Study Polar Function Graphs in AP Precalculus with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: What symmetry test indicates a polar graph is symmetric about the line θ = π 2 \theta=\frac{\pi}{2} θ = 2 π ? Answer: Replace θ \theta θ with π − θ \pi-\theta π − θ ; equation unchanged. Reflection across y y y -axis supplements angle to π \pi π .
Flashcard 2: What is the polar equation of the line through the origin making angle α \alpha α with the positive x x x -axis? Answer: θ = α \theta=\alpha θ = α . Constant angle creates a ray from the origin.
Flashcard 3: What equation relates r r r , x x x , and y y y for polar graphs in the plane? Answer: r 2 = x 2 + y 2 r^2=x^2+y^2 r 2 = x 2 + y 2 . Squaring both sides of r = x 2 + y 2 r=\sqrt{x^2+y^2} r = x 2 + y 2 gives this identity.
Flashcard 4: Convert the polar point ( r , θ ) = ( 2 , π 3 ) (r,\theta)=(2,\frac{\pi}{3}) ( r , θ ) = ( 2 , 3 π ) to Cartesian coordinates. Answer: ( x , y ) = ( 1 , 3 ) (x,y)=(1,\sqrt{3}) ( x , y ) = ( 1 , 3 ) . x = 2 cos ( π 3 ) = 1 x=2\cos(\frac{\pi}{3})=1 x = 2 cos ( 3 π ) = 1 , y = 2 sin ( π 3 ) = 3 y=2\sin(\frac{\pi}{3})=\sqrt{3} y = 2 sin ( 3 π ) = 3 .
Flashcard 5: Convert the Cartesian point ( x , y ) = ( − 3 , 1 ) (x,y)=(-\sqrt{3},1) ( x , y ) = ( − 3 , 1 ) to polar with r > 0 r>0 r > 0 and 0 ≤ θ < 2 π 0\le\theta<2\pi 0 ≤ θ < 2 π . Answer: ( r , θ ) = ( 2 , 5 π 6 ) (r,\theta)=(2,\frac{5\pi}{6}) ( r , θ ) = ( 2 , 6 5 π ) . r = 3 + 1 = 2 r=\sqrt{3+1}=2 r = 3 + 1 = 2 ; θ \theta θ in Q2 where tan θ = − 1 3 \tan\theta=-\frac{1}{\sqrt{3}} tan θ = − 3 1 .
Flashcard 6: What is the equation relating r r r , x x x , and y y y for polar coordinates? Answer: r 2 = x 2 + y 2 r^2=x^2+y^2 r 2 = x 2 + y 2 . Pythagorean theorem relates radius to Cartesian coordinates.
Flashcard 7: For a rose r = a sin ( n θ ) r=a\sin(n\theta) r = a sin ( n θ ) with even n n n , how many petals are graphed? Answer: 2 n 2n 2 n petals. Even n n n requires full period [ 0 , 2 π ] [0,2\pi] [ 0 , 2 π ] to trace all petals.
Flashcard 8: What symmetry test indicates a polar graph is symmetric about the polar axis? Answer: Replace θ \theta θ with − θ -\theta − θ ; equation unchanged. Reflection across x x x -axis negates angle.
Flashcard 9: What is the polar axis in the polar coordinate system? Answer: The positive x x x -axis. The reference line from which angles θ \theta θ are measured.
Flashcard 10: How many petals does the rose curve r = a cos ( n θ ) r=a\cos(n\theta) r = a cos ( n θ ) have when n n n is even? Answer: 2 n 2n 2 n petals. Even n n n requires two periods to trace all petals.
Flashcard 11: What is the polar-to-Cartesian conversion formula for x x x and y y y ? Answer: x = r cos ( θ ) , y = r sin ( θ ) x=r\cos(\theta),\ y=r\sin(\theta) x = r cos ( θ ) , y = r sin ( θ ) . Uses trigonometric definitions on the unit circle extended to radius r r r .
Flashcard 12: What circle does r = 2 cos ( θ ) r=2\cos(\theta) r = 2 cos ( θ ) represent (center and radius)? Answer: Center ( 1 , 0 ) (1,0) ( 1 , 0 ) , radius 1 1 1 . Complete the square: ( x − 1 ) 2 + y 2 = 1 (x-1)^2+y^2=1 ( x − 1 ) 2 + y 2 = 1 from x 2 + y 2 = 2 x x^2+y^2=2x x 2 + y 2 = 2 x .
