Calculus Help Finding the Arc Length












1












$begingroup$


So I'm having trouble with this problem:




Let $$x=frac{y^2+2y}{8}-ln(y+1)$$
Find the arclength for $0leq yleq 2$.




My work. I know the Arc length formula is $(1+ (x'^2))^{1/2}$ in this case but when I take $x'$ I get $$x' = (y/4 + 1/4 - 1/(y+1))$$ and when I square that I get something way too hectic and adding $1$ to all of that makes it even more of a mess. Am I missing something here?






calculus integration definite-integrals arc-length







share|cite|improve this question











$endgroup$












  • $begingroup$
    $ln3+1$ comes from what Robert Z suggests
    $endgroup$
    – Sameer Baheti
    Dec 17 '18 at 6:38
















1












$begingroup$


So I'm having trouble with this problem:




Let $$x=frac{y^2+2y}{8}-ln(y+1)$$
Find the arclength for $0leq yleq 2$.




My work. I know the Arc length formula is $(1+ (x'^2))^{1/2}$ in this case but when I take $x'$ I get $$x' = (y/4 + 1/4 - 1/(y+1))$$ and when I square that I get something way too hectic and adding $1$ to all of that makes it even more of a mess. Am I missing something here?






calculus integration definite-integrals arc-length







share|cite|improve this question











$endgroup$












  • $begingroup$
    $ln3+1$ comes from what Robert Z suggests
    $endgroup$
    – Sameer Baheti
    Dec 17 '18 at 6:38














1












1








1





$begingroup$


So I'm having trouble with this problem:




Let $$x=frac{y^2+2y}{8}-ln(y+1)$$
Find the arclength for $0leq yleq 2$.




My work. I know the Arc length formula is $(1+ (x'^2))^{1/2}$ in this case but when I take $x'$ I get $$x' = (y/4 + 1/4 - 1/(y+1))$$ and when I square that I get something way too hectic and adding $1$ to all of that makes it even more of a mess. Am I missing something here?






calculus integration definite-integrals arc-length







share|cite|improve this question











$endgroup$




So I'm having trouble with this problem:




Let $$x=frac{y^2+2y}{8}-ln(y+1)$$
Find the arclength for $0leq yleq 2$.




My work. I know the Arc length formula is $(1+ (x'^2))^{1/2}$ in this case but when I take $x'$ I get $$x' = (y/4 + 1/4 - 1/(y+1))$$ and when I square that I get something way too hectic and adding $1$ to all of that makes it even more of a mess. Am I missing something here?






calculus integration definite-integrals arc-length




calculus integration definite-integrals arc-length






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 7:20









Martin Rosenau

1,1661310




1,1661310










asked Dec 17 '18 at 5:53









krauser126krauser126

474




474












  • $begingroup$
    $ln3+1$ comes from what Robert Z suggests
    $endgroup$
    – Sameer Baheti
    Dec 17 '18 at 6:38


















  • $begingroup$
    $ln3+1$ comes from what Robert Z suggests
    $endgroup$
    – Sameer Baheti
    Dec 17 '18 at 6:38
















$begingroup$
$ln3+1$ comes from what Robert Z suggests
$endgroup$
– Sameer Baheti
Dec 17 '18 at 6:38




$begingroup$
$ln3+1$ comes from what Robert Z suggests
$endgroup$
– Sameer Baheti
Dec 17 '18 at 6:38










1 Answer
1






active

oldest

votes


















1












$begingroup$

You are going in the right direction. However note that after squaring and adding $1$ a perfect square turns out:
$$1+(x'(y))^2=1+left(frac{y+1}{4}-frac{1}{y+1}right)^2
=frac{(y+1)^2}{16}+frac{1}{(y+1)^2}+frac{1}{2}=left(frac{t}{4}+frac{1}{t}right)^2.$$
where $t=y+1$. Now integrate the square root of the right-hand side for $t$ in $(1,3)$ and you will get the result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So what is the final result?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:45










  • $begingroup$
    Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:48










  • $begingroup$
    "completing the square doesn't work here" why?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:49










  • $begingroup$
    Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:50










  • $begingroup$
    @krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:52













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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043581%2fcalculus-help-finding-the-arc-length%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









1












$begingroup$

You are going in the right direction. However note that after squaring and adding $1$ a perfect square turns out:
$$1+(x'(y))^2=1+left(frac{y+1}{4}-frac{1}{y+1}right)^2
=frac{(y+1)^2}{16}+frac{1}{(y+1)^2}+frac{1}{2}=left(frac{t}{4}+frac{1}{t}right)^2.$$
where $t=y+1$. Now integrate the square root of the right-hand side for $t$ in $(1,3)$ and you will get the result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So what is the final result?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:45










