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Calculus II Question

2

$begingroup$

Find the length of the following parametric curve.

$$x(t)=5+6t^4 ,quad y(t)=5+4t^6 ,qquad 0  ≤  t  ≤  2.$$

I used the formula
$$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
And I found
$$frac23cdot 17^3/2+4-frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.

My steps:
$$left(fracdxdtright) = 24cdot t^3 $$
$$left(fracdydtright) = 24cdot t^5 $$
$$int_0^2sqrtleft(24cdot t^3right)^2+left(24cdot t^5right)^2dt$$
$$int_0^2sqrtleft(576cdot t^6right)+left(576cdot t^10right)dt$$
$$int_0^2sqrtleft(576cdot t^6right) cdot left(1+t^4right)dt$$
$$24+int_0^2sqrtleft(t^6right) cdot left(1+t^4right)dt$$

$$frac23cdot 17^3/2+4-frac23$$

calculus integration

share|cite|improve this question

edited 56 mins ago

rash

585116

asked 1 hour ago

curiousengcuriouseng

134

New contributor

curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
  $endgroup$
  – Ross Millikan
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Isn't there a square root missing in your length formula?
  $endgroup$
  – John Wayland Bales
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
  $endgroup$
  – David Peterson
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
  $endgroup$
  – John Omielan
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @JohnOmielan that’s exactly what’s wrong
  $endgroup$
  – Shalop
  1 hour ago
  

  
  
  
  

 | 
show 5 more comments

2

$begingroup$

Find the length of the following parametric curve.

$$x(t)=5+6t^4 ,quad y(t)=5+4t^6 ,qquad 0  ≤  t  ≤  2.$$

I used the formula
$$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
And I found
$$frac23cdot 17^3/2+4-frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.

My steps:
$$left(fracdxdtright) = 24cdot t^3 $$
$$left(fracdydtright) = 24cdot t^5 $$
$$int_0^2sqrtleft(24cdot t^3right)^2+left(24cdot t^5right)^2dt$$
$$int_0^2sqrtleft(576cdot t^6right)+left(576cdot t^10right)dt$$
$$int_0^2sqrtleft(576cdot t^6right) cdot left(1+t^4right)dt$$
$$24+int_0^2sqrtleft(t^6right) cdot left(1+t^4right)dt$$

$$frac23cdot 17^3/2+4-frac23$$

calculus integration

share|cite|improve this question

edited 56 mins ago

rash

585116

asked 1 hour ago

curiousengcuriouseng

134

New contributor

curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
  $endgroup$
  – Ross Millikan
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Isn't there a square root missing in your length formula?
  $endgroup$
  – John Wayland Bales
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
  $endgroup$
  – David Peterson
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
  $endgroup$
  – John Omielan
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @JohnOmielan that’s exactly what’s wrong
  $endgroup$
  – Shalop
  1 hour ago
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

Find the length of the following parametric curve.

$$x(t)=5+6t^4 ,quad y(t)=5+4t^6 ,qquad 0  ≤  t  ≤  2.$$

I used the formula
$$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
And I found
$$frac23cdot 17^3/2+4-frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.

My steps:
$$left(fracdxdtright) = 24cdot t^3 $$
$$left(fracdydtright) = 24cdot t^5 $$
$$int_0^2sqrtleft(24cdot t^3right)^2+left(24cdot t^5right)^2dt$$
$$int_0^2sqrtleft(576cdot t^6right)+left(576cdot t^10right)dt$$
$$int_0^2sqrtleft(576cdot t^6right) cdot left(1+t^4right)dt$$
$$24+int_0^2sqrtleft(t^6right) cdot left(1+t^4right)dt$$

$$frac23cdot 17^3/2+4-frac23$$

calculus integration

share|cite|improve this question

edited 56 mins ago

rash

585116

asked 1 hour ago

curiousengcuriouseng

134

New contributor

curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 56 mins ago

rash

585116

asked 1 hour ago

curiousengcuriouseng

134

New contributor

curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 56 mins ago

rash

585116

asked 1 hour ago

curiousengcuriouseng

134

New contributor

curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 56 mins ago

rash

585116

edited 56 mins ago

rash

585116

asked 1 hour ago

curiousengcuriouseng

134

New contributor

curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

curiousengcuriouseng

134

2 Answers
2

active

oldest

votes

2

$begingroup$

Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$

Which gives us:

$$int_0^2 24sqrtt^6+t^10dt$$

Which, when integrated, gives us: $$68sqrt17-4$$

I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.

