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



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






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



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






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 45 mins 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 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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    active

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    2 Answers
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    active

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    $begingroup$

    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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            Knooppunt Holsloot

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            Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
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            Altaar (religie)

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            Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
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            Gregoriusmis

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            Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
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