Displacement Equation in a Vector Field












1












$begingroup$


While building a program for a college level Physics course, I have a vector field representing force such that $vec F = F_x(x,y) hat{i} +F_y(x,y) hat{j}$. I am asked to create a program which shows the path of a mass $m$ with an initial velocity vector ($vec v_o$) and position vector ($vec x_o$) under the influence of the field.



My solution is to make the following steps:



$vec F = mfrac{partial^2vec x}{partial t^2}$ and as such $frac{vec F}{m} = frac{partial^2vec x}{partial t^2}$



By integration of each side, I find that the path would be



$vec x = frac{vec F Delta t^2}{2m} +vec v_o Delta t +vec x_o$ Which is the same as the typical displacement equation.



My question is as follows; Is this intuitive use of calculus correct considering that $vec F$ is position dependent, and thus time-dependent as well?






calculus vector-spaces physics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are $F_x$ and $F_y$ constants?
    $endgroup$
    – Botond
    Dec 7 '18 at 15:33










  • $begingroup$
    @Botond, No, $vec F = F_x(x,y) hat{i} + F_y(x,y) hat{j}$
    $endgroup$
    – BooleanDesigns
    Dec 7 '18 at 15:35
















1












$begingroup$


While building a program for a college level Physics course, I have a vector field representing force such that $vec F = F_x(x,y) hat{i} +F_y(x,y) hat{j}$. I am asked to create a program which shows the path of a mass $m$ with an initial velocity vector ($vec v_o$) and position vector ($vec x_o$) under the influence of the field.



My solution is to make the following steps:



$vec F = mfrac{partial^2vec x}{partial t^2}$ and as such $frac{vec F}{m} = frac{partial^2vec x}{partial t^2}$



By integration of each side, I find that the path would be



$vec x = frac{vec F Delta t^2}{2m} +vec v_o Delta t +vec x_o$ Which is the same as the typical displacement equation.



My question is as follows; Is this intuitive use of calculus correct considering that $vec F$ is position dependent, and thus time-dependent as well?






calculus vector-spaces physics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are $F_x$ and $F_y$ constants?
    $endgroup$
    – Botond
    Dec 7 '18 at 15:33










  • $begingroup$
    @Botond, No, $vec F = F_x(x,y) hat{i} + F_y(x,y) hat{j}$
    $endgroup$
    – BooleanDesigns
    Dec 7 '18 at 15:35














1












1








1





$begingroup$


While building a program for a college level Physics course, I have a vector field representing force such that $vec F = F_x(x,y) hat{i} +F_y(x,y) hat{j}$. I am asked to create a program which shows the path of a mass $m$ with an initial velocity vector ($vec v_o$) and position vector ($vec x_o$) under the influence of the field.



My solution is to make the following steps:



$vec F = mfrac{partial^2vec x}{partial t^2}$ and as such $frac{vec F}{m} = frac{partial^2vec x}{partial t^2}$



By integration of each side, I find that the path would be



$vec x = frac{vec F Delta t^2}{2m} +vec v_o Delta t +vec x_o$ Which is the same as the typical displacement equation.



My question is as follows; Is this intuitive use of calculus correct considering that $vec F$ is position dependent, and thus time-dependent as well?






calculus vector-spaces physics







share|cite|improve this question











$endgroup$




While building a program for a college level Physics course, I have a vector field representing force such that $vec F = F_x(x,y) hat{i} +F_y(x,y) hat{j}$. I am asked to create a program which shows the path of a mass $m$ with an initial velocity vector ($vec v_o$) and position vector ($vec x_o$) under the influence of the field.



My solution is to make the following steps:



$vec F = mfrac{partial^2vec x}{partial t^2}$ and as such $frac{vec F}{m} = frac{partial^2vec x}{partial t^2}$



By integration of each side, I find that the path would be



$vec x = frac{vec F Delta t^2}{2m} +vec v_o Delta t +vec x_o$ Which is the same as the typical displacement equation.