share|cite|improve this answer

answered 1 hour ago

Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost

527217

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
  $endgroup$
  – curiouseng
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @curiouseng You are very welcome, regards!
  $endgroup$
  – Bertrand Wittgenstein's Ghost
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.

Line 5 is correct.

Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.

You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
$$beginalign*
24 int_t=0^2 sqrtt^6(1+t^4) , dt
&= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_u=1^17 sqrtu , du \
&= 6 left[frac2u^3/23 right]_u=0^17 \
&= 4 (17^3/2 - 1) \
&= 68 sqrt17 - 4.
endalign*$$

share|cite|improve this answer

answered 35 mins ago

heropupheropup

64.8k764103

$endgroup$

add a comment | 

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2

$begingroup$

Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$

Which gives us:

$$int_0^2 24sqrtt^6+t^10dt$$

Which, when integrated, gives us: $$68sqrt17-4$$

I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.

share|cite|improve this answer

answered 1 hour ago

Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost

527217

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
  $endgroup$
  – curiouseng
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @curiouseng You are very welcome, regards!
  $endgroup$
  – Bertrand Wittgenstein's Ghost
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$

Which gives us:

$$int_0^2 24sqrtt^6+t^10dt$$

Which, when integrated, gives us: $$68sqrt17-4$$

I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.

share|cite|improve this answer

answered 1 hour ago

Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost

527217

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
  $endgroup$
  – curiouseng
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @curiouseng You are very welcome, regards!
  $endgroup$
  – Bertrand Wittgenstein's Ghost
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$

Which gives us:

$$int_0^2 24sqrtt^6+t^10dt$$

Which, when integrated, gives us: $$68sqrt17-4$$

I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.

share|cite|improve this answer

answered 1 hour ago

Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost

527217

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost

527217

answered 1 hour ago

Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost

527217

answered 1 hour ago

Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost

527217

2

$begingroup$

Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.

Line 5 is correct.

Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.

You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
$$beginalign*
24 int_t=0^2 sqrtt^6(1+t^4) , dt
&= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_u=1^17 sqrtu , du \
&= 6 left[frac2u^3/23 right]_u=0^17 \
&= 4 (17^3/2 - 1) \
&= 68 sqrt17 - 4.
endalign*$$

share|cite|improve this answer

answered 35 mins ago

heropupheropup

64.8k764103

$endgroup$

add a comment | 

2

$begingroup$

Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.

Line 5 is correct.

Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.

You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
$$beginalign*
24 int_t=0^2 sqrtt^6(1+t^4) , dt
&= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_u=1^17 sqrtu , du \
&= 6 left[frac2u^3/23 right]_u=0^17 \
&= 4 (17^3/2 - 1) \
&= 68 sqrt17 - 4.
endalign*$$

share|cite|improve this answer

answered 35 mins ago

heropupheropup

64.8k764103

$endgroup$

add a comment | 

$begingroup$

Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.

Line 5 is correct.

Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.

You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
$$beginalign*
24 int_t=0^2 sqrtt^6(1+t^4) , dt
&= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_u=1^17 sqrtu , du \
&= 6 left[frac2u^3/23 right]_u=0^17 \
&= 4 (17^3/2 - 1) \
&= 68 sqrt17 - 4.
endalign*$$

share|cite|improve this answer

answered 35 mins ago

heropupheropup

64.8k764103

$endgroup$

share|cite|improve this answer

answered 35 mins ago

heropupheropup

64.8k764103

answered 35 mins ago

heropupheropup

64.8k764103

answered 35 mins ago

heropupheropup

64.8k764103