11.3 Notes.Pptx

11.3

Polar Functions

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall What you’ll learn about… …and why n Polar Coordinates Polar equations enable us to n Polar Curves define some interesting and n Slopes of Polar Curves important curves that would n Areas Enclosed by Polar Curves be difficult or impossible to n A Small Polar Gallery define in the form y = f(x).

2 2 2 x = r cosθ r = x + y

y y = r sinθ tanθ = x

Find rectangular coordinates for the point with the polar coordinate:

⎛ π ⎞ a. ⎜4, ⎟ ⎝ 2 ⎠ a. (0, 4) b.(−3,π ) b.(3, 0) ⎛ 5π ⎞ c.⎜16, ⎟ ⎝ 6 ⎠ c.(−8 3,8) ⎛ π ⎞ d.⎜− 2,− ⎟ d. −1,1 ⎝ 4 ⎠ ( )

F ind two different sets of polar coordinates for the point with the rectangular coordinate: A point has infinitely many sets a.(0, 4) of polar coordinates: b.(−3,3) a.(1,0),(1,2π ) c.(0,−4) ⎛ 3π ⎞ ⎛ π ⎞ b.⎜3 2, ⎟,⎜−3 2,− ⎟ d.(1, 3) ⎝ 4 ⎠ ⎝ 4 ⎠ ⎛ π ⎞ ⎛ 3π ⎞ c.⎜4,− ⎟,⎜4, ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ π ⎞ ⎛ 4π ⎞ d.⎜2, ⎟,⎜−2, ⎟ ⎝ 3 ⎠ ⎝ 3 ⎠

The polar graph of r = f (θ ) is the curve defined parametrically by: x = r cosθ = f (θ )cosθ y = r sinθ = f (θ )sinθ

F ind the slope of the rose curve r = 2sin 3θ at the point where θ = π / 6.

d 2sin 3θ sinθ dy dy / dθ ( ) = = dθ = − 3 π π d dx θ = dx / dθ θ = 6 6 (2sin 3θ cosθ ) π d θ = θ 6

x = 2sin3θ cosθ y = 2sin3θ sinθ dy 6cos3θ sinθ + 2sin3θ cosθ = dx 6cos3θ cosθ − 2sin3θ sinθ 3cos3θ sinθ +sin3θ cosθ = 3cos3θ cosθ −sin3θ sinθ ⎛ 1 ⎞ ⎛ 3 ⎞ 3(0)⎜ ⎟+1⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ = ⎛ 3 ⎞ ⎛ 1 ⎞ = − 3 0⎜ ⎟−1⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠

The area of the region between the origin and the curve r = f (θ ) for α ≤ θ ≤ β is

2 β 1β 1 β 1 β 1 2 AA r 2 rd2d f ( f)( d) .d . == ∫α∫ θ θ= =∫α ∫( θ( )θ θ) θ α2 2 2α 2

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F ind the area of the region in the plane enclosed by the cardioid r = 2(1+ cosθ ). 2π 2π 1 1 2 A = ∫ r2 dθ = ∫ ⋅ 4(1+ cosθ) dθ 0 2 0 2 2π = ∫ 2(1+ 2cosθ + cos2 θ)dθ 0 2π = ∫ (3+ 4cosθ + cos2θ)dθ 0 2π ⎡ sin2θ ⎤ = ⎢3θ + 4sinθ + ⎥ = 6π − 0 = 6π ⎣ 2 ⎦0

The area of the region between r1 (θ ) and r2 (θ ) for α ≤ θ ≤ β is

β 2 β 2 β β 1 1 2 β 1 1 2 β 1 1 AA = r (r ()θ)d dθ − r ( r)(θd) dθ = r 2 rr22 −dr2. dθ. = ∫α∫ ( 2( θ2 ) ) θ − ∫α ∫ ( 1 θ( 1 ) )θ = ∫α ∫( 2 − 21 ) θ1 α2 2 2α 2 2 α 2 ( )

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Find the area of the region lying between the inner and outer loops of the limacon r = 1− 2sinθ

The inner loop begins and ends when r = 0. € 1 1− 2sinθ = 0 ⇒ sinθ = 2 π 5π ⇒ θ = , 6 6

