Tangent at the pole for the equation $r = 2(1 - sintheta)$ The 2019 Stack Overflow Developer Survey Results Are InTangent Line for Polar Curve R=$3-3sintheta $To prove $(sintheta + csctheta)^2 + (costheta +sectheta)^2 ge 9$Tangent undefined for polar curves ($r^2=a^2sin(stheta)$)?find the equations of the tangents at the pole.Solving $sin(theta) + cos(theta) = -1$ using the T-FormulaVertical tangent line for r = 1 + cos($theta$)Symmetry of $r = sin (2 theta)$ regarding polar axis, pole and line $theta = fracpi2$Why is $intfracsin (3theta + pi )+1415 times fracsin (3theta - pi )+1415,dtheta$ near double the area of $r=1$ with area $pi?$Find the equations of the lines tangent to $r=4sin(3theta)$ at the pole.Identify the symmetries and sketch the curve $r=sin (theta/2)$
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Tangent at the pole for the equation $r = 2(1 - sintheta)$
The 2019 Stack Overflow Developer Survey Results Are InTangent Line for Polar Curve R=$3-3sintheta $To prove $(sintheta + csctheta)^2 + (costheta +sectheta)^2 ge 9$Tangent undefined for polar curves ($r^2=a^2sin(stheta)$)?find the equations of the tangents at the pole.Solving $sin(theta) + cos(theta) = -1$ using the T-FormulaVertical tangent line for r = 1 + cos($theta$)Symmetry of $r = sin (2 theta)$ regarding polar axis, pole and line $theta = fracpi2$Why is $intfracsin (3theta + pi )+1415 times fracsin (3theta - pi )+1415,dtheta$ near double the area of $r=1$ with area $pi?$Find the equations of the lines tangent to $r=4sin(3theta)$ at the pole.Identify the symmetries and sketch the curve $r=sin (theta/2)$
$begingroup$
I was asked to find the tangents at the pole for the following equation: $r=2(1-sintheta)$.
I understand that the requirements for tangency at the pole are $f(theta)=0$ and $f'(theta) neq 0$. I set $0=2(1-sinpi)$ and got $theta= fracpi2$. But when I plugged that into the derivative, I got $f'(theta)=0$. Why is that? Am I solving it wrong? (By the way, my $f'$ was $f'(theta)=-2costheta$). Thank You!
calculus algebra-precalculus polar-coordinates
edited Mar 30 at 10:13
N. F. Taussig
45.1k103358
asked Mar 30 at 2:05
Matthew CherryMatthew Cherry
62
$endgroup$
add a comment |
$begingroup$
I was asked to find the tangents at the pole for the following equation: $r=2(1-sintheta)$.
I understand that the requirements for tangency at the pole are $f(theta)=0$ and $f'(theta) neq 0$. I set $0=2(1-sinpi)$ and got $theta= fracpi2$. But when I plugged that into the derivative, I got $f'(theta)=0$. Why is that? Am I solving it wrong? (By the way, my $f'$ was $f'(theta)=-2costheta$). Thank You!
calculus algebra-precalculus polar-coordinates
edited Mar 30 at 10:13
N. F. Taussig
45.1k103358
asked Mar 30 at 2:05
Matthew CherryMatthew Cherry
62
$endgroup$
$begingroup$
What is f??????
$endgroup$
– William Elliot
Mar 30 at 2:26
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Mar 30 at 9:10
$begingroup$
You meant to write $2(1 - sintheta) = 0 implies theta = fracpi2$.
$endgroup$
– N. F. Taussig
Mar 30 at 9:23
add a comment |
$begingroup$
I was asked to find the tangents at the pole for the following equation: $r=2(1-sintheta)$.
I understand that the requirements for tangency at the pole are $f(theta)=0$ and $f'(theta) neq 0$. I set $0=2(1-sinpi)$ and got $theta= fracpi2$. But when I plugged that into the derivative, I got $f'(theta)=0$. Why is that? Am I solving it wrong? (By the way, my $f'$ was $f'(theta)=-2costheta$). Thank You!
calculus algebra-precalculus polar-coordinates
edited Mar 30 at 10:13
N. F. Taussig
45.1k103358
asked Mar 30 at 2:05
Matthew CherryMatthew Cherry
62
$endgroup$
I was asked to find the tangents at the pole for the following equation: $r=2(1-sintheta)$.
