A biquadratic function $y=ax^4+bx^3+cx^2+dx+e(ane 0)$ touches $y=px+q$ at $alpha,beta(alpha<beta)$.












0














A biquadratic function $y=ax^4+bx^3+cx^2+dx+e(ane 0)$ touches $y=px+q$ at $alpha,beta(alpha<beta)$.Area of the region bounded by the graphs is $frac{a(alpha-beta)^5}{lambda}$.Find $lambda.$




At $alpha,beta$,the slope of line$(p)=$derivative of biquadratic function$=4aalpha^3+3balpha^2+2calpha+d=4abeta^3+3bbeta^2+2cbeta+d$


I dont know how to solve further.






calculus algebra-precalculus definite-integrals







share|cite|improve this question






















  • To find $q$, note that a touching point is on both curves. $y(alpha)=aalpha^4+balpha^3+calpha^2+dalpha+e=palpha+q$. Then all you need is to write the integral $int_alpha^beta(ax^4+bx^3+cx^2+dx+e-px-q)dx$
    – Andrei
    Nov 26 at 15:38
















0














A biquadratic function $y=ax^4+bx^3+cx^2+dx+e(ane 0)$ touches $y=px+q$ at $alpha,beta(alpha<beta)$.Area of the region bounded by the graphs is $frac{a(alpha-beta)^5}{lambda}$.Find $lambda.$




At $alpha,beta$,the slope of line$(p)=$derivative of biquadratic function$=4aalpha^3+3balpha^2+2calpha+d=4abeta^3+3bbeta^2+2cbeta+d$


I dont know how to solve further.






calculus algebra-precalculus definite-integrals







share|cite|improve this question






















  • To find $q$, note that a touching point is on both curves. $y(alpha)=aalpha^4+balpha^3+calpha^2+dalpha+e=palpha+q$. Then all you need is to write the integral $int_alpha^beta(ax^4+bx^3+cx^2+dx+e-px-q)dx$
    – Andrei
    Nov 26 at 15:38














0












0








0







A biquadratic function $y=ax^4+bx^3+cx^2+dx+e(ane 0)$ touches $y=px+q$ at $alpha,beta(alpha<beta)$.Area of the region bounded by the graphs is $frac{a(alpha-beta)^5}{lambda}$.Find $lambda.$




At $alpha,beta$,the slope of line$(p)=$derivative of biquadratic function$=4aalpha^3+3balpha^2+2calpha+d=4abeta^3+3bbeta^2+2cbeta+d$


I dont know how to solve further.






calculus algebra-precalculus definite-integrals







share|cite|improve this question













A biquadratic function $y=ax^4+bx^3+cx^2+dx+e(ane 0)$ touches $y=px+q$ at $alpha,beta(alpha<beta)$.Area of the region bounded by the graphs is $frac{a(alpha-beta)^5}{lambda}$.Find $lambda.$




At $alpha,beta$,the slope of line$(p)=$derivative of biquadratic function$=4aalpha^3+3balpha^2+2calpha+d=4abeta^3+3bbeta^2+2cbeta+d$


I dont know how to solve further.






calculus algebra-precalculus definite-integrals




calculus algebra-precalculus definite-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 at 15:03









user984325

248112




248112












  • To find $q$, note that a touching point is on both curves. $y(alpha)=aalpha^4+balpha^3+calpha^2+dalpha+e=palpha+q$. Then all you need is to write the integral $int_alpha^beta(ax^4+bx^3+cx^2+dx+e-px-q)dx$
    – Andrei
    Nov 26 at 15:38


















  • To find $q$, note that a touching point is on both curves. $y(alpha)=aalpha^4+balpha^3+calpha^2+dalpha+e=palpha+q$. Then all you need is to write the integral $int_alpha^beta(ax^4+bx^3+cx^2+dx+e-px-q)dx$
    – Andrei
    Nov 26 at 15:38
















To find $q$, note that a touching point is on both curves. $y(alpha)=aalpha^4+balpha^3+calpha^2+dalpha+e=palpha+q$. Then all you need is to write the integral $int_alpha^beta(ax^4+bx^3+cx^2+dx+e-px-q)dx$
– Andrei
Nov 26 at 15:38




To find $q$, note that a touching point is on both curves. $y(alpha)=aalpha^4+balpha^3+calpha^2+dalpha+e=palpha+q$. Then all you need is to write the integral $int_alpha^beta(ax^4+bx^3+cx^2+dx+e-px-q)dx$
– Andrei
Nov 26 at 15:38










