Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
Then does my answer address your concern? – copper.hat Nov 12 '12 at 18:58

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).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.