Etydhv

Rounding a number extracted by jq to limit the decimal points

What are the components of a legend (in the sense of a tale, not a figure legend)?

Effects of ~10atm pressure on engine design

Can't find the release for this wiring harness connector

Unexpected Netflix account registered to my Gmail address - any way it could be a hack attempt?

What's the difference between "за ... от" and "в ... от"?

correct spelling of "carruffel" (fuzz, hustle, all that jazz)

Is the expression "To think you would stoop so low" often misused?

Was this character’s old age look CGI or make-up?

Why are solar panels kept tilted?

Why did the metro bus stop at each railway crossing, despite no warning indicating a train was coming?

Safety when modifying old electrical work

Is Germany still exporting arms to countries involved in Yemen?

Why is a set not a partition of itself?

Is 12 minutes connection in Bristol Temple Meads long enough?

Anabelian geometry ~ higher category theory

Is there anything special about -1 (0xFFFFFFFF) regarding ADC?

Conditional probability - sum of dice is even given that at least one is a five

If current results hold, Man City would win PL title

Tikz draw contour without some edges, and fill

How can I answer high-school writing prompts without sounding weird and fake?

How do employ ' ("prime") in math mode at the correct depth?

declared variable inside void setup is forgotten in void loop

How to cope with regret and shame about not fully utilizing opportunities during PhD?

Why do I get two different answers when solving for arclength?

Integration problem. Two different answers.What am I missing when solving this integral with trigonometric substition?Arclength of parametric curveWhy are these two answers different?Indefinite integral vs definite integral: Why the different answers?Different answers for integral of $sin^3x$Equality of tw0 arclengthsA definite integral with two different answersLoophole? I'm getting 2 different answers when solving a differential equation in 2 different methodsWhy are these two ways of measuring the length of the groove in a phonograph record different?

3

$begingroup$

I am given that $fracdxdt=8t*cos(t)$ and $fracdydt=8t*sin(t)$. I tried solving for the arclength from t=0 to t=1.

Method 1:
Arclength = $int_0^1 sqrt(fracdxdt)^2+(fracdydt)^2 dx$ = 4.

Method 2:

Arclength = $int_0^1 sqrt1+(fracdydx)^2 dx$. However, when I solve using method 2, I get 1.22619, when the answer should be 4. What is causing this difference?

integration arc-length

share|cite|improve this question

asked 40 mins ago

JayJay

755

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  About your second method: Can you tell us under what conditions it is guaranteed to work? [You can click "edit" below your question to add your answer to this question.] Also: exactly how did you compute $dy/dx$?
  $endgroup$
  – John Hughes
  35 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your expressions are equally. The problem is likely to be with your steps. Edit your question to show how you solve each of them
  $endgroup$
  – Tojrah
  34 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$.
  $endgroup$
  – MathIsFun
  33 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

I am given that $fracdxdt=8t*cos(t)$ and $fracdydt=8t*sin(t)$. I tried solving for the arclength from t=0 to t=1.

Method 1:
Arclength = $int_0^1 sqrt(fracdxdt)^2+(fracdydt)^2 dx$ = 4.

Method 2:

Arclength = $int_0^1 sqrt1+(fracdydx)^2 dx$. However, when I solve using method 2, I get 1.22619, when the answer should be 4. What is causing this difference?

integration arc-length

share|cite|improve this question

asked 40 mins ago

JayJay

755

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  About your second method: Can you tell us under what conditions it is guaranteed to work? [You can click "edit" below your question to add your answer to this question.] Also: exactly how did you compute $dy/dx$?
  $endgroup$
  – John Hughes
  35 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your expressions are equally. The problem is likely to be with your steps. Edit your question to show how you solve each of them
  $endgroup$
  – Tojrah
  34 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$.
  $endgroup$
  – MathIsFun
  33 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

I am given that $fracdxdt=8t*cos(t)$ and $fracdydt=8t*sin(t)$. I tried solving for the arclength from t=0 to t=1.

Method 1:
Arclength = $int_0^1 sqrt(fracdxdt)^2+(fracdydt)^2 dx$ = 4.

Method 2:

Arclength = $int_0^1 sqrt1+(fracdydx)^2 dx$. However, when I solve using method 2, I get 1.22619, when the answer should be 4. What is causing this difference?

integration arc-length

share|cite|improve this question

asked 40 mins ago

JayJay

755

$endgroup$

share|cite|improve this question

asked 40 mins ago

JayJay

755

share|cite|improve this question

asked 40 mins ago

JayJay

755

asked 40 mins ago

JayJay

755

asked 40 mins ago

JayJay

755

3 Answers
3

active

oldest

votes

5

$begingroup$

Your second formula applies when you see $y$ as a function of $x$; you don't say how you found $dy/dx$.

