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
$$int_0^2sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
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:
$$left(frac{dx}{dt}right) = 24cdot t^3 $$
$$left(frac{dy}{dt}right) = 24cdot t^5 $$
$$int_0^2sqrt{left(24cdot t^3right)^2+left(24cdot t^5right)^2}dt$$
$$int_0^2sqrt{left(576cdot t^6right)+left(576cdot t^10right)}dt$$
$$int_0^2sqrt{left(576cdot t^6right) cdot left(1+t^4right)}dt$$
$$24+int_0^2sqrt{left(t^6right) cdot left(1+t^4right)}dt$$



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










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  • 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
    51 mins ago






  • 1




    $begingroup$
    @JohnOmielan that’s exactly what’s wrong
    $endgroup$
    – Shalop
    49 mins ago
















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
$$int_0^2sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
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:
$$left(frac{dx}{dt}right) = 24cdot t^3 $$
$$left(frac{dy}{dt}right) = 24cdot t^5 $$
$$int_0^2sqrt{left(24cdot t^3right)^2+left(24cdot t^5right)^2}dt$$
$$int_0^2sqrt{left(576cdot t^6right)+left(576cdot t^10right)}dt$$
$$int_0^2sqrt{left(576cdot t^6right) cdot left(1+t^4right)}dt$$
$$24+int_0^2sqrt{left(t^6right) cdot left(1+t^4right)}dt$$



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










share|cite|improve this question









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
    51 mins ago






  • 1




    $begingroup$
    @JohnOmielan that’s exactly what’s wrong
    $endgroup$
    – Shalop
    49 mins ago














2












2








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
$$int_0^2sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
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:
$$left(frac{dx}{dt}right) = 24cdot t^3 $$
$$left(frac{dy}{dt}right) = 24cdot t^5 $$
$$int_0^2sqrt{left(24cdot t^3right)^2+left(24cdot t^5right)^2}dt$$
$$int_0^2sqrt{left(576cdot t^6right)+left(576cdot t^10right)}dt$$
$$int_0^2sqrt{left(576cdot t^6right) cdot left(1+t^4right)}dt$$
$$24+int_0^2sqrt{left(t^6right) cdot left(1+t^4right)}dt$$



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










share|cite|improve this question









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.







$endgroup$




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
$$int_0^2sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
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:
$$left(frac{dx}{dt}right) = 24cdot t^3 $$
$$left(frac{dy}{dt}right) = 24cdot t^5 $$
$$int_0^2sqrt{left(24cdot t^3right)^2+left(24cdot t^5right)^2}dt$$
$$int_0^2sqrt{left(576cdot t^6right)+left(576cdot t^10right)}dt$$
$$int_0^2sqrt{left(576cdot t^6right) cdot left(1+t^4right)}dt$$
$$24+int_0^2sqrt{left(t^6right) cdot left(1+t^4right)}dt$$



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







calculus integration






share|cite|improve this question









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









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




share|cite|improve this question








edited 45 mins ago









rash

585116




585116






New contributor




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Check out our Code of Conduct.









asked 1 hour ago









curiousengcuriouseng

134




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.





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.






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








  • 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
    51 mins ago






  • 1




    $begingroup$
    @JohnOmielan that’s exactly what’s wrong
    $endgroup$
    – Shalop
    49 mins ago














  • 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
    51 mins ago






  • 1




    $begingroup$
    @JohnOmielan that’s exactly what’s wrong
    $endgroup$
    – Shalop
    49 mins ago








3




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






$begingroup$
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




1




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




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




1




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




$begingroup$
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




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
51 mins ago




$begingroup$
@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
51 mins ago




1




1




$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
49 mins ago




$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
49 mins ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

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



Which gives us:



$$int_0^2 24sqrt{t^6+t^{10}}dt$$



Which, when integrated, gives us: $$68sqrt{17}-4$$



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









$endgroup$













  • $begingroup$
    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
    52 mins ago










  • $begingroup$
    @curiouseng You are very welcome, regards!
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    52 mins ago



















2












$begingroup$

Line 4 should read $$int_{t=0}^2 sqrt{576 t^6 + 576 t^{10}} , dt.$$ This is a typesetting error.



Line 5 is correct.



Line 6 should read $$24 int_{t=0}^2 sqrt{t^6 (1+t^4)} , dt.$$ 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:
$$begin{align*}
24 int_{t=0}^2 sqrt{t^6(1+t^4)} , dt
&= 24 int_{t=0}^2 t^3 sqrt{1+t^4} , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_{u=1}^{17} sqrt{u} , du \
&= 6 left[frac{2u^{3/2}}{3} right]_{u=0}^{17} \
&= 4 (17^{3/2} - 1) \
&= 68 sqrt{17} - 4.
end{align*}$$






share|cite|improve this answer









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    2 Answers
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    2 Answers
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    active

