tough question in Line integral / Multivariable Calculas












2












$begingroup$



Let $C$ be a curve in the $(x − y)$-plane. For every point $(x, y)$ of $C$
let $u(x, y)$ denote the unit vector in the direction of the tangent line
to $C$ at $(x, y)$. Let $S$ be the surface obtained by taking the union of
all straight line segments connecting $(1, 2, 3)$ to points of $C$. Express
the area of $S$ as an integral of the first type, on the curve $C$, of some
function of $x$ and $y$. (hint: try to use the function $u(x, y)$.)




really Hard question , i couldn't understand how to use the fact that line integral will help here since i don't have a function $f(x,y)$ to calculate $ int_{C} f(x(t),y(t))sqrt{x'(t)^2 + y'(t)^2 } dt$ also i can parameterize $S$ like that :
$S=:k(1-x(t),2-y(t),3) , kin[0,1]$ where $(x(t),y(t),0)$ is the curve $C$



and why $u(x,y)$ is given here.
Unit vector






integration multivariable-calculus







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Let $C$ be a curve in the $(x − y)$-plane. For every point $(x, y)$ of $C$
    let $u(x, y)$ denote the unit vector in the direction of the tangent line
    to $C$ at $(x, y)$. Let $S$ be the surface obtained by taking the union of
    all straight line segments connecting $(1, 2, 3)$ to points of $C$. Express
    the area of $S$ as an integral of the first type, on the curve $C$, of some
    function of $x$ and $y$. (hint: try to use the function $u(x, y)$.)




    really Hard question , i couldn't understand how to use the fact that line integral will help here since i don't have a function $f(x,y)$ to calculate $ int_{C} f(x(t),y(t))sqrt{x'(t)^2 + y'(t)^2 } dt$ also i can parameterize $S$ like that :
    $S=:k(1-x(t),2-y(t),3) , kin[0,1]$ where $(x(t),y(t),0)$ is the curve $C$



    and why $u(x,y)$ is given here.
    Unit vector






    integration multivariable-calculus







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      2



      $begingroup$



      Let $C$ be a curve in the $(x − y)$-plane. For every point $(x, y)$ of $C$
      let $u(x, y)$ denote the unit vector in the direction of the tangent line
      to $C$ at $(x, y)$. Let $S$ be the surface obtained by taking the union of
      all straight line segments connecting $(1, 2, 3)$ to points of $C$. Express
      the area of $S$ as an integral of the first type, on the curve $C$, of some
      function of $x$ and $y$. (hint: try to use the function $u(x, y)$.)




      really Hard question , i couldn't understand how to use the fact that line integral will help here since i don't have a function $f(x,y)$ to calculate $ int_{C} f(x(t),y(t))sqrt{x'(t)^2 + y'(t)^2 } dt$ also i can parameterize $S$ like that :
      $S=:k(1-x(t),2-y(t),3) , kin[0,1]$ where $(x(t),y(t),0)$ is the curve $C$



      and why $u(x,y)$ is given here.
      Unit vector






      integration multivariable-calculus







      share|cite|improve this question











      $endgroup$





      Let $C$ be a curve in the $(x − y)$-plane. For every point $(x, y)$ of $C$
      let $u(x, y)$ denote the unit vector in the direction of the tangent line
      to $C$ at $(x, y)$. Let $S$ be the surface obtained by taking the union of
      all straight line segments connecting $(1, 2, 3)$ to points of $C$. Express
      the area of $S$ as an integral of the first type, on the curve $C$, of some
      function of $x$ and $y$. (hint: try to use the function $u(x, y)$.)




      really Hard question , i couldn't understand how to use the fact that line integral will help here since i don't have a function $f(x,y)$ to calculate $ int_{C} f(x(t),y(t))sqrt{x'(t)^2 + y'(t)^2 } dt$ also i can parameterize $S$ like that :
      $S=:k(1-x(t),2-y(t),3) , kin[0,1]$ where $(x(t),y(t),0)$ is the curve $C$



      and why $u(x,y)$ is given here.
      Unit vector






      integration multivariable-calculus




      integration multivariable-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 7:04







      Mather

















      asked Jan 4 at 21:21









      Mather Mather

      3888




      3888






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          $c(t) = (x(t),y(t), 0)\
          frac {dc}{dt} = u(t)$



