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**[SOLVED] Trajectory using gradient and differential equations**

## Homework Statement

A heat-seeking particle is located at the point P on a flat metal plate whose temperature

at a point (x, y) is T(x, y). Find parametric equations for the trajectory of the particle if

it moves continuously in the direction of maximum temperature increase.

T(x, y) = 5 − 4x[tex]^{2}[/tex] − y[tex]^{2}[/tex]; P(1, 4).

## Homework Equations

velocity vector = [tex]\left \langle[/tex]x'(t), [tex]\right y'(t)\rangle[/tex]

[tex]\nabla T(x,y)[/tex] = [tex]\left\langle[/tex][tex]T_{x}[/tex], [tex]\right T_{y}\rangle[/tex]

## The Attempt at a Solution

Let x = x(t), y = y(t) be parametric equations for the trajectory of the particle.

We want the particle to move continuously in the direction of maximum temperature

increase. Equivalently, we want the velocity vector [tex]\left \langle[/tex]x'(t), [tex]\right y'(t)\rangle[/tex] to point in the direction

of [tex]\nabla T(x(t),y(t))[/tex]. Since [tex]\nabla T(x,y)[/tex]= [tex]\left \langle[/tex] -8x, [tex]\right -2y\rangle[/tex], the previous condition will be satisfied if

[tex]\left \langle[/tex]x'(t), [tex]\right y'(t)\rangle[/tex] = [tex]\left \langle[/tex] -8x(t), [tex]\right -2y(t)\rangle[/tex] for all t. To find such functions x(t) and y(t), we must

solve the differential equations

x'(t) = -8x(t), y'(t) = -2y(t)

but this is where I get stuck...I know that to determine solution, we need the initial condition of x(0) = 1, y(0) = 4. But as to solve the differential equations, I have no idea. This is from my multivariable differential calculus class, and I think they assume that I know how to do this... sadly I do not. Any help would be greatly appreciated... thanks