My question is as follows; Is this intuitive use of calculus correct considering that $vec F$ is position dependent, and thus time-dependent as well?






calculus vector-spaces physics




calculus vector-spaces physics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 15:36







BooleanDesigns

















asked Dec 7 '18 at 15:31









BooleanDesignsBooleanDesigns

84




84












  • $begingroup$
    Are $F_x$ and $F_y$ constants?
    $endgroup$
    – Botond
    Dec 7 '18 at 15:33










  • $begingroup$
    @Botond, No, $vec F = F_x(x,y) hat{i} + F_y(x,y) hat{j}$
    $endgroup$
    – BooleanDesigns
    Dec 7 '18 at 15:35


















  • $begingroup$
    Are $F_x$ and $F_y$ constants?
    $endgroup$
    – Botond
    Dec 7 '18 at 15:33










  • $begingroup$
    @Botond, No, $vec F = F_x(x,y) hat{i} + F_y(x,y) hat{j}$
    $endgroup$
    – BooleanDesigns
    Dec 7 '18 at 15:35
















$begingroup$
Are $F_x$ and $F_y$ constants?
$endgroup$
– Botond
Dec 7 '18 at 15:33




$begingroup$
Are $F_x$ and $F_y$ constants?
$endgroup$
– Botond
Dec 7 '18 at 15:33












$begingroup$
@Botond, No, $vec F = F_x(x,y) hat{i} + F_y(x,y) hat{j}$
$endgroup$
– BooleanDesigns
Dec 7 '18 at 15:35




$begingroup$
@Botond, No, $vec F = F_x(x,y) hat{i} + F_y(x,y) hat{j}$
$endgroup$
– BooleanDesigns
Dec 7 '18 at 15:35










2 Answers
2






active

oldest

votes


















0












$begingroup$

Note that Newton's equation is not
$$vec{F}=m frac{partial^2 vec{r}}{partial t^2}$$
But
$$vec{F}=m frac{mathrm{d}^2 vec{r}}{mathrm{d}t^2}$$
So you have two equations:
$$F_x(x(t),y(t))=m ddot{x}$$
$$F_y(x(t),y(t))=m ddot{y}$$
So your solution is not correct, because you are treating $F$ as a constant.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    From a numeric standpoint (i.e. computation) If I use my solution over an iteration algorithm (Euler's Method for example), would it still yield the same curve as the parametric solution to the system of equations that would be derived from your solutions?
    $endgroup$
    – BooleanDesigns
    Dec 7 '18 at 15:46










  • $begingroup$
    @BooleanDesigns It won't, of couse. But for a nice $F$ and small $Delta t$s, it will be quite close to that.
    $endgroup$
    – Botond
    Dec 7 '18 at 15:51










  • $begingroup$
    After some revisions to the method I was using originally, I've decided to run Euler's Method to estimate the correct curves in the following way: Given that there is some $vec F_x$ and some $vec F_y$ for all points in the field, It would follow that $ddot{x} = frac{vec F_x}{m}$ and the same would be true for $y$. From this, $dot{x} = frac{vec F_x Delta t}{m} + dot{x_0}$. As such, the estimates should show that for some $Delta t$, the estimates should be $x=frac{vec F_x Delta t^2}{m} + vec v_xo Delta t + x_o$ and the same should apply to $y$. Is this a correct method?
    $endgroup$
    – BooleanDesigns
    Dec 8 '18 at 22:05












  • $begingroup$
    @BooleanDesigns My homework was quite similar to yours, so I will tell you what did I do. I "separeted" the second order differential equations (Is it the correct word?) into first order differential equations: $dot{x}=v_x$, $dot{v_x}=f_x$, and so on. Then, I made a function, with the initial conditions as the input, which did something like this with every equation: $x=x_0+v_{x0}*dt$, $v=v_0+f(x)*dt$ and iterated it: It called itself, with the new values as initial values. Is it understandable?
    $endgroup$
    – Botond
    Dec 8 '18 at 23:09












  • $begingroup$
    Upon implementing your new method, they appear to be the same (or at least very close) for about 98% of cases, but in extremely complex configurations, my algorithm yields slight variation. As to why my method is not equivalent, I'm not quite sure yet; but your method is definitely the correct one, so thank you.
    $endgroup$
    – BooleanDesigns
    Dec 9 '18 at 18:09



















