Polar coordinates - parallel to initial line question

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I have the polar curve $r=asqrtsin(2theta)$, where $a neq 0$ (graph shown above). I want to find the points where the tangent to the curve is parallel to the initial line, $theta = 0$.

I have written $y=rsintheta$ and worked out the solutions to $fracdydtheta = 0$. I get to the stage where one of my solutions is $sin theta = 0$ and another solution which leads to $theta = -fracpi3$.

My question is why do we get the solution $sintheta = 0$. I don't see how we have a tangent line parallel to the initial line at $theta = fracpi2$, for example... So if someone can share a light on the origin of this solution, that would be helpful.

Thanks

polar-coordinates

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asked Jul 16 at 16:11

PhysicsMathsLove

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up vote
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I have the polar curve $r=asqrtsin(2theta)$, where $a neq 0$ (graph shown above). I want to find the points where the tangent to the curve is parallel to the initial line, $theta = 0$.

I have written $y=rsintheta$ and worked out the solutions to $fracdydtheta = 0$. I get to the stage where one of my solutions is $sin theta = 0$ and another solution which leads to $theta = -fracpi3$.

My question is why do we get the solution $sintheta = 0$. I don't see how we have a tangent line parallel to the initial line at $theta = fracpi2$, for example... So if someone can share a light on the origin of this solution, that would be helpful.

Thanks

polar-coordinates

share|cite|improve this question

asked Jul 16 at 16:11

PhysicsMathsLove

1,070313

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I have the polar curve $r=asqrtsin(2theta)$, where $a neq 0$ (graph shown above). I want to find the points where the tangent to the curve is parallel to the initial line, $theta = 0$.

I have written $y=rsintheta$ and worked out the solutions to $fracdydtheta = 0$. I get to the stage where one of my solutions is $sin theta = 0$ and another solution which leads to $theta = -fracpi3$.

My question is why do we get the solution $sintheta = 0$. I don't see how we have a tangent line parallel to the initial line at $theta = fracpi2$, for example... So if someone can share a light on the origin of this solution, that would be helpful.

Thanks

polar-coordinates

share|cite|improve this question

asked Jul 16 at 16:11

PhysicsMathsLove

1,070313

I have the polar curve $r=asqrtsin(2theta)$, where $a neq 0$ (graph shown above). I want to find the points where the tangent to the curve is parallel to the initial line, $theta = 0$.

I have written $y=rsintheta$ and worked out the solutions to $fracdydtheta = 0$. I get to the stage where one of my solutions is $sin theta = 0$ and another solution which leads to $theta = -fracpi3$.

My question is why do we get the solution $sintheta = 0$. I don't see how we have a tangent line parallel to the initial line at $theta = fracpi2$, for example... So if someone can share a light on the origin of this solution, that would be helpful.

Thanks

polar-coordinates

share|cite|improve this question

asked Jul 16 at 16:11

PhysicsMathsLove

1,070313

share|cite|improve this question

share|cite|improve this question

asked Jul 16 at 16:11

PhysicsMathsLove

1,070313

asked Jul 16 at 16:11

PhysicsMathsLove

1,070313

asked Jul 16 at 16:11

PhysicsMathsLove

1,070313

1,070313

add a comment | 

add a comment | 

2 Answers
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$(asinthetasqrtsin2theta)'=a(costhetasin2theta+sinthetacos2theta)/sqrtsin2theta=asin3theta/sqrtsin2theta=0$

So solutions as you said are $theta_1=0, theta_2=pi/3, theta_3=pi, theta_4=4pi/3$ , no solutions at $pi/2$ (maybe you misspelled). Solutions in $2pi/3$ and $5pi/3$ are excluded cos curve isn't defined there.

You see that curve reaches 4 times the point $(x,y)=(0,0)$, once when $theta=0$, second when $theta=pi/2$, then when $theta=pi$ and when $theta=3pi/2$

In the first and 3rd case curve is parallel to x axis, in the 2nd and 4th case it is parallel to y axis.

Hope this clears it for you...

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answered Jul 16 at 18:44

Djura Marinkov

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First, your $r(theta)$ is only valid for $theta in [0,pi/2]cup [pi, 3pi/2]$, as you need $sin(2theta)geq 0$, so I'm not sure where you're getting $-pi/3 equiv 5pi/3$ as a solution.

Next, we have $y = sqrt2 (sin theta)^3/2 (cos theta)^1/2,$ so
$dy/dtheta$ vanishes when
$$3 (sin theta)^1/2(cos theta)^3/2 - (sin theta)^5/2(costheta)^-1/2 = 0.$$

Two solutions are $thetain0,pi$ and two non-solutions (since the derivative diverges) are $thetainpi/2, 3pi/2$. Away from these special values we can multiply by $(cos theta)^1/2(sintheta)^-1/2$ to yield
$$3cos^2theta - sin^2theta = 0$$
or
$$sin(theta) = pm fracsqrt32,$$
which in $[0,pi/2]cup [pi, 3pi/2]$ has solutions $theta = pi/3$ and $theta = 4pi/3$. Therefore
$$theta in 0,pi, pi/3, 4pi/3$$
which makes sense given the plot.

