0 Differentiating this leads to cos θ = 0 or cos 2 θ = 2323 Giving the points ( 2, ) and the pole. d d θ (r cos θ) = 0

11 Summary of key points: This PowerPoint produced by R.Collins ; Updated Apr x 2 + y 2 = r 2 y = r sin θx = r cos θ tan θ = yxyx Relationship between Polar and Cartesian Equations Tangents parallel to the initial line: d d θ (r sin θ) = 0 Tangents perpendicular to the initial line: d d θ (r cos θ) = 0


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