This example was in my book and I am curious as to where the time dependent quantity $k$ came from and why it is the same for both the $x$ and $y$ relationships.
Example 5 Heat seeking particle at location $(2,-3)$ on metal plate whose temperature at $(x,y)$ is
$$ T( x , y ) = 20 − 4 x^2 − y^2 $$
Find the path of the particle as it continuously moves in the direction of maximum temperature increase.
Solution: Let the path be represented by
$$ r ( t ) = x ( t ) \hat{i} + y ( t ) \hat{j} $$
A tangent vector at each point (x(t),y(t)) is given by the derivative of r
$$ r ′ ( t ) = x ′ ( t ) \hat{i} + y ′ ( t ) \hat{j} $$
The gradient of T is $$ \nabla T ( x , y ) = − 8 x \hat{i} − 2 y \hat{j} $$
The particle seeks maximum temperature increase, so the directions of
$$ r ′ ( t ) = x ′ ( t ) \hat{i} + y ′ ( t ) \hat{j} $$
and
$$ \nabla T ( x , y ) = − 8 x \hat{i} − 2 y \hat{j} $$
are the same, giving: $$ − 8 x = k \frac{d x}{d t}\\ − 2 y = k \frac{d y}{d t} $$
where k depends on t.
Solve for $dt/k$ and equate the results: $$ d x/ − 8 x = d y/− 2 y $$
The solution to this differential equation is $x = C y^4$.
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$\begingroup$You know the two vectors $\nabla T$ and $r' (t)$ are always proportional. So call that proportionality $k(t)$. It does depend on time because all you know is the directions match up at all times, but there is no information about the speed of the particle (A bit unrealistic, but okay). It has to be the same for both $x$ and $y$ because otherwise $\nabla T$ and $r'(t)$ would be pointing in different directions.
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