1












$begingroup$

Your approach is close to the explicit Euler method. You can also have a look at other numerical methods for ordinary differential equations. Some of them are beneficial if your problem has a particular structure, for instance it is Hamiltonian; in that case symplectic integrators might be preferred.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Note that Newton's equation is not
    $$vec{F}=m frac{partial^2 vec{r}}{partial t^2}$$
    But
    $$vec{F}=m frac{mathrm{d}^2 vec{r}}{mathrm{d}t^2}$$
    So you have two equations:
    $$F_x(x(t),y(t))=m ddot{x}$$
    $$F_y(x(t),y(t))=m ddot{y}$$
    So your solution is not correct, because you are treating $F$ as a constant.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      From a numeric standpoint (i.e. computation) If I use my solution over an iteration algorithm (Euler's Method for example), would it still yield the same curve as the parametric solution to the system of equations that would be derived from your solutions?
      $endgroup$
      – BooleanDesigns
      Dec 7 '18 at 15:46










    • $begingroup$
      @BooleanDesigns It won't, of couse. But for a nice $F$ and small $Delta t$s, it will be quite close to that.
      $endgroup$
      – Botond
      Dec 7 '18 at 15:51










    • $begingroup$
      After some revisions to the method I was using originally, I've decided to run Euler's Method to estimate the correct curves in the following way: Given that there is some $vec F_x$ and some $vec F_y$ for all points in the field, It would follow that $ddot{x} = frac{vec F_x}{m}$ and the same would be true for $y$. From this, $dot{x} = frac{vec F_x Delta t}{m} + dot{x_0}$. As such, the estimates should show that for some $Delta t$, the estimates should be $x=frac{vec F_x Delta t^2}{m} + vec v_xo Delta t + x_o$ and the same should apply to $y$. Is this a correct method?
      $endgroup$
      – BooleanDesigns
      Dec 8 '18 at 22:05












    • $begingroup$
      @BooleanDesigns My homework was quite similar to yours, so I will tell you what did I do. I "separeted" the second order differential equations (Is it the correct word?) into first order differential equations: $dot{x}=v_x$, $dot{v_x}=f_x$, and so on. Then, I made a function, with the initial conditions as the input, which did something like this with every equation: $x=x_0+v_{x0}*dt$, $v=v_0+f(x)*dt$ and iterated it: It called itself, with the new values as initial values. Is it understandable?
      $endgroup$
      – Botond
      Dec 8 '18 at 23:09












    • $begingroup$
      Upon implementing your new method, they appear to be the same (or at least very close) for about 98% of cases, but in extremely complex configurations, my algorithm yields slight variation. As to why my method is not equivalent, I'm not quite sure yet; but your method is definitely the correct one, so thank you.
      $endgroup$
      – BooleanDesigns
      Dec 9 '18 at 18:09
















    0












    $begingroup$

    Note that Newton's equation is not
    $$vec{F}=m frac{partial^2 vec{r}}{partial t^2}$$
    But
    $$vec{F}=m frac{mathrm{d}^2 vec{r}}{mathrm{d}t^2}$$
    So you have two equations:
    $$F_x(x(t),y(t))=m ddot{x}$$
    $$F_y(x(t),y(t))=m ddot{y}$$
    So your solution is not correct, because you are treating $F$ as a constant.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      From a numeric standpoint (i.e. computation) If I use my solution over an iteration algorithm (Euler's Method for example), would it still yield the same curve as the parametric solution to the system of equations that would be derived from your solutions?
      $endgroup$
      – BooleanDesigns
      Dec 7 '18 at 15:46










    • $begingroup$
      @BooleanDesigns It won't, of couse. But for a nice $F$ and small $Delta t$s, it will be quite close to that.
      $endgroup$
      – Botond
      Dec 7 '18 at 15:51










