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What is this length of this curve using a calculator?

1

$begingroup$

I have this question:

Set up an integral that represents the length of the curve.Use a calculator to find the length correct to 4 places.

$$y^2 = lnx$$ and $-1 leq y leq 1$

so implicitly differentiating:

$$2y fracdydx = frac1x$$

$$fracdydx = frac1x2y$$

and $y = sqrtlnx$

So the curve is

$$int_-1^1 sqrt1+(frac12xlnx)^2$$

$$int_-1^1 sqrt1+(frac14x^2lnx^2)$$

Is that right? I can just plug that into a CAD right?

Wolfram exceeded the time allotted... did I set this up incorrectly?

integration

share|cite|improve this question

edited Mar 19 at 21:38

Jwan622

asked Mar 19 at 21:33

Jwan622Jwan622

2,33411632

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure, but shouldn't that be a $ln(x)$ instead of a $ln^2(x)$ in the last line, because you are squaring the square root of the log?
  $endgroup$
  – Seth
  Mar 19 at 22:00
  

  
  
  
  

add a comment | 

1

$begingroup$

I have this question:

Set up an integral that represents the length of the curve.Use a calculator to find the length correct to 4 places.

$$y^2 = lnx$$ and $-1 leq y leq 1$

so implicitly differentiating:

$$2y fracdydx = frac1x$$

$$fracdydx = frac1x2y$$

and $y = sqrtlnx$

So the curve is

$$int_-1^1 sqrt1+(frac12xlnx)^2$$

$$int_-1^1 sqrt1+(frac14x^2lnx^2)$$

Is that right? I can just plug that into a CAD right?

Wolfram exceeded the time allotted... did I set this up incorrectly?

integration

share|cite|improve this question

edited Mar 19 at 21:38

Jwan622

asked Mar 19 at 21:33

Jwan622Jwan622

2,33411632

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure, but shouldn't that be a $ln(x)$ instead of a $ln^2(x)$ in the last line, because you are squaring the square root of the log?
  $endgroup$
  – Seth
  Mar 19 at 22:00
  

  
  
  
  

add a comment | 

$begingroup$

I have this question:

Set up an integral that represents the length of the curve.Use a calculator to find the length correct to 4 places.

$$y^2 = lnx$$ and $-1 leq y leq 1$

so implicitly differentiating:

$$2y fracdydx = frac1x$$

$$fracdydx = frac1x2y$$

and $y = sqrtlnx$

So the curve is

$$int_-1^1 sqrt1+(frac12xlnx)^2$$

$$int_-1^1 sqrt1+(frac14x^2lnx^2)$$

Is that right? I can just plug that into a CAD right?

Wolfram exceeded the time allotted... did I set this up incorrectly?

integration

share|cite|improve this question

edited Mar 19 at 21:38

Jwan622

asked Mar 19 at 21:33

Jwan622Jwan622

2,33411632

$endgroup$

share|cite|improve this question

edited Mar 19 at 21:38

Jwan622

asked Mar 19 at 21:33

Jwan622Jwan622

2,33411632

share|cite|improve this question

edited Mar 19 at 21:38

Jwan622

asked Mar 19 at 21:33

Jwan622Jwan622

2,33411632

asked Mar 19 at 21:33

Jwan622Jwan622

2,33411632

asked Mar 19 at 21:33

Jwan622Jwan622

2,33411632

2 Answers
2

active

oldest

votes

2

$begingroup$

The limits of integration in the OP are not correct. In fact, inasmuch as $log(x)$ is not defined for $xle 0$, the integral in the OP is also not defined.

To proceed correctly, we note that $y=sqrtlog(x)$ for $yin[0,1]$ ($xin [1,e]$) and $y=-sqrtlog(x)$ for $yin [-1,0]$ ($xin [1,e]$). Then, we see that

$$beginalign
textLength of Curve&=2int_1^e sqrt1+left(fracdydxright)^2,dx\\
&=2int_1^e sqrt1+frac14x^2log(x),dx\\
&approx. 4.25523282937328
endalign$$

