The defining exponential function formulas for the hyperbolic functions give

First calculate the curl:By Stokes theorem, use the positively oriented equator $\mathbf{r}(\varphi )=(a\cos \varphi ,a\sin \varphi ,0)$, $0\le \varphi \le 2\pi$. Its tangent is $a(-\sin \varphi ,\cos \varphi ,0)$, while the restricted field is $(0,3a^2\cos \varphi \sin \varphi -a\cos \varphi ,0)$. ThusThe first term integrates to zero over a full period and the second uses ${\int }_0^{2\pi }{\cos }^2\varphi d\varphi =\pi$.

One can also calculate directly on the curved surface. With spherical coordinates $0\le \theta \le \pi /2$, $0\le \varphi \le 2\pi$,Substitution gives the scalar integrand $a^3{\sin }^2\theta \cos \theta \sin \varphi -a^2\sin \theta \cos \theta$. Its azimuthal sine term vanishes, and the rest gives $-2\pi a^2{\int }_0^{\pi /2}\sin \theta \cos \theta d\theta =-\pi a^2$, confirming the orientation and the same result.

For a smooth oriented surface $S$ with positively oriented boundary $C=\partial S$, Stokes theorem statesHere $d\mathbf{S}=\mathbf{n}dA$ uses the chosen unit normal vector, while $d\mathbf{r}$ is tangent to the directed boundary. The boundary orientation follows the right-hand rule: curling the fingers along $C$ makes the thumb point in the normal direction. The field must be continuously differentiable on a neighbourhood of the surface, with the usual piecewise-smooth allowance for its boundary.

For the upper hemisphere with outward normal, the induced equatorial direction is counterclockwise as viewed from positive $z$.

The upper branch leaves the origin horizontally and curves upward, growing like $x^{5/2}$. The lower branch shares the horizontal tangent, rises to its small maximum at $x=16/25$, then crosses the axis at $x=1$ and falls toward $-\infty$. Its concavity changes at $x=64/225$, before the maximum. A separate zoom makes those nearby lower-branch features visible.

For $x>0$, differentiate the two explicit branches:The plus branch is strictly increasing and strictly convex for $x>0$. Its only zero is $x=0$, also its one-sided global minimum; it has no interior extremum or inflection point.

The minus branch rises initially, then has $y_{-}^{\prime }=0$ at $\sqrt{x}=4/5$, changing from positive to negative. Its second derivative changes sign at $\sqrt{x}=8/15$. Its zeros follow from $y_{-}=x^2(1-\sqrt{x})$. ConsequentlyThe origin is a one-sided local minimum of both graph branches, because their domain ends there; it is not an interior critical point. The minus branch has no global minimum since it tends to $-\infty$ as $x\to \infty$. Its maximum and inflection heights, useful for the sketch, are $256/3125$ and $28672/759375$, respectively.

A real square on the left requires $x^5\ge 0$, so $x\ge 0$. Taking the two square roots gives exactly the two real graph branchesThey meet only at the origin. The local coordinate change $Y=y-x^2$ turns the equation into $Y^2=x^5$, the plane cusp of type (2,5); both branches share the tangent $y=0$ there.

For the specified polynomial,The first equation $F_y=0$ forces $y=x^2$; substitution into $F_x=0$ then gives $x=0$, and hence $y=0$. It also satisfies the curve equation. Therefore the unique singular point is

Along a differentiable graph satisfying the implicit equation, the multivariable chain rule givesThus, wherever $F_y\ne 0$,The implicit function theorem supplies such a local graph under that nonvanishing condition. At a singular point both numerator and denominator vanish, so this division cannot determine its tangent.

Along a change in volume at fixed temperature, the first law of thermodynamics givesUse the Helmholtz Maxwell relation derived in (a) to replace the entropy derivative. This provesFor the stated Van der Waals equation, temperature-independent $a,b$ give $P_T{|}_V=R/(V-b)$. Hence $U_V{|}_T=a/V^2$. Integrating with respect to $V$ yieldsThe dilute-limit condition holds at every fixed $T$ and forces $h(T)=cT$, soThe result is the internal energy of a van der Waals gas in the fixed-amount convention of the question. Its volume derivative is positive even though the attractive energy correction itself is negative.

For fixed amount of substance, take the internal energy $U=U(S,V)$. The first law of thermodynamics identifies $T=U_S{|}_V$ and $-P=U_V{|}_S$. Equality of mixed second derivatives givesFor the Helmholtz free energy, subtract the product differential $d(TS)$ from the first law of thermodynamics to obtain $dF=-SdT-PdV$. Its natural variables are $(T,V)$, so $F_T=-S$ and $F_V=-P$. Exactness therefore givesFor the Gibbs free energy, the product rule gives $dG=-SdT+VdP$. Thus $G_T=-S$, $G_P=V$, andFor the enthalpy, $dH=TdS+VdP$, giving $H_S=T$, $H_P=V$ andThese are the four Maxwell relations, each following from exactness in the appropriate pair of natural variables.

Assume the components of $\mathbf{F}$ have continuous second partial derivatives. Expanding the divergence of its curl givesEach positive term cancels a negative term because the corresponding mixed derivatives commute. Consequently

Put $c=\sqrt{a^2+b^2}$ and denote the distances to $(-c,0)$ and $(c,0)$ by $d_{-}$ and $d_{+}$. Direct expansion using $c^2=a^2+b^2$ givesSince $c>a>0$ and $\cosh t\ge 1$, both expressions inside the final square roots are positive. Hencefor every point of this right branch. This is the focal-distance identity for a hyperbola; on the left branch the signed difference would reverse, while its absolute value would remain $2a$.

Solving for $y$ gives $y=\pm (b/a)\sqrt{x^2-a^2}$ with $x\ge a$. For large $x$, the square root is $x-a^2/(2x)+O(x^{-3})$, so the two asymptotes areThe curve lies between these lines on the right side and has its nearest point, with vertical tangent, at $(a,0)$. As $t$ increases it travels from the lower unbounded end through that vertex to the upper unbounded end. The left hyperbola branch is not part of the given curve.

The hyperbolic functions identity eliminates $t$ from the parametric curve:The restriction is essential: $\cosh t\ge 1$, so the parametrization traces only the right branch of the full hyperbola. Conversely, every point on that right branch is reached by taking $t=\mathrm{arsinh}(y/b)$.

The Euclidean distance is $D(x,y)=\sqrt{x^2+y^2}$. Along the parametric curve,This is at least $a^2$, with equality exactly when $t=0$. Therefore

Differentiate $y^{\prime }=-y+3x$ and use the other equation:Since $3x=y^{\prime }+y$, this becomes $y^{\prime \prime }-3y^{\prime }+2y=0$. Its roots are one and two, so $y=A{e}^t+B{e}^{2t}$. The given data imply $y^{\prime }(0)=-2+3=1$, hence $A+B=2$ and $A+2B=1$. Thus $A=3$, $B=-1$, givingRecover the other dependent variable from $x=(y^{\prime }+y)/3$:These expressions satisfy both original equations and the two initial values.