I understand that the requirements for tangency at the pole are $f(theta)=0$ and $f'(theta) neq 0$. I set $0=2(1-sinpi)$ and got $theta= fracpi2$. But when I plugged that into the derivative, I got $f'(theta)=0$. Why is that? Am I solving it wrong? (By the way, my $f'$ was $f'(theta)=-2costheta$). Thank You!
calculus algebra-precalculus polar-coordinates
calculus algebra-precalculus polar-coordinates
edited Mar 30 at 10:13
N. F. Taussig
45.1k103358
asked Mar 30 at 2:05
Matthew CherryMatthew Cherry
62
edited Mar 30 at 10:13
N. F. Taussig
45.1k103358
asked Mar 30 at 2:05
Matthew CherryMatthew Cherry
62
edited Mar 30 at 10:13
N. F. Taussig
45.1k103358
edited Mar 30 at 10:13
N. F. Taussig
45.1k103358
edited Mar 30 at 10:13
N. F. Taussig
45.1k103358
45.1k103358
asked Mar 30 at 2:05
Matthew CherryMatthew Cherry
62
asked Mar 30 at 2:05
Matthew CherryMatthew Cherry
62
asked Mar 30 at 2:05
Matthew CherryMatthew Cherry
62
62
$begingroup$
What is f??????
$endgroup$
– William Elliot
Mar 30 at 2:26
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Mar 30 at 9:10
$begingroup$
You meant to write $2(1 - sintheta) = 0 implies theta = fracpi2$.
$endgroup$
– N. F. Taussig
Mar 30 at 9:23
add a comment |
$begingroup$
What is f??????
$endgroup$
– William Elliot
Mar 30 at 2:26
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Mar 30 at 9:10
$begingroup$
You meant to write $2(1 - sintheta) = 0 implies theta = fracpi2$.
$endgroup$
– N. F. Taussig
Mar 30 at 9:23
$begingroup$
What is f??????
$endgroup$
– William Elliot
Mar 30 at 2:26
$begingroup$
What is f??????
$endgroup$
– William Elliot
Mar 30 at 2:26
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Mar 30 at 9:10
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Mar 30 at 9:10
$begingroup$
You meant to write $2(1 - sintheta) = 0 implies theta = fracpi2$.
$endgroup$
– N. F. Taussig
Mar 30 at 9:23
$begingroup$
You meant to write $2(1 - sintheta) = 0 implies theta = fracpi2$.
$endgroup$
– N. F. Taussig
Mar 30 at 9:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You calculated the derivative incorrectly. The derivative is
$$fracdydx = fracdfracdydthetadfracdxdtheta$$
where
beginalign*
x & = rcostheta\
y & = rsintheta
endalign*
By the product rule,
beginalign*
fracdxdtheta & = r'costheta - rsintheta\
fracdydtheta & = r'sintheta + rcostheta
endalign*
where
$$r' = fracdrdtheta$$
Hence,
$$fracdydx = fracr'sintheta + rcosthetar'costheta - rsintheta$$
We were given the function $r(theta) = 2(1 - sintheta)$. At the pole, $r = 0$, so we obtain
beginalign*
2(1 - sintheta) & = 0\
1 - sintheta & = 0\
1 & = sintheta\
fracpi2 & = theta
endalign*
Notice that at the pole, since $r = 0$,
beginalign*
fracdydx & = fracr'sintheta + rcosthetar'costheta - rsintheta\
& = fracr'sinthetar'costheta\
& = tantheta
endalign*
Observe that
beginalign*
lim_theta to fracpi2^- tantheta & = infty\
lim_theta to fracpi2^+ tantheta & = -infty
endalign*
Thus, the function $r(theta) = 2(1 - sintheta)$ has a vertical tangent at the pole, as can be seen from viewing its graph, which is a cardioid.
answered Mar 30 at 10:10
N. F. TaussigN. F. Taussig
45.1k103358
$endgroup$
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
You calculated the derivative incorrectly. The derivative is
$$fracdydx = fracdfracdydthetadfracdxdtheta$$
where
beginalign*
x & = rcostheta\
y & = rsintheta
endalign*
By the product rule,
beginalign*
fracdxdtheta & = r'costheta - rsintheta\
fracdydtheta & = r'sintheta + rcostheta
endalign*
where
$$r' = fracdrdtheta$$
Hence,
$$fracdydx = fracr'sintheta + rcosthetar'costheta - rsintheta$$
We were given the function $r(theta) = 2(1 - sintheta)$. At the pole, $r = 0$, so we obtain
beginalign*
2(1 - sintheta) & = 0\
1 - sintheta & = 0\
1 & = sintheta\
fracpi2 & = theta
endalign*
Notice that at the pole, since $r = 0$,
beginalign*
fracdydx & = fracr'sintheta + rcosthetar'costheta - rsintheta\
& = fracr'sinthetar'costheta\
& = tantheta
endalign*
Observe that
beginalign*
lim_theta to fracpi2^- tantheta & = infty\
lim_theta to fracpi2^+ tantheta & = -infty
endalign*
Thus, the function $r(theta) = 2(1 - sintheta)$ has a vertical tangent at the pole, as can be seen from viewing its graph, which is a cardioid.
