For a parametric curve with $dx/dt\ne 0$, the chain rule givesThe specified point has $\cos t=\sin t=1/\sqrt{2}$, hence $t=\pi /4$ in the given traversal. Therefore

Eliminating the parameter gives $x^2+y^2/4=1$, an ellipse centered at the origin with horizontal semiaxis one and vertical semiaxis two. At $t=0$ the particle is at $(1,0)$ with velocity $(0,2)$, so it travels counterclockwise, reaching $(0,2),(-1,0),(0,-2)$ successively and returning after $2\pi$. The right panel of the graph figure shows this direction.

Conditioned on being inside the circle, the point is a uniform random point in a disk of radius $1/2$. The annulus from $r$ to $r+dr$ has area $2\pi rdr+O(d{r}^2)$. Dividing by the disk's area givesThe radial probability density function is $8r$ on $[0,1/2]$ and integrates to one. Without conditioning, the corresponding annulus probability for a point in the original square is $2\pi rdr+O(d{r}^2)$.

For a uniform distribution on the unit square, probability is the ratio of areas. The incircle has radius $1/2$, so

Within this polynomial family, $u(0,0)=0$ already holds and $u(1,0)=1+\beta =0$ gives $\beta =-1$. Thus one solution isTwo specified point values alone do not give uniqueness among all solutions of the heat equation.

The proposed function has partial derivatives $u_t=1$ and $u_{xx}=2\alpha$. Substitution in the heat equation requires $1=\alpha$, henceThe coefficient $\beta$ remains arbitrary.

The product rule for divergence gives $\nabla \cdot (r^2\mathbf{r})=\nabla (r^2)\cdot \mathbf{r}+r^2\nabla \cdot \mathbf{r}$. Since $\nabla \cdot \mathbf{r}=3$, the result is

Differentiate the sum of squares in Cartesian coordinates:

The integrand is an odd function, because $x$ is odd and $\cos x$ is even. Its integral over the symmetric interval is therefore

The derivative of $\tan x$ is $1/{\cos }^{2}x$. The fundamental theorem of calculus gives

For a separated harmonic function $V=F(x)G(y)$, Laplace's equation gives $F^{\prime \prime }/F=-G^{\prime \prime }/G=-k^2$. Bounded periodic dependence in $x$ has $F=A\cos (kx)+B\sin (kx)$; decay on each half-plane selects $G=e^{-k|y|}$ for $k>0$, when the potential is matched continuously at the sheet. These are the separated decaying modes. General decaying solutions can be superpositions of them; decay alone does not force a single value of $k$.

The prescribed cosine density selects $k=1$ and $V=C\cos x{e}^{-|y|}$. Its normal derivative jump is $-2C\cos x$. Matching the stated jump $2\pi {\Sigma }_0\cos x$ givesThis is the gravitational potential of a sinusoidal sheet in the normalization of the question. It is continuous at the sheet and decays on both sides. Since ${\Sigma }_0\cos x$ changes sign, it can physically describe a Fourier component of a density perturbation rather than an everywhere positive isolated mass sheet.

For $c>0$ and $l>0$, the zero-endpoint Dirichlet boundary conditions select the Fourier sine basis functions $\sin (n\pi x/l)$. Separation of variables gives temporal factors with angular frequencies ${\omega }_n=n\pi c/l$. Zero initial displacement removes the temporal cosine terms, givingDifferentiating at $t=0$ and using Fourier orthogonality givesThe evaluation property of the given initial velocity means $v(x)=\delta (x-l/2)$ in the sense of a Dirac delta, an idealized impulsively struck fixed-end string. Thus $a_n=2\sin (n\pi /2)/(n\pi c)$: every even mode vanishes, while the lowest two excited modes have $n=1,3$. Their ordinary frequencies and amplitude ratio areThe signed coefficient ratio is $a_3/a_1=-1/3$. The impulse solution is interpreted as a distributional weak solution, rather than one with a smooth initial velocity.

The identity $\nabla \cdot (\mathbf{a}\times \mathbf{v})=\mathbf{v}\cdot (\nabla \times \mathbf{a})-\mathbf{a}\cdot (\nabla \times \mathbf{v})$ reduces to $-\mathbf{a}\cdot (\nabla \times \mathbf{v})$ when $\mathbf{a}$ is constant. In components, this follows by differentiating ${\epsilon }_{ijk}{a}_j{v}_k$ and using the antisymmetry of the Levi-Civita symbol. The divergence theorem and the cyclic scalar triple product identity now giveSince $\mathbf{a}$ is arbitrary,The order of the cross product determines the minus sign, as in the vector curl form of the divergence theorem.

Since $\mathbf{a}$ is constant, componentwise differentiation gives $\nabla \cdot (\mathbf{a}\varphi )={\sum }_i{a}_i{\partial }_i\varphi =\mathbf{a}\cdot \nabla \varphi$. Applying the divergence theorem to this vector field givesThis holds for every constant vector $\mathbf{a}$, so the vector integrals agree. This is the vector gradient form of the divergence theorem. With $\varphi =z$, the gradient is the constant unit vector ${\mathbf{e}}_z$. Therefore

For a bounded region $V$ with piecewise smooth closed boundary $S$, outward unit normal vector $\mathbf{n}$ and continuously differentiable vector field $\mathbf{F}$, the divergence theorem states ${\int }_S\mathbf{F}\cdot \mathbf{n}dS={\int }_V\nabla \cdot \mathbf{F}dV$. Here $dS$ is scalar surface area, $dV$ is volume, and $d\mathbf{S}=\mathbf{n}dS$ is the oriented area element.

For $\mathbf{F}=\mathbf{r}$, its divergence is three. The ellipsoid has semiaxes $a,a,b$, so its volume is $4\pi a^{2}b/3$, obtained by scaling the unit ball. Consequently