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Convert $x^2 + y^2 = 2ax$ to a polar equation, where $a > 0$.

Is $r=2cos(theta)$ a one-petal polar function?Test for symmetry for polar graphsConvert $y^2 = 4(x + 1)$ to a polar equationellipse polar co-ordinate conversionParabola equation from cartesian to polar representationConverting polar equation to cartesian coordinatesConverting Polar Equations into Cartesian equationsConverting an equation from Cartesian to Polar form?Converting cartesian rectangular equation to it's corresponding polar equationVerify my work : Polar equation to Cartesian Equation

1

$begingroup$

This isn't homework exactly, but rather I'm teaching a class on precalculus and I'm a little stuck on a question from the textbook.

Convert $x^2 + y^2 = 2ax$ to a polar equation, where $a>0$.

This is incredibly simple to do: note that $x^2 + y^2 = r^2$, and $x = rcostheta$. Then the equation is simply: $$r^2 = 2arcostheta.$$ However, the answer in the back of the textbook is $$r = 2acostheta.$$

Here's my question: Why can we divide by $r$? What happens if $r=0$? Not for every $theta$ will $r=0$, so why is it valid to simply divide by it?

$r$ is the equation of a circle with radius $a$ whose center lies on the $x$-axis, and graphically the pole $r=0$ exists on the graph. So what gives?

algebra-precalculus proof-verification polar-coordinates

share|cite|improve this question

asked Nov 8 '17 at 1:13

Decaf-MathDecaf-Math

3,422926

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe... $r^2=2a rcos(theta) to r(r-2acos(theta))=0 to r=0vee r=2acos(theta)$ with $cos(theta)ge 0$.
  $endgroup$
  – Ixion
  Nov 8 '17 at 1:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that if $r=0$ then we get the point $(0,0)$, the same that we get when $cos(theta)=0$.
  $endgroup$
  – Ixion
  Nov 8 '17 at 1:48
  

  
  
  
  

add a comment | 

1

$begingroup$

This isn't homework exactly, but rather I'm teaching a class on precalculus and I'm a little stuck on a question from the textbook.

Convert $x^2 + y^2 = 2ax$ to a polar equation, where $a>0$.

This is incredibly simple to do: note that $x^2 + y^2 = r^2$, and $x = rcostheta$. Then the equation is simply: $$r^2 = 2arcostheta.$$ However, the answer in the back of the textbook is $$r = 2acostheta.$$

Here's my question: Why can we divide by $r$? What happens if $r=0$? Not for every $theta$ will $r=0$, so why is it valid to simply divide by it?

$r$ is the equation of a circle with radius $a$ whose center lies on the $x$-axis, and graphically the pole $r=0$ exists on the graph. So what gives?

algebra-precalculus proof-verification polar-coordinates

share|cite|improve this question

asked Nov 8 '17 at 1:13

Decaf-MathDecaf-Math

3,422926

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe... $r^2=2a rcos(theta) to r(r-2acos(theta))=0 to r=0vee r=2acos(theta)$ with $cos(theta)ge 0$.
  $endgroup$
  – Ixion
  Nov 8 '17 at 1:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that if $r=0$ then we get the point $(0,0)$, the same that we get when $cos(theta)=0$.
  $endgroup$
  – Ixion
  Nov 8 '17 at 1:48
  

  
  
  
  

add a comment | 

$begingroup$

This isn't homework exactly, but rather I'm teaching a class on precalculus and I'm a little stuck on a question from the textbook.

Convert $x^2 + y^2 = 2ax$ to a polar equation, where $a>0$.

This is incredibly simple to do: note that $x^2 + y^2 = r^2$, and $x = rcostheta$. Then the equation is simply: $$r^2 = 2arcostheta.$$ However, the answer in the back of the textbook is $$r = 2acostheta.$$

Here's my question: Why can we divide by $r$? What happens if $r=0$? Not for every $theta$ will $r=0$, so why is it valid to simply divide by it?