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    2












    $begingroup$

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



    Which gives us:



    $$int_0^2 24sqrt{t^6+t^{10}}dt$$



    Which, when integrated, gives us: $$68sqrt{17}-4$$



    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









    $endgroup$













    • $begingroup$
      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
      52 mins ago










    • $begingroup$
      @curiouseng You are very welcome, regards!
      $endgroup$
      – Bertrand Wittgenstein's Ghost
      52 mins ago
















    2












    $begingroup$

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



    Which gives us:



    $$int_0^2 24sqrt{t^6+t^{10}}dt$$



    Which, when integrated, gives us: $$68sqrt{17}-4$$



    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









    $endgroup$













    • $begingroup$
      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
      52 mins ago










    • $begingroup$
      @curiouseng You are very welcome, regards!
      $endgroup$
      – Bertrand Wittgenstein's Ghost
      52 mins ago














    2












    2








    2





    $begingroup$

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



    Which gives us:



    $$int_0^2 24sqrt{t^6+t^{10}}dt$$



    Which, when integrated, gives us: $$68sqrt{17}-4$$



    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









    $endgroup$



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



    Which gives us:



    $$int_0^2 24sqrt{t^6+t^{10}}dt$$



    Which, when integrated, gives us: $$68sqrt{17}-4$$



    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












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost

    527217




    527217












    • $begingroup$
      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
      52 mins ago










    • $begingroup$
      @curiouseng You are very welcome, regards!
      $endgroup$
      – Bertrand Wittgenstein's Ghost
      52 mins ago


















    • $begingroup$
      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
      52 mins ago










    • $begingroup$
      @curiouseng You are very welcome, regards!
      $endgroup$
      – Bertrand Wittgenstein's Ghost
      52 mins ago
















    $begingroup$
    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
    52 mins ago




    $begingroup$
    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
    52 mins ago












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




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











    2












    $begingroup$

    Line 4 should read $$int_{t=0}^2 sqrt{576 t^6 + 576 t^{10}} , dt.$$ This is a typesetting error.



    Line 5 is correct.



    Line 6 should read $$24 int_{t=0}^2 sqrt{t^6 (1+t^4)} , dt.$$ 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:
    $$begin{align*}
    24 int_{t=0}^2 sqrt{t^6(1+t^4)} , dt
    &= 24 int_{t=0}^2 t^3 sqrt{1+t^4} , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
    &= 6 int_{u=1}^{17} sqrt{u} , du \
    &= 6 left[frac{2u^{3/2}}{3} right]_{u=0}^{17} \
    &= 4 (17^{3/2} - 1) \
    &= 68 sqrt{17} - 4.
    end{align*}$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Line 4 should read $$int_{t=0}^2 sqrt{576 t^6 + 576 t^{10}} , dt.$$ This is a typesetting error.



      Line 5 is correct.



      Line 6 should read $$24 int_{t=0}^2 sqrt{t^6 (1+t^4)} , dt.$$ 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:
      $$begin{align*}
      24 int_{t=0}^2 sqrt{t^6(1+t^4)} , dt
      &= 24 int_{t=0}^2 t^3 sqrt{1+t^4} , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
      &= 6 int_{u=1}^{17} sqrt{u} , du \
      &= 6 left[frac{2u^{3/2}}{3} right]_{u=0}^{17} \
      &= 4 (17^{3/2} - 1) \
      &= 68 sqrt{17} - 4.
      end{align*}$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Line 4 should read $$int_{t=0}^2 sqrt{576 t^6 + 576 t^{10}} , dt.$$ This is a typesetting error.



        Line 5 is correct.



        Line 6 should read $$24 int_{t=0}^2 sqrt{t^6 (1+t^4)} , dt.$$ 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:
        $$begin{align*}
        24 int_{t=0}^2 sqrt{t^6(1+t^4)} , dt
        &= 24 int_{t=0}^2 t^3 sqrt{1+t^4} , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
        &= 6 int_{u=1}^{17} sqrt{u} , du \
        &= 6 left[frac{2u^{3/2}}{3} right]_{u=0}^{17} \
        &= 4 (17^{3/2} - 1) \
        &= 68 sqrt{17} - 4.
        end{align*}$$






        share|cite|improve this answer









        $endgroup$



        Line 4 should read $$int_{t=0}^2 sqrt{576 t^6 + 576 t^{10}} , dt.$$ This is a typesetting error.



        Line 5 is correct.



        Line 6 should read $$24 int_{t=0}^2 sqrt{t^6 (1+t^4)} , dt.$$ 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:
        $$begin{align*}
        24 int_{t=0}^2 sqrt{t^6(1+t^4)} , dt
        &= 24 int_{t=0}^2 t^3 sqrt{1+t^4} , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
        &= 6 int_{u=1}^{17} sqrt{u} , du \
        &= 6 left[frac{2u^{3/2}}{3} right]_{u=0}^{17} \
        &= 4 (17^{3/2} - 1) \
        &= 68 sqrt{17} - 4.
        end{align*}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 24 mins ago









        heropupheropup

        64.8k764103




        64.8k764103






















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