    • $begingroup$
      After some revisions to the method I was using originally, I've decided to run Euler's Method to estimate the correct curves in the following way: Given that there is some $vec F_x$ and some $vec F_y$ for all points in the field, It would follow that $ddot{x} = frac{vec F_x}{m}$ and the same would be true for $y$. From this, $dot{x} = frac{vec F_x Delta t}{m} + dot{x_0}$. As such, the estimates should show that for some $Delta t$, the estimates should be $x=frac{vec F_x Delta t^2}{m} + vec v_xo Delta t + x_o$ and the same should apply to $y$. Is this a correct method?
      $endgroup$
      – BooleanDesigns
      Dec 8 '18 at 22:05












    • $begingroup$
      @BooleanDesigns My homework was quite similar to yours, so I will tell you what did I do. I "separeted" the second order differential equations (Is it the correct word?) into first order differential equations: $dot{x}=v_x$, $dot{v_x}=f_x$, and so on. Then, I made a function, with the initial conditions as the input, which did something like this with every equation: $x=x_0+v_{x0}*dt$, $v=v_0+f(x)*dt$ and iterated it: It called itself, with the new values as initial values. Is it understandable?
      $endgroup$
      – Botond
      Dec 8 '18 at 23:09












    • $begingroup$
      Upon implementing your new method, they appear to be the same (or at least very close) for about 98% of cases, but in extremely complex configurations, my algorithm yields slight variation. As to why my method is not equivalent, I'm not quite sure yet; but your method is definitely the correct one, so thank you.
      $endgroup$
      – BooleanDesigns
      Dec 9 '18 at 18:09














    0












    0








    0





    $begingroup$

    Note that Newton's equation is not
    $$vec{F}=m frac{partial^2 vec{r}}{partial t^2}$$
    But
    $$vec{F}=m frac{mathrm{d}^2 vec{r}}{mathrm{d}t^2}$$
    So you have two equations:
    $$F_x(x(t),y(t))=m ddot{x}$$
    $$F_y(x(t),y(t))=m ddot{y}$$
    So your solution is not correct, because you are treating $F$ as a constant.






    share|cite|improve this answer









    $endgroup$



    Note that Newton's equation is not
    $$vec{F}=m frac{partial^2 vec{r}}{partial t^2}$$
    But
    $$vec{F}=m frac{mathrm{d}^2 vec{r}}{mathrm{d}t^2}$$
    So you have two equations:
    $$F_x(x(t),y(t))=m ddot{x}$$
    $$F_y(x(t),y(t))=m ddot{y}$$
    So your solution is not correct, because you are treating $F$ as a constant.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 7 '18 at 15:44









    BotondBotond

    5,7782732




    5,7782732












    • $begingroup$
      From a numeric standpoint (i.e. computation) If I use my solution over an iteration algorithm (Euler's Method for example), would it still yield the same curve as the parametric solution to the system of equations that would be derived from your solutions?
      $endgroup$
      – BooleanDesigns
      Dec 7 '18 at 15:46










    • $begingroup$
      @BooleanDesigns It won't, of couse. But for a nice $F$ and small $Delta t$s, it will be quite close to that.
      $endgroup$
      – Botond
      Dec 7 '18 at 15:51










    • $begingroup$
      After some revisions to the method I was using originally, I've decided to run Euler's Method to estimate the correct curves in the following way: Given that there is some $vec F_x$ and some $vec F_y$ for all points in the field, It would follow that $ddot{x} = frac{vec F_x}{m}$ and the same would be true for $y$. From this, $dot{x} = frac{vec F_x Delta t}{m} + dot{x_0}$. As such, the estimates should show that for some $Delta t$, the estimates should be $x=frac{vec F_x Delta t^2}{m} + vec v_xo Delta t + x_o$ and the same should apply to $y$. Is this a correct method?
      $endgroup$
      – BooleanDesigns
      Dec 8 '18 at 22:05












