Total Derivative of Vector Function of Time

Question

I'm reading Baraff's '98 paper: http://run.usc.edu/cs599-s10/cloth/baraff-witkin98.pdf

He defines condition vector functions that equal zero $C(x(t))=0$. An energy function is defined using a vector function:

$$ E_c(x)=\frac{1}{2}k*C(x(t))^TC(x(t)) $$

The confusion comes from equation (7). (7) comes from computing the partial derivative of $\frac{\partial E_c(x)}{\partial x}$. Applying the chain rule, I imagine it would look something like:

$$ \frac{\partial E_c(x)}{\partial x}=k*C(x(t))*\frac{dx}{dt} $$

Could anyone clear up how to obtain the correct result in equation (7)? I must be miss-applying the chain rule, but am still a bit new to vector calculus (especially in regards to total vs partial derivatives).

For convenience, here's equation (7):

$$ f_i=-\partial \frac{E_c}{\partial x_i}=-k\frac{\partial C(x)}{\partial x_i}C(x) $$

Here's a screenshot of the relevant section from the above paper:

Answer 1

If $E(x) = { 1\over 2} \|C(x)\|^2$ then $DE(x) = C^T(x) DC(x)$, or ${\partial E(x)\over \partial x} = C^T(x) {\partial C(x)\over \partial x} $.

Hence ${\partial E(x)\over \partial x_i} = {\partial E(x)\over \partial x} e_i = C^T(x) {\partial C(x)\over \partial x}e_i = C^T(x) {\partial C(x)\over \partial x_i} = \sum_k [C(x)]_k [{\partial C(x)\over \partial x_i}]_k$.

Answer 2

If you write $\mathbf C(\mathbf x)$ as a vector $$ \mathbf C(\mathbf x)=\begin{pmatrix}C_1(\mathbf x)\\\vdots\\C_p(\mathbf x)\end{pmatrix}$$ the energy is given by $E_{C}=\frac k2\sum_{j=1}^pC_j(\mathbf x)^2$. From there, this is pretty straightforward. $$f_i=-\frac{\partial E_C}{\partial x_i}=-k\sum_{j=1}^p\frac{\partial C_j}{\partial x_i}C_j.$$