Determining the path of a particle in a field

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Integration can be used to determine the path of a particle in a field if the field is properly defined and initial conditions are provided. A vector field must have components, so a function like F(x,y) = x + y is not a valid vector field. Instead, a valid vector field would be expressed as F = xi + yj, allowing for the application of Newton's equations of motion to derive the particle's path. If the field represents potential energy, the force is derived from the negative gradient of that potential. Understanding these principles is essential for accurately modeling particle trajectories in a given field.
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Homework Statement
using integration to determine paths of travel
Relevant Equations
F(x,y) = x+y [for example]
This is not a specific homework question, but more of a general query.

If provided with a simple vector field indicating forces (for example, an electrical field), can you use integration to determine the path of a particle placed in that field, if also provided with some initial conditions? Let's say this is a simple 2D plane where the field follows something fairly simple, like F(x,y) = x+ y. Can this be done? Can anyone point me in the right direction to learn more about this?

Consider this as an attempt to change a grid of vectors into a set of field lines that, at the same time, show the path of a set particles within that field
 
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A vector field has components. Your F does not.
Apart from that, the answer to your question is yes: you integrate Newton's equations of motion to get the path.
(That's how the SUVAT equations come about).
 
Your field, F= x+ y, is NOT a "vector field" because x+ y is not a vector!

If you had something like F= xi+ yj then, since Force= mass* acceleration, the acceleration would be \frac{d^2x}{dt^2}= x and \frac{d^2y}{dt^2}= y. The "characteristic equation" of both is r^2= 1 and So that x(t)= Ae^t+ Be^{-t} and y(t)= Ce^t+ De^{-t}.

Another possibility is that F is not a force vector but a potential energy field. In that case, the force is given by the negative of the gradient of the potential energy field. With potential energy x+ y, the force vector is -i- j. Then \frac{d^2x}{dt^2}= -1and \frac{d^2y}{dt^2}= -1so that, integrating twice, x(t)= -\frac{t^2}{2}+ At+ B and y(t)= -\frac{t^2}{2}+ Ct+ D.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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