Given a surface $z=f(x,y)$ we need to find the change in temperature $t(x,y,z)$ in the direction of $(a,b)$ at point $(x_0,y_0)$.
My current way of thinking is finding the tangent plane of $f(x,y)$ at $(x_0,y_0)$ using $h = f(x_0, y_0) + f_x(x_0, y_0) * (x-x_0) + f_y(x_0, y_0) * (y-y_0)$
I then proceed to find the directional vector by doing: $v = (x_0+a, y_0+b,h(x_0+a, y_0+b)) - (x_0, y_0, f(x_0, y_0))$
And normalizing it:
$v = v/||v||$
I then dot the gradient of t with v.
My question is how correct is this and if not where have I gone wrong ?
EDIT:: Fixed a mistake
I may be missing the point, but it seems like the problem is to find the derivative of the function $\phi(x,y) = t(x,y,f(x,y))$? (I cannot tell if $z=f(x,y)$ describes a functional relationship or a constraint.)
Assuming it is a functional relationship:
$\frac{\partial \phi(x,y)}{\partial x} = \frac{\partial t(x,y,f(x,y))}{\partial x}+\frac{\partial t(x,y,f(x,y))}{\partial z}\frac{\partial f(x,y)}{\partial x}$
$\frac{\partial \phi(x,y)}{\partial y} = \frac{\partial t(x,y,f(x,y))}{\partial y}+\frac{\partial t(x,y,f(x,y))}{\partial z}\frac{\partial f(x,y)}{\partial y}$
The rate of change of $\phi$ in the direction $(a,b)$ is just given by $\frac{\partial \phi(x,y)}{\partial x} a + \frac{\partial \phi(x,y)}{\partial y} b$, where the partials are given above. You may wish to normalize the pair $(a,b)$ by dividing by length, but this depends on what you want to do with the gradient.
If it describes a constraint, then I need more information about $f$ (non-zero partials, for example).