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Calculus II Question



The Next CEO of Stack OverflowLength of an AstroidUnderstanding this calculus simplificationIntegration problem: $int x^{2} -x 4^{-x^{2}} dx$Finding the parametric form of a standard equationApplication of “twice the integral” even if the function is not graphically even?Find the length of the parametric curveFind the exact length of the parametric curve(Not sure what I'm doing wrong)Calculus 2 moments question.The length of a parametric curveParametric curve length - calculus












3












$begingroup$


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



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






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.







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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
    3 hours ago








  • 1




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






  • 1




    $begingroup$
    @JohnOmielan that’s exactly what’s wrong
    $endgroup$
    – Shalop
    2 hours ago
















3












$begingroup$


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



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






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.







$endgroup$








  • 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
    3 hours ago








  • 1




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






  • 1




    $begingroup$
    @JohnOmielan that’s exactly what’s wrong
    $endgroup$
    – Shalop
    2 hours ago














3












3








3





$begingroup$


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



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






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.







$endgroup$




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




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 2 hours ago









rash

595116




595116






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 3 hours ago









curiousengcuriouseng

235




235




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
    3 hours ago








  • 1




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






  • 1




    $begingroup$
    @JohnOmielan that’s exactly what’s wrong
    $endgroup$
    – Shalop
    2 hours 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
    3 hours ago








  • 1




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






  • 1




    $begingroup$
    @JohnOmielan that’s exactly what’s wrong
    $endgroup$
    – Shalop
    2 hours 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
3 hours 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
3 hours ago






1




1




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




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




1




1




$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
2 hours ago




$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
2 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

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
    2 hours ago










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



















3












$begingroup$

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









$endgroup$














    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    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
      2 hours ago










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
















    3












    $begingroup$

    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
      2 hours ago










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














    3












    3








    3





    $begingroup$

    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 2 hours ago









    Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost

    537217




    537217












    • $begingroup$
      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
      2 hours ago










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










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












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




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











    3












    $begingroup$

    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









    $endgroup$


















      3












      $begingroup$

      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









      $endgroup$
















        3












        3








        3





        $begingroup$

        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 1 hour ago









        heropupheropup

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