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edited Jul 16 at 18:34

answered Jul 16 at 18:09

user7530

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

$(asinthetasqrtsin2theta)'=a(costhetasin2theta+sinthetacos2theta)/sqrtsin2theta=asin3theta/sqrtsin2theta=0$

So solutions as you said are $theta_1=0, theta_2=pi/3, theta_3=pi, theta_4=4pi/3$ , no solutions at $pi/2$ (maybe you misspelled). Solutions in $2pi/3$ and $5pi/3$ are excluded cos curve isn't defined there.

You see that curve reaches 4 times the point $(x,y)=(0,0)$, once when $theta=0$, second when $theta=pi/2$, then when $theta=pi$ and when $theta=3pi/2$

In the first and 3rd case curve is parallel to x axis, in the 2nd and 4th case it is parallel to y axis.

Hope this clears it for you...

share|cite|improve this answer

answered Jul 16 at 18:44

Djura Marinkov

2,3661816

add a comment | 

up vote
1
down vote



accepted

$(asinthetasqrtsin2theta)'=a(costhetasin2theta+sinthetacos2theta)/sqrtsin2theta=asin3theta/sqrtsin2theta=0$

So solutions as you said are $theta_1=0, theta_2=pi/3, theta_3=pi, theta_4=4pi/3$ , no solutions at $pi/2$ (maybe you misspelled). Solutions in $2pi/3$ and $5pi/3$ are excluded cos curve isn't defined there.

You see that curve reaches 4 times the point $(x,y)=(0,0)$, once when $theta=0$, second when $theta=pi/2$, then when $theta=pi$ and when $theta=3pi/2$

In the first and 3rd case curve is parallel to x axis, in the 2nd and 4th case it is parallel to y axis.

Hope this clears it for you...

share|cite|improve this answer

answered Jul 16 at 18:44

Djura Marinkov

2,3661816

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

$(asinthetasqrtsin2theta)'=a(costhetasin2theta+sinthetacos2theta)/sqrtsin2theta=asin3theta/sqrtsin2theta=0$

So solutions as you said are $theta_1=0, theta_2=pi/3, theta_3=pi, theta_4=4pi/3$ , no solutions at $pi/2$ (maybe you misspelled). Solutions in $2pi/3$ and $5pi/3$ are excluded cos curve isn't defined there.

You see that curve reaches 4 times the point $(x,y)=(0,0)$, once when $theta=0$, second when $theta=pi/2$, then when $theta=pi$ and when $theta=3pi/2$

In the first and 3rd case curve is parallel to x axis, in the 2nd and 4th case it is parallel to y axis.

Hope this clears it for you...

share|cite|improve this answer

answered Jul 16 at 18:44

Djura Marinkov

2,3661816

$(asinthetasqrtsin2theta)'=a(costhetasin2theta+sinthetacos2theta)/sqrtsin2theta=asin3theta/sqrtsin2theta=0$

So solutions as you said are $theta_1=0, theta_2=pi/3, theta_3=pi, theta_4=4pi/3$ , no solutions at $pi/2$ (maybe you misspelled). Solutions in $2pi/3$ and $5pi/3$ are excluded cos curve isn't defined there.

You see that curve reaches 4 times the point $(x,y)=(0,0)$, once when $theta=0$, second when $theta=pi/2$, then when $theta=pi$ and when $theta=3pi/2$

In the first and 3rd case curve is parallel to x axis, in the 2nd and 4th case it is parallel to y axis.

Hope this clears it for you...

share|cite|improve this answer

answered Jul 16 at 18:44

Djura Marinkov

2,3661816

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 18:44

Djura Marinkov

2,3661816

answered Jul 16 at 18:44

Djura Marinkov

2,3661816

answered Jul 16 at 18:44

Djura Marinkov

2,3661816

2,3661816

add a comment | 

add a comment | 

up vote
0
down vote

First, your $r(theta)$ is only valid for $theta in [0,pi/2]cup [pi, 3pi/2]$, as you need $sin(2theta)geq 0$, so I'm not sure where you're getting $-pi/3 equiv 5pi/3$ as a solution.

