# change in t(x,y,z) in the direction (a,b) on f(x,y)

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

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Finding the tangent plane is the right way to go, but you need a unit vector in the direction of travel along the surface. You have defined $v$ to be a scalar, so you can't dot $v$ with $\nabla t$. –  icurays1 Nov 12 '12 at 18:49
Sorry, I edited the question –  Kassym Dorsel Nov 12 '12 at 18:53
Does $z=f(x,y)$ represent a constraint on $(x,y)$ or a function? –  copper.hat Nov 12 '12 at 18:56
It's a function. –  Kassym Dorsel Nov 12 '12 at 18:57

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).

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This only gives me the rate of change. Right ? ie the gradient. I would still need to dot that with a unit directional vector to find the change in the specified direction (a,b) –  Kassym Dorsel Nov 12 '12 at 19:02
Correct. I have added a comment to that effect in the answer. –  copper.hat Nov 12 '12 at 19:10
This gives me the same answer as my original solution if I don't normalize my directional vector. Even if I normalize (a,b) I'm still missing the normalization of the change of z –  Kassym Dorsel Nov 12 '12 at 19:27
By using the functional relationship $z=f(x,y)$ you have removed $z$ from consideration. The only variables left are $(x,y)$. –  copper.hat Nov 12 '12 at 19:29