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For positive wavenumber, the condition gives $\omega =C{k}^{-n/2}$ on the positive-frequency branch. The group velocity and phase velocity are consequently related byThe required positive ratio $v_g/v_p=2$ gives $\boxed{n=-4}$. Isotropy makes these velocities parallel to the wave vector when this coefficient is positive. Identifying group velocity with energy transport assumes the usual lossless narrowband wave packet regime.

The two planes bound $z-1\le y\le 1-z$, and the elliptical cross-section has $-1\le z\le 1$. Under $x=2r\cos \theta$, $z=r\sin \theta$, the absolute Jacobian determinant for $(x,z)$ is $2r$. ThereforeAs a check, the cross-sectional ellipse has area $2\pi$, while the part of the thickness proportional to $z$ integrates to zero by reflection. The mean thickness is therefore two.

The triangular support can be written $0\le x\le \pi /2$, $\pi /2-x\le y\le \pi /2$. The density is nonnegative there, and the normalization of a probability distribution requiresHence $\boxed{A=2}$. Multiplying the joint probability density by the required function cancels $\sin x$ on the interior. Its expectation isThe singularity at $x=0$ lies at a support vertex of zero probability; the displayed integral establishes integrability. Assigning any value to the random function at that single vertex does not change the expectation.

Choose the normal pointing away from the origin, equivalently the outward normal of the first-octant tetrahedron. Parametrize the face by $\mathbf{r}=(x,y,2-x/2-y/2)$ on $D=\{x,y\ge 0:x+y\le 4\}$. Its oriented area element isThus the surface integral isThe divergence theorem checks this result: $\nabla \cdot \mathbf{F}=1$, the tetrahedron's volume is $4\cdot 4\cdot 2/6=16/3$, and each coordinate-plane face has zero flux. The opposite orientation gives $-16/3$; a signed flux requires an orientation.

Put $h=x-1/\sqrt{2}$ and $k=y-1/(2\sqrt{2})$. At the expansion point the exponent is one and the gradient is $(\sqrt{2}e,\sqrt{2}e)$. The first-order multivariable Taylor expansion isThe directional error of a first-order Taylor approximation is determined by the omitted terms, not just the equal first partial derivatives. The Hessian matrix at the expansion point is $e\left(\begin{array}{cc}4& 2\\ 2& 2\end{array}\right)$, soFor equal small steps along the coordinate axes, the leading error is $2e{h}^2$ along $x$ and $e{k}^2$ along $y$. The error grows faster along the $x$ axis, although the first-order change in the function is equal along the two axes.

For an exact differential $Pdx+Qdy$, equality of mixed partial derivatives requires $P_y=Q_x$. The matching pair is $P=f_1$, $Q=f_3$, sinceThe other pairings do not give the same partial derivatives on an open region. Integrating $f_1$ with respect to $x$ gives $\ln x\sin y+e^x{\cos }^{2}y+C(y)$. Its $y$ derivative agrees with $f_3$ precisely when $C^{\prime }(y)=0$. HenceThis also proves existence, rather than relying only on the necessary mixed-partial derivative test.

Use logarithmic differentiation on the positive function: $\ln f=x^{-1}\ln (x^2+2)$. The product rule and chain rule giveThe positive base permits this real definition on both components of the indicated domain. There is no need to treat a variable exponent as a constant.