Sec 5 Symmetry in polar coordinate

Definitions

Symmetry about the Polar Axis The curve is symmetric about the polar axis ( the x-axis)if replacing the point (r, θ) by either the point (r, - θ) or an equivalent polar representation of it results into an equation equivalent to the original equation

Symmetry about the Polar Axis The curve is symmetric about the line θ=π/2 ( the y-axis) if replacing the point (r, θ) by either the point (r, π - θ) or an equivalent polar representation of it, such as (-r, - θ) results into an equation equivalent to the original equation.

Symmetry about the Pole. The curve is symmetric about the pole (the origin) if replacing the point (r, θ) by either the point (- r, θ) or an equivalent polar representation of it, such as ( r, π+θ) results into an equation equivalent to the original equation.

Examples Test all type of symmetry for each of the following functions r = f(θ) = cos4θ r = f(θ) = cos 2θ + 5 r = f(θ) = sin2θ

Solutions

1. r = f(θ) = cos4θ a. f(-θ) = cos(-4θ) = cos4θ = f(θ), which means that, the curve is symmetric about the polar axis b. f(π - θ) = cos( (4(π - θ) ) = cos( (4π - 4θ ) = cos(-4θ) = cos4θ = f(θ) which means that, the curve is symmetric about the y-axis c. Replacing the point (r, θ) by the point ( -r, θ) does not result into an equation equivalent to the original one. However, replacing the point (r, θ) by the equivalent representation ( r, π + θ) in the original equation, we get: r = cos[4(π + θ)] = cos4θ Which is equivalent to the original equation r = cos4θ. This show that the curve is symmetric about the pole.

The rose curve r = cos4θ exhibits all of the three types of polar symmetry

Example 2. r = f(θ) = cos 2θ + 5 Do it!

What types of polar symmetry does the curve r = cos 2θ + 5 have?

3. r = f(θ) = sin2θ Replacing the point (r, θ) by the point (r, - θ) does not result into an equation equivalent to the original one. This, however this does not indicate a lack of symmetry about the polar axis. The point ( -r, π – θ) is equivalent polar representation for the same point with the polar representation (r, - θ) Replacing the point (r, θ) by ( -r, π – θ) in the original equation, we get: - r = f(π – θ) = sin[2(π – θ)] = sin[2π –2θ] = sin (–2θ) = - sin2θ Which is equivalent to the original equation r = sin2θ. This show that the curve is symmetric about the polar axis. Replacing the point (r, θ) by the point (r, π – θ) does not result into an equation equivalent to the original one. This however this does not indicate a lack of symmetry about the y-axis. The point ( -r, – θ) is equivalent polar representation for the same point with the polar representation (r, π – θ). Replacing the point (r, θ) by ( -r, – θ) in the original equation, we get: - r = sin(– 2θ) = - sin2θ Which is equivalent to the original equation r = sin2θ. This show that the curve is symmetric about the y-axis. Replacing the point (r, θ) by the point ( -r, θ) does not result into an equation equivalent to the original one. However, replacing the point (r, θ) by the equivalent representation ( r, π + θ) in the original equation, we get: r = sin[2(π + θ)] = sin2θ Which is equivalent to the original equation r = sin2θ. This show that the curve is symmetric about the pole.

The rose curve r = sin2θ exhibits all of the three types of polar symmetry

Sec 6 Points of Intersection

Consider the curves r = f(θ) & r = g(θ) To find the points of intersection of these two curves, we find the simultaneous solution for these equations, and at the same time remind ourselves of the following: The simultaneous solution may fail to yields all points of intersection. For instance the pole (origin) has different representations. On one curve it might have one or more representation, while on the other curve a completely distinct representation or set of representations from those on the first curve. Thus although it might be a common point, there is no common representation for it that satisfies both equation simultaneously. A solution to both equations may cease to be as such, when replaced by another representation for the same point.

Examples Find all points of intersection of the given two curves: r = 4, r = 8sinθ r = 8cosθ, r = 8sinθ r = cos2θ, r = sinθ

Solutions

1. r = 4 & r = 8sinθ Solving the equations simultaneously, we get: sinθ = 1/2 → θ = π/6, θ = 5π/6 Thus the points are (4, π/6 ) and (4, 5π/6 ) Graph the two curves and indicate the points of intersection!

The intersection of the curves r = 4 & r = 8sinθ

2. r = 8cosθ & r = 8sinθ Solving the equations simultaneously, we get: tan θ = 1 → θ = π/4, θ = 5π/4, which correspond to the representations ( 4√2, π/4 ) and ( - 4√2, 5π/4 ). These representations represent the same point. Why? What's the Cartesian coordinates of this point. Graph the curves and notice that they intersect also at the pole. Thus the curves intersect at two point: The pole and the point ( 4√2, π/4 ).

The intersection of the curves r = 8cosθ & r = 8sinθ

3. r = cos2θ, r = sinθ Solving the equations simultaneously, we get: cos2θ = sinθ → 1 – 2sin 2 θ = sinθ → 2sin 2 θ + sinθ – 1 = 0 → ( 2sinθ – 1 ) ( sin θ + 1) = 0 → sin θ = 1/2 Or sin θ = -1 → θ = π/6, θ = 5π/6 Or θ = 3π/2 → The point of intersection are: ( 1/2, π/6 ) and ( 1/2, 5π/6 ) and ( -1, 3π/2) Notice that while ( -1, 3π/2) is a representation of the same point who has also the representation ( 1, π/2), the latter satisfies r = sinθ but not r = cos2θ. Graph the two curves and indicate the points of intersection.

The intersection of the curves r = cos2θ & r = sinθ

Sec 7 Arch Length

Let r = f(θ), and let dr/dθ be continuous on [θ 1, θ 2 ]. Then, the arc length L of the curve from θ = θ 1 to θ = θ 2 is: Provided no part of the graph is traced more than once on the interval [θ1, θ2 ].

Examples Find the length of the curve : r = 2 – 2cosθ r = 2 + 2cosθ

Solutions

1. r = 2 – 2cosθ

r = 2 – 2cosθ

2. r = 2 + 2cosθ

Another Method

2. r = 2 + 2cosθ

Sec 8 Area

Let r = f(θ) and 0 < θ 2 – θ 1 ≤ 2π Let r = f(θ) be continuous and either f(θ) ≥ 0 or f(θ) ≤ 0 on [θ 1, θ 2 ] Then the area A of the region enclosed by the curve r = f(θ) and the lines θ = θ 1 and θ = θ 2 is:

Examples Find the area enclosed by : The curve r = 1 – cosθ, the positive x- axis, the y-axis. The curve r = cos2θ The curve r = 4 + 4cosθ, but outside the circle r=6

Solutions

1. The area enclosed by r = 1 – cosθ, the positive x-axis and the y-axis

2. The area enclosed by the rose curve r = cos2θ

1. The area enclosed by r = 4 +4 cosθ, but outside the circle r = 6

The area A is the difference of the area A 1 enclosed by r = 4 + 4cosθ and the lines θ=-π/3 and θ =π/3 and the area A 2 enclosed by r = 6 and the same lines. How do we know that? Thus,