  • $begingroup$
    Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:48










  • $begingroup$
    "completing the square doesn't work here" why?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:49










  • $begingroup$
    Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:50










  • $begingroup$
    @krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:52


















1












$begingroup$

You are going in the right direction. However note that after squaring and adding $1$ a perfect square turns out:
$$1+(x'(y))^2=1+left(frac{y+1}{4}-frac{1}{y+1}right)^2
=frac{(y+1)^2}{16}+frac{1}{(y+1)^2}+frac{1}{2}=left(frac{t}{4}+frac{1}{t}right)^2.$$
where $t=y+1$. Now integrate the square root of the right-hand side for $t$ in $(1,3)$ and you will get the result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So what is the final result?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:45










  • $begingroup$
    Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:48










  • $begingroup$
    "completing the square doesn't work here" why?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:49










  • $begingroup$
    Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:50










  • $begingroup$
    @krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:52
















1












1








1





$begingroup$

You are going in the right direction. However note that after squaring and adding $1$ a perfect square turns out:
$$1+(x'(y))^2=1+left(frac{y+1}{4}-frac{1}{y+1}right)^2
=frac{(y+1)^2}{16}+frac{1}{(y+1)^2}+frac{1}{2}=left(frac{t}{4}+frac{1}{t}right)^2.$$
where $t=y+1$. Now integrate the square root of the right-hand side for $t$ in $(1,3)$ and you will get the result.






share|cite|improve this answer











$endgroup$



You are going in the right direction. However note that after squaring and adding $1$ a perfect square turns out:
$$1+(x'(y))^2=1+left(frac{y+1}{4}-frac{1}{y+1}right)^2
=frac{(y+1)^2}{16}+frac{1}{(y+1)^2}+frac{1}{2}=left(frac{t}{4}+frac{1}{t}right)^2.$$
where $t=y+1$. Now integrate the square root of the right-hand side for $t$ in $(1,3)$ and you will get the result.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 6:36

























answered Dec 17 '18 at 6:15









Robert ZRobert Z

98.6k1068139




98.6k1068139












  • $begingroup$
    So what is the final result?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:45










  • $begingroup$
    Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:48










  • $begingroup$
    "completing the square doesn't work here" why?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:49










  • $begingroup$
    Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:50










  • $begingroup$
    @krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:52




















  • $begingroup$
    So what is the final result?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:45










  • $begingroup$
    Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:48










  • $begingroup$
    "completing the square doesn't work here" why?
    $endgroup$
    – Robert Z
    Dec 17 '18 at 6:49










  • $begingroup$
    Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
    $endgroup$
    – krauser126
    Dec 17 '18 at 6:50










  • $begingroup$
    @krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
    $endgroup$
    – Dylan
    Dec 17 '18 at 6:52


















$begingroup$
So what is the final result?
$endgroup$
– Robert Z
Dec 17 '18 at 6:45




$begingroup$
So what is the final result?
$endgroup$
– Robert Z
Dec 17 '18 at 6:45












$begingroup$
Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
$endgroup$
– krauser126
Dec 17 '18 at 6:48




$begingroup$
Thank you. I will try it now. I was just wondering, is there a method in order to figure out how to factor this into a perfect square? (completing the square doesn't work here so how did you manage to figure it out?
$endgroup$
– krauser126
Dec 17 '18 at 6:48












$begingroup$
"completing the square doesn't work here" why?
$endgroup$
– Robert Z
Dec 17 '18 at 6:49




$begingroup$
"completing the square doesn't work here" why?
$endgroup$
– Robert Z
Dec 17 '18 at 6:49












$begingroup$
Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
$endgroup$
– krauser126
Dec 17 '18 at 6:50




$begingroup$
Wouldn't that only work in something of the form Ax^2 + bx + c? Also, I got 1 + ln(3) as the answer.
$endgroup$
– krauser126
Dec 17 '18 at 6:50












$begingroup$
@krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
$endgroup$
– Dylan
Dec 17 '18 at 6:52






$begingroup$
@krauser126 Use $(a+b)^2 = a^2 + b^2 + 2ab$. In this case $ab=frac14$
$endgroup$
– Dylan
Dec 17 '18 at 6:52




















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043581%2fcalculus-help-finding-the-arc-length%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

How do I know what Microsoft account the skydrive app is syncing to?

Image
up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
Read more

When does type information flow backwards in C++?

Image
up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
Read more

Grease: Live!

Image
A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
Read more