2 Answers
2






active

oldest

votes


















3














Let $f(x)$ be the vertical distance between the two functions at $x$. Then $f(x)$ is a quartic polynomial with leading coefficient $a$, and $f(alpha)=f'(alpha)=f(beta)=f'(beta)=0$. So $f$ has two double roots, which accounts for all its roots; that is,
$$
f(x)=a(x-alpha)^2(x-beta)^2
$$
Moreover, $int_alpha^beta f(x) , dx$ is the area you're looking for.






share|cite|improve this answer





















  • Very intuitive solution
    – user984325
    Nov 27 at 8:16



















0














One way to do this is as follows: set $f(x) = ax^4+bx^3+cx^2+dx+e$, $l(x)=px+q$. Then solve the system
$$f(alpha) = l(alpha),quad f(beta)=l(beta)$$
to get $p$ and $q$ in terms of the other variables. You can substitute this value of $p$ into
$$f'(alpha) = p,quad f'(beta)=p$$
and solve for $b, c$ in terms of $a$, $alpha$, and $beta$. Put all of this together and integrate (after substitution)
$$int_alpha^beta (f(x)-l(x)),dx.$$
If this is a homework problem, either there's an easier way or your teacher is a sadist.






share|cite|improve this answer





















    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%2f3014434%2fa-biquadratic-function-y-ax4bx3cx2dxea-ne-0-touches-y-pxq-at-alp%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    Let $f(x)$ be the vertical distance between the two functions at $x$. Then $f(x)$ is a quartic polynomial with leading coefficient $a$, and $f(alpha)=f'(alpha)=f(beta)=f'(beta)=0$. So $f$ has two double roots, which accounts for all its roots; that is,
    $$
    f(x)=a(x-alpha)^2(x-beta)^2
    $$
    Moreover, $int_alpha^beta f(x) , dx$ is the area you're looking for.






    share|cite|improve this answer





















    • Very intuitive solution
      – user984325
      Nov 27 at 8:16
















    3














    Let $f(x)$ be the vertical distance between the two functions at $x$. Then $f(x)$ is a quartic polynomial with leading coefficient $a$, and $f(alpha)=f'(alpha)=f(beta)=f'(beta)=0$. So $f$ has two double roots, which accounts for all its roots; that is,
    $$
    f(x)=a(x-alpha)^2(x-beta)^2
    $$
    Moreover, $int_alpha^beta f(x) , dx$ is the area you're looking for.






    share|cite|improve this answer





















    • Very intuitive solution
      – user984325
      Nov 27 at 8:16














    3












    3








    3






    Let $f(x)$ be the vertical distance between the two functions at $x$. Then $f(x)$ is a quartic polynomial with leading coefficient $a$, and $f(alpha)=f'(alpha)=f(beta)=f'(beta)=0$. So $f$ has two double roots, which accounts for all its roots; that is,
    $$
    f(x)=a(x-alpha)^2(x-beta)^2
    $$
    Moreover, $int_alpha^beta f(x) , dx$ is the area you're looking for.






    share|cite|improve this answer












    Let $f(x)$ be the vertical distance between the two functions at $x$. Then $f(x)$ is a quartic polynomial with leading coefficient $a$, and $f(alpha)=f'(alpha)=f(beta)=f'(beta)=0$. So $f$ has two double roots, which accounts for all its roots; that is,
    $$
    f(x)=a(x-alpha)^2(x-beta)^2
    $$
    Moreover, $int_alpha^beta f(x) , dx$ is the area you're looking for.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 26 at 17:00









    Micah

    29.6k1363105




    29.6k1363105












    • Very intuitive solution
      – user984325
      Nov 27 at 8:16


















    • Very intuitive solution
      – user984325
      Nov 27 at 8:16
















    Very intuitive solution
    – user984325
    Nov 27 at 8:16




    Very intuitive solution
    – user984325
    Nov 27 at 8:16











    0














    One way to do this is as follows: set $f(x) = ax^4+bx^3+cx^2+dx+e$, $l(x)=px+q$. Then solve the system
    $$f(alpha) = l(alpha),quad f(beta)=l(beta)$$
    to get $p$ and $q$ in terms of the other variables. You can substitute this value of $p$ into
    $$f'(alpha) = p,quad f'(beta)=p$$
    and solve for $b, c$ in terms of $a$, $alpha$, and $beta$. Put all of this together and integrate (after substitution)
    $$int_alpha^beta (f(x)-l(x)),dx.$$
    If this is a homework problem, either there's an easier way or your teacher is a sadist.






