Find the Arc length of the parametric curve
$begingroup$
$$x=6t-6sint$$
$$y=6-6cost$$
Find the arc length of the parametric curve
$$Arc length = int_{0}^{2pi} sqrt{(6-6cost)^2+(6sint)^2}dt\
=int_{0}^{2pi} sqrt{36-72cost+36cos^2t+36sin^2t}dt \
=int_{0}^{2pi} 6 sqrt{1-2cost+cos^2t+sin^2t}dt\
=int_{0}^{2pi} 6sqrt{2-2cost}dt\
=int_{0}^{2pi} 6sqrt{2}sqrt{1-cost}dt\
= 6sqrt{2}int_{0}^{2pi}sqrt{1-cost}dt\
=6sqrt{2}int_{0}^{2pi}sqrt{1-cost}frac{sqrt{(1+cost)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sqrt{(1-cos^2t)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sint}{sqrt{(1+cost)}} dt\$$
Let $$u =1+cost$$ $$du=-sint$$
$$=-6sqrt{2}int_{0}^{2pi}frac{1}{sqrt{u}}\
=-12sqrt{2}[u^frac{1}{2}]\
=0$$
But the answer is 48.
calculus
asked Dec 15 '18 at 16:41
EvianEvian
84
$endgroup$
add a comment |
$begingroup$
$$x=6t-6sint$$
$$y=6-6cost$$
Find the arc length of the parametric curve
$$Arc length = int_{0}^{2pi} sqrt{(6-6cost)^2+(6sint)^2}dt\
=int_{0}^{2pi} sqrt{36-72cost+36cos^2t+36sin^2t}dt \
=int_{0}^{2pi} 6 sqrt{1-2cost+cos^2t+sin^2t}dt\
=int_{0}^{2pi} 6sqrt{2-2cost}dt\
=int_{0}^{2pi} 6sqrt{2}sqrt{1-cost}dt\
= 6sqrt{2}int_{0}^{2pi}sqrt{1-cost}dt\
=6sqrt{2}int_{0}^{2pi}sqrt{1-cost}frac{sqrt{(1+cost)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sqrt{(1-cos^2t)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sint}{sqrt{(1+cost)}} dt\$$
Let $$u =1+cost$$ $$du=-sint$$
$$=-6sqrt{2}int_{0}^{2pi}frac{1}{sqrt{u}}\
=-12sqrt{2}[u^frac{1}{2}]\
=0$$
But the answer is 48.
calculus
asked Dec 15 '18 at 16:41
EvianEvian
84
$endgroup$
1
$begingroup$
the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves
$endgroup$
– G Cab
Dec 15 '18 at 16:47
add a comment |
$begingroup$
$$x=6t-6sint$$
$$y=6-6cost$$
Find the arc length of the parametric curve
$$Arc length = int_{0}^{2pi} sqrt{(6-6cost)^2+(6sint)^2}dt\
=int_{0}^{2pi} sqrt{36-72cost+36cos^2t+36sin^2t}dt \
=int_{0}^{2pi} 6 sqrt{1-2cost+cos^2t+sin^2t}dt\
=int_{0}^{2pi} 6sqrt{2-2cost}dt\
=int_{0}^{2pi} 6sqrt{2}sqrt{1-cost}dt\
= 6sqrt{2}int_{0}^{2pi}sqrt{1-cost}dt\
=6sqrt{2}int_{0}^{2pi}sqrt{1-cost}frac{sqrt{(1+cost)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sqrt{(1-cos^2t)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sint}{sqrt{(1+cost)}} dt\$$
Let $$u =1+cost$$ $$du=-sint$$
$$=-6sqrt{2}int_{0}^{2pi}frac{1}{sqrt{u}}\
=-12sqrt{2}[u^frac{1}{2}]\
=0$$
But the answer is 48.
