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

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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|edit

asked 58 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
  53 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
  52 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$.
  $endgroup$
  – MathIsFun
  51 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|edit

asked 58 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
  53 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
  52 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The interval $t=0$ to $1$ does not correspond to the interval $x=0$ to $1$.
  $endgroup$
  – MathIsFun
  51 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|edit

asked 58 mins ago

JayJay

755

$endgroup$

share|cite|edit

asked 58 mins ago

JayJay

755

share|cite|edit

asked 58 mins ago

JayJay

755

asked 58 mins ago

JayJay

755

asked 58 mins ago

JayJay

755

3 Answers
3

active

oldest

votes

4

$begingroup$

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 43 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 39 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 32 mins ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

42.9k42061

$endgroup$

add a comment | 

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4

$begingroup$

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 43 mins ago

Martin ArgeramiMartin Argerami

131k1284186

$endgroup$

add a comment | 

4

$begingroup$

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 43 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 43 mins ago

Martin ArgeramiMartin Argerami

131k1284186

$endgroup$

share|cite|improve this answer

answered 43 mins ago

Martin ArgeramiMartin Argerami

131k1284186

answered 43 mins ago

Martin ArgeramiMartin Argerami

131k1284186

answered 43 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 39 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 39 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 39 mins ago

Ak19Ak19

1,38410

$endgroup$

share|cite|improve this answer

answered 39 mins ago

Ak19Ak19

1,38410

answered 39 mins ago

Ak19Ak19

1,38410

answered 39 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 32 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 32 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 32 mins ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

42.9k42061

$endgroup$

share|cite|improve this answer

answered 32 mins ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

42.9k42061

answered 32 mins ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

42.9k42061

answered 32 mins ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

42.9k42061