    • $begingroup$
      @BooleanDesigns My homework was quite similar to yours, so I will tell you what did I do. I "separeted" the second order differential equations (Is it the correct word?) into first order differential equations: $dot{x}=v_x$, $dot{v_x}=f_x$, and so on. Then, I made a function, with the initial conditions as the input, which did something like this with every equation: $x=x_0+v_{x0}*dt$, $v=v_0+f(x)*dt$ and iterated it: It called itself, with the new values as initial values. Is it understandable?
      $endgroup$
      – Botond
      Dec 8 '18 at 23:09












    • $begingroup$
      Upon implementing your new method, they appear to be the same (or at least very close) for about 98% of cases, but in extremely complex configurations, my algorithm yields slight variation. As to why my method is not equivalent, I'm not quite sure yet; but your method is definitely the correct one, so thank you.
      $endgroup$
      – BooleanDesigns
      Dec 9 '18 at 18:09


















    • $begingroup$
      From a numeric standpoint (i.e. computation) If I use my solution over an iteration algorithm (Euler's Method for example), would it still yield the same curve as the parametric solution to the system of equations that would be derived from your solutions?
      $endgroup$
      – BooleanDesigns
      Dec 7 '18 at 15:46










    • $begingroup$
      @BooleanDesigns It won't, of couse. But for a nice $F$ and small $Delta t$s, it will be quite close to that.
      $endgroup$
      – Botond
      Dec 7 '18 at 15:51










    • $begingroup$
      After some revisions to the method I was using originally, I've decided to run Euler's Method to estimate the correct curves in the following way: Given that there is some $vec F_x$ and some $vec F_y$ for all points in the field, It would follow that $ddot{x} = frac{vec F_x}{m}$ and the same would be true for $y$. From this, $dot{x} = frac{vec F_x Delta t}{m} + dot{x_0}$. As such, the estimates should show that for some $Delta t$, the estimates should be $x=frac{vec F_x Delta t^2}{m} + vec v_xo Delta t + x_o$ and the same should apply to $y$. Is this a correct method?
      $endgroup$
      – BooleanDesigns
      Dec 8 '18 at 22:05












    • $begingroup$
      @BooleanDesigns My homework was quite similar to yours, so I will tell you what did I do. I "separeted" the second order differential equations (Is it the correct word?) into first order differential equations: $dot{x}=v_x$, $dot{v_x}=f_x$, and so on. Then, I made a function, with the initial conditions as the input, which did something like this with every equation: $x=x_0+v_{x0}*dt$, $v=v_0+f(x)*dt$ and iterated it: It called itself, with the new values as initial values. Is it understandable?
      $endgroup$
      – Botond
      Dec 8 '18 at 23:09












    • $begingroup$
      Upon implementing your new method, they appear to be the same (or at least very close) for about 98% of cases, but in extremely complex configurations, my algorithm yields slight variation. As to why my method is not equivalent, I'm not quite sure yet; but your method is definitely the correct one, so thank you.
      $endgroup$
      – BooleanDesigns
      Dec 9 '18 at 18:09
















    $begingroup$
    From a numeric standpoint (i.e. computation) If I use my solution over an iteration algorithm (Euler's Method for example), would it still yield the same curve as the parametric solution to the system of equations that would be derived from your solutions?
    $endgroup$
    – BooleanDesigns
    Dec 7 '18 at 15:46




    $begingroup$
    From a numeric standpoint (i.e. computation) If I use my solution over an iteration algorithm (Euler's Method for example), would it still yield the same curve as the parametric solution to the system of equations that would be derived from your solutions?
    $endgroup$
    – BooleanDesigns
    Dec 7 '18 at 15:46












    $begingroup$
    @BooleanDesigns It won't, of couse. But for a nice $F$ and small $Delta t$s, it will be quite close to that.
    $endgroup$
    – Botond
    Dec 7 '18 at 15:51




    $begingroup$
    @BooleanDesigns It won't, of couse. But for a nice $F$ and small $Delta t$s, it will be quite close to that.
    $endgroup$
    – Botond
    Dec 7 '18 at 15:51