          The parmeterization of $S$ should be



          $S = (k +(1-k) x,2 k + (1-k) y, 3k)$



          or



          $((1-k) +k x,2(1- k) + k) y(t), 3(1-k))$



          with



          $0le k le 1$



          $dS = |frac {partial S}{dk} times frac {partial S}{dt}|$



          $frac {partial S}{dk} = (1-x, 2-y,3) = (1,2,3) - c(t)$



          $frac {partial S}{dt} = (1-k)u(t)$



          $int_0^1 (1-k) dk int_0^t |(1,2,3)times u(t) - c(t)times u(t)| dt$



          $frac 12 int_0^t |(1,2,3)times u(t) - c(t)times u(t)| dt$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you this is what I tried to do , I didn’t understand one thing , $ u(x(t),y(t)) $ is a unit vector shouldn’t we multiply this Vector function with the length of the tangent line to the curve to get the surface area ? or it doesn’t matter and if it doesn’t why
            $endgroup$
            – Mather
            Jan 5 at 7:00





















          0












          $begingroup$

          This is a very non-rigorous derivation.



          Consider a point $C(t) = (x(t),y(t),0) in C$ and a second point infinitesimally close to it:
          $$C(t+dt) = (x(t+dt),y(t+dt),0) = (x(t)+dot{x}(t),dt,y(t)+dot{y}(t),dt,0)$$
          We need to calculate the infinitesimal area $dA$ of the triangle with vertices $C(t),C(t+dt)$ and $(1,2,3)$.



          The area is given by
          begin{align}
          dA &= frac12 |(C(t+dt) - C(t)) times ((1,2,3) - C(t))| \
          &= frac12 |(dot{x},dt,dot{y},dt,0) times (1-x,2-y,3)|\
          &= frac{dt}2 |(dot{x},dot{y},0) times (1-x(t),2-y(t),3)|\
          &= frac{dt}2 |(3dot{y},-3dot{x},(2-y)dot{x}+(x-1)dot{y})|\
          end{align}



          so $$A = int_C,dA = frac12int_{t}|(3dot{y}(t),-3dot{x}(t),(2-y(t))dot{x}(t)+(x(t)-1)dot{y}(t))|,dt$$






          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%2f3062126%2ftough-question-in-line-integral-multivariable-calculas%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









            2












            $begingroup$

            $c(t) = (x(t),y(t), 0)\
            frac {dc}{dt} = u(t)$



            The parmeterization of $S$ should be



            $S = (k +(1-k) x,2 k + (1-k) y, 3k)$



            or



            $((1-k) +k x,2(1- k) + k) y(t), 3(1-k))$



            with



            $0le k le 1$



            $dS = |frac {partial S}{dk} times frac {partial S}{dt}|$



            $frac {partial S}{dk} = (1-x, 2-y,3) = (1,2,3) - c(t)$



            $frac {partial S}{dt} = (1-k)u(t)$



            $int_0^1 (1-k) dk int_0^t |(1,2,3)times u(t) - c(t)times u(t)| dt$



            $frac 12 int_0^t |(1,2,3)times u(t) - c(t)times u(t)| dt$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you this is what I tried to do , I didn’t understand one thing , $ u(x(t),y(t)) $ is a unit vector shouldn’t we multiply this Vector function with the length of the tangent line to the curve to get the surface area ? or it doesn’t matter and if it doesn’t why
              $endgroup$
              – Mather
              Jan 5 at 7:00


















            2












            $begingroup$

            $c(t) = (x(t),y(t), 0)\
            frac {dc}{dt} = u(t)$



            The parmeterization of $S$ should be



            $S = (k +(1-k) x,2 k + (1-k) y, 3k)$



            or



            $((1-k) +k x,2(1- k) + k) y(t), 3(1-k))$



            with



            $0le k le 1$



            $dS = |frac {partial S}{dk} times frac {partial S}{dt}|$



            $frac {partial S}{dk} = (1-x, 2-y,3) = (1,2,3) - c(t)$



            $frac {partial S}{dt} = (1-k)u(t)$



            $int_0^1 (1-k) dk int_0^t |(1,2,3)times u(t) - c(t)times u(t)| dt$



            $frac 12 int_0^t |(1,2,3)times u(t) - c(t)times u(t)| dt$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you this is what I tried to do , I didn’t understand one thing , $ u(x(t),y(t)) $ is a unit vector shouldn’t we multiply this Vector function with the length of the tangent line to the curve to get the surface area ? or it doesn’t matter and if it doesn’t why
              $endgroup$
              – Mather
              Jan 5 at 7:00
















