path followed by a light-seeking particle ?

Question

The intensity of light is given by I(x,y)=A-2*x^2-y^2. Find the path followed by a light-seeking particle that originates at poin (-2,1).

Answer 1

I(x, y) = A-2x²-y²

Let (x, y) denote the coordinates of the particle. You're not clear about what sense the particle is light-seeking. Do you mean that it's velocity is proportional to the gradient of the intensity, or that it's acceleration is proportional to the intensity? I will assume that "light-seeking particle" means that the velocity of the particle will be equal to the gradient of the intensity of light - if you mean the acceleration, the solution will be different (and you will need to make an additional assumption to find a unique solution, such as the particle starting from rest). In either case, we must first find the gradient.

∇I = (-4x, -2y)

Now, we can write this as a system of differential equations:

∂x/∂t = -4x
∂y/∂t = -2y

Fortunately, each differential equation involves only one variable, so we can solve them independently. Rewriting them:

∂x/∂t / x = -4
∂y/∂t / y = -2

Integrating both sides:

ln |x| = -4t+C₁
ln |y| = -2t+C₂

Exponentiating:

x = K₁e^(-4t)
y = K₂e^(-2t)

All that remains is to find the values of K₁ and K₂. But we know that at t=0, x=-2 and y=1, so a quick substitution yields that K₁=-2 and K₂=1. Therefore, the path of the particle is:

x(t) = -2e^(-4t)
y(t) = e^(-2t)

If you meant for the acceleration of the particle to be proportional to the gradient of the intensity, then you would instead solve the differential equations ∂²x/∂t²=-4x and ∂²y/∂t²=-2y.