$r$ is the equation of a circle with radius $a$ whose center lies on the $x$-axis, and graphically the pole $r=0$ exists on the graph. So what gives?

algebra-precalculus proof-verification polar-coordinates

share|cite|improve this question

asked Nov 8 '17 at 1:13

Decaf-MathDecaf-Math

3,422926

$endgroup$

share|cite|improve this question

asked Nov 8 '17 at 1:13

Decaf-MathDecaf-Math

3,422926

share|cite|improve this question

asked Nov 8 '17 at 1:13

Decaf-MathDecaf-Math

3,422926

asked Nov 8 '17 at 1:13

Decaf-MathDecaf-Math

3,422926

asked Nov 8 '17 at 1:13

Decaf-MathDecaf-Math

3,422926

2 Answers
2

active

oldest

votes

1

$begingroup$

$r=0$ covers only one point, namely the pole. Can $r=0$ satisfy the new equation? Yes, if $theta=fracpi2$. Therefore, you haven't changed the set of points that satisfy the equation.

share|cite|improve this answer

edited Mar 21 at 8:45

Shailesh

3,97792134

answered Mar 21 at 5:35

user656351user656351

111

$endgroup$

add a comment | 

0

$begingroup$

Well, I admit they do not give enough detail.
$$ (x-a)^2 + y^2 = a^2. $$
With $a > 0,$ this is tangent to the vertical axis at the origin. The polar expression $r = 2 a cos theta$ covers the circle twice, once with $ r$ positive while $; - pi/2 < theta < pi / 2,$ then again with $r$ negative while, say, $; pi/2 < theta < 3 pi / 2.$ At the odd multiples of $pi/2,$ we get the origin because $r=0.$

I guess you need to decide whether your students will be able to deal with negative $r.$

share|cite|improve this answer

edited Nov 8 '17 at 1:40

answered Nov 8 '17 at 1:26

Will JagyWill Jagy

104k5102201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, I mean how do I go from $r^2 = 2arcostheta$ to $r = 2acostheta$? The obvious answer would be to divide by $r$, but I'm just concerned for when $r=0$: how do I know I'm allowed to do that? You claim that $r$ is positive for $thetainleft[-piover2,piover2right]$, but if $theta = -piover2$ or $theta = piover2$, then $costheta = 0$, which $0$ is neither positive nor negative.
  $endgroup$
  – Decaf-Math
  Nov 8 '17 at 1:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Decaf-Math I had originally strict inequality signs, I put them back now. I agree that "dividing by $r$" is not an adequate description. However, the curve $r = 2a cos theta$ actually does cover the entire circle.
  $endgroup$
  – Will Jagy
  Nov 8 '17 at 1:36
  

  
  
  
  

add a comment | 

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1

$begingroup$

$r=0$ covers only one point, namely the pole. Can $r=0$ satisfy the new equation? Yes, if $theta=fracpi2$. Therefore, you haven't changed the set of points that satisfy the equation.

share|cite|improve this answer

edited Mar 21 at 8:45

Shailesh

3,97792134

answered Mar 21 at 5:35

user656351user656351

111

$endgroup$

add a comment | 

1

$begingroup$

$r=0$ covers only one point, namely the pole. Can $r=0$ satisfy the new equation? Yes, if $theta=fracpi2$. Therefore, you haven't changed the set of points that satisfy the equation.

share|cite|improve this answer

edited Mar 21 at 8:45

Shailesh

3,97792134

answered Mar 21 at 5:35

user656351user656351

111

$endgroup$

add a comment | 

$begingroup$

$r=0$ covers only one point, namely the pole. Can $r=0$ satisfy the new equation? Yes, if $theta=fracpi2$. Therefore, you haven't changed the set of points that satisfy the equation.

share|cite|improve this answer

edited Mar 21 at 8:45

Shailesh

3,97792134

answered Mar 21 at 5:35

user656351user656351

111

$endgroup$

share|cite|improve this answer

edited Mar 21 at 8:45

Shailesh

3,97792134

answered Mar 21 at 5:35

user656351user656351

111

edited Mar 21 at 8:45

Shailesh

3,97792134

edited Mar 21 at 8:45

Shailesh

3,97792134

answered Mar 21 at 5:35

user656351user656351

111

answered Mar 21 at 5:35

user656351user656351

111

0

$begingroup$

Well, I admit they do not give enough detail.
$$ (x-a)^2 + y^2 = a^2. $$
With $a > 0,$ this is tangent to the vertical axis at the origin. The polar expression $r = 2 a cos theta$ covers the circle twice, once with $ r$ positive while $; - pi/2 < theta < pi / 2,$ then again with $r$ negative while, say, $; pi/2 < theta < 3 pi / 2.$ At the odd multiples of $pi/2,$ we get the origin because $r=0.$

