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

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?I am getting two different answers for a basic integration problemDifferent 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?

8

$begingroup$

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

Method 1:
$$textArclength = int_0^1 sqrtleft(fracdxdtright)^2+left(fracdydtright)^2 dx = 4.$$

Method 2:

$$textArclength = int_0^1 sqrt1+left(fracdydxright)^2 dx.$$ 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

edited May 13 at 10:43

David K

56.6k345126

asked May 13 at 1:33

JayJay

1006

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  May 13 at 1:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  May 13 at 1:39
  

  
  
  
  


• 
  
  
  

  
  15
  

  
  
  
  
  
  

  
  $begingroup$
  The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$.
  $endgroup$
  – MathIsFun
  May 13 at 1:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A previous edit corrected the $dx$ in the first formula to $dt$ (among many other changes, mostly unnecessary). This error is a significant part of the question, so I restored it. The appropriate way to deal with an error like this is to inform the OP about it in an answer (as someone did) rather than stealthily "fixing" the problem.
  $endgroup$
  – David K
  May 13 at 10:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why is it a significant part of the question? It is not, if OP had integrated the first formulare w.r.t. to $x$, then his result would depend on $t$.
  $endgroup$
  – infinitezero
  May 13 at 14:40
  

  
  
  
  

add a comment | 

8

$begingroup$

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

Method 1:
$$textArclength = int_0^1 sqrtleft(fracdxdtright)^2+left(fracdydtright)^2 dx = 4.$$

Method 2:

$$textArclength = int_0^1 sqrt1+left(fracdydxright)^2 dx.$$ 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

edited May 13 at 10:43

David K

56.6k345126

asked May 13 at 1:33

JayJay

1006

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  May 13 at 1:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  May 13 at 1:39
  

  
  
  
  


• 
  
  
  

  
  15
  

  
  
  
  
  
  

  
  $begingroup$
  The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$.
  $endgroup$
  – MathIsFun
  May 13 at 1:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A previous edit corrected the $dx$ in the first formula to $dt$ (among many other changes, mostly unnecessary). This error is a significant part of the question, so I restored it. The appropriate way to deal with an error like this is to inform the OP about it in an answer (as someone did) rather than stealthily "fixing" the problem.
  $endgroup$
  – David K
  May 13 at 10:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why is it a significant part of the question? It is not, if OP had integrated the first formulare w.r.t. to $x$, then his result would depend on $t$.
  $endgroup$
  – infinitezero
  May 13 at 14:40
  

  
  
  
  

add a comment | 

$begingroup$

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

Method 1:
$$textArclength = int_0^1 sqrtleft(fracdxdtright)^2+left(fracdydtright)^2 dx = 4.$$

Method 2:

$$textArclength = int_0^1 sqrt1+left(fracdydxright)^2 dx.$$ 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

edited May 13 at 10:43

David K

56.6k345126

asked May 13 at 1:33

JayJay

1006

$endgroup$

share|cite|improve this question

edited May 13 at 10:43

David K

56.6k345126

asked May 13 at 1:33

JayJay

1006

share|cite|improve this question

edited May 13 at 10:43

David K

56.6k345126

asked May 13 at 1:33

JayJay

1006

edited May 13 at 10:43

David K

56.6k345126

edited May 13 at 10:43

David K

56.6k345126

asked May 13 at 1:33

JayJay

1006

asked May 13 at 1:33

JayJay

1006

3 Answers
3

active

oldest

votes

17

$begingroup$

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 May 13 at 1:49

Martin ArgeramiMartin Argerami

132k1285188

$endgroup$

add a comment | 

3

$begingroup$

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 May 13 at 1:52

Ak19Ak19

2,160212

$endgroup$

add a comment | 

3

$begingroup$

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 May 13 at 1:59

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.1k42162

$endgroup$

add a comment | 

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17

$begingroup$

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 May 13 at 1:49

Martin ArgeramiMartin Argerami

132k1285188

$endgroup$

add a comment | 

17

$begingroup$

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 May 13 at 1:49

Martin ArgeramiMartin Argerami

132k1285188

$endgroup$

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 May 13 at 1:49

Martin ArgeramiMartin Argerami

132k1285188

$endgroup$

share|cite|improve this answer

answered May 13 at 1:49

Martin ArgeramiMartin Argerami

132k1285188

answered May 13 at 1:49

Martin ArgeramiMartin Argerami

132k1285188

answered May 13 at 1:49

Martin ArgeramiMartin Argerami

132k1285188

3

$begingroup$

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 May 13 at 1:52

Ak19Ak19

2,160212

$endgroup$

add a comment | 

3

$begingroup$

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 May 13 at 1:52

Ak19Ak19

2,160212

$endgroup$

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 May 13 at 1:52

Ak19Ak19

2,160212

$endgroup$

share|cite|improve this answer

answered May 13 at 1:52

Ak19Ak19

2,160212

answered May 13 at 1:52

Ak19Ak19

2,160212

answered May 13 at 1:52

Ak19Ak19

2,160212

3

$begingroup$

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 May 13 at 1:59

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.1k42162

$endgroup$

add a comment | 

3

$begingroup$

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 May 13 at 1:59

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.1k42162

$endgroup$

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 May 13 at 1:59

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.1k42162

$endgroup$

share|cite|improve this answer

answered May 13 at 1:59

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.1k42162

answered May 13 at 1:59

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.1k42162

answered May 13 at 1:59

Mohammad Riazi-KermaniMohammad Riazi-Kermani

44.1k42162