tough question in Line integral / Multivariable Calculas












2















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





























    2















    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



























      2












      2








      2


      2






      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
















      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

      1777




      1777






















          2 Answers
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          active

          oldest

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          2














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



          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





















          • 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
            – Mather
            Jan 5 at 7:00





















          0














          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





















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            2 Answers
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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














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



            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





















            • 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
              – Mather
              Jan 5 at 7:00


















            2














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



            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





















            • 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
              – Mather
              Jan 5 at 7:00
















            2












            2








            2






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



            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












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



            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

            44.2k31854




            44.2k31854












            • 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
              – Mather
              Jan 5 at 7:00




















            • 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
              – Mather
              Jan 5 at 7:00


















            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
            – Mather
            Jan 5 at 7:00






            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
            – Mather
            Jan 5 at 7:00













            0














            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


























              0














              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
























                0












                0








                0






                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












                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

                27k62446




                27k62446






























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