Flashcard 13: What is the symmetry test for a polar equation using the pole (origin)? Answer: Replace r r r with − r -r − r or θ \theta θ with θ + π \theta+\pi θ + π ; same implies symmetry. Points symmetric about pole are π \pi π radians apart.
Flashcard 14: Identify an equivalent coordinate to ( r , θ ) (r,\theta) ( r , θ ) using a 2 π 2\pi 2 π angle change. Answer: ( r , θ + 2 π k ) (r,\theta+2\pi k) ( r , θ + 2 πk ) for any integer k k k . Adding multiples of 2 π 2\pi 2 π to θ \theta θ gives the same point.
Flashcard 15: Find the polar intercept angles where r = a cos θ r=a\cos\theta r = a cos θ crosses the pole (origin). Answer: θ = π 2 , 3 π 2 \theta=\frac{\pi}{2},\ \frac{3\pi}{2} θ = 2 π , 2 3 π . r = 0 r=0 r = 0 when cos θ = 0 \cos\theta=0 cos θ = 0 , at odd multiples of π 2 \frac{\pi}{2} 2 π .
Flashcard 16: Find the polar intercept angles where r = a sin θ r=a\sin\theta r = a sin θ crosses the pole (origin). Answer: θ = 0 , π \theta=0,\ \pi θ = 0 , π . r = 0 r=0 r = 0 when sin θ = 0 \sin\theta=0 sin θ = 0 , at multiples of π \pi π .
Flashcard 17: Identify an equivalent coordinate to ( r , θ ) (r,\theta) ( r , θ ) using a negative radius. Answer: ( − r , θ + π ) (-r,\theta+\pi) ( − r , θ + π ) . Negative r r r reflects the point through the pole.
Flashcard 18: Identify the equivalent polar coordinate of ( r , θ ) = ( 3 , π 6 ) (r,\theta)=(3,\frac{\pi}{6}) ( r , θ ) = ( 3 , 6 π ) using a negative radius. Answer: ( − 3 , 7 π 6 ) (-3,\frac{7\pi}{6}) ( − 3 , 6 7 π ) . Add π \pi π to angle and negate radius for equivalent point.
Flashcard 19: What symmetry test indicates a polar graph is symmetric about the pole (origin)? Answer: Replace r r r with − r -r − r (or θ \theta θ with θ + π \theta+\pi θ + π ); unchanged. Point and its opposite have same location.
Flashcard 20: Identify the period of r = cos θ r=\cos\theta r = cos θ and r = sin θ r=\sin\theta r = sin θ as polar functions. Answer: Period = 2 π =2\pi = 2 π . Basic trig functions complete one cycle in 2 π 2\pi 2 π .
Flashcard 21: What is the conversion from polar to Cartesian coordinates for a point ( r , θ ) (r,\theta) ( r , θ ) ? Answer: x = r cos θ , y = r sin θ x=r\cos\theta,\ y=r\sin\theta x = r cos θ , y = r sin θ . Uses trigonometric definitions where x x x is horizontal and y y y is vertical projection.
Flashcard 22: Which condition determines a limacon r = a + b cos ( θ ) r=a+b\cos(\theta) r = a + b cos ( θ ) has an inner loop? Answer: Inner loop occurs when ∣ a ∣ < ∣ b ∣ |a|<|b| ∣ a ∣ < ∣ b ∣ . Inner loop forms when r r r becomes negative for some θ \theta θ .
Flashcard 23: What circle does r = 4 sin ( θ ) r=4\sin(\theta) r = 4 sin ( θ ) represent (center and radius)? Answer: Center ( 0 , 2 ) (0,2) ( 0 , 2 ) , radius 2 2 2 . Complete the square: x 2 + ( y − 2 ) 2 = 4 x^2+(y-2)^2=4 x 2 + ( y − 2 ) 2 = 4 from x 2 + y 2 = 4 y x^2+y^2=4y x 2 + y 2 = 4 y .
Flashcard 24: What is the symmetry test for a polar equation using the x x x -axis (polar axis)? Answer: Replace θ \theta θ with − θ -\theta − θ ; same equation implies symmetry. Points symmetric about polar axis have angles θ \theta θ and − θ -\theta − θ .
Flashcard 25: Identify the period of r = cos ( n θ ) r=\cos(n\theta) r = cos ( n θ ) and r = sin ( n θ ) r=\sin(n\theta) r = sin ( n θ ) for integer n ≠ 0 n\neq 0 n = 0 . Answer: Period = 2 π ∣ n ∣ =\frac{2\pi}{|n|} = ∣ n ∣ 2 π . Coefficient n n n compresses period by factor of ∣ n ∣ |n| ∣ n ∣ .
