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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
asked Jul 16 at 16:11
PhysicsMathsLove
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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
asked Jul 16 at 16:11
PhysicsMathsLove
1,070313
add a comment |Â
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
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
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
asked Jul 16 at 16:11
PhysicsMathsLove
1,070313
1,070313
add a comment |Â
add a comment |Â
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...
answered Jul 16 at 18:44
Djura Marinkov
2,3661816
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up vote
0
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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.
answered Jul 16 at 18:09
user7530
33.4k558109
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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...
answered Jul 16 at 18:44
Djura Marinkov
2,3661816
add a comment |Â
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...
answered Jul 16 at 18:44
Djura Marinkov
2,3661816
add a comment |Â
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...
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...
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
answered Jul 16 at 18:44
Djura Marinkov
2,3661816
2,3661816
add a comment |Â
add a comment |Â
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.
answered Jul 16 at 18:09
user7530
33.4k558109
add a comment |Â
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.
answered Jul 16 at 18:09
user7530
33.4k558109
add a comment |Â
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.
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.
answered Jul 16 at 18:09
user7530
33.4k558109
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 |Â
add a comment |Â
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