Tangent at the pole for the equation $r = 2(1 - sintheta)$Tangent 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 $r=sin (theta/2)$
cryptic clue: mammal sounds like relative consumer (8)
What to wear for invited talk in Canada
How to deal with fear of taking dependencies
Denied boarding due to overcrowding, Sparpreis ticket. What are my rights?
2015 GMC truck check engine light no longer on
Weird behaviour when using querySelector
Does the average primeness of natural numbers tend to zero?
Why Is Death Allowed In the Matrix?
Are white and non-white police officers equally likely to kill black suspects?
LM317 - Calculate dissipation due to voltage drop
Set up public ip on server
Could a US political party gain complete control over the government by removing checks & balances?
Wild Shape Centaur Into a Giant Elk: do their Charges stack?
Symmetry in quantum mechanics
How would photo IDs work for shapeshifters?
Prime joint compound before latex paint?
I am not able to install anything in ubuntu
Some basic questions on halt and move in Turing machines
How do you conduct xenoanthropology after first contact?
My colleague's body is amazing
Can I interfere when another PC is about to be attacked?
Can a planet have a different gravitational pull depending on its location in orbit around its sun?
How to handle columns with categorical data and many unique values
Why is an old chain unsafe?
Tangent at the pole for the equation $r = 2(1 - sintheta)$
Tangent 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
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3167834%2ftangent-at-the-pole-for-the-equation-r-21-sin-theta%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3167834%2ftangent-at-the-pole-for-the-equation-r-21-sin-theta%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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