Flashcard 26: What is the polar equation of the y y y -axis? Answer: θ = π 2 \theta=\frac{\pi}{2} θ = 2 π . Vertical line through origin has angle 90 ° 90° 90° from positive x x x -axis.
Flashcard 27: What is the polar equation of the x x x -axis? Answer: θ = 0 \theta=0 θ = 0 . Horizontal line through origin has angle 0 ° 0° 0° from positive x x x -axis.
Flashcard 28: What is the graph type of r = a r=a r = a for a > 0 a>0 a > 0 in polar coordinates? Answer: Circle centered at the pole with radius a a a . Constant r r r means all points are equidistant from pole.
Flashcard 29: What is the maximum value of r r r for r = a cos θ r=a\cos\theta r = a cos θ (or r = a sin θ r=a\sin\theta r = a sin θ ) with a > 0 a>0 a > 0 ? Answer: r max = a r_{\max}=a r m a x = a . Cosine/sine maximum is 1, scaled by coefficient a a a .
Flashcard 30: Identify the polar equation of a circle centered at the origin with radius a > 0 a>0 a > 0 . Answer: r = a r=a r = a . Constant radius from origin defines a circle.
Flashcard 31: What is the graph type of r = a ( 1 + cos ( θ ) ) r=a(1+\cos(\theta)) r = a ( 1 + cos ( θ )) for a > 0 a>0 a > 0 ? Answer: Cardioid (a limacon with a cusp). Heart-shaped curve with cusp at pole when θ = π \theta=\pi θ = π .
Flashcard 32: What is the symmetry test for a polar equation using the y y y -axis (θ = π 2 \theta=\frac{\pi}{2} θ = 2 π line)? Answer: Replace θ \theta θ with π − θ \pi-\theta π − θ ; same equation implies symmetry. Points symmetric about y y y -axis have supplementary angles from polar axis.
Flashcard 33: What is the Cartesian form of the polar equation r = 2 sin ( θ ) r=2\sin(\theta) r = 2 sin ( θ ) ? Answer: x 2 + y 2 = 2 y x^2+y^2=2y x 2 + y 2 = 2 y . Substitute r sin ( θ ) = y r\sin(\theta)=y r sin ( θ ) = y and r 2 = x 2 + y 2 r^2=x^2+y^2 r 2 = x 2 + y 2 to convert.
Flashcard 34: What is the Cartesian form of the polar equation r = 2 cos ( θ ) r=2\cos(\theta) r = 2 cos ( θ ) ? Answer: x 2 + y 2 = 2 x x^2+y^2=2x x 2 + y 2 = 2 x . Substitute r cos ( θ ) = x r\cos(\theta)=x r cos ( θ ) = x and r 2 = x 2 + y 2 r^2=x^2+y^2 r 2 = x 2 + y 2 to convert.
Flashcard 35: For a rose r = a cos ( n θ ) r=a\cos(n\theta) r = a cos ( n θ ) with odd n n n , how many petals are graphed? Answer: n n n petals. Odd n n n traces the complete rose in one period [ 0 , π ] [0,\pi] [ 0 , π ] .
Flashcard 36: What is the conversion from Cartesian to polar coordinates for a point ( x , y ) (x,y) ( x , y ) ? Answer: r = x 2 + y 2 , tan θ = y x r=\sqrt{x^2+y^2},\ \tan\theta=\frac{y}{x} r = x 2 + y 2 , tan θ = x y . Distance formula gives r r r ; angle found from slope ratio.
Flashcard 37: What is the pole in the polar coordinate system? Answer: The origin ( 0 , 0 ) (0,0) ( 0 , 0 ) . The fixed point from which all distances r r r are measured.
Flashcard 38: How many petals does the rose curve r = a cos ( n θ ) r=a\cos(n\theta) r = a cos ( n θ ) have when n n n is odd? Answer: n n n petals. Odd n n n traces all petals in one period.
Flashcard 39: What is the Cartesian-to-polar conversion formula for r r r in terms of x x x and y y y ? Answer: r = x 2 + y 2 r=\sqrt{x^2+y^2} r = x 2 + y 2 . Apply the Pythagorean theorem to find distance from origin.
Flashcard 40: For a rose r = a cos ( n θ ) r=a\cos(n\theta) r = a cos ( n θ ) , what is the maximum radius (petal length)? Answer: Maximum r = ∣ a ∣ r=|a| r = ∣ a ∣ . Petals reach farthest when cos ( n θ ) = ± 1 \cos(n\theta)=\pm 1 cos ( n θ ) = ± 1 .