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

This question is based on section 6 of the paper Kriging and splines with derivative information.

A parametric curve $\phi(u)$ in three dimensions is deformed by the function $f$ to a new curve $\psi(u) = f[ \phi(u)]$.

As a result, for any parameter $u$, a point $\mathbf{s} = \phi(u)$ is mapped to a new point $\mathbf{s}^\prime = \psi(u)$ and the gradient $\mathbf{t} = \dot{\phi}(u)$ is mapped to a new gradient $\mathbf{t}^\prime = \dot{\psi}(u)$.

I am trying to express the gradient $\mathbf{t}^\prime$ in terms of $\mathbf{t}$. Here is my attempt.

Starting from the expression $\psi(u) = f[ \phi(u)]$, by the chain rule, for the $x$ component of $\psi(u)$

$$\frac{\partial \psi}{\partial x} = \frac{\partial f}{\partial \phi}\frac{\partial \phi}{\partial x} \mbox{ } (\ast)$$

where $\psi$ and $\phi$ are taken to mean $\psi(u)$ and $\phi(u)$ respectively. Because

$$\frac{\partial \phi}{\partial x} = \mathbf{t}_x \mbox{ }\mbox{ }\mbox{ and }\mbox{ }\mbox{ } \frac{\partial \psi}{\partial x} = \mathbf{t}_x^\prime$$

Equation $(\ast)$ becomes

$$\mathbf{t}_x^\prime = \frac{\partial f}{\partial \phi} \mathbf{t}_x$$

By a similar argument,

$$\mathbf{t}_y^\prime = \frac{\partial f}{\partial \phi} \mathbf{t}_y \mbox{ }\mbox{ }\mbox{ }\mbox{ and }\mbox{ }\mbox{ }\mbox{ } \mathbf{t}_z^\prime = \frac{\partial f}{\partial \phi} \mathbf{t}_z$$

The problem is that I don't know what to make of $\frac{\partial f}{\partial \phi}$. From what I understand from the paper this term is supposed to be a component of the gradient $\nabla f$. Unfortunately, I can't explain why this is so.

share|cite|improve this question
The function $f$ is a mapping $\mathbb{R}^3 \to \mathbb{R}^3$, so the derivative $D f(u)$ is a linear mapping $\mathbb{R}^3 \to \mathbb{R}^3$. The mapping you want is $t' = D f(u) t$. – copper.hat Jun 23 '12 at 9:03
You have very strange notations but have you tried the chain rule? – Mercy King Jun 23 '12 at 9:35
@copper.hat I understand that both mappings are from $\mathbb{R}^3 \rightarrow \mathbb{R}^3$ (thanks). Unfortunately, I don't understand why this means that the mapping that I seek has the form $\mathbf{t}^\prime = D f(u) \mathbf{t}$. – Olumide Jun 24 '12 at 0:09
@Mercy I have applied the chain rule, as mentioned in my post. I what other way should I apply the chain rule? PS: on my notation, I'm avoiding shorthand notation for now and am trying to explicit as possible. – Olumide Jun 24 '12 at 0:11
@Mercy I've fixed a small problem with to the notation. – Olumide Jun 24 '12 at 0:18
up vote 1 down vote accepted

This is not an answer, but doesn't fit easily into comments. I find your notation a bit confusing, for example, I would write $t_x = \frac{\partial \phi_x(u)}{\partial u}$ instead of what you have above.

The componentwise expression of the chain rule in the comments above is as follows (note that $\psi = f \circ \phi$):

Let $f(x,y,z) = \begin{bmatrix} f_x(x,y,z) \\ f_y(x,y,z) \\ f_z(x,y,z) \end{bmatrix}$, $\phi(u) = \begin{bmatrix} \phi_x(u) \\ \phi_y(u) \\ \phi_z(u) \end{bmatrix}$, $\psi(u) = \begin{bmatrix} f_x(\phi_x(u), \phi_y(u), \phi_z(u)) \\ f_y(\phi_x(u), \phi_y(u), \phi_z(u)) \\ f_z(\phi_x(u), \phi_y(u), \phi_z(u)) \end{bmatrix}$.

Then using the chain rule componentwise you have (I have suppressed the arguments to simplify):

$$ \begin{bmatrix} \frac{\partial \psi_x}{\partial u} \\ \frac{\partial \psi_y}{\partial u} \\ \frac{\partial \psi_z}{\partial u} \end{bmatrix} = \begin{bmatrix} \frac{\partial f_x}{\partial x}\frac{\partial \phi_x}{\partial u} + \frac{\partial f_x}{\partial y}\frac{\partial \phi_y}{\partial u} + \frac{\partial f_x}{\partial z}\frac{\partial \phi_z}{\partial u} \\ \frac{\partial f_y}{\partial x}\frac{\partial \phi_x}{\partial u} + \frac{\partial f_y}{\partial y}\frac{\partial \phi_y}{\partial u} + \frac{\partial f_y}{\partial z}\frac{\partial \phi_z}{\partial u} \\ \frac{\partial f_z}{\partial x}\frac{\partial \phi_x}{\partial u} + \frac{\partial f_z}{\partial y}\frac{\partial \phi_y}{\partial u} + \frac{\partial f_z}{\partial z}\frac{\partial \phi_z}{\partial u} \end{bmatrix} = \begin{bmatrix} \frac{\partial f_x}{\partial x} & \frac{\partial f_x}{\partial y} & \frac{\partial f_x}{\partial z} \\ \frac{\partial f_y}{\partial x} & \frac{\partial f_y}{\partial y} & \frac{\partial f_y}{\partial z} & \\ \frac{\partial f_z}{\partial x} & \frac{\partial f_z}{\partial y} & \frac{\partial f_z}{\partial z} & \end{bmatrix} \begin{bmatrix} \frac{\partial \phi_x}{\partial u} \\ \frac{\partial \phi_y}{\partial u} \\ \frac{\partial \phi_z}{\partial u} \end{bmatrix}. $$ The above can be written more succinctly as $D \psi(u) = Df(\phi(u)) \, D\phi(u)$.

The vector $t$ is given by $t = \begin{bmatrix} \frac{\partial \phi_x}{\partial u} \\ \frac{\partial \phi_y}{\partial u} \\ \frac{\partial \phi_z}{\partial u} \end{bmatrix}$, and similarly for $t'$ ($\phi$ replaced by $\psi$).

share|cite|improve this answer
Of course $f$ is a vector-valued function $(f_x , f_y, f_z)^T$. It just didn't occur to me that the components of the tangent could be/are defined as $\left[\frac{\partial_x \phi}{\partial u} , \frac{\partial \phi_y}{\partial u} , \frac{\partial \phi_z}{\partial u}\right]^T$. – Olumide Jun 24 '12 at 16:52

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.