Next, we have $y = sqrt2 (sin theta)^3/2 (cos theta)^1/2,$ so
$dy/dtheta$ vanishes when
$$3 (sin theta)^1/2(cos theta)^3/2 - (sin theta)^5/2(costheta)^-1/2 = 0.$$

Two solutions are $thetain0,pi$ and two non-solutions (since the derivative diverges) are $thetainpi/2, 3pi/2$. Away from these special values we can multiply by $(cos theta)^1/2(sintheta)^-1/2$ to yield
$$3cos^2theta - sin^2theta = 0$$
or
$$sin(theta) = pm fracsqrt32,$$
which in $[0,pi/2]cup [pi, 3pi/2]$ has solutions $theta = pi/3$ and $theta = 4pi/3$. Therefore
$$theta in 0,pi, pi/3, 4pi/3$$
which makes sense given the plot.

share|cite|improve this answer

edited Jul 16 at 18:34

answered Jul 16 at 18:09

user7530

33.4k558109

add a comment | 

up vote
0
down vote

First, your $r(theta)$ is only valid for $theta in [0,pi/2]cup [pi, 3pi/2]$, as you need $sin(2theta)geq 0$, so I'm not sure where you're getting $-pi/3 equiv 5pi/3$ as a solution.

Next, we have $y = sqrt2 (sin theta)^3/2 (cos theta)^1/2,$ so
$dy/dtheta$ vanishes when
$$3 (sin theta)^1/2(cos theta)^3/2 - (sin theta)^5/2(costheta)^-1/2 = 0.$$

Two solutions are $thetain0,pi$ and two non-solutions (since the derivative diverges) are $thetainpi/2, 3pi/2$. Away from these special values we can multiply by $(cos theta)^1/2(sintheta)^-1/2$ to yield
$$3cos^2theta - sin^2theta = 0$$
or
$$sin(theta) = pm fracsqrt32,$$
which in $[0,pi/2]cup [pi, 3pi/2]$ has solutions $theta = pi/3$ and $theta = 4pi/3$. Therefore
$$theta in 0,pi, pi/3, 4pi/3$$
which makes sense given the plot.

share|cite|improve this answer

edited Jul 16 at 18:34

answered Jul 16 at 18:09

user7530

33.4k558109

add a comment | 

up vote
0
down vote

up vote
0
down vote

First, your $r(theta)$ is only valid for $theta in [0,pi/2]cup [pi, 3pi/2]$, as you need $sin(2theta)geq 0$, so I'm not sure where you're getting $-pi/3 equiv 5pi/3$ as a solution.

Next, we have $y = sqrt2 (sin theta)^3/2 (cos theta)^1/2,$ so
$dy/dtheta$ vanishes when
$$3 (sin theta)^1/2(cos theta)^3/2 - (sin theta)^5/2(costheta)^-1/2 = 0.$$

Two solutions are $thetain0,pi$ and two non-solutions (since the derivative diverges) are $thetainpi/2, 3pi/2$. Away from these special values we can multiply by $(cos theta)^1/2(sintheta)^-1/2$ to yield
$$3cos^2theta - sin^2theta = 0$$
or
$$sin(theta) = pm fracsqrt32,$$
which in $[0,pi/2]cup [pi, 3pi/2]$ has solutions $theta = pi/3$ and $theta = 4pi/3$. Therefore
$$theta in 0,pi, pi/3, 4pi/3$$
which makes sense given the plot.

share|cite|improve this answer

edited Jul 16 at 18:34

answered Jul 16 at 18:09

user7530

33.4k558109

First, your $r(theta)$ is only valid for $theta in [0,pi/2]cup [pi, 3pi/2]$, as you need $sin(2theta)geq 0$, so I'm not sure where you're getting $-pi/3 equiv 5pi/3$ as a solution.

Next, we have $y = sqrt2 (sin theta)^3/2 (cos theta)^1/2,$ so
$dy/dtheta$ vanishes when
$$3 (sin theta)^1/2(cos theta)^3/2 - (sin theta)^5/2(costheta)^-1/2 = 0.$$

Two solutions are $thetain0,pi$ and two non-solutions (since the derivative diverges) are $thetainpi/2, 3pi/2$. Away from these special values we can multiply by $(cos theta)^1/2(sintheta)^-1/2$ to yield
$$3cos^2theta - sin^2theta = 0$$
or
$$sin(theta) = pm fracsqrt32,$$
which in $[0,pi/2]cup [pi, 3pi/2]$ has solutions $theta = pi/3$ and $theta = 4pi/3$. Therefore
$$theta in 0,pi, pi/3, 4pi/3$$
which makes sense given the plot.

share|cite|improve this answer

edited Jul 16 at 18:34

answered Jul 16 at 18:09

user7530

33.4k558109

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 18:34

edited Jul 16 at 18:34

edited Jul 16 at 18:34

answered Jul 16 at 18:09

user7530

33.4k558109

answered Jul 16 at 18:09

user7530

33.4k558109

answered Jul 16 at 18:09

user7530

33.4k558109

33.4k558109

add a comment | 

add a comment | 

 

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