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Find the area of the region lying between the inner and outer loops of the limacon r = 1− 2sinθ The inner loop is traced as θ increases from π/6 to 5π/6. 1 5π/6 2 1 5π/6 A = (1− 2sinθ) dθ = 1− 4sinθ + 4sin2 θ dθ 2 ∫ π/6 2 ∫ π/6 ( ) 1 5π/6 ⎛ ⎛1− cos2θ ⎞⎞ 1 5π/6 € = 1− 4sinθ + 4 dθ = 1− 4sinθ + 2 1− cos2θ dθ ∫ π/6 ⎜ ⎜ ⎟⎟ ∫ π/6 ( ( )) 2 ⎝ ⎝ €2 ⎠⎠ 2 1 5π/6 = (3− 4sinθ − 2cos2θ)dθ 2 ∫ π/6 1 5π/6 = [3θ + 4cosθ −sin2θ] 2 π/6 ⎡⎛ ⎛ ⎞ ⎛ ⎞⎞ ⎛ ⎛ ⎞ ⎞⎤ 1 5π − 3 3 π 3 3 1 ⎡ ⎤ = ⎢⎜ + 4⎜ ⎟−⎜− ⎟⎟−⎜ + 4⎜ ⎟− ⎟⎥ = ⎣2π − 3 3⎦ 2⎣⎢⎝ 2 ⎝ 2 ⎠ ⎝ 2 ⎠⎠ ⎝ 2 ⎝ 2 ⎠ 2 ⎠⎦⎥ 2 3 3 = π − 2 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall Slide 10- 19 Area Between Polar Curves

Find the area of the region lying between the inner and outer loops of the limacon r = 1− 2sinθ The outer loop is traced as θ increases from 5π/6 to 13π/6. 1 13π/6 2 1 13π/6 A = (1− 2sinθ) dθ = 1− 4sinθ + 4sin2 θ dθ 2 ∫ 5π/6 2 ∫ 5π/6 ( ) 1 13π/6 ⎛ ⎛1− cos2θ ⎞⎞ 1 13π/6 € = 1− 4sinθ + 4 dθ = 1− 4sinθ + 2 1− cos2θ dθ ∫ 5 /6 ⎜ ⎜ ⎟⎟ ∫ 5 /6 ( ( )) 2 π ⎝ ⎝ 2€ ⎠⎠ 2 π 1 13π/6 = (3− 4sinθ − 2cos2θ)dθ 2 ∫ 5π/6 1 13π/6 = [3θ + 4cosθ −sin2θ] 2 5π/6 ⎡⎛ ⎛ ⎞ ⎛ ⎞⎞ ⎛ ⎛ ⎞ ⎛ ⎞⎞⎤ 1 13π 3 3 5π 3 3 1 ⎡ ⎤ = ⎢⎜ + 4⎜ ⎟−⎜ ⎟⎟−⎜ + 4⎜− ⎟−⎜− ⎟⎟⎥ = ⎣4π + 3 3⎦ 2⎣⎢⎝ 2 ⎝ 2 ⎠ ⎝ 2 ⎠⎠ ⎝ 2 ⎝ 2 ⎠ ⎝ 2 ⎠⎠⎦⎥ 2 3 3 = 2π + 2 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall Slide 10- 20 Area Between Polar Curves

Find the area of the region lying between the inner and outer loops of the limacon r = 1− 2sinθ

The area between the inner and outer loops will be € 3 3 ⎛ 3 3 ⎞ 2π + − ⎜ π − ⎟ = π + 3 3 ≈ 8.34 2 ⎜ 2 ⎟ ⎝ ⎠

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Y ou may use a graphing calculator to solve the following problems. 1. Which of the following is equal to the area of the region inside the polar curve r = 2cosθ and outside the polar curve r = cosθ? / 2 π / 2π 2 2 ((A)A) 3 3∫0 cocoss θθdθdθ ∫0 π 2 π 2 ((B)B) 3 3∫0 cocoss θθddθθ ∫0 33 π / 2π / 2 2 (C) cos 2 d (C) ∫0∫ cosθθ dθθ € 22 0 π / 2π / 2 € ((D)D) 3 3∫0 cocossθθdθdθ ∫0 (E) 3 π cπos d (E) 3∫0 cosθθ dθθ € ∫0