    $begingroup$
    After some revisions to the method I was using originally, I've decided to run Euler's Method to estimate the correct curves in the following way: Given that there is some $vec F_x$ and some $vec F_y$ for all points in the field, It would follow that $ddot{x} = frac{vec F_x}{m}$ and the same would be true for $y$. From this, $dot{x} = frac{vec F_x Delta t}{m} + dot{x_0}$. As such, the estimates should show that for some $Delta t$, the estimates should be $x=frac{vec F_x Delta t^2}{m} + vec v_xo Delta t + x_o$ and the same should apply to $y$. Is this a correct method?
    $endgroup$
    – BooleanDesigns
    Dec 8 '18 at 22:05






    $begingroup$
    After some revisions to the method I was using originally, I've decided to run Euler's Method to estimate the correct curves in the following way: Given that there is some $vec F_x$ and some $vec F_y$ for all points in the field, It would follow that $ddot{x} = frac{vec F_x}{m}$ and the same would be true for $y$. From this, $dot{x} = frac{vec F_x Delta t}{m} + dot{x_0}$. As such, the estimates should show that for some $Delta t$, the estimates should be $x=frac{vec F_x Delta t^2}{m} + vec v_xo Delta t + x_o$ and the same should apply to $y$. Is this a correct method?
    $endgroup$
    – BooleanDesigns
    Dec 8 '18 at 22:05














    $begingroup$
    @BooleanDesigns My homework was quite similar to yours, so I will tell you what did I do. I "separeted" the second order differential equations (Is it the correct word?) into first order differential equations: $dot{x}=v_x$, $dot{v_x}=f_x$, and so on. Then, I made a function, with the initial conditions as the input, which did something like this with every equation: $x=x_0+v_{x0}*dt$, $v=v_0+f(x)*dt$ and iterated it: It called itself, with the new values as initial values. Is it understandable?
    $endgroup$
    – Botond
    Dec 8 '18 at 23:09






    $begingroup$
    @BooleanDesigns My homework was quite similar to yours, so I will tell you what did I do. I "separeted" the second order differential equations (Is it the correct word?) into first order differential equations: $dot{x}=v_x$, $dot{v_x}=f_x$, and so on. Then, I made a function, with the initial conditions as the input, which did something like this with every equation: $x=x_0+v_{x0}*dt$, $v=v_0+f(x)*dt$ and iterated it: It called itself, with the new values as initial values. Is it understandable?
    $endgroup$
    – Botond
    Dec 8 '18 at 23:09














    $begingroup$
    Upon implementing your new method, they appear to be the same (or at least very close) for about 98% of cases, but in extremely complex configurations, my algorithm yields slight variation. As to why my method is not equivalent, I'm not quite sure yet; but your method is definitely the correct one, so thank you.
    $endgroup$
    – BooleanDesigns
    Dec 9 '18 at 18:09




    $begingroup$
    Upon implementing your new method, they appear to be the same (or at least very close) for about 98% of cases, but in extremely complex configurations, my algorithm yields slight variation. As to why my method is not equivalent, I'm not quite sure yet; but your method is definitely the correct one, so thank you.
    $endgroup$
    – BooleanDesigns
    Dec 9 '18 at 18:09











    1












    $begingroup$

    Your approach is close to the explicit Euler method. You can also have a look at other numerical methods for ordinary differential equations. Some of them are beneficial if your problem has a particular structure, for instance it is Hamiltonian; in that case symplectic integrators might be preferred.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Your approach is close to the explicit Euler method. You can also have a look at other numerical methods for ordinary differential equations. Some of them are beneficial if your problem has a particular structure, for instance it is Hamiltonian; in that case symplectic integrators might be preferred.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Your approach is close to the explicit Euler method. You can also have a look at other numerical methods for ordinary differential equations. Some of them are beneficial if your problem has a particular structure, for instance it is Hamiltonian; in that case symplectic integrators might be preferred.






        share|cite|improve this answer









        $endgroup$



        Your approach is close to the explicit Euler method. You can also have a look at other numerical methods for ordinary differential equations. Some of them are beneficial if your problem has a particular structure, for instance it is Hamiltonian; in that case symplectic integrators might be preferred.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 15:40









        FedericoFederico

        5,034514




        5,034514






























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