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



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






calculus integration







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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
    2 hours 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
















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



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






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
    2 hours 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














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



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






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 1 hour ago









rash

585116




585116






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 2 hours ago









curiousengcuriouseng

135




135




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
    2 hours 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














  • 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
    2 hours 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








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
2 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
2 hours 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
1 hour 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
1 hour ago




1




1




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




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










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





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    2 Answers
    2






    active

    oldest

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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
      1 hour ago










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










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










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










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












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




    $begingroup$
    @curiouseng You are very welcome, regards!
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    1 hour 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 44 mins ago









        heropupheropup

        64.8k764103




        64.8k764103






















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            الفوسفات في المغرب

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            قطار ينقل فوسفات من منطقة خريبكة متجه نحو الدار البيضاء. ديار عمال بمصانع الفوسفات بمدينة آسفي. المحتوى هنا ينقصه الاستشهاد بمصادر. يرجى إيراد مصادر موثوق بها. أي معلومات غير موثقة يمكن التشكيك بها وإزالتها. (أغسطس 2015) الفوسفات في المغرب هو ثالث مصدر عالمي للفوسفات، ويتوفر المغرب على %75 من الاحتياطي العالمي  [بحاجة لمصدر] ويتوزع بنسب متفاوتة بين عدة مناطق أساسية وهي: وادي زم بن جرير بوكراع خريبكة اليوسفية أولاد عبدون كنتور واد الذهب مستقلة. هذه الأخيرة لم تفتح بعد للاستغلال. الصادرات الحامض الفوسفوري P205 تمتص الهند %50 من ا لصادرات الإجمالية آسيا (دون روسيا)%59 أوروبا %30 أمريكا اللاتينية %11 أسمدة صلبة أوروبا 38 % أمريكا اللاتينية %30 آسيا (دون روسيا) 22 % أستراليا 9 % بوابة المغرب هذه بذرة مقالة عن علم الاقتصاد أو موضوع متعلق به بحاجة للتوسيع. شارك في تحريرها. This page is only for reference, If you need detailed information, please check here
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            Four equal circles intersect: What is the area of the small shaded portion and its height

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            3 $begingroup$ In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this. geometry circle share | cite | improve this question edited 2 hours ago Abdulhameed asked 2 hours ago ...
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            جامعة ليفربول

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            جامعة ليفربول University of Liverpool الأسماء السابقة كلية ليفربول الجامعية (1882-1884) جامعة فيكتوريا في ليفربول (1884-1903) معلومات التأسيس 1882 (منذ 137 سنة) الإنتماءات رابطة الجامعات الأوروبية مجموعة راسل جامعات القرميد الأحمر المنحة المالية 143.5 مليون £ النوع جامعة عامة الكليات 6 الموقع الجغرافي إحداثيات 53°24′22″N 2°58′01″W  /  53.406°N 2.967°W  / 53.406; -2.967   المدينة ليفربول المكان   إنجلترا البلد   المملكة المتحدة رقم الهاتف 7942000 151 (44+) إحصاءات الطلاب 17,405 (سنة 2008) عضوية مجموعة رسل [1] أورسيد [2]   طلاب الدراسات العليا 3,860 الموقع www.liverpool.ac.uk تعديل   جامعة ليفربول مبنى فكتوريا جامعة ليفربول (بالإنجليزية: The University of Liverpool )) إحدى الجامعات البريطانية العريقة، تقع في مدينة ليفربول على الساحل الشمالي الغربي من أنجلترا . وهي عضو في مجموعة راسل للجامعات وتعتبر من ضمن جامعات الطوب الأحمر الستة. محتويات ...
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