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Y ou may use a graphing calculator to solve the following problems. 1. Which of the following is equal to the area of the region inside the polar curve r = 2cosθ and outside the polar curve r = cosθ? π / 2π / 2 2 2 ((A)A) 3 3∫0 cocoss θθdθdθ ∫0 π 2 (B) 3 cπos 2 d (B) 3∫0 cosθθ dθθ ∫0 3 π / 2 (C) 3 πc/ 2os2 2 d (C) ∫0∫ cosθθ dθθ € 22 0 π / 2π / 2 € ((D)D) 3 3∫0 cocossθθdθdθ ∫0 (E) 3 π cπos d (E) 3∫0 cosθθ dθθ € ∫0

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2 . For what values of t does the curve given by the parametric 3 2 44 22 eequations quations x = t − t −−11 aandnd yy = t ++22tt −−88tt hhaveave aa vverticalertical ttangent?angent? (A) 0 only (B) 1 only € (C) 0 and 2/3 only (D) 0, 2/3, and 1 (E) no value

2 . For what values of t does the curve given by the parametric 3 2 4 2 eequations quations x = t 3 − t 2 −−11 aandnd yy = t 4 ++22tt2 −−88tt hhaveave aa vverticalertical ttangent?angent? (A) 0 only (B) 1 only € (C) 0 and 2/3 only (D) 0, 2/3, and 1 (E) no value

3 . The length of the path described by the parametric equations xx == t 2 and y = t from t = 0 to t = 4 is given by which integral?

4 4 ((A)A) ∫0 4t4+t +1d1tdt ∫0 4 2 (B) 2 4 t 2 1dt (B) 2∫0 t++1dt € ∫0 4 2 (C) ∫0 4 2t +1dt € (C) 2t 2 +1dt 4∫0 2 (D) ∫0 4t +1dt 4 2 € (D) 4 4t +1dt ∫0 2 (E) 2π ∫0 4t +1dt 4 € (E) 2 4t 2 +1dt ∫0 €

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3 . The length of the path described by the parametric equations x = t 2 and y = t from t = 0 to t = 4 is given by which integral?

4 4 ((A)A) ∫0 4t4+t +1d1tdt ∫0 4 2 (B) 2 4 t 2 1dt (B) 2∫0 t++1dt ∫0 4 2 (C) ∫0 4 2t +1dt € (C) 2t 2 +1dt 4∫0 2 (D) ∫0 4t +1dt 4 2 € (D) 4 4t +1dt ∫0 2 (E) 2π ∫0 4t +1dt 4 € (E) 2 4t 2 +1dt ∫0 €

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1 . Find the component form of a vector with magnitude 4 and direction angle 30! . 2. Find the area of a 30! sector of a circle of radius 6. 3. Find the area of a sector if a circle of radius 8 that has a central angle of π /8 radians. 4. Find the rectangular equation of a circle of radius 5 centered at the origin.

1 . Find the component form of a vector with magnitude 4 and direction angle 30! . 2 3,2 2. Find the area of a 30! sector of a circle of radius 6. 3π 3. Find the area of a sector if a circle of radius 8 that has a central angle of π /8 radians. 4π 4. Find the rectangular equation of a circle of radius 5 centered at the origin. x 2 + y 2 = 25

G iven x = 3cost, y = 5sin t, 0 ≤ t ≤ 2π. 5. Find dy / dx. 6. Find the slope of the curve at t = 2. 7. Find the points on the curve where the slope is zero. 8. Find the points on the curve where the slope is undefined. 9. Find the length of the curve from t = 0 to t = π.

G iven x = 3cost, y = 5sin t, 0 ≤ t ≤ 2π. 5. Find dy / dx. − 5/ 3 cot t 6. Find the slope of the curve at t = 2. − 5/ 3 cot 2 7. Find the points on the curve where the slope is zero. (0,5) and (0, − 5) 8. Find the points on the curve where the slope is undefined. (3,0) and ( − 3,0) 9. Find the length of the curve from t = 0 to t = π. ≈ 12.763