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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.

The common homogeneous solution is $C_1\cos 2x+C_2\sin 2x$. For forcing $\sin x$, a trial $a\sin x$ gives $3a\sin x=\sin x$, so the first general solution isFor forcing $\sin 2x$, the frequency produces resonance in a differential equation. A trial $ax\cos 2x$ gives $y_p^{\prime \prime }+4y_p=-4a\sin 2x$, so $a=-1/4$ and the second solution isThe superposition principle permits adding the two particular solutions. The sum-forcing equation therefore hasOnly one pair of homogeneous constants is needed, because the two original homogeneous families coincide.

The characteristic equation has roots zero and three, giving $y_h=C_1+C_2{e}^{3x}$. Since the forcing is quadratic and the homogeneous solution contains a constant, use the polynomial trial $y_p=A{x}^3+B{x}^2+Cx$. Substitution givesMatching coefficients gives $A=-1/3$, $B=-1/3$, and $C=-8/9$. Therefore

The tangent half-angle substitution $t=\tan (x/2)$ gives $\sin x=2t/(1+t^2)$ and $dx=2dt/(1+t^2)$. The endpoints become zero and one, soUse completing the square, $t^2+t+1=(t+1/2{)}^2+3/4$, and integrate:Thus both integrals equal $\boxed{\pi /(3\sqrt{3})}$.