Path of a heat-seeking particle

[MathChalenged]

Where does the k in the following example come from and why is it the same in both equations?

This is an example from my textbook (pg 937, Calculus, 8th, Larson, Hostetler Edwards). After they set the particle's path's derivative equal to the temperature function's gradient (in large font), they insert a k.

Example 5
Heat seeking particle at location (2,-3) on metal plate who temp at (x,y) is
$T(x,y)=20-4x^2-y^2$

Find the path of the particle as it continuously moves in the direction of maximum temperature increase.

Solution: Let the path be represented by
$\mathbf{r}(t)=x(t) \mathbf{\hat{i}} + y(t) \mathbf{\hat{j}}$

A tangent vector at each point (x(t),y(t)) is given by the derivative of r
$\mathbf{r}'(t)=x'(t) \mathbf{\hat{i}} + y'(t) \mathbf{\hat{j}}$

The gradient of T is
$\nabla T(x,y)=-8x \mathbf{\hat{i}} -2y \mathbf{\hat{j}}$

The particle seeks maximum temperature increase, so the directions of $r'(t)=x'(t) \mathbf{\hat{i}} + y'(t) \mathbf{\hat{j}}$ and $\nabla T(x,y)=-8x \mathbf{\hat{i}} - 2y \mathbf{\hat{j}}$ are the same, giving:
$-8x=k \frac{dx}{dt} \text{ and } -2y=k \frac{dy}{dt}$

where k depends on t. Solve for dt/k and equate the results:
$\frac{dx}{-8x}=\frac{dy}{-2y}$

The solution to this differential equation is $x=Cy^4$.

The particle starts at (2,-3), therefore C=2/81 and the path of the heat-seeking particle is
$x=\frac{2}{81}y^4$

 

[grey_earl]

You already answered your question yourself:

... the directions of $r'(t)=x'(t) \mathbf{\hat{i}} + y'(t) \mathbf{\hat{j}}$ and $\nabla T(x,y)=-8x \mathbf{\hat{i}} - 2y \mathbf{\hat{j}}$ are the same ...

You don't know how fast the particle moves, so you don't know x'(t) (which is the velocity in direction x) and y'(t). You only know that it goes in that direction, so you just insert an arbitrary velocity k and calculate. Since you only want to find the curve x(y), the speed with which it is traversed doesn't matter - in the final answer there is no k.

 

[MathChalenged]

Thanks.