Playing a bit loose with differentials, we have
$$
fracdydx=fracfracdydtfracdxdt=frac8tsin t8tcos t=tan t.
$$
Then
$$
sqrt1+left(fracdydx right)^2,dx=sqrt1+tan ^2 t ,dx=frac1cos t,dx
=frac1cos t,8t,cos t,dt=8t,dt.
$$
So your second integral is (now the limits are on $t$, note that we don't easily know the limits on $x$)
$$
int_0^18t,dt = 4.
$$

share|cite|improve this answer

answered 24 mins ago

Martin ArgeramiMartin Argerami

131k1284186

$endgroup$

add a comment | 

1

$begingroup$

Your first method requires a change. (It is $dt$ not $dx$)

$$I = int^1_0 sqrt(fracdxdt)^2 + (fracdydt)^2dt = int^1_0 sqrt(8t)^2(cos^2t + sin^2t)dt =int^1_0 8tdt = 4[t^2]^1_0 = 4$$

Now, for the 2nd method.

It is actually an equivalence of the first one. It can be deduced like this.

$$int sqrt(fracdxdt)^2 + (fracdydt)^2dt = int fracdxdtsqrt1 + frac(fracdydt)^2(fracdxdt)^2dt = intsqrt1+(fracdydx)^2dx$$

So, the second method also yields 4.

share|cite|improve this answer

answered 21 mins ago

Ak19Ak19

1,38410

$endgroup$

add a comment | 

0

$begingroup$

The second method should give you the correct answer as well.

Note that $$ sqrt1+(fracdydx)^2 dx =sqrt 1+tan^2(t)(8tcos(t))dt$$

so the arc length is $$int _0^1 sqrt 1+tan^2(t)(8tcos(t))dt = int _0^1 8tdt=4$$

share|cite|improve this answer

answered 14 mins ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

42.9k42061

$endgroup$

add a comment | 

Your Answer

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

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3223929%2fwhy-do-i-get-two-different-answers-when-solving-for-arclength%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

5

$begingroup$

Your second formula applies when you see $y$ as a function of $x$; you don't say how you found $dy/dx$.

Playing a bit loose with differentials, we have
$$
fracdydx=fracfracdydtfracdxdt=frac8tsin t8tcos t=tan t.
$$
Then
$$
sqrt1+left(fracdydx right)^2,dx=sqrt1+tan ^2 t ,dx=frac1cos t,dx
=frac1cos t,8t,cos t,dt=8t,dt.
$$
So your second integral is (now the limits are on $t$, note that we don't easily know the limits on $x$)
$$
int_0^18t,dt = 4.
$$

share|cite|improve this answer

answered 24 mins ago

Martin ArgeramiMartin Argerami

131k1284186

$endgroup$

add a comment | 

5

$begingroup$

Your second formula applies when you see $y$ as a function of $x$; you don't say how you found $dy/dx$.

Playing a bit loose with differentials, we have
$$
fracdydx=fracfracdydtfracdxdt=frac8tsin t8tcos t=tan t.
$$
Then
$$
sqrt1+left(fracdydx right)^2,dx=sqrt1+tan ^2 t ,dx=frac1cos t,dx
=frac1cos t,8t,cos t,dt=8t,dt.
$$
So your second integral is (now the limits are on $t$, note that we don't easily know the limits on $x$)
$$
int_0^18t,dt = 4.
$$

share|cite|improve this answer

answered 24 mins ago

Martin ArgeramiMartin Argerami

131k1284186

$endgroup$

add a comment | 

$begingroup$

Your second formula applies when you see $y$ as a function of $x$; you don't say how you found $dy/dx$.

