Question 1

For a  ≠
eq=  0, the polar curve  r=a(3−2cos⁡θ) r = a ( 3 - 2 \cos 	heta ) r=a(3−2cosθ)   is a

Accepted Answer

A)  Circle
B)  Cardioid
C)  Limaçon with inner loop
D)  Dimpled limaçon
E)  Convex limaçon

Question 2

The area of the surface generated by revolving the curve  x(t) =cos⁡t,y(t) =2+sin⁡tx ( t )  = \cos t , \quad y ( t )  = 2 + \sin tx(t) =cost,y(t) =2+sint  with  t∈[0,2π]t \in [ 0,2 \pi ]t∈[0,2π]  about the x-axis is

Accepted Answer

A)    4π24 \pi ^ { 2 }4π2 
B)    8π28 \pi ^ { 2 }8π2 
C)    12π212 \pi ^ { 2 }12π2 
D)    16π216 \pi ^ { 2 }16π2 
E)    2π22 \pi ^ { 2 }2π2

Question 3

For a  ≠
eq=  0, the polar curve  r=a(1−3sin⁡θ) r = a ( 1 - 3 \sin 	heta ) r=a(1−3sinθ)   is a

Accepted Answer

A)  Circle
B)  Cardioid
C)  Limaçon with inner loop
D)  Dimpled limaçon
E)  Convex limaçon

Question 4

For a ≠
eq=  0, the polar curve  r=asin⁡3θr = a \sin 3 	hetar=asin3θ  is a

Accepted Answer

A)  Straight line
B)  Circle
C)  Three-petal rose
D)  Cardioid
E)  Limaçon

Question 5

The polar equation that corresponds to the rectangular equation xy = 1 is

Accepted Answer

A)    r2=−tan⁡θsec⁡θr ^ { 2 } = - 	an 	heta \sec 	hetar2=−tanθsecθ 
B)    r2=tan⁡θsec⁡θr ^ { 2 } = 	an 	heta \sec 	hetar2=tanθsecθ 
C)    r2=csc⁡θsec⁡θr ^ { 2 } = \csc 	heta \sec 	hetar2=cscθsecθ 
D)    r2=−csc⁡θsec⁡θr ^ { 2 } = - \csc 	heta \sec 	hetar2=−cscθsecθ 
E)    r2=cos⁡θr ^ { 2 } = \cos 	hetar2=cosθ

Question 6

The rectangular equation of the conic section  r=11+sin⁡θr = \frac { 1 } { 1 + \sin 	heta }r=1+sinθ1​  is

Accepted Answer

A)    y2=1−2xy ^ { 2 } = 1 - 2 xy2=1−2x 
B)    y2=2x+1y ^ { 2 } = 2 x + 1y2=2x+1 
C)    x2=1−2yx ^ { 2 } = 1 - 2 yx2=1−2y 
D)    x2=2y+1x ^ { 2 } = 2 y + 1x2=2y+1 
E)    y2=1−xy ^ { 2 } = 1 - xy2=1−x

Question 7

The rectangular equation of the plane curve, with parametric equations  x(t) =3−t,y(t) =t+3x ( t )  = 3 - t , y ( t )  = t + 3x(t) =3−t,y(t) =t+3  with  t∈(−∞,∞) ,t \in ( - \infty , \infty ) ,t∈(−∞,∞) ,  is

Accepted Answer

A)    x−y=6x - y = 6x−y=6 
B)    x+y=−6x + y = - 6x+y=−6 
C)    x−y=−6x - y = - 6x−y=−6 
D)    x+y=6x + y = 6x+y=6 
E)    x+y=−3x + y = - 3x+y=−3

Question 8

The conic section  r=11+sin⁡θr = \frac { 1 } { 1 + \sin 	heta }r=1+sinθ1​  is

Accepted Answer

A)  A straight line
B)  A circle
C)  A parabola
D)  An ellipse
E)  A hyperbola

Question 9

For a > 0, the polar curve  r=−acos⁡2θr = - a \cos 2 	hetar=−acos2θ  is a

Accepted Answer

A)  Circle
B)  Two-petal rose
C)  Four-petal rose
D)  Limaçon
E)  Lemniscate

Question 10

The rectangular equation of the conic section  r=21+2cos⁡θr = \frac { 2 } { 1 + 2 \cos 	heta }r=1+2cosθ2​  is

Accepted Answer

A)    2y2=1−2x2 y ^ { 2 } = 1 - 2 x2y2=1−2x 
B)    x2−3y2−8x=4x ^ { 2 } - 3 y ^ { 2 } - 8 x = 4x2−3y2−8x=4 
C)    3x2−y2−8y=43 x ^ { 2 } - y ^ { 2 } - 8 y = 43x2−y2−8y=4 
D)    x2+3y2+8y=4x ^ { 2 } + 3 y ^ { 2 } + 8 y = 4x2+3y2+8y=4 
E)    3x2−y2−8x=−43 x ^ { 2 } - y ^ { 2 } - 8 x = - 43x2−y2−8x=−4

Question 11

The rectangular equation of the plane curve, with parametric equations  x(t) =4sin⁡t,y(t) =3cos⁡t,x ( t )  = 4 \sin t , y ( t )  = 3 \cos t,x(t) =4sint,y(t) =3cost,  is

Accepted Answer

A)    x29+y216=1\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 16 } = 19x2​+16y2​=1 
B)    x216+y29=1\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 116x2​+9y2​=1 
C)    x216−y29=1\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 116x2​−9y2​=1 
D)    −x216+y29=1- \frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1−16x2​+9y2​=1 
E)    16x2+9y2=116 x ^ { 2 } + 9 y ^ { 2 } = 116x2+9y2=1

Question 12

The rectangular equation of the plane curve, with parametric equations  x(t) =csc⁡t,y(t) =cot⁡t,x ( t )  = \csc t , y ( t )  = \cot t,x(t) =csct,y(t) =cott,  is

Accepted Answer

A)    y=11−x2y = \frac { 1 } { 1 - x ^ { 2 } }y=1−x21​ 
B)    y=1x2+1y = \frac { 1 } { x ^ { 2 } + 1 }y=x2+11​ 
C)    x2−y2=1x ^ { 2 } - y ^ { 2 } = 1x2−y2=1 
D)    x2+y2=1x ^ { 2 } + y ^ { 2 } = 1x2+y2=1 
E)    y=−1x2+1y = - \frac { 1 } { x ^ { 2 } + 1 }y=−x2+11​

Question 13

The rectangular equation of the plane curve, with parametric equations  x(t) =2t+5,y(t) =4t−7,x ( t )  = 2 t + 5 , y ( t )  = 4 t - 7,x(t) =2t+5,y(t) =4t−7,  is

Accepted Answer

A)    y=4x−3y = 4 x - 3y=4x−3 
B)    y=2x−7y = 2 x - 7y=2x−7 
C)    y=2x−17y = 2 x - 17y=2x−17 
D)    y=4x+3y = 4 x + 3y=4x+3 
E)    y=2x−2y = 2 x - 2y=2x−2

Question 14

The area of the surface generated by revolving the curve  x(t) =t,y=3t2x ( t )  = t , \quad y = 3 t ^ { 2 }x(t) =t,y=3t2  with  t∈[0,2]t \in [ 0,2 ]t∈[0,2]  about the y-axis is

Accepted Answer

A)    π(145145−1) 60\frac { \pi ( 145 \sqrt { 145 } - 1 )  } { 60 }60π(145145c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−1) ​ 
B)    π(145145−1) 54\frac { \pi ( 145 \sqrt { 145 } - 1 )  } { 54 }54π(145145c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−1) ​ 
C)    π(145145−1) 48\frac { \pi ( 145 \sqrt { 145 } - 1 )  } { 48 }48π(145145c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−1) ​ 
D)    π(145145−1) 42\frac { \pi ( 145 \sqrt { 145 } - 1 )  } { 42 }42π(145145c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−1) ​ 
E)    π(145145−1) 36\frac { \pi ( 145 \sqrt { 145 } - 1 )  } { 36 }36π(145145c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−1) ​

Question 15

If A is the area of the intersection of the regions inside  r=−2cos⁡θr = - 2 \cos 	hetar=−2cosθ  and outside  r=1,r = 1,r=1,  then A is

Accepted Answer

A)    6π−166 \pi - 166π−16 
B)    33−π3 \sqrt { 3 } - \pi33c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−π 
C)    4π−336\frac { 4 \pi - 3 \sqrt { 3 } } { 6 }64π−33c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​ 
D)    π+1−3\pi + 1 - \sqrt { 3 }π+1−3c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​ 
E)    8−π4\frac { 8 - \pi } { 4 }48−π​

Question 16

The rectangular equation that corresponds to the polar equation  r2=cos⁡θr ^ { 2 } = \cos 	hetar2=cosθ  is

Accepted Answer

A)    (x2+y2) 32=x\left( x ^ { 2 } + y ^ { 2 } \right)  ^ { \frac { 3 } { 2 } } = x(x2+y2) 23​=x 
B)    (x2+y2) 32=y\left( x ^ { 2 } + y ^ { 2 } \right)  ^ { \frac { 3 } { 2 } } = y(x2+y2) 23​=y 
C)    (x2−y2) 32=x\left( x ^ { 2 } - y ^ { 2 } \right)  ^ { \frac { 3 } { 2 } } = x(x2−y2) 23​=x 
D)    (x2−y2) 32=y\left( x ^ { 2 } - y ^ { 2 } \right)  ^ { \frac { 3 } { 2 } } = y(x2−y2) 23​=y 
E)    (−x2+y2) 32=x\left( - x ^ { 2 } + y ^ { 2 } \right)  ^ { \frac { 3 } { 2 } } = x(−x2+y2) 23​=x

Question 17

The polar equation that corresponds to the rectangular equation  2x+y=42 x + y = 42x+y=4  is

Accepted Answer

A)    r=42cos⁡θ+sin⁡θr = \frac { 4 } { 2 \cos 	heta + \sin 	heta }r=2cosθ+sinθ4​ 
B)    r=42cos⁡θ−sin⁡θr = \frac { 4 } { 2 \cos 	heta - \sin 	heta }r=2cosθ−sinθ4​ 
C)    r=4cos⁡θ+2sin⁡θr = \frac { 4 } { \cos 	heta + 2 \sin 	heta }r=cosθ+2sinθ4​ 
D)    r=4cos⁡θ−2sin⁡θr = \frac { 4 } { \cos 	heta - 2 \sin 	heta }r=cosθ−2sinθ4​ 
E)    r=−42cos⁡θ+sin⁡θr = \frac { - 4 } { 2 \cos 	heta + \sin 	heta }r=2cosθ+sinθ−4​

Question 18

Let  x(t) =2t3,y(t) =3t2x ( t )  = 2 t ^ { 3 } , y ( t )  = 3 t ^ { 2 }x(t) =2t3,y(t) =3t2  be the parametric equations of a curve. Then  dydx\frac { d y } { d x }dxdy​  is

Accepted Answer

A)    −1t- \frac { 1 } { t }−t1​ 
B)    2t\frac { 2 } { t }t2​ 
C)    t22\frac { t ^ { 2 } } { 2 }2t2​ 
D)    1t\frac { 1 } { t }t1​ 
E)    −t2- \frac { t } { 2 }−2t​

Question 19

If A is the area of the region bounded by  r=2+2cos⁡θ,r = 2 + 2 \cos 	heta,r=2+2cosθ,  then A is

Accepted Answer

A)    8π8 \pi8π 
B)    6π6 \pi6π 
C)    4π4 \pi4π 
D)    2π2 \pi2π 
E)    π\piπ

Question 20

The rectangular equation of the plane curve, with parametric equations  x(t) =sin⁡3t,y(t) =cos⁡23tx ( t )  = \sin 3 t , y ( t )  = \cos ^ { 2 } 3 tx(t) =sin3t,y(t) =cos23t  with  t∈[0,π6],t \in \left[ 0 , \frac { \pi } { 6 } \right],t∈[0,6π​],  is

Accepted Answer

A)    y=1−x2y = 1 - x ^ { 2 }y=1−x2 
B)    y=x2−1y = x ^ { 2 } - 1y=x2−1 
C)    y=x2+1y = x ^ { 2 } + 1y=x2+1 
D)    y=−1−x2y = - \sqrt { 1 - x ^ { 2 } }y=−1−x2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​ 
E)    y=1−x2y = \sqrt { 1 - x ^ { 2 } }y=1−x2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​

Exam 10: Parametric Equations; Polar Equations

The area of the surface generated by revolving the curve x(t) =t,y=3t2x ( t ) = t , \quad y = 3 t ^ { 2 }x(t) =t,y=3t2 with t∈[0,2]t \in [ 0,2 ]t∈[0,2] about the y-axis is

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The rectangular equation of the plane curve, with parametric equations x(t) =csc⁡t,y(t) =cot⁡t,x ( t ) = \csc t , y ( t ) = \cot t,x(t) =csct,y(t) =cott, is

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For a > 0, the polar curve r=−acos⁡2θr = - a \cos 2 \thetar=−acos2θ is a

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Let x(t) =2t3,y(t) =3t2x ( t ) = 2 t ^ { 3 } , y ( t ) = 3 t ^ { 2 }x(t) =2t3,y(t) =3t2 be the parametric equations of a curve. Then dydx\frac { d y } { d x }dxdy​ is

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The area of the surface generated by revolving the curve x(t) =cos⁡t,y(t) =2+sin⁡tx ( t ) = \cos t , \quad y ( t ) = 2 + \sin tx(t) =cost,y(t) =2+sint with t∈[0,2π]t \in [ 0,2 \pi ]t∈[0,2π] about the x-axis is

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For a ≠\neq= 0, the polar curve r=a(3−2cos⁡θ) r = a ( 3 - 2 \cos \theta ) r=a(3−2cosθ) is a

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For a ≠\neq= 0, the polar curve r=a(1−3sin⁡θ) r = a ( 1 - 3 \sin \theta ) r=a(1−3sinθ) is a

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The conic section r=11+sin⁡θr = \frac { 1 } { 1 + \sin \theta }r=1+sinθ1​ is

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The rectangular equation of the plane curve, with parametric equations x(t) =2t+5,y(t) =4t−7,x ( t ) = 2 t + 5 , y ( t ) = 4 t - 7,x(t) =2t+5,y(t) =4t−7, is

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For a ≠\neq= 0, the polar curve r=asin⁡3θr = a \sin 3 \thetar=asin3θ is a

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The rectangular equation that corresponds to the polar equation r2=cos⁡θr ^ { 2 } = \cos \thetar2=cosθ is

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If A is the area of the intersection of the regions inside r=−2cos⁡θr = - 2 \cos \thetar=−2cosθ and outside r=1,r = 1,r=1, then A is

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The polar equation that corresponds to the rectangular equation 2x+y=42 x + y = 42x+y=4 is

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The rectangular equation of the conic section r=11+sin⁡θr = \frac { 1 } { 1 + \sin \theta }r=1+sinθ1​ is

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The polar equation that corresponds to the rectangular equation xy = 1 is

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If A is the area of the region bounded by r=2+2cos⁡θ,r = 2 + 2 \cos \theta,r=2+2cosθ, then A is

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The rectangular equation of the conic section r=21+2cos⁡θr = \frac { 2 } { 1 + 2 \cos \theta }r=1+2cosθ2​ is

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The rectangular equation of the plane curve, with parametric equations x(t) =sin⁡3t,y(t) =cos⁡23tx ( t ) = \sin 3 t , y ( t ) = \cos ^ { 2 } 3 tx(t) =sin3t,y(t) =cos23t with t∈[0,π6],t \in \left[ 0 , \frac { \pi } { 6 } \right],t∈[0,6π​], is

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The rectangular equation of the plane curve, with parametric equations x(t) =3−t,y(t) =t+3x ( t ) = 3 - t , y ( t ) = t + 3x(t) =3−t,y(t) =t+3 with t∈(−∞,∞) ,t \in ( - \infty , \infty ) ,t∈(−∞,∞) , is

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The rectangular equation of the plane curve, with parametric equations x(t) =4sin⁡t,y(t) =3cos⁡t,x ( t ) = 4 \sin t , y ( t ) = 3 \cos t,x(t) =4sint,y(t) =3cost, is

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The area of the surface generated by revolving the curve $x ( t ) = t , \quad y = 3 t ^ { 2 }$ with $t \in [ 0,2 ]$ about the y-axis is

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The rectangular equation of the plane curve, with parametric equations $x ( t ) = \csc t , y ( t ) = \cot t,$ is

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For a > 0, the polar curve $r = - a \cos 2 \theta$ is a

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Let $x ( t ) = 2 t ^ { 3 } , y ( t ) = 3 t ^ { 2 }$ be the parametric equations of a curve. Then $\frac { d y } { d x }$ is

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The area of the surface generated by revolving the curve $x ( t ) = \cos t , \quad y ( t ) = 2 + \sin t$ with $t \in [ 0,2 \pi ]$ about the x-axis is

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For a $\neq$ 0, the polar curve $r = a ( 3 - 2 \cos \theta )$ is a

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For a $\neq$ 0, the polar curve $r = a ( 1 - 3 \sin \theta )$ is a

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The conic section $r = \frac { 1 } { 1 + \sin \theta }$ is

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The rectangular equation of the plane curve, with parametric equations $x ( t ) = 2 t + 5 , y ( t ) = 4 t - 7,$ is

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For a $\neq$ 0, the polar curve $r = a \sin 3 \theta$ is a

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The rectangular equation that corresponds to the polar equation $r ^ { 2 } = \cos \theta$ is

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If A is the area of the intersection of the regions inside $r = - 2 \cos \theta$ and outside $r = 1,$ then A is

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The polar equation that corresponds to the rectangular equation $2 x + y = 4$ is

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The rectangular equation of the conic section $r = \frac { 1 } { 1 + \sin \theta }$ is

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If A is the area of the region bounded by $r = 2 + 2 \cos \theta,$ then A is

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The rectangular equation of the conic section $r = \frac { 2 } { 1 + 2 \cos \theta }$ is

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The rectangular equation of the plane curve, with parametric equations $x ( t ) = \sin 3 t , y ( t ) = \cos ^ { 2 } 3 t$ with $t \in \left[ 0 , \frac { \pi } { 6 } \right],$ is

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The rectangular equation of the plane curve, with parametric equations $x ( t ) = 3 - t , y ( t ) = t + 3$ with $t \in ( - \infty , \infty ) ,$ is

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The rectangular equation of the plane curve, with parametric equations $x ( t ) = 4 \sin t , y ( t ) = 3 \cos t,$ is

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