    share|cite|improve this answer


























      0














      One way to do this is as follows: set $f(x) = ax^4+bx^3+cx^2+dx+e$, $l(x)=px+q$. Then solve the system
      $$f(alpha) = l(alpha),quad f(beta)=l(beta)$$
      to get $p$ and $q$ in terms of the other variables. You can substitute this value of $p$ into
      $$f'(alpha) = p,quad f'(beta)=p$$
      and solve for $b, c$ in terms of $a$, $alpha$, and $beta$. Put all of this together and integrate (after substitution)
      $$int_alpha^beta (f(x)-l(x)),dx.$$
      If this is a homework problem, either there's an easier way or your teacher is a sadist.






      share|cite|improve this answer
























        0












        0








        0






        One way to do this is as follows: set $f(x) = ax^4+bx^3+cx^2+dx+e$, $l(x)=px+q$. Then solve the system
        $$f(alpha) = l(alpha),quad f(beta)=l(beta)$$
        to get $p$ and $q$ in terms of the other variables. You can substitute this value of $p$ into
        $$f'(alpha) = p,quad f'(beta)=p$$
        and solve for $b, c$ in terms of $a$, $alpha$, and $beta$. Put all of this together and integrate (after substitution)
        $$int_alpha^beta (f(x)-l(x)),dx.$$
        If this is a homework problem, either there's an easier way or your teacher is a sadist.






        share|cite|improve this answer












        One way to do this is as follows: set $f(x) = ax^4+bx^3+cx^2+dx+e$, $l(x)=px+q$. Then solve the system
        $$f(alpha) = l(alpha),quad f(beta)=l(beta)$$
        to get $p$ and $q$ in terms of the other variables. You can substitute this value of $p$ into
        $$f'(alpha) = p,quad f'(beta)=p$$
        and solve for $b, c$ in terms of $a$, $alpha$, and $beta$. Put all of this together and integrate (after substitution)
        $$int_alpha^beta (f(x)-l(x)),dx.$$
        If this is a homework problem, either there's an easier way or your teacher is a sadist.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 at 16:44









        rogerl

        17.4k22746




        17.4k22746






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3014434%2fa-biquadratic-function-y-ax4bx3cx2dxea-ne-0-touches-y-pxq-at-alp%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

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more

            Province de Neuquén

            Image
            Province de Neuquén Provincia del Neuquén .mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Héraldique Drapeau Localisation de la province de Neuquén Administration Pays   Argentine Capitale Neuquén Gouverneur Omar Gutiérrez ISO 3166-2 AR-Q Démographie Gentilé Neuquino/a Population 550 344  hab. (2010) Densité 5,8  hab./km 2 Géographie Superficie 94 078  km 2 Liens Site web http://www.neuquen.gov.ar modifier   La province de Neuquén est une province d'Argentine, qui se trouve à l'extrémité nord-ouest de la Patagonie argentine. Avec une superficie de 94 378  km 2 , elle est la plus petite province des régions australes du pays. Sa géographie s'abîme sur l'imposante Cordillère des Andes. Sommaire 1 Histoire 1.1 Avant la découverte par les Espagnols 1.2 ...
            Read more

            Journaliste

            Image
            Pour les articles homonymes, voir Journaliste (homonymie). Journaliste Journalistes Appellation Journaliste Secteur d'activité Média - Communication Professions voisines Journaliste sportif - Reporter-photographe - Éditorialiste Codes CITP 2451, 2642 IDEO (France) 10197 ROME (France) E1106 modifier - modifier le code - modifier Wikidata Un journaliste est une personne qui recueille, écrit ou distribue des informations. L'activité d'un journaliste est le journalisme. Un journaliste peut travailler sur des sujets généraux ou se spécialiser dans certaines questions. La plupart des journalistes ont tendance à se spécialiser et, en collaborant avec d'autres journalistes, rédigent un article ou mettent en forme un reportage qui couvre l'actualité. Par exemple, un journaliste sportif couvre les nouvelles dans le monde du sport, mais ce journaliste peut aussi faire partie d'un journal qui couvre d'autres su...
            Read more