calculus parametric curve length The Next CEO of Stack OverflowFind the length of the...

What is the result of assigning to std::vector<T>::begin()?

Rotate a column

Inappropriate reference requests from Journal reviewers

Bold, vivid family

Would this house-rule that treats advantage as a +1 to the roll instead (and disadvantage as -1) and allows them to stack be balanced?

Anatomically Correct Strange Women In Ponds Distributing Swords

How to solve a differential equation with a term to a power?

Do I need to enable Dev Hub in my PROD Org?

Is micro rebar a better way to reinforce concrete than rebar?

What was the first Unix version to run on a microcomputer?

What does convergence in distribution "in the Gromov–Hausdorff" sense mean?

How do scammers retract money, while you can’t?

Example of a Mathematician/Physicist whose Other Publications during their PhD eclipsed their PhD Thesis

Which tube will fit a -(700 x 25c) wheel?

Can I equip Skullclamp on a creature I am sacrificing?

Preparing Indesign booklet with .psd graphics for print

Why does standard notation not preserve intervals (visually)

If/When UK leaves the EU, can a future goverment conduct a referendum to join the EU?

How to add tiny 0.5A 120V load to very remote split phase 240v 3 wire well house

How do I go from 300 unfinished/half written blog posts, to published posts?

What is ( CFMCC ) on ILS approach chart?

Why has the US not been more assertive in confronting Russia in recent years?

Why do we use the plural of movies in this phrase "We went to the movies last night."?

Written every which way



calculus parametric curve length



The Next CEO of Stack OverflowFind the length of the parametric curve (Difficult)Parametric Curve Tangent EquationsParametric 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 curveTransforming quadratic parametric curve to implicit formLength of a parametric curve formula: What does the integral represent?












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 :( Can anyone show me the steps?










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
    4 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 :( Can anyone show me the steps?










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
    4 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 :( Can anyone show me the steps?










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 :( Can anyone show me the steps?







calculus parametric






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









Matt A Pelto

2,667621




2,667621










asked 5 hours ago









McAMcA

224




224












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




















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


















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






$begingroup$
Is this $$x=5+frac{9}{2}t^3,y=4+3t^{9/2}$$?
$endgroup$
– Dr. Sonnhard Graubner
4 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.
Check out our Code of Conduct.






$endgroup$





















    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









      $endgroup$














        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3167507%2fcalculus-parametric-curve-length%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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 4 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 4 hours ago

























                    answered 4 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 4 hours ago









                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                            78.2k42867




                            78.2k42867






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3167507%2fcalculus-parametric-curve-length%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                Anexo:Material bélico de la Fuerza Aérea de Chile Índice Aeronaves Defensa...

                                Always On Availability groups resolving state after failover - Remote harden of transaction...

                                update json value to null Announcing the arrival of Valued Associate #679: Cesar Manara ...