answered Mar 30 at 10:10
N. F. TaussigN. F. Taussig
45.1k103358
$endgroup$
add a comment |
$begingroup$
You calculated the derivative incorrectly. The derivative is
$$fracdydx = fracdfracdydthetadfracdxdtheta$$
where
beginalign*
x & = rcostheta\
y & = rsintheta
endalign*
By the product rule,
beginalign*
fracdxdtheta & = r'costheta - rsintheta\
fracdydtheta & = r'sintheta + rcostheta
endalign*
where
$$r' = fracdrdtheta$$
Hence,
$$fracdydx = fracr'sintheta + rcosthetar'costheta - rsintheta$$
We were given the function $r(theta) = 2(1 - sintheta)$. At the pole, $r = 0$, so we obtain
beginalign*
2(1 - sintheta) & = 0\
1 - sintheta & = 0\
1 & = sintheta\
fracpi2 & = theta
endalign*
Notice that at the pole, since $r = 0$,
beginalign*
fracdydx & = fracr'sintheta + rcosthetar'costheta - rsintheta\
& = fracr'sinthetar'costheta\
& = tantheta
endalign*
Observe that
beginalign*
lim_theta to fracpi2^- tantheta & = infty\
lim_theta to fracpi2^+ tantheta & = -infty
endalign*
Thus, the function $r(theta) = 2(1 - sintheta)$ has a vertical tangent at the pole, as can be seen from viewing its graph, which is a cardioid.
answered Mar 30 at 10:10
N. F. TaussigN. F. Taussig
45.1k103358
$endgroup$
add a comment |
$begingroup$
You calculated the derivative incorrectly. The derivative is
$$fracdydx = fracdfracdydthetadfracdxdtheta$$
where
beginalign*
x & = rcostheta\
y & = rsintheta
endalign*
By the product rule,
beginalign*
fracdxdtheta & = r'costheta - rsintheta\
fracdydtheta & = r'sintheta + rcostheta
endalign*
where
$$r' = fracdrdtheta$$
Hence,
$$fracdydx = fracr'sintheta + rcosthetar'costheta - rsintheta$$
We were given the function $r(theta) = 2(1 - sintheta)$. At the pole, $r = 0$, so we obtain
beginalign*
2(1 - sintheta) & = 0\
1 - sintheta & = 0\
1 & = sintheta\
fracpi2 & = theta
endalign*
Notice that at the pole, since $r = 0$,
beginalign*
fracdydx & = fracr'sintheta + rcosthetar'costheta - rsintheta\
& = fracr'sinthetar'costheta\
& = tantheta
endalign*
Observe that
beginalign*
lim_theta to fracpi2^- tantheta & = infty\
lim_theta to fracpi2^+ tantheta & = -infty
endalign*
Thus, the function $r(theta) = 2(1 - sintheta)$ has a vertical tangent at the pole, as can be seen from viewing its graph, which is a cardioid.
answered Mar 30 at 10:10
N. F. TaussigN. F. Taussig
45.1k103358
$endgroup$
You calculated the derivative incorrectly. The derivative is
$$fracdydx = fracdfracdydthetadfracdxdtheta$$
where
beginalign*
x & = rcostheta\
y & = rsintheta
endalign*
By the product rule,
beginalign*
fracdxdtheta & = r'costheta - rsintheta\
fracdydtheta & = r'sintheta + rcostheta
endalign*
where
$$r' = fracdrdtheta$$
Hence,
$$fracdydx = fracr'sintheta + rcosthetar'costheta - rsintheta$$
We were given the function $r(theta) = 2(1 - sintheta)$. At the pole, $r = 0$, so we obtain
beginalign*
2(1 - sintheta) & = 0\
1 - sintheta & = 0\
1 & = sintheta\
fracpi2 & = theta
endalign*
Notice that at the pole, since $r = 0$,
beginalign*
fracdydx & = fracr'sintheta + rcosthetar'costheta - rsintheta\
& = fracr'sinthetar'costheta\
& = tantheta
endalign*
Observe that
beginalign*
lim_theta to fracpi2^- tantheta & = infty\
lim_theta to fracpi2^+ tantheta & = -infty
endalign*
Thus, the function $r(theta) = 2(1 - sintheta)$ has a vertical tangent at the pole, as can be seen from viewing its graph, which is a cardioid.
answered Mar 30 at 10:10
N. F. TaussigN. F. Taussig
45.1k103358
answered Mar 30 at 10:10
N. F. TaussigN. F. Taussig
45.1k103358
answered Mar 30 at 10:10
N. F. TaussigN. F. Taussig
45.1k103358
answered Mar 30 at 10:10
N. F. TaussigN. F. Taussig
45.1k103358
45.1k103358
add a comment |
add a comment |
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$begingroup$
What is f??????
$endgroup$
– William Elliot
Mar 30 at 2:26
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Mar 30 at 9:10
$begingroup$
You meant to write $2(1 - sintheta) = 0 implies theta = fracpi2$.
$endgroup$
– N. F. Taussig
Mar 30 at 9:23