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Section 10.3 Polar Coordinates and Functions Ruipeng Shen March 10, 12 1 Polar Coordinates Pole and Polar Axis We choose a point in the plane that is called the pole (or origin) and is labeled O. Then we draw a ray (half-line) starting at O called the polar axis. This axis is usually drawn horizontally to the right and corresponds to the positive x-axis in Cartesian coordinates. Polar Coordinates If P is any other point in the plane. Let r be the distance from O to P and let θ be the angle (usually measured in radians) from the polar axis to the line OP as in Figure 1. Then the point P is represented by the ordered pair (r, θ). P(r,θ) P(r,θ+2𝛑) r r θ+2𝛑 θ O r O O -θ r P(-r,θ) P(r,-θ) Figure 1: Polar Coordinates 1 θ Negative Values of θ or r The angle is considered positive if the rotation from the polar axis to OP is in the counter-clockwise direction, negative if the rotation is in the clockwise direction. The polar coordinates (−r, θ) and (r, θ) represent two points that are symmetric about the origin. Remark 1. A single point may have a lot of different representations of polar coordinates. For example, the polar coordinates (r, θ + 2π), (−r, θ + π) represent the same point as (r, θ). We usually use positive radius and angle between 0 and 2π. Example 2. Plot the points where the polar coordinates are given. (2, 5π/4) Solution (2, −π/4) (3, π/2) See the figure 2. P2(3,𝛑/2) O P1(2,5𝛑/4) P3(2,-𝛑/4) Figure 2: Example 2 - Polar Coordinates Switch between two coordinate systems By basic definition of trigonometric functions, we can convert polar coordinates into Cartesian coordinates x = r cos θ y = r sin θ We can also convert Cartesian coordinates into polar coordinates y r 2 = x2 + y 2 tan θ = x Example 3. Convert the point (2, π/3) from polar coordinates to Cartesian coordinates Solution We have x = 2 cos π = 1; 3 y = 2 sin √ Thus the Cartesian coordinates are (1, 3). π √ = 3. 3 Example 4. Represent the point with Cartesian coordinates (1, −1) in term of polar coordinates. 2 Solution Let us choose r to be positive; we can calculate r= p 12 + (−1)2 = √ 2; tan θ = −1 = −1 1 Since√the point is in the fourth quadrant, we can choose θ = 7π/4. Thus the polar coordinates √ are ( 2, 7π/4). One can also use ( 2, −π/4). 2 Polar Curves The graph of a polar equation r = f (θ) or more generally F (r, θ) = 0, consists of all points P that have at least one polar representation (r, θ) whose coordinates satisfy the equation. Example 5. What curve is represented by the polar equation r = 2? How about θ = 1? Solution This can be understood by considering the geometrical meaning of polar coordinates. The curves are the circle centered at the origin with radius 2 and the line y = (tan 1)x, respectively. Example 6. Let us consider the curves with a polar equation r = 2 cos θ. Find a Cartesian equation and sketch the curve. Solution Using the identities x = r cos θ and r2 = x2 + y 2 , we have r = 2 cos θ ⇐⇒ r2 = 2r cos θ ⇐⇒ x2 + y 2 = 2x ⇐⇒ (x − 1)2 + y 2 = 1 This is a circle centered at (1, 0) with a radius 1. r=2cosθ O θ 2 Figure 3: Circle: Example 6 Example 7 (Cardioid). Sketch the curve r = 1 − sin θ. Solution Let us first sketch the graph of r = 1 − sin θ in the Cartesian coordinates (on a r-θ plane). This enables us to read at a glance the values of r that correspond to increasing values of θ. 3 2 1 -1 0 1 2 3 4 5 6 7 8 Figure 4: Graph of r = 1 − sin θ in the r-θ coordinate system 0.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 -0.5 r = 1- sin θ -1 -1.5 -2 -2.5 Figure 5: Cardioid: polar equation r = 1 − sin θ Example 8. Sketch the curve r = cos 2θ. Solution Again we first sketch the graph of r = cos 2θ in the Cartesian coordinates, then sketch the curve. 1 -1 0 1 2 3 4 5 6 7 8 -1 Figure 6: Graph of r = cos 2θ in the r-θ coordinate system 4 1.5 θ=3π/4 θ=π/4 1 0.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 r = cos 2θ -0.5 θ=5π/4 θ=7π/4 -1 -1.5 Figure 7: Graph of the polar equation r = cos 2θ 3 Tangents to Polar Curves To find a tangent line to a polar curve r = f (θ), we regard θ as a parameter and write its parametric equation as x = r cos θ = f (θ) cos θ; y = r sin θ = f (θ) sin θ. Thus the slope of the parametric curve is given by dy dr sin θ + r cos θ dy f 0 (θ) sin θ + f (θ) cos θ dθ dθ = . = = 0 dx dr dx f (θ) cos θ − f (θ) sin θ cos θ − r sin θ dθ dθ Remark 9. In particular, if f (θ0 ) = 0, then we have dy = tan θ0 . dx Example 10. Consider the cardioid r = 1 − sin θ in the Figure 5. (a) Find the formula for the slope of the curve in term of θ. (b) Find the points where the tangent line is horizontal or vertical. 5 (1) 2.5 Solution (a) By the formula (1), we have dr sin θ + r cos θ dy (− cos θ) sin θ + (1 − sin θ) cos θ = dθ = dr dx (− cos θ) cos θ − (1 − sin θ) sin θ cos θ − r sin θ dθ cos θ(1 − 2 sin θ) = −(1 − sin θ)(1 + 2 sin θ) (b) In order to seek horizontal or vertical tangents, we solve the equation 1 π 3π π 5π dy = cos θ(1 − 2 sin θ) = 0 =⇒ cos θ = 0 or sin θ = =⇒ θ = , , , dθ 2 2 2 6 6 dx 1 π 7π 11π = − (1 − sin θ)(1 + 2 sin θ) = 0 =⇒ sin θ = 1 or sin θ = − =⇒ θ = , , dθ 2 2 6 6 Thus we have θ π 6 π 2 5π 6 7π 6 3π 2 11π 6 y 0 (t) = 0? Yes Yes Yes No Yes No x0 (t) = 0? No Yes No Yes No Yes Tangent Horizontal Indeterminate Horizontal Vertical Horizontal Vertical Finally when θ = π/2, the radius r = 0. Thus the tangent line is vertical, according to Remark 9. In summary, the tangent is horizontal at the points (1/2, π/6), (1/2, 5π/6) and (2, 3π/2); the tangent is vertical at the points (3/2, 7π/6), (3/2, 11π/6) and the pole. 6 0.5π 0.5 -2 -1.5 π -1 -0.5 0 0.5 -0.5 -11.5π -1.5 -2 -2.5 Figure 8: Cardioid in Polar Coordinates 7 1 1.5 2 2.5