Playing a bit loose with differentials, we have
$$
fracdydx=fracfracdydtfracdxdt=frac8tsin t8tcos t=tan t.
$$
Then
$$
sqrt1+left(fracdydx right)^2,dx=sqrt1+tan ^2 t ,dx=frac1cos t,dx
=frac1cos t,8t,cos t,dt=8t,dt.
$$
So your second integral is (now the limits are on $t$, note that we don't easily know the limits on $x$)
$$
int_0^18t,dt = 4.
$$

share|cite|improve this answer

answered 24 mins ago

Martin ArgeramiMartin Argerami

131k1284186

$endgroup$

share|cite|improve this answer

answered 24 mins ago

Martin ArgeramiMartin Argerami

131k1284186

answered 24 mins ago

Martin ArgeramiMartin Argerami

131k1284186

answered 24 mins ago

Martin ArgeramiMartin Argerami

131k1284186

1

$begingroup$

Your first method requires a change. (It is $dt$ not $dx$)

$$I = int^1_0 sqrt(fracdxdt)^2 + (fracdydt)^2dt = int^1_0 sqrt(8t)^2(cos^2t + sin^2t)dt =int^1_0 8tdt = 4[t^2]^1_0 = 4$$

Now, for the 2nd method.

It is actually an equivalence of the first one. It can be deduced like this.

$$int sqrt(fracdxdt)^2 + (fracdydt)^2dt = int fracdxdtsqrt1 + frac(fracdydt)^2(fracdxdt)^2dt = intsqrt1+(fracdydx)^2dx$$

So, the second method also yields 4.

share|cite|improve this answer

answered 21 mins ago

Ak19Ak19

1,38410

$endgroup$

add a comment | 

1

$begingroup$

Your first method requires a change. (It is $dt$ not $dx$)

$$I = int^1_0 sqrt(fracdxdt)^2 + (fracdydt)^2dt = int^1_0 sqrt(8t)^2(cos^2t + sin^2t)dt =int^1_0 8tdt = 4[t^2]^1_0 = 4$$

Now, for the 2nd method.

It is actually an equivalence of the first one. It can be deduced like this.

$$int sqrt(fracdxdt)^2 + (fracdydt)^2dt = int fracdxdtsqrt1 + frac(fracdydt)^2(fracdxdt)^2dt = intsqrt1+(fracdydx)^2dx$$

So, the second method also yields 4.

share|cite|improve this answer

answered 21 mins ago

Ak19Ak19

1,38410

$endgroup$

add a comment | 

$begingroup$

Your first method requires a change. (It is $dt$ not $dx$)

$$I = int^1_0 sqrt(fracdxdt)^2 + (fracdydt)^2dt = int^1_0 sqrt(8t)^2(cos^2t + sin^2t)dt =int^1_0 8tdt = 4[t^2]^1_0 = 4$$

Now, for the 2nd method.

It is actually an equivalence of the first one. It can be deduced like this.

$$int sqrt(fracdxdt)^2 + (fracdydt)^2dt = int fracdxdtsqrt1 + frac(fracdydt)^2(fracdxdt)^2dt = intsqrt1+(fracdydx)^2dx$$

So, the second method also yields 4.

share|cite|improve this answer

answered 21 mins ago

Ak19Ak19

1,38410

$endgroup$

share|cite|improve this answer

answered 21 mins ago

Ak19Ak19

1,38410

answered 21 mins ago

Ak19Ak19

1,38410

answered 21 mins ago

Ak19Ak19

1,38410

0

$begingroup$

The second method should give you the correct answer as well.

Note that $$ sqrt1+(fracdydx)^2 dx =sqrt 1+tan^2(t)(8tcos(t))dt$$

so the arc length is $$int _0^1 sqrt 1+tan^2(t)(8tcos(t))dt = int _0^1 8tdt=4$$

share|cite|improve this answer

answered 14 mins ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

42.9k42061

$endgroup$

add a comment | 

0

$begingroup$

The second method should give you the correct answer as well.

Note that $$ sqrt1+(fracdydx)^2 dx =sqrt 1+tan^2(t)(8tcos(t))dt$$

so the arc length is $$int _0^1 sqrt 1+tan^2(t)(8tcos(t))dt = int _0^1 8tdt=4$$

share|cite|improve this answer

answered 14 mins ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

42.9k42061

$endgroup$

add a comment | 

$begingroup$

The second method should give you the correct answer as well.

Note that $$ sqrt1+(fracdydx)^2 dx =sqrt 1+tan^2(t)(8tcos(t))dt$$

so the arc length is $$int _0^1 sqrt 1+tan^2(t)(8tcos(t))dt = int _0^1 8tdt=4$$

share|cite|improve this answer

answered 14 mins ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

42.9k42061

$endgroup$

share|cite|improve this answer

answered 14 mins ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

42.9k42061

answered 14 mins ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

42.9k42061

answered 14 mins ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

42.9k42061