Parametric curve length - calculus












3












$begingroup$


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







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$endgroup$












  • $begingroup$
    Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    6 hours ago


















3












$begingroup$


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











$endgroup$












  • $begingroup$
    Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    6 hours ago
















3












3








3





$begingroup$


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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 35 mins ago









Peter Mortensen

565310




565310










asked 6 hours ago









McAMcA

204




204












  • $begingroup$
    Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    6 hours ago




















  • $begingroup$
    Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    6 hours ago


















$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
6 hours ago






$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
6 hours ago












3 Answers
3






active

oldest

votes


















2












$begingroup$

Apply the formula for arc length, we get
$$
int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
$$
Then we make the change of variable $v=t^3+1$ to get
$$
int_1^9 frac 9 2 sqrt{v} dv = 78.
$$






share|cite|improve this answer








New contributor




EagleToLearn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$





















    2












    $begingroup$

    begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}



    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 $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
      $$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Apply the formula for arc length, we get
        $$
        int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
        $$
        Then we make the change of variable $v=t^3+1$ to get
        $$
        int_1^9 frac 9 2 sqrt{v} dv = 78.
        $$






        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
          $$
          int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
          $$
          Then we make the change of variable $v=t^3+1$ to get
          $$
          int_1^9 frac 9 2 sqrt{v} dv = 78.
          $$






          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
            $$
            int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
            $$
            Then we make the change of variable $v=t^3+1$ to get
            $$
            int_1^9 frac 9 2 sqrt{v} dv = 78.
            $$






            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
            $$
            int_0^2 frac{27{{t}^{2}},sqrt{{{t}^{3}}+1}}{2} dt
            $$
            Then we make the change of variable $v=t^3+1$ to get
            $$
            int_1^9 frac 9 2 sqrt{v} dv = 78.
            $$







            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 6 hours ago









            EagleToLearnEagleToLearn

            233




            233




            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












                $begingroup$

                begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}



                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 $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}



                  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 $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}



                    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 $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.






                    share|cite|improve this answer











                    $endgroup$



                    begin{aligned}L&=int_0^2 sqrt{frac{729}4t^4+frac{729}4t^7}dt\&=int_0^2sqrt{frac{729}4t^4(1+t^3)}dt\&=frac{27}2int_0^2t^2(1+t^3)^{frac12}dt\&=3(1+t^3)^{frac32}big]_0^2end{aligned}



                    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 $f(t)=frac{27}2t^2(1+t^3)^{frac12}$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 5 hours ago

























                    answered 6 hours ago









                    Matt A PeltoMatt A Pelto

                    2,667621




                    2,667621























                        1












                        $begingroup$

                        You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
                        $$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
                          $$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
                            $$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$






                            share|cite|improve this answer









                            $endgroup$



                            You must use the formula $$int_{0}^{2}sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2}dt$$
                            $$dx=frac{9}{2}3t^2dt$$ and $$dy=3cdot frac{9}{2}t^{7/2}dt$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 6 hours ago









                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                            78.2k42867




                            78.2k42867






























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