Line Integral Work Done [closed]












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I have a problem when comes to question 2, I don't know how to put this into parametric form. And I am not sure if this is a parabola. Thanks in advance.
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closed as off-topic by Saad, Abcd, Shailesh, Lee David Chung Lin, user91500 Jan 12 at 9:58


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    $begingroup$


    I have a problem when comes to question 2, I don't know how to put this into parametric form. And I am not sure if this is a parabola. Thanks in advance.
    The question










    share|cite|improve this question









    $endgroup$



    closed as off-topic by Saad, Abcd, Shailesh, Lee David Chung Lin, user91500 Jan 12 at 9:58


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Abcd, Shailesh, Lee David Chung Lin, user91500

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      0












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      0





      $begingroup$


      I have a problem when comes to question 2, I don't know how to put this into parametric form. And I am not sure if this is a parabola. Thanks in advance.
      The question










      share|cite|improve this question









      $endgroup$




      I have a problem when comes to question 2, I don't know how to put this into parametric form. And I am not sure if this is a parabola. Thanks in advance.
      The question







      integration parametric line-integrals






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      asked Jan 11 at 22:51









      joshjosh

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      closed as off-topic by Saad, Abcd, Shailesh, Lee David Chung Lin, user91500 Jan 12 at 9:58


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Abcd, Shailesh, Lee David Chung Lin, user91500

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Saad, Abcd, Shailesh, Lee David Chung Lin, user91500 Jan 12 at 9:58


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Abcd, Shailesh, Lee David Chung Lin, user91500

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
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          $begingroup$

          If $y=4x^2$ then the $y$-coordinate of 4 times the square of the $x$-coordinate.



          Let the $x$-coordinate be $t$, then the $y$-coordinate of $4t^2$.



          Your path is given by $gamma(t) = (t,4t^2)$, where $0 le t le 1$.



          The curve $y=4x^2$ is indeed a parabola, with focus $(x,y)=(0,frac{1}{16})$ and directrix $y=-frac{1}{16}$.






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            1 Answer
            1






            active

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            1 Answer
            1






            active

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            active

            oldest

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            active

            oldest

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            1












            $begingroup$

            If $y=4x^2$ then the $y$-coordinate of 4 times the square of the $x$-coordinate.



            Let the $x$-coordinate be $t$, then the $y$-coordinate of $4t^2$.



            Your path is given by $gamma(t) = (t,4t^2)$, where $0 le t le 1$.



            The curve $y=4x^2$ is indeed a parabola, with focus $(x,y)=(0,frac{1}{16})$ and directrix $y=-frac{1}{16}$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              If $y=4x^2$ then the $y$-coordinate of 4 times the square of the $x$-coordinate.



              Let the $x$-coordinate be $t$, then the $y$-coordinate of $4t^2$.



              Your path is given by $gamma(t) = (t,4t^2)$, where $0 le t le 1$.



              The curve $y=4x^2$ is indeed a parabola, with focus $(x,y)=(0,frac{1}{16})$ and directrix $y=-frac{1}{16}$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                If $y=4x^2$ then the $y$-coordinate of 4 times the square of the $x$-coordinate.



                Let the $x$-coordinate be $t$, then the $y$-coordinate of $4t^2$.



                Your path is given by $gamma(t) = (t,4t^2)$, where $0 le t le 1$.



                The curve $y=4x^2$ is indeed a parabola, with focus $(x,y)=(0,frac{1}{16})$ and directrix $y=-frac{1}{16}$.






                share|cite|improve this answer











                $endgroup$



                If $y=4x^2$ then the $y$-coordinate of 4 times the square of the $x$-coordinate.



                Let the $x$-coordinate be $t$, then the $y$-coordinate of $4t^2$.



                Your path is given by $gamma(t) = (t,4t^2)$, where $0 le t le 1$.



                The curve $y=4x^2$ is indeed a parabola, with focus $(x,y)=(0,frac{1}{16})$ and directrix $y=-frac{1}{16}$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 11 at 23:18

























                answered Jan 11 at 23:11









                Fly by NightFly by Night

                25.9k32978




                25.9k32978















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