On each of the two radial legs, $\mathbf{F}=\mathbf{a}\times \mathbf{r}$ is perpendicular to the tangent. Each contribution is zero, giving $\boxed{\int \mathbf{F}\cdot d\mathbf{r}=0}$. This differs from the straight-path answer, confirming path dependence.

For $\mathbf{G}$, use $\mathbf{r}=(s,0,0)$ with $s$ decreasing from one to zero, followed by $\mathbf{r}=(0,s,0)$ with $s$ increasing from zero to one. The two contributions areThey cancel: $\boxed{\int \mathbf{G}\cdot d\mathbf{r}=0}$. This agreement is a consequence of the conservative vector field property already proved, rather than evidence sufficient by itself to establish it.

Parametrize the segment by $\mathbf{r}(t)=(1-t,t,0)$, $0\le t\le 1$. Its tangent is $(-1,1,0)$ and $\mathbf{F}=(-t,1-t,t-1)$, so $\mathbf{F}\cdot {\mathbf{r}}^{\prime }=1$. Hence its line integral isFor the other line integral, $|\mathbf{r}{|}^2=(1-t{)}^2+t^2$ and $\mathbf{r}\cdot {\mathbf{r}}^{\prime }=2t-1$. ThusThis also equals the change in $\Phi$, by the fundamental theorem for line integrals.

For the specified coordinate vectors,Choose the orthogonal matrix representing a quarter-turn,Then $S^{T}S=I$ and direct matrix multiplication gives $S^{T}LS=\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)=M$. Thus these two shear mappings are orthogonally similar.

Since $M^T=I-\mathbf{a}{\mathbf{b}}^T=I-N$, nilpotence already gives $(I+N)(I-N)=I$. Explicitly, in index notation,Thus $\boxed{L{M}^T=I}$, or equivalently $M^T=L^{-1}$. Here ${\delta }_{ij}$ is the Kronecker delta.

Let $N=\mathbf{a}{\mathbf{b}}^T$. The orthogonality condition gives $N^2=\mathbf{a}({\mathbf{b}}^T\mathbf{a}){\mathbf{b}}^T=0$: this is the nilpotent part of a transvection when both vectors are nonzero. In two dimensions,Expanding its determinant, the fourth-degree terms cancel:These formulas also cover either vector being zero.

The Bernoulli differential equation becomes linear after $z=1/y$: multiplying by $-y^{-2}$ gives $z^{\prime }-2z=-x^2$. A quadratic particular solution is $z_p=x^2/2+x/2+1/4$, while the homogeneous solution is $C{e}^{2x}$. The initial condition gives $z(0)=1$ and $C=3/4$. HenceThe denominator is positive for all real $x$, since $2x^2+2x+1=2(x+1/2{)}^2+1/2$. Thus this initial value problem has the displayed solution on the entire real line. The separate solution $y=0$ does not satisfy its initial condition.

For $\mathbf{u}=(1,1{)}^T$ and $\mathbf{v}=(1,-1{)}^T$, the outer products areTheir inner product is zero and both squared norms are two. Thus $P_u=\mathbf{u}{\mathbf{u}}^T/2$ and $P_v=\mathbf{v}{\mathbf{v}}^T/2$ are orthogonal spectral projectors, with $P_u+P_v=I$. Since $A\mathbf{u}=5\mathbf{u}$ and $A\mathbf{v}=\mathbf{v}$, the eigenvalues are $\lambda =5$, $\mu =1$, andMultiplying these decompositions uses $P_u{P}_v=0$ and $P_u^2=P_u$, $P_v^2=P_v$, giving the identity matrix.

Direct matrix multiplication givesConsequently $\boxed{A^2-6A+5I=0}$, the Cayley-Hamilton theorem for its characteristic polynomial $t^2-6t+5$. Rearranging gives $A(6I-A)=5I$, so

The upper boundary follows $y=x$ until its intersection with $y=1/x$ at $(1,1)$, then follows the hyperbola to $(2,1/2)$. The lower boundary is $y=0$. Therefore the two vertical-slice descriptions are $0\le x\le 1$, $0\le y\le x$, and $1\le x\le 2$, $0\le y\le 1/x$.

