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The symmetric and antisymmetric parts of a matrix areIndeed $S+A=M$, while the transpose identities $(M^T{)}^T=M$ give $S^T=S$ and $A^T=-A$. Thus $S$ is a symmetric matrix and $A$ is an antisymmetric matrix. The decomposition is also unique: a matrix that is both symmetric and antisymmetric must equal its own negative and hence vanish.

Put $\mathbf{W}=\nabla \times \mathbf{F}$ and $\mathbf{v}=\dot{\mathbf{r}}$. All vector fields and their spatial partial derivatives below are evaluated at $\mathbf{r}(t)$. The product rule for a dot product givesBecause the vector field has no explicit time dependence, the multivariable chain rule gives $d\mathbf{W}/dt=J_{\mathbf{W}}\dot{\mathbf{r}}$, where the Jacobian matrix has entries $(J_{\mathbf{W}}{)}_{ij}={\partial }_j{W}_i$. ConsequentlyThe first term measures the change of the curl sampled as the point moves; the second measures the change in the velocity vector used in the dot product.

To give the result explicitly in $P,Q,R$, define the directional derivative operator $L=\dot{x}{\partial }_x+\dot{y}{\partial }_y+\dot{z}{\partial }_z$. Sincethe same expression isFor example $L(R_y-Q_z)=\dot{x}(R_{yx}-Q_{zx})+\dot{y}(R_{yy}-Q_{zy})+\dot{z}(R_{yz}-Q_{zz})$; the other two terms expand in the same way. Thus only the first and second spatial partial derivatives of $P,Q,R$, together with the velocity and acceleration of the parametric curve, occur.

Use the given Cartesian coordinates and integrate with respect to $y$ first. Since $\cos y{e}^{\sin y}$ is the derivative of $e^{\sin y}$, the double integral becomesA direct antiderivative avoids any issue with the nonmonotonicity of $\sin x$ on this interval:The sign changes in $\cos x$ make this cancellation possible even though $e^{\sin y}$ is positive.

Substitute the partial derivatives obtained above. The first-order terms cancel exactly:The transformed partial differential equation is thereforeIn these characteristic coordinates, the mixed partial derivative is the only surviving term.