# Computing the gradient of the function $\psi(u) = f[ \phi(u)]$

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.

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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

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 $\psi$).

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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