calculus
asked Dec 15 '18 at 16:41
EvianEvian
84
$endgroup$
$$x=6t-6sint$$
$$y=6-6cost$$
Find the arc length of the parametric curve
$$Arc length = int_{0}^{2pi} sqrt{(6-6cost)^2+(6sint)^2}dt\
=int_{0}^{2pi} sqrt{36-72cost+36cos^2t+36sin^2t}dt \
=int_{0}^{2pi} 6 sqrt{1-2cost+cos^2t+sin^2t}dt\
=int_{0}^{2pi} 6sqrt{2-2cost}dt\
=int_{0}^{2pi} 6sqrt{2}sqrt{1-cost}dt\
= 6sqrt{2}int_{0}^{2pi}sqrt{1-cost}dt\
=6sqrt{2}int_{0}^{2pi}sqrt{1-cost}frac{sqrt{(1+cost)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sqrt{(1-cos^2t)}}{sqrt{(1+cost)}} dt\
=6sqrt{2}int_{0}^{2pi}frac{sint}{sqrt{(1+cost)}} dt\$$
Let $$u =1+cost$$ $$du=-sint$$
$$=-6sqrt{2}int_{0}^{2pi}frac{1}{sqrt{u}}\
=-12sqrt{2}[u^frac{1}{2}]\
=0$$
But the answer is 48.
calculus
calculus
asked Dec 15 '18 at 16:41
EvianEvian
84
asked Dec 15 '18 at 16:41
EvianEvian
84
edited Dec 15 '18 at 16:49
Evian
asked Dec 15 '18 at 16:41
EvianEvian
84
asked Dec 15 '18 at 16:41
EvianEvian
84
asked Dec 15 '18 at 16:41
EvianEvian
84
84
1
$begingroup$
the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves
$endgroup$
– G Cab
Dec 15 '18 at 16:47
add a comment |
1
$begingroup$
the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves
$endgroup$
– G Cab
Dec 15 '18 at 16:47
1
1
$begingroup$
the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves
$endgroup$
– G Cab
Dec 15 '18 at 16:47
$begingroup$
the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves
$endgroup$
– G Cab
Dec 15 '18 at 16:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Leaving the coefficient $6$ on the side, we have
$$s=int_0^{2pi}sqrt{(1-cos t)^2+(sin t)^2},dt=int_0^{2pi}sqrt{2-2cos t},dt=2int_0^{2pi}left|sinfrac t2right|,dt.$$
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k675227
$endgroup$
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%2f3041693%2ffind-the-arc-length-of-the-parametric-curve%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
$begingroup$
Leaving the coefficient $6$ on the side, we have
$$s=int_0^{2pi}sqrt{(1-cos t)^2+(sin t)^2},dt=int_0^{2pi}sqrt{2-2cos t},dt=2int_0^{2pi}left|sinfrac t2right|,dt.$$
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
Leaving the coefficient $6$ on the side, we have
$$s=int_0^{2pi}sqrt{(1-cos t)^2+(sin t)^2},dt=int_0^{2pi}sqrt{2-2cos t},dt=2int_0^{2pi}left|sinfrac t2right|,dt.$$
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
Leaving the coefficient $6$ on the side, we have
$$s=int_0^{2pi}sqrt{(1-cos t)^2+(sin t)^2},dt=int_0^{2pi}sqrt{2-2cos t},dt=2int_0^{2pi}left|sinfrac t2right|,dt.$$
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k675227
$endgroup$
Leaving the coefficient $6$ on the side, we have
$$s=int_0^{2pi}sqrt{(1-cos t)^2+(sin t)^2},dt=int_0^{2pi}sqrt{2-2cos t},dt=2int_0^{2pi}left|sinfrac t2right|,dt.$$
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k675227
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k675227
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k675227
answered Dec 15 '18 at 16:54
Yves DaoustYves Daoust
128k675227
128k675227
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%2f3041693%2ffind-the-arc-length-of-the-parametric-curve%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
1
$begingroup$
the arc length is $ds^2=dx^2+dy^2=(x'^2+y'^2)dt^2$ : the derivatives and not the functions themselves
$endgroup$
– G Cab
Dec 15 '18 at 16:47