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Parametric curve length - calculus [on hold]



The Next CEO of Stack OverflowDetect “cusp” in parametric curveFind the length of the parametric curve (Difficult)Parametric curve parametriced by lengthCompute the length of a parametric curve.Arc Length parametric curveSampling a curve (parametric)Arc Length with Parametric EquationsFind the length of the parametric curveDetermine the length of the Parametric Curve given by the set of parametric equations.Length of a parametric curve formula: What does the integral represent?










2












$begingroup$


Find the length of the following parametric curve.



$x = 5 + frac92 t^3$, $y = 4 + 3 t^frac92$, $0 leq t leq 2$.



I used integration and after some point I got lost :( What are the steps?






calculus parametric







share|cite|improve this question











$endgroup$



put on hold as off-topic by Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    Is this $$x=5+frac92t^3,y=4+3t^9/2$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    2 days ago
















2












$begingroup$


Find the length of the following parametric curve.



$x = 5 + frac92 t^3$, $y = 4 + 3 t^frac92$, $0 leq t leq 2$.



I used integration and after some point I got lost :( What are the steps?






calculus parametric







share|cite|improve this question











$endgroup$



put on hold as off-topic by Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    Is this $$x=5+frac92t^3,y=4+3t^9/2$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    2 days ago














2












2








2





$begingroup$


Find the length of the following parametric curve.



$x = 5 + frac92 t^3$, $y = 4 + 3 t^frac92$, $0 leq t leq 2$.



I used integration and after some point I got lost :( What are the steps?






calculus parametric







share|cite|improve this question











$endgroup$




Find the length of the following parametric curve.



$x = 5 + frac92 t^3$, $y = 4 + 3 t^frac92$, $0 leq t leq 2$.



I used integration and after some point I got lost :( What are the steps?






calculus parametric




calculus parametric






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Peter Mortensen

565310




565310










asked 2 days ago









McAMcA

264




264




put on hold as off-topic by Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.







put on hold as off-topic by Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Travis, Jyrki Lahtonen, Leucippus, Eevee Trainer, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.











  • $begingroup$
    Is this $$x=5+frac92t^3,y=4+3t^9/2$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    2 days ago

















  • $begingroup$
    Is this $$x=5+frac92t^3,y=4+3t^9/2$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    2 days ago
















$begingroup$
Is this $$x=5+frac92t^3,y=4+3t^9/2$$?
$endgroup$
– Dr. Sonnhard Graubner
2 days ago





$begingroup$
Is this $$x=5+frac92t^3,y=4+3t^9/2$$?
$endgroup$
– Dr. Sonnhard Graubner
2 days ago











3 Answers
3






active

oldest

votes


















2












$begingroup$

You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
$$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    Apply the formula for arc length, we get
    $$
    int_0^2 frac27t^2,sqrtt^3+12 dt
    $$
    Then we make the change of variable $v=t^3+1$ to get
    $$
    int_1^9 frac 9 2 sqrtv dv = 78.
    $$






    share|cite|improve this answer








    New contributor




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






    $endgroup$




















      2












      $begingroup$

      beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned



      Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^frac32$ is an antiderivative of $f(t)=frac272t^2(1+t^3)^frac12$.






      share|cite|improve this answer











      $endgroup$



















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
        $$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$






        share|cite|improve this answer









        $endgroup$

















          2












          $begingroup$

          You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
          $$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$






          share|cite|improve this answer









          $endgroup$















            2












            2








            2





            $begingroup$

            You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
            $$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$






            share|cite|improve this answer









            $endgroup$



            You must use the formula $$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
            $$dx=frac923t^2dt$$ and $$dy=3cdot frac92t^7/2dt$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            78.4k42867




            78.4k42867





















                2












                $begingroup$

                Apply the formula for arc length, we get
                $$
                int_0^2 frac27t^2,sqrtt^3+12 dt
                $$
                Then we make the change of variable $v=t^3+1$ to get
                $$
                int_1^9 frac 9 2 sqrtv dv = 78.
                $$






                share|cite|improve this answer








                New contributor




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






                $endgroup$

















                  2












                  $begingroup$

                  Apply the formula for arc length, we get
                  $$
                  int_0^2 frac27t^2,sqrtt^3+12 dt
                  $$
                  Then we make the change of variable $v=t^3+1$ to get
                  $$
                  int_1^9 frac 9 2 sqrtv dv = 78.
                  $$






                  share|cite|improve this answer








                  New contributor




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






                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    Apply the formula for arc length, we get
                    $$
                    int_0^2 frac27t^2,sqrtt^3+12 dt
                    $$
                    Then we make the change of variable $v=t^3+1$ to get
                    $$
                    int_1^9 frac 9 2 sqrtv dv = 78.
                    $$






                    share|cite|improve this answer








                    New contributor




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






                    $endgroup$



                    Apply the formula for arc length, we get
                    $$
                    int_0^2 frac27t^2,sqrtt^3+12 dt
                    $$
                    Then we make the change of variable $v=t^3+1$ to get
                    $$
                    int_1^9 frac 9 2 sqrtv dv = 78.
                    $$







                    share|cite|improve this answer








                    New contributor




                    EagleToLearn 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 answer



                    share|cite|improve this answer






                    New contributor




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









                    answered 2 days ago









                    EagleToLearnEagleToLearn

                    534




                    534




                    New contributor




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





                    New contributor





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






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





















                        2












                        $begingroup$

                        beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned



                        Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^frac32$ is an antiderivative of $f(t)=frac272t^2(1+t^3)^frac12$.






                        share|cite|improve this answer











                        $endgroup$

















                          2












                          $begingroup$

                          beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned



                          Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^frac32$ is an antiderivative of $f(t)=frac272t^2(1+t^3)^frac12$.






                          share|cite|improve this answer











                          $endgroup$















                            2












                            2








                            2





                            $begingroup$

                            beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned



                            Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^frac32$ is an antiderivative of $f(t)=frac272t^2(1+t^3)^frac12$.






                            share|cite|improve this answer











                            $endgroup$



                            beginalignedL&=int_0^2 sqrtfrac7294t^4+frac7294t^7dt\&=int_0^2sqrtfrac7294t^4(1+t^3)dt\&=frac272int_0^2t^2(1+t^3)^frac12dt\&=3(1+t^3)^frac32big]_0^2endaligned



                            Made the leap from the third line to the fourth line by recognizing that $F(t)=3(1+t^3)^frac32$ is an antiderivative of $f(t)=frac272t^2(1+t^3)^frac12$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 2 days ago

























                            answered 2 days ago









                            Matt A PeltoMatt A Pelto

                            2,667621




                            2,667621













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