Integrating with respect to $y$ first reduces the double integral toUsing the antiderivatives from part (a),

Let $I={\int }_0^{\infty }{e}^{-x^2}dx$. Positivity allows the squared integral to be written as a first-quadrant double integral. In polar coordinates the Jacobian determinant is $r$, soUsing integration by parts, with the boundary term $x{e}^{-x^2}$ vanishing at both endpoints,The substitution $u=x^2$ also gives the two antiderivativesThese values also follow from the Gaussian moment recurrence and can be checked by differentiation.

At an interior local extremum of a differentiable function, $f^{\prime }(x)=0$. The second-derivative test classifies a stationary point: a positive second derivative gives a strict local minimum, and a negative second derivative gives a strict local maximum. A zero second derivative is inconclusive: $x^4$, $-x^4$, and $x^3$ at zero illustrate a minimum, a maximum, and neither.

HereThus $f^{\prime \prime }(1/3)=-8$ and $f^{\prime \prime }(3)=8$. Substitution into $f$ gives

Expand the constant right-edge value in a Fourier sine series. Its coefficients areThe superposition principle combines the normalized modes, givingThis is a harmonic function in a rectangle with one constant boundary value, with the coordinate roles interchanged. The series and its derivatives converge uniformly on compact subsets of $0<x<\pi$, so it satisfies Laplace's equation there. It vanishes on the other three edges and approaches one at every interior point of the right edge. At the two right corners the one-sided prescribed values disagree; no continuous extension can satisfy both. The stated open-edge condition avoids that incompatibility, and isolated corner assignments do not change the bounded interior Dirichlet problem.

For the first family, vanishing at $y=\pi$ requires $\sin (k\pi )=0$, so $k=n$ is a positive integer. The second family would require $\sinh (k\pi )=0$, impossible for real $k>0$; the mode $xy$ also fails unless its coefficient is zero. Conveniently normalized surviving modes areAt $x=\pi$, these become the Fourier sine series basis functions $\sin (ny)$.

The zero value at $x=0$ removes the even factor in $x$, and the zero value at $y=0$ removes the even factor in $y$. Up to multiplicative constants, the nonzero separated modes areThe zero-separation case also supplies $\boxed{xy}$. Each is a harmonic function and vanishes on both axes.

Factor $e^{i\varphi /2}$ from the numerator and denominator:For $0<\varphi <\pi$, the tangent ranges over all positive real values. Thus $\boxed{\mathrm{Re}w=0,\mathrm{Im}w>0}$. Using the principal complex logarithm,The real part ranges over the entire real line, while the imaginary part is fixed: the locus is the horizontal line $\mathrm{Im}\log w=\pi /2$. Any other fixed logarithm branch gives the parallel line $\mathrm{Im}\log w=\pi /2+2\pi k$ for one integer $k$; allowing every branch gives the union of those lines.

Euler's formula and its version with $-\theta$ giveSquaring $e^{i\theta }=\cos \theta +i\sin \theta$ and comparing real and imaginary parts with $e^{2i\theta }$ yields the double-angle formulas

Complex conjugation reverses the argument, reciprocation also inverts the modulus, and multiplication by $-1$ adds $\pi$ to the argument. With all arguments chosen in $[0,2\pi )$ and $0<\varphi <\pi /2$,Thus $z$ is in the first quadrant, $z^{\ast }$ and $z^{-1}$ lie on the same fourth-quadrant ray at radii two and one-half, and $-z$ is diametrically opposite $z$.

Differentiate both algebraic constraints at fixed $y$ or fixed $x$:The multivariable chain rule for $\varphi (x,y)=\psi (z,u)$ therefore givesDivide the first identity by $x$ and the second by $y$, and add. The ${\psi }_u$ terms cancel, leavingThe Jacobian determinant of $(x,y)\mapsto (z,u)$ is $-3xy/(zu)$, so the calculation is valid on real branches with $xyzu\ne 0$. This explicitly supplies the local independence assumption underlying the change of coordinates.