---

Alternatively, we have $x=e^y^2$ so the

$$beginalign
textLength of Curve&=int_-1^1 sqrt1+left(fracdxdyright)^2,dy\\
&=2int_0^1 sqrt1+4y^2e^2y^2,dy\\
& approx 4.25523282937328
endalign$$

share|cite|improve this answer

edited Mar 20 at 16:07

answered Mar 19 at 21:52

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you change the limit of integration and forget to multiply by 2 at the end?
  $endgroup$
  – Jwan622
  Mar 19 at 22:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you took the integral but forgot to multiply it by 2 right?
  $endgroup$
  – Jwan622
  Mar 20 at 15:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jwan622 Ah, now I understand. Yes, I did forget the factor of $2$. I've edited accordingly.
  $endgroup$
  – Mark Viola
  Mar 20 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think I misunderstood something about what limits of integration to use... if the arc length formula that I use uses dy, then the limits of integration have to be for y. Conversely, if the arc length formula I use uses dx, then the limits have to be for x right?
  $endgroup$
  – Jwan622
  Mar 22 at 19:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jwan622 Yes, that is correct. And that is, in fact, what you will see in the two approaches that I presented herein.
  $endgroup$
  – Mark Viola
  Mar 22 at 19:15
  
  

  
  
  
  

add a comment | 

1

$begingroup$

What would make this easier is to integrate with respect to $y$, and not $x$. You can re-write this as $x=e^y^2$. Now $fracdxdy=e^y^2.2y$. Hence, the length of the curve can be written as $$int_-1^1 sqrt4y^2e^2y^2+1dy$$ You can now plug this into the calculator.

share|cite|improve this answer

edited Mar 19 at 21:58

answered Mar 19 at 21:45

Anju GeorgeAnju George

988

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkViola- Ah if you meant the $4y^2$ factor, I have now added it
  $endgroup$
  – Anju George
  Mar 19 at 21:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you get to $x = e^y^2$ again? What's the rule? YOu can just take everything as an exponent to e?
  $endgroup$
  – Jwan622
  Mar 19 at 22:11
  

  
  
  
  

add a comment | 

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2

$begingroup$

The limits of integration in the OP are not correct. In fact, inasmuch as $log(x)$ is not defined for $xle 0$, the integral in the OP is also not defined.

To proceed correctly, we note that $y=sqrtlog(x)$ for $yin[0,1]$ ($xin [1,e]$) and $y=-sqrtlog(x)$ for $yin [-1,0]$ ($xin [1,e]$). Then, we see that

$$beginalign
textLength of Curve&=2int_1^e sqrt1+left(fracdydxright)^2,dx\\
&=2int_1^e sqrt1+frac14x^2log(x),dx\\
&approx. 4.25523282937328
endalign$$

---

Alternatively, we have $x=e^y^2$ so the

$$beginalign
textLength of Curve&=int_-1^1 sqrt1+left(fracdxdyright)^2,dy\\
&=2int_0^1 sqrt1+4y^2e^2y^2,dy\\
& approx 4.25523282937328
endalign$$

share|cite|improve this answer

edited Mar 20 at 16:07

answered Mar 19 at 21:52

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you change the limit of integration and forget to multiply by 2 at the end?
  $endgroup$
  – Jwan622
  Mar 19 at 22:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you took the integral but forgot to multiply it by 2 right?
  $endgroup$
  – Jwan622
  Mar 20 at 15:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jwan622 Ah, now I understand. Yes, I did forget the factor of $2$. I've edited accordingly.
  $endgroup$
  – Mark Viola
  Mar 20 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think I misunderstood something about what limits of integration to use... if the arc length formula that I use uses dy, then the limits of integration have to be for y. Conversely, if the arc length formula I use uses dx, then the limits have to be for x right?
  $endgroup$
  – Jwan622
  Mar 22 at 19:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jwan622 Yes, that is correct. And that is, in fact, what you will see in the two approaches that I presented herein.
  $endgroup$
  – Mark Viola
  Mar 22 at 19:15
  
  

  
  
  
  

add a comment | 

2

$begingroup$

The limits of integration in the OP are not correct. In fact, inasmuch as $log(x)$ is not defined for $xle 0$, the integral in the OP is also not defined.

To proceed correctly, we note that $y=sqrtlog(x)$ for $yin[0,1]$ ($xin [1,e]$) and $y=-sqrtlog(x)$ for $yin [-1,0]$ ($xin [1,e]$). Then, we see that

$$beginalign
textLength of Curve&=2int_1^e sqrt1+left(fracdydxright)^2,dx\\
&=2int_1^e sqrt1+frac14x^2log(x),dx\\
&approx. 4.25523282937328
endalign$$

