Multivariable Calculus

Question 1

(a) [6 marks] Let $\mathbf{F}=(P,Q,R)$ with $P=\frac{-y}{x^2+y^2}+2xz$, $Q=\frac{x}{x^2+y^2}$, $R=x^2+\cos z$.

Curl components:

• $(\nabla\times\mathbf F)_x = R_y - Q_z = 0 - 0 = 0$.
• $(\nabla\times\mathbf F)_y = P_z - R_x = 2x - 2x = 0$.
• $(\nabla\times\mathbf F)_z = Q_x - P_y$. Here $Q_x=\frac{(x^2+y^2)-x(2x)}{(x^2+y^2)^2}=\frac{y^2-x^2}{(x^2+y^2)^2}$ and $P_y=\frac{-(x^2+y^2)+y(2y)}{(x^2+y^2)^2}=\frac{y^2-x^2}{(x^2+y^2)^2}$. So $Q_x-P_y=0$.

Thus $\nabla\times\mathbf F=\mathbf 0$ (4). Divergence: $\nabla\cdot\mathbf F = P_x+Q_y+R_z$. $P_x=\frac{2xy}{(x^2+y^2)^2}+2z$, $Q_y=\frac{-2xy}{(x^2+y^2)^2}$, $R_z=-\sin z$. Sum $=2z-\sin z$ (2).

(b) [9 marks] The domain $\mathbb R^3\setminus\{z\text{-axis}\}$ is not simply connected (loops around the $z$-axis cannot contract to a point) (3). Poincaré's lemma / the "curl-free $\Rightarrow$ conservative" implication requires a simply connected domain; it fails here (2). The vortex part $\mathbf G$ is the gradient of $\theta=\operatorname{atan2}(y,x)$, which is multivalued (jumps by $2\pi$ around a loop). The remaining part $\mathbf F-\mathbf G=(2xz,\,0,\,x^2+\cos z)$ is conservative with potential $\phi = x^2 z + \sin z + \text{const}$ Check: $\phi_x=2xz$, $\phi_y=0$, $\phi_z=x^2+\cos z$. ✓ (4)

(c) [7 marks] Parametrize $\mathbf r(t)=(2\cos t, 2\sin t, 1)$, $t\in[0,2\pi]$, $\mathbf r'(t)=(-2\sin t, 2\cos t, 0)$. The gradient part $\nabla\phi$ contributes $0$ around the closed loop (path-independent, closed curve). Only $\mathbf G$ contributes. On $C$: $x^2+y^2=4$, so $\mathbf G=(-\tfrac{y}{4},\tfrac{x}{4},0)=(-\tfrac{\sin t}{2},\tfrac{\cos t}{2},0)$. $\mathbf G\cdot\mathbf r' = (-\tfrac{\sin t}{2})(-2\sin t)+(\tfrac{\cos t}{2})(2\cos t)=\sin^2 t+\cos^2 t=1$. $\oint_C \mathbf F\cdot d\mathbf r=\int_0^{2\pi}1\,dt=2\pi.$ (marks: parametrization 2, drop gradient part 2, integrand 2, value 1)

(d) [4 marks] Any surface $S$ bounded by $C$ (which encircles the $z$-axis) must cross the $z$-axis, where $\mathbf F$ (and $\nabla\times\mathbf F$) is undefined (2). Stokes' theorem requires $\mathbf F$ to be $C^1$ on an open set containing $S$; this fails because $S$ contains a singular point (2). Hence Stokes cannot be applied and the direct value $2\pi\neq0$ stands.

(e) [8 marks] Generalized Stokes: $\int_{\partial M}\omega=\int_M d\omega$.

• Green ($1$-form on a plane region): $M$ = 2D region, $\partial M$ = its boundary curve, $\omega=P\,dx+Q\,dy$, $d\omega=(Q_x-P_y)\,dx\wedge dy$ (2).
• Stokes (classical): $M$ = oriented surface in $\mathbb R^3$, $\partial M$ = boundary curve, $\omega=\mathbf F\cdot d\mathbf r$ ($1$-form), $d\omega=(\nabla\times\mathbf F)\cdot d\mathbf S$ (2).
• Divergence: $M$ = solid region $E$, $\partial M$ = closed surface, $\omega=$ the $2$-form dual to $\mathbf F$ (flux form), $d\omega=(\nabla\cdot\mathbf F)\,dV$ (2). Unifying idea: the exterior derivative $d$ generalizes grad/curl/div, and orientation of the boundary is induced from $M$; each theorem equates a boundary integral to an integral of the derivative over the interior (2).

Question 2

(a) [9 marks] $\nabla U=(1,2,1)$, $\nabla g=(2x,2y,8z)$. Equations: $1=2\lambda x,\;2=2\lambda y,\;1=8\lambda z$ $\Rightarrow x=\frac{1}{2\lambda}, y=\frac{1}{\lambda}, z=\frac{1}{8\lambda}$ (3). Constraint: $x^2+y^2+4z^2=12$: $\frac{1}{4\lambda^2}+\frac{1}{\lambda^2}+4\cdot\frac{1}{64\lambda^2}=12$ $\Rightarrow \frac{1}{\lambda^2}\left(\frac14+1+\frac{1}{16}\right)=12$. Sum $=\frac{4+16+1}{16}=\frac{21}{16}$. So $\frac{21}{16\lambda^2}=12\Rightarrow \lambda^2=\frac{21}{192}=\frac{7}{64}\Rightarrow\lambda=\pm\frac{\sqrt7}{8}$ (4). Points: with $\lambda=\frac{\sqrt7}{8}$: $x=\frac{4}{\sqrt7}, y=\frac{8}{\sqrt7}, z=\frac{1}{\sqrt7}$; with $\lambda=-\frac{\sqrt7}{8}$: negatives thereof (2).

(b) [6 marks] $U=x+2y+z$. At the $\lambda>0$ point: $U=\frac{4}{\sqrt7}+\frac{16}{\sqrt7}+\frac{1}{\sqrt7}=\frac{21}{\sqrt7}=3\sqrt7$ (2). At the $\lambda<0$ point: $U=-3\sqrt7$ (1). Global max $=3\sqrt7\approx7.937$; global min $=-3\sqrt7$ (2). Existence: $U$ is continuous and the ellipsoid $S$ is compact (closed & bounded), so by the Extreme Value Theorem the max and min are attained (1).

(c) [10 marks] Now $g=x^2+y^2+4z^2-12$, $h=z$. On $z=0$: $g$ reduces to circle $x^2+y^2=12$. $\nabