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Is $r^2-4rcos(theta)=14$ an equation of a circle or cylinder?
Is $r=2cos(theta)$ a one-petal polar function?General Cartesian/Rectangular Equation for Polar Rose ($r=sin(ktheta)$)Calculate Area of SurfaceSketch $r=cos(5 theta)$? $r$ as a function of $theta$ in cartesian coordinatesTo prove $(sintheta + csctheta)^2 + (costheta +sectheta)^2 ge 9$How to convert $r = 2scos (theta+t)$ into Cartesian coordinates?How can this be the surface area of an intersection of cone and cylinder?Question on polar coordinates and cartesian coordinatesHow to sketch $lnleft(sqrtx^2+y^2right)=-arctanleft(fracyxright)$ in polar coordinates?Show that $ sqrt2 + sqrt2 + sqrt2 + 2 cos 8theta = 2 cos theta$
$begingroup$
A question asks to identify the surface of the polar equation
beginequationtag1
r^2-4rcos(theta)=14.
endequation
I converted $(1)$ into Cartesian coordinates:
beginequationtag2
(x-2)^2+y^2 = (3sqrt2)^2.
endequation
I would have thought $(2)$ represents the equation of a cylinder, because the question asks to identify a surface.
algebra-precalculus graphing-functions surfaces
asked Mar 14 at 16:02
GurjinderGurjinder
552417
$endgroup$
add a comment |
$begingroup$
A question asks to identify the surface of the polar equation
beginequationtag1
r^2-4rcos(theta)=14.
endequation
I converted $(1)$ into Cartesian coordinates:
beginequationtag2
(x-2)^2+y^2 = (3sqrt2)^2.
endequation
I would have thought $(2)$ represents the equation of a cylinder, because the question asks to identify a surface.
algebra-precalculus graphing-functions surfaces
asked Mar 14 at 16:02
GurjinderGurjinder
552417
$endgroup$
$begingroup$
in how many dimensions?
$endgroup$
– J. W. Tanner
Mar 14 at 16:07
1
$begingroup$
In $mathbbR^2$ it's a circle. In $mathbbR^3$ it's a cylinder (using cylindrical coordinates $(x,y,z)=(rcostheta,rsintheta,z)$).
$endgroup$
– Barry Cipra
Mar 14 at 16:08
$begingroup$
@BarryCipra Okay. So is the answer dependent on the context of the question?
$endgroup$
– Gurjinder
Mar 14 at 16:11
$begingroup$
@Gurjinder, yes, exactly.
$endgroup$
– Barry Cipra
Mar 14 at 16:12
$begingroup$
@BarryCipra Cool, thanks.
$endgroup$
– Gurjinder
Mar 14 at 16:13
add a comment |
$begingroup$
A question asks to identify the surface of the polar equation
beginequationtag1
r^2-4rcos(theta)=14.
endequation
I converted $(1)$ into Cartesian coordinates:
beginequationtag2
(x-2)^2+y^2 = (3sqrt2)^2.
endequation
I would have thought $(2)$ represents the equation of a cylinder, because the question asks to identify a surface.
algebra-precalculus graphing-functions surfaces
asked Mar 14 at 16:02
GurjinderGurjinder
552417
$endgroup$
A question asks to identify the surface of the polar equation
beginequationtag1
r^2-4rcos(theta)=14.
endequation
I converted $(1)$ into Cartesian coordinates:
beginequationtag2
(x-2)^2+y^2 = (3sqrt2)^2.
endequation
I would have thought $(2)$ represents the equation of a cylinder, because the question asks to identify a surface.
algebra-precalculus graphing-functions surfaces
algebra-precalculus graphing-functions surfaces
asked Mar 14 at 16:02
GurjinderGurjinder
552417
asked Mar 14 at 16:02
GurjinderGurjinder
552417
asked Mar 14 at 16:02
GurjinderGurjinder
552417
asked Mar 14 at 16:02
GurjinderGurjinder
552417
asked Mar 14 at 16:02
GurjinderGurjinder
552417
552417
$begingroup$
in how many dimensions?
$endgroup$
– J. W. Tanner
Mar 14 at 16:07
1
$begingroup$
In $mathbbR^2$ it's a circle. In $mathbbR^3$ it's a cylinder (using cylindrical coordinates $(x,y,z)=(rcostheta,rsintheta,z)$).
$endgroup$
– Barry Cipra
Mar 14 at 16:08
$begingroup$
@BarryCipra Okay. So is the answer dependent on the context of the question?
$endgroup$
– Gurjinder
Mar 14 at 16:11
$begingroup$
@Gurjinder, yes, exactly.
$endgroup$
– Barry Cipra
Mar 14 at 16:12
$begingroup$
@BarryCipra Cool, thanks.
$endgroup$
– Gurjinder
Mar 14 at 16:13
add a comment |
$begingroup$
in how many dimensions?
$endgroup$
– J. W. Tanner
Mar 14 at 16:07
1
$begingroup$
In $mathbbR^2$ it's a circle. In $mathbbR^3$ it's a cylinder (using cylindrical coordinates $(x,y,z)=(rcostheta,rsintheta,z)$).
$endgroup$
– Barry Cipra
Mar 14 at 16:08
$begingroup$
@BarryCipra Okay. So is the answer dependent on the context of the question?
$endgroup$
– Gurjinder
Mar 14 at 16:11
$begingroup$
@Gurjinder, yes, exactly.
$endgroup$
– Barry Cipra
Mar 14 at 16:12
$begingroup$
@BarryCipra Cool, thanks.
$endgroup$
– Gurjinder
Mar 14 at 16:13
$begingroup$
in how many dimensions?
$endgroup$
– J. W. Tanner
Mar 14 at 16:07
$begingroup$
in how many dimensions?
$endgroup$
– J. W. Tanner
Mar 14 at 16:07
1
1
$begingroup$
In $mathbbR^2$ it's a circle. In $mathbbR^3$ it's a cylinder (using cylindrical coordinates $(x,y,z)=(rcostheta,rsintheta,z)$).
$endgroup$
– Barry Cipra
Mar 14 at 16:08
$begingroup$
In $mathbbR^2$ it's a circle. In $mathbbR^3$ it's a cylinder (using cylindrical coordinates $(x,y,z)=(rcostheta,rsintheta,z)$).
$endgroup$
– Barry Cipra
Mar 14 at 16:08
$begingroup$
@BarryCipra Okay. So is the answer dependent on the context of the question?
$endgroup$
– Gurjinder
Mar 14 at 16:11
$begingroup$
@BarryCipra Okay. So is the answer dependent on the context of the question?
$endgroup$
– Gurjinder
Mar 14 at 16:11
$begingroup$
@Gurjinder, yes, exactly.
$endgroup$
– Barry Cipra
Mar 14 at 16:12
$begingroup$
@Gurjinder, yes, exactly.
$endgroup$
– Barry Cipra
Mar 14 at 16:12
$begingroup$
@BarryCipra Cool, thanks.
$endgroup$
– Gurjinder
Mar 14 at 16:13
$begingroup$
@BarryCipra Cool, thanks.
$endgroup$
– Gurjinder
Mar 14 at 16:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is of course a circle. General formula for the circle is $$(x-p)^2+(y-q)^2 =r^2$$
where $O=(p,q)$ is a center and $r$ radius. At your case $O=(2,0)$ and $r=3sqrt2$
answered Mar 14 at 16:05
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
2
$begingroup$
I think we should recognize that this is also the equation of a cylinder whose axis of symmetry is parallel to the $z$ axis
$endgroup$
– WW1
Mar 14 at 16:08
add a comment |
$begingroup$
$$beginequationtag1
r^2-4rcos(theta)=14.
