geometry and motion- particle moving along a path

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ellemay96
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Assume the pressure in some region of the plane is given byp(x,y)=(7/12) x−2y+x^2 +3y^2.
Assume a particle initially at (x,y) = (1,1) moves along a path that is always tangent to ∇p, moving inthe direction of −∇p. Find the path of the particle and its final position as t → ∞.

I know the equation that you use is r'(t)=−∇p(r(t)).
I found ∇p=(-1+2x,-2+6y). However, I am confused as to how to work out what r(t) is. Any help is appreciated
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atsruser
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Brief thoughts:

You have:

\nabla p = (f_x, g_y)

and the particle has some time dependent position vector \bold{r}(t) which we can split into x-y components like so:

\bold{r}(t) = (x(t), y(t))

So \bold{r}'(t) = -\nabla p = -(f_x, g_y) \Rightarrow (x'(t), y'(t)) = -(f_x,g_y) which gives you two 1D separable equations to solve:

x'(t) = \frac{dx}{dt} = -f_x

y'(t) = \frac{dy}{dt} = -g_y

from which you can find the time dependence of the x and y components.
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ellemay96
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(Original post by atsruser)
Brief thoughts:

You have:

\nabla p = (f_x, g_y)

and the particle has some time dependent position vector \bold{r}(t) which we can split into x-y components like so:

\bold{r}(t) = (x(t), y(t))

So \bold{r}'(t) = -\nabla p = -(f_x, g_y) \Rightarrow (x'(t), y'(t)) = -(f_x,g_y) which gives you two 1D separable equations to solve:

x'(t) = \frac{dx}{dt} = -f_x

y'(t) = \frac{dy}{dt} = -g_y

from which you can find the time dependence of the x and y components.
I got 1-2x=dx/dt and 2-6y=dy/dt.
I solved these ode's to get t=-0.5ln(1-2x) and t=-(1/6)ln(1-3y). Does this seem correct?
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atsruser
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(Original post by ellemay96)
I got 1-2x=dx/dt and 2-6y=dy/dt.
I solved these ode's to get t=-0.5ln(1-2x) and t=-(1/6)ln(1-3y). Does this seem correct?
They look plausible (I haven't worked through the details) but

1. what happened to the 7/12 factor?

2. you need to invert these equations to figure out what happens to x and y as t tends to infinity.
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