I guess you need to decide whether your students will be able to deal with negative $r.$

share|cite|improve this answer

edited Nov 8 '17 at 1:40

answered Nov 8 '17 at 1:26

Will JagyWill Jagy

104k5102201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, I mean how do I go from $r^2 = 2arcostheta$ to $r = 2acostheta$? The obvious answer would be to divide by $r$, but I'm just concerned for when $r=0$: how do I know I'm allowed to do that? You claim that $r$ is positive for $thetainleft[-piover2,piover2right]$, but if $theta = -piover2$ or $theta = piover2$, then $costheta = 0$, which $0$ is neither positive nor negative.
  $endgroup$
  – Decaf-Math
  Nov 8 '17 at 1:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Decaf-Math I had originally strict inequality signs, I put them back now. I agree that "dividing by $r$" is not an adequate description. However, the curve $r = 2a cos theta$ actually does cover the entire circle.
  $endgroup$
  – Will Jagy
  Nov 8 '17 at 1:36
  

  
  
  
  

add a comment | 

0

$begingroup$

Well, I admit they do not give enough detail.
$$ (x-a)^2 + y^2 = a^2. $$
With $a > 0,$ this is tangent to the vertical axis at the origin. The polar expression $r = 2 a cos theta$ covers the circle twice, once with $ r$ positive while $; - pi/2 < theta < pi / 2,$ then again with $r$ negative while, say, $; pi/2 < theta < 3 pi / 2.$ At the odd multiples of $pi/2,$ we get the origin because $r=0.$

I guess you need to decide whether your students will be able to deal with negative $r.$

share|cite|improve this answer

edited Nov 8 '17 at 1:40

answered Nov 8 '17 at 1:26

Will JagyWill Jagy

104k5102201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, I mean how do I go from $r^2 = 2arcostheta$ to $r = 2acostheta$? The obvious answer would be to divide by $r$, but I'm just concerned for when $r=0$: how do I know I'm allowed to do that? You claim that $r$ is positive for $thetainleft[-piover2,piover2right]$, but if $theta = -piover2$ or $theta = piover2$, then $costheta = 0$, which $0$ is neither positive nor negative.
  $endgroup$
  – Decaf-Math
  Nov 8 '17 at 1:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Decaf-Math I had originally strict inequality signs, I put them back now. I agree that "dividing by $r$" is not an adequate description. However, the curve $r = 2a cos theta$ actually does cover the entire circle.
  $endgroup$
  – Will Jagy
  Nov 8 '17 at 1:36
  

  
  
  
  

add a comment | 

$begingroup$

Well, I admit they do not give enough detail.
$$ (x-a)^2 + y^2 = a^2. $$
With $a > 0,$ this is tangent to the vertical axis at the origin. The polar expression $r = 2 a cos theta$ covers the circle twice, once with $ r$ positive while $; - pi/2 < theta < pi / 2,$ then again with $r$ negative while, say, $; pi/2 < theta < 3 pi / 2.$ At the odd multiples of $pi/2,$ we get the origin because $r=0.$

I guess you need to decide whether your students will be able to deal with negative $r.$

share|cite|improve this answer

edited Nov 8 '17 at 1:40

answered Nov 8 '17 at 1:26

Will JagyWill Jagy

104k5102201

$endgroup$

share|cite|improve this answer

edited Nov 8 '17 at 1:40

answered Nov 8 '17 at 1:26

Will JagyWill Jagy

104k5102201

answered Nov 8 '17 at 1:26

Will JagyWill Jagy

104k5102201

answered Nov 8 '17 at 1:26

Will JagyWill Jagy

104k5102201