endequation$$
$$x^2+y^2-4x=14$$
$$(x-2)^2+y^2=18$$
$$(x-2)^2+y^2=(3sqrt2)^2$$
answered Mar 14 at 16:09
E.H.EE.H.E
15.9k11968
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is of course a circle. General formula for the circle is $$(x-p)^2+(y-q)^2 =r^2$$
where $O=(p,q)$ is a center and $r$ radius. At your case $O=(2,0)$ and $r=3sqrt2$
answered Mar 14 at 16:05
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
2
$begingroup$
I think we should recognize that this is also the equation of a cylinder whose axis of symmetry is parallel to the $z$ axis
$endgroup$
– WW1
Mar 14 at 16:08
add a comment |
$begingroup$
This is of course a circle. General formula for the circle is $$(x-p)^2+(y-q)^2 =r^2$$
where $O=(p,q)$ is a center and $r$ radius. At your case $O=(2,0)$ and $r=3sqrt2$
answered Mar 14 at 16:05
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
2
$begingroup$
I think we should recognize that this is also the equation of a cylinder whose axis of symmetry is parallel to the $z$ axis
$endgroup$
– WW1
Mar 14 at 16:08
add a comment |
$begingroup$
This is of course a circle. General formula for the circle is $$(x-p)^2+(y-q)^2 =r^2$$
where $O=(p,q)$ is a center and $r$ radius. At your case $O=(2,0)$ and $r=3sqrt2$
answered Mar 14 at 16:05
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
This is of course a circle. General formula for the circle is $$(x-p)^2+(y-q)^2 =r^2$$
where $O=(p,q)$ is a center and $r$ radius. At your case $O=(2,0)$ and $r=3sqrt2$
answered Mar 14 at 16:05
Maria MazurMaria Mazur
47.9k1260120
answered Mar 14 at 16:05
Maria MazurMaria Mazur
47.9k1260120
answered Mar 14 at 16:05
Maria MazurMaria Mazur
47.9k1260120
answered Mar 14 at 16:05
Maria MazurMaria Mazur
47.9k1260120
47.9k1260120
2
$begingroup$
I think we should recognize that this is also the equation of a cylinder whose axis of symmetry is parallel to the $z$ axis
$endgroup$
– WW1
Mar 14 at 16:08
add a comment |
2
$begingroup$
I think we should recognize that this is also the equation of a cylinder whose axis of symmetry is parallel to the $z$ axis
$endgroup$
– WW1
Mar 14 at 16:08
2
2
$begingroup$
I think we should recognize that this is also the equation of a cylinder whose axis of symmetry is parallel to the $z$ axis
$endgroup$
– WW1
Mar 14 at 16:08
$begingroup$
I think we should recognize that this is also the equation of a cylinder whose axis of symmetry is parallel to the $z$ axis
$endgroup$
– WW1
Mar 14 at 16:08
add a comment |
$begingroup$
$$beginequationtag1
r^2-4rcos(theta)=14.
endequation$$
$$x^2+y^2-4x=14$$
$$(x-2)^2+y^2=18$$
$$(x-2)^2+y^2=(3sqrt2)^2$$
answered Mar 14 at 16:09
E.H.EE.H.E
15.9k11968
$endgroup$
add a comment |
$begingroup$
$$beginequationtag1
r^2-4rcos(theta)=14.
endequation$$
$$x^2+y^2-4x=14$$
$$(x-2)^2+y^2=18$$
$$(x-2)^2+y^2=(3sqrt2)^2$$
answered Mar 14 at 16:09
E.H.EE.H.E
15.9k11968
$endgroup$
add a comment |
$begingroup$
$$beginequationtag1
r^2-4rcos(theta)=14.
endequation$$
$$x^2+y^2-4x=14$$
$$(x-2)^2+y^2=18$$
$$(x-2)^2+y^2=(3sqrt2)^2$$
answered Mar 14 at 16:09
E.H.EE.H.E
15.9k11968
$endgroup$
$$beginequationtag1
r^2-4rcos(theta)=14.
endequation$$
$$x^2+y^2-4x=14$$
$$(x-2)^2+y^2=18$$
$$(x-2)^2+y^2=(3sqrt2)^2$$
answered Mar 14 at 16:09
E.H.EE.H.E
15.9k11968
answered Mar 14 at 16:09
E.H.EE.H.E
15.9k11968
answered Mar 14 at 16:09
E.H.EE.H.E
15.9k11968
answered Mar 14 at 16:09
E.H.EE.H.E
15.9k11968
15.9k11968
add a comment |
add a comment |
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YGU9bPvC5bd,eYL2ul m1FxFlky
$begingroup$
in how many dimensions?
$endgroup$
– J. W. Tanner
Mar 14 at 16:07
1
$begingroup$
In $mathbbR^2$ it's a circle. In $mathbbR^3$ it's a cylinder (using cylindrical coordinates $(x,y,z)=(rcostheta,rsintheta,z)$).
$endgroup$
– Barry Cipra
Mar 14 at 16:08
$begingroup$
@BarryCipra Okay. So is the answer dependent on the context of the question?
$endgroup$
– Gurjinder
Mar 14 at 16:11
$begingroup$
@Gurjinder, yes, exactly.
$endgroup$
– Barry Cipra
Mar 14 at 16:12
$begingroup$
@BarryCipra Cool, thanks.
$endgroup$
– Gurjinder
Mar 14 at 16:13