Polar Coordinates, Area in Polar Coordinates

Lecture 32 18.01 Fall 2006 Lecture 32: Polar Co-ordinates, Area in Polar Co-ordinates Polar Coordinates r θ Figure 1: Polar Co-ordinates. In polar coordinates, we specify an object’s position in terms of its distance r from the origin and the angle θ that the ray from the origin to the point makes with respect to the x-axis. Example 1. What are the polar coordinates for the point specified by (1, −1) in rectangular coordinates? r (1,-1) Figure 2: Rectangular Co-ordinates to Polar Co-ordinates. r = θ = � 12 + (−1)2 = π − 4 √ 2 In most cases, we use the convention that r ≥ 0 and 0 ≤ θ ≤ 2π. But another common convention is to say r ≥ 0 and −π ≤ θ ≤ π. All values of θ and even negative values of r can be used. 1 Lecture 32 18.01 Fall 2006 r y θ x Figure 3: Rectangular Co-ordinates to Polar Co-ordinates. Regardless of whether we allow positive or negative values of r or θ, what is always true is: x = r cos θ and y = r sin θ √ 3π For instance, x = 1, y = −1 can be represented by r = − 2, θ = : 4 √ 3π 1 = x = − 2 cos 4 and √ 3π − 1 = y = − 2 sin 4 Example 2. Consider a circle of radius a with its center at x = a, y = 0. We want to find an equation that relates r to θ. (a,0) Figure 4: Circle of radius a with center at x = a, y = 0. 2 Lecture 32 18.01 Fall 2006 We know the equation for the circle in rectangular coordinates is (x − a)2 + y 2 = a2 Start by plugging in: x = r cos θ and y = r sin θ This gives us (r cos θ − a)2 + (r sin θ)2 = a2 r2 cos2 θ − 2arcosθ + a2 + r2 sin2 θ = a2 r2 − 2ar cos θ = 0 r = 2a cos θ π π The range of 0 ≤ θ ≤ traces out the top half of the circle, while − ≤ θ ≤ 0 traces out the bottom 2 2 half. Let’s graph this. y θ = π/4 r θ θ=0 (a,0) x Figure 5: r = 2a cos θ, −π/2 ≤ θ ≤ π/2. At θ = 0, r = 2a =⇒ x = 2a, y = 0 √ π π At θ = , r = 2a cos = a 2 4 4 The main issue is finding the range of θ tracing the circle once. In this case, θ = − θ = π 2 −π π <θ< . 2 2 π (down) 2 (up) π 3π <θ< . When θ = π, r = 2a cos π = 2a(−1) = −2a. The 2 2 π 3π radius points “backwards”. In the range < θ < , the same circle is traced out a second time. 2 2 Weird range (avoid this one): 3 Lecture 32 18.01 Fall 2006 r=f(θ) Figure 6: Using polar co-ordinates to find area of a generic function. Area in Polar Coordinates Since radius is a function of angle (r = f (θ)), we will integrate with respect to θ. The question is: what, exactly, should we integrate? � θ2 ?? dθ θ1 Let’s look at a very small slice of this region: rdθ dθ r Figure 7: Approximate slice of area in polar coordinates. This infinitesimal slice is approximately a right triangle. To find its area, we take: Area of slice ≈ 1 1 (base) (height) = r(r dθ) 2 2 So, � θ2 Total Area = θ1 4 1 2 r dθ 2 Lecture 32 18.01 Fall 2006 π π < θ < (the circle in Figure 5). 2 2 � π/2 � π/2 1 A = area = (2a cos θ)2 dθ = 2a2 cos2 θ dθ −π/2 2 −π/2 Example 3. r = 2a cos θ, and − 1 1 + cos 2θ, we can rewrite this as 2 2 � π/2 � A = area = (1 + cos 2θ) dθ = a2 Because cos2 θ = −π/2 = πa2 + π/2 dθ + a2 −π/2 � π/2 cos 2θ dθ −π/2 0 �π/2 1 1 ��� � � � sin 2θ� = πa2 + [sin π − sin(−π)] � 2 2�� −π/2 A = area = πa2 Example 4: Circle centered at the Origin. r=a Figure 8: Example 4: Circle centered at the origin x = r cos θ; y = r sin θ x + y 2 = r2 cos2 θ + r2 sin2 θ = r2 2 The circle is x2 + y 2 = a2 , so r = a and x = a cos θ; y = a sin θ � A= 0 2π 1 2 1 a dθ = a2 · 2π = πa2 . 2 2 5 Lecture 32 18.01 Fall 2006 Example 5: A Ray. In this case, θ = b. θ=b Figure 9: Example 5: The ray θ = b, 0 ≤ r < ∞. The range of r is 0 ≤ r < ∞; x = r cos b; y = r sin b. Example 6: Finding the Polar Formula, based on the Cartesian Formula y 1/sin θ 1 θ x Figure 10: Example 6: Cartesian Form to Polar Form Consider, in cartesian coordinates, the line y = 1. To find the polar coordinate equation, plug in y = r sin θ and x = r cos θ and solve for r. r sin θ = 1 =⇒ r = 1 sin θ 6 with 0<θ<π Lecture 32 18.01 Fall 2006 Example 7: Going back to (x, y) coordinates from r = f (θ). Start with 1 r= . 1 1 + 2 sin θ Hence, r+ Plug in r = � x2 + y 2 : � � x2 + y 2 = 1 − y 2 r sin θ = 1 2 y =1 2 � y �2 y2 x2 + y 2 = 1 − =1−y+ 2 4 x2 + y 2 + =⇒ Finally, 3y 2 +y =1 4 This is an equation for an ellipse, with the origin at one focus. x2 + Useful conversion formulas: r= � x2 + y 2 and θ = tan−1 �y� x Example 8: A Rose r = cos(2θ) The graph looks a bit like a flower: r<0 π/4 r>0 1 r>0 r<0 Figure 11: Example 8: Rose For the first “petal” − π π <θ< 4 4 Note: Next lecture is Lecture 34 as Lecture 33 is Exam 4. 7 -π/4