---

Alternatively, we have $x=e^y^2$ so the

$$beginalign
textLength of Curve&=int_-1^1 sqrt1+left(fracdxdyright)^2,dy\\
&=2int_0^1 sqrt1+4y^2e^2y^2,dy\\
& approx 4.25523282937328
endalign$$

share|cite|improve this answer

edited Mar 20 at 16:07

answered Mar 19 at 21:52

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you change the limit of integration and forget to multiply by 2 at the end?
  $endgroup$
  – Jwan622
  Mar 19 at 22:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you took the integral but forgot to multiply it by 2 right?
  $endgroup$
  – Jwan622
  Mar 20 at 15:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jwan622 Ah, now I understand. Yes, I did forget the factor of $2$. I've edited accordingly.
  $endgroup$
  – Mark Viola
  Mar 20 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think I misunderstood something about what limits of integration to use... if the arc length formula that I use uses dy, then the limits of integration have to be for y. Conversely, if the arc length formula I use uses dx, then the limits have to be for x right?
  $endgroup$
  – Jwan622
  Mar 22 at 19:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Jwan622 Yes, that is correct. And that is, in fact, what you will see in the two approaches that I presented herein.
  $endgroup$
  – Mark Viola
  Mar 22 at 19:15
  
  

  
  
  
  

add a comment | 

$begingroup$

The limits of integration in the OP are not correct. In fact, inasmuch as $log(x)$ is not defined for $xle 0$, the integral in the OP is also not defined.

To proceed correctly, we note that $y=sqrtlog(x)$ for $yin[0,1]$ ($xin [1,e]$) and $y=-sqrtlog(x)$ for $yin [-1,0]$ ($xin [1,e]$). Then, we see that

$$beginalign
textLength of Curve&=2int_1^e sqrt1+left(fracdydxright)^2,dx\\
&=2int_1^e sqrt1+frac14x^2log(x),dx\\
&approx. 4.25523282937328
endalign$$

---

Alternatively, we have $x=e^y^2$ so the

$$beginalign
textLength of Curve&=int_-1^1 sqrt1+left(fracdxdyright)^2,dy\\
&=2int_0^1 sqrt1+4y^2e^2y^2,dy\\
& approx 4.25523282937328
endalign$$

share|cite|improve this answer

edited Mar 20 at 16:07

answered Mar 19 at 21:52

Mark ViolaMark Viola

134k1278176

$endgroup$

share|cite|improve this answer

edited Mar 20 at 16:07

answered Mar 19 at 21:52

Mark ViolaMark Viola

134k1278176

answered Mar 19 at 21:52

Mark ViolaMark Viola

134k1278176

answered Mar 19 at 21:52

Mark ViolaMark Viola

134k1278176

1

$begingroup$

What would make this easier is to integrate with respect to $y$, and not $x$. You can re-write this as $x=e^y^2$. Now $fracdxdy=e^y^2.2y$. Hence, the length of the curve can be written as $$int_-1^1 sqrt4y^2e^2y^2+1dy$$ You can now plug this into the calculator.

share|cite|improve this answer

edited Mar 19 at 21:58

answered Mar 19 at 21:45

Anju GeorgeAnju George

988

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkViola- Ah if you meant the $4y^2$ factor, I have now added it
  $endgroup$
  – Anju George
  Mar 19 at 21:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you get to $x = e^y^2$ again? What's the rule? YOu can just take everything as an exponent to e?
  $endgroup$
  – Jwan622
  Mar 19 at 22:11
  

  
  
  
  

add a comment | 

1

$begingroup$

What would make this easier is to integrate with respect to $y$, and not $x$. You can re-write this as $x=e^y^2$. Now $fracdxdy=e^y^2.2y$. Hence, the length of the curve can be written as $$int_-1^1 sqrt4y^2e^2y^2+1dy$$ You can now plug this into the calculator.

share|cite|improve this answer

edited Mar 19 at 21:58

answered Mar 19 at 21:45

Anju GeorgeAnju George

988

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkViola- Ah if you meant the $4y^2$ factor, I have now added it
  $endgroup$
  – Anju George
  Mar 19 at 21:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you get to $x = e^y^2$ again? What's the rule? YOu can just take everything as an exponent to e?
  $endgroup$
  – Jwan622
  Mar 19 at 22:11
  

  
  
  
  

add a comment | 

$begingroup$

What would make this easier is to integrate with respect to $y$, and not $x$. You can re-write this as $x=e^y^2$. Now $fracdxdy=e^y^2.2y$. Hence, the length of the curve can be written as $$int_-1^1 sqrt4y^2e^2y^2+1dy$$ You can now plug this into the calculator.

share|cite|improve this answer

edited Mar 19 at 21:58

answered Mar 19 at 21:45

Anju GeorgeAnju George

988

$endgroup$

share|cite|improve this answer

edited Mar 19 at 21:58

answered Mar 19 at 21:45

Anju GeorgeAnju George

988

answered Mar 19 at 21:45

Anju GeorgeAnju George

988

answered Mar 19 at 21:45

Anju GeorgeAnju George

988