            2












            2








            2





            $begingroup$

            $c(t) = (x(t),y(t), 0)\
            frac {dc}{dt} = u(t)$



            The parmeterization of $S$ should be



            $S = (k +(1-k) x,2 k + (1-k) y, 3k)$



            or



            $((1-k) +k x,2(1- k) + k) y(t), 3(1-k))$



            with



            $0le k le 1$



            $dS = |frac {partial S}{dk} times frac {partial S}{dt}|$



            $frac {partial S}{dk} = (1-x, 2-y,3) = (1,2,3) - c(t)$



            $frac {partial S}{dt} = (1-k)u(t)$



            $int_0^1 (1-k) dk int_0^t |(1,2,3)times u(t) - c(t)times u(t)| dt$



            $frac 12 int_0^t |(1,2,3)times u(t) - c(t)times u(t)| dt$






            share|cite|improve this answer









            $endgroup$



            $c(t) = (x(t),y(t), 0)\
            frac {dc}{dt} = u(t)$



            The parmeterization of $S$ should be



            $S = (k +(1-k) x,2 k + (1-k) y, 3k)$



            or



            $((1-k) +k x,2(1- k) + k) y(t), 3(1-k))$



            with



            $0le k le 1$



            $dS = |frac {partial S}{dk} times frac {partial S}{dt}|$



            $frac {partial S}{dk} = (1-x, 2-y,3) = (1,2,3) - c(t)$



            $frac {partial S}{dt} = (1-k)u(t)$



            $int_0^1 (1-k) dk int_0^t |(1,2,3)times u(t) - c(t)times u(t)| dt$



            $frac 12 int_0^t |(1,2,3)times u(t) - c(t)times u(t)| dt$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 4 at 21:59









            Doug MDoug M

            45.3k31954




            45.3k31954












            • $begingroup$
              Thank you this is what I tried to do , I didn’t understand one thing , $ u(x(t),y(t)) $ is a unit vector shouldn’t we multiply this Vector function with the length of the tangent line to the curve to get the surface area ? or it doesn’t matter and if it doesn’t why
              $endgroup$
              – Mather
              Jan 5 at 7:00




















            • $begingroup$
              Thank you this is what I tried to do , I didn’t understand one thing , $ u(x(t),y(t)) $ is a unit vector shouldn’t we multiply this Vector function with the length of the tangent line to the curve to get the surface area ? or it doesn’t matter and if it doesn’t why
              $endgroup$
              – Mather
              Jan 5 at 7:00


















            $begingroup$
            Thank you this is what I tried to do , I didn’t understand one thing , $ u(x(t),y(t)) $ is a unit vector shouldn’t we multiply this Vector function with the length of the tangent line to the curve to get the surface area ? or it doesn’t matter and if it doesn’t why
            $endgroup$
            – Mather
            Jan 5 at 7:00






            $begingroup$
            Thank you this is what I tried to do , I didn’t understand one thing , $ u(x(t),y(t)) $ is a unit vector shouldn’t we multiply this Vector function with the length of the tangent line to the curve to get the surface area ? or it doesn’t matter and if it doesn’t why
            $endgroup$
            – Mather
            Jan 5 at 7:00













            0












            $begingroup$

            This is a very non-rigorous derivation.



            Consider a point $C(t) = (x(t),y(t),0) in C$ and a second point infinitesimally close to it:
            $$C(t+dt) = (x(t+dt),y(t+dt),0) = (x(t)+dot{x}(t),dt,y(t)+dot{y}(t),dt,0)$$
            We need to calculate the infinitesimal area $dA$ of the triangle with vertices $C(t),C(t+dt)$ and $(1,2,3)$.



            The area is given by
            begin{align}
            dA &= frac12 |(C(t+dt) - C(t)) times ((1,2,3) - C(t))| \
            &= frac12 |(dot{x},dt,dot{y},dt,0) times (1-x,2-y,3)|\
            &= frac{dt}2 |(dot{x},dot{y},0) times (1-x(t),2-y(t),3)|\
            &= frac{dt}2 |(3dot{y},-3dot{x},(2-y)dot{x}+(x-1)dot{y})|\
            end{align}



            so $$A = int_C,dA = frac12int_{t}|(3dot{y}(t),-3dot{x}(t),(2-y(t))dot{x}(t)+(x(t)-1)dot{y}(t))|,dt$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              This is a very non-rigorous derivation.



              Consider a point $C(t) = (x(t),y(t),0) in C$ and a second point infinitesimally close to it:
              $$C(t+dt) = (x(t+dt),y(t+dt),0) = (x(t)+dot{x}(t),dt,y(t)+dot{y}(t),dt,0)$$
              We need to calculate the infinitesimal area $dA$ of the triangle with vertices $C(t),C(t+dt)$ and $(1,2,3)$.



              The area is given by
              begin{align}
              dA &= frac12 |(C(t+dt) - C(t)) times ((1,2,3) - C(t))| \
              &= frac12 |(dot{x},dt,dot{y},dt,0) times (1-x,2-y,3)|\
              &= frac{dt}2 |(dot{x},dot{y},0) times (1-x(t),2-y(t),3)|\
              &= frac{dt}2 |(3dot{y},-3dot{x},(2-y)dot{x}+(x-1)dot{y})|\
              end{align}



              so $$A = int_C,dA = frac12int_{t}|(3dot{y}(t),-3dot{x}(t),(2-y(t))dot{x}(t)+(x(t)-1)dot{y}(t))|,dt$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                This is a very non-rigorous derivation.



                Consider a point $C(t) = (x(t),y(t),0) in C$ and a second point infinitesimally close to it:
                $$C(t+dt) = (x(t+dt),y(t+dt),0) = (x(t)+dot{x}(t),dt,y(t)+dot{y}(t),dt,0)$$
                We need to calculate the infinitesimal area $dA$ of the triangle with vertices $C(t),C(t+dt)$ and $(1,2,3)$.



                The area is given by
                begin{align}
                dA &= frac12 |(C(t+dt) - C(t)) times ((1,2,3) - C(t))| \
                &= frac12 |(dot{x},dt,dot{y},dt,0) times (1-x,2-y,3)|\
                &= frac{dt}2 |(dot{x},dot{y},0) times (1-x(t),2-y(t),3)|\
                &= frac{dt}2 |(3dot{y},-3dot{x},(2-y)dot{x}+(x-1)dot{y})|\
                end{align}



                so $$A = int_C,dA = frac12int_{t}|(3dot{y}(t),-3dot{x}(t),(2-y(t))dot{x}(t)+(x(t)-1)dot{y}(t))|,dt$$






                share|cite|improve this answer









                $endgroup$



                This is a very non-rigorous derivation.



                Consider a point $C(t) = (x(t),y(t),0) in C$ and a second point infinitesimally close to it:
                $$C(t+dt) = (x(t+dt),y(t+dt),0) = (x(t)+dot{x}(t),dt,y(t)+dot{y}(t),dt,0)$$
                We need to calculate the infinitesimal area $dA$ of the triangle with vertices $C(t),C(t+dt)$ and $(1,2,3)$.



                The area is given by
                begin{align}
                dA &= frac12 |(C(t+dt) - C(t)) times ((1,2,3) - C(t))| \
                &= frac12 |(dot{x},dt,dot{y},dt,0) times (1-x,2-y,3)|\
                &= frac{dt}2 |(dot{x},dot{y},0) times (1-x(t),2-y(t),3)|\
                &= frac{dt}2 |(3dot{y},-3dot{x},(2-y)dot{x}+(x-1)dot{y})|\
                end{align}



                so $$A = int_C,dA = frac12int_{t}|(3dot{y}(t),-3dot{x}(t),(2-y(t))dot{x}(t)+(x(t)-1)dot{y}(t))|,dt$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 22:15









                mechanodroidmechanodroid

                28.2k62548




                28.2k62548






























                    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%2f3062126%2ftough-question-in-line-integral-multivariable-calculas%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

                    Ellipse (mathématiques)

                    Image
                    Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                    Read more

                    Quarter-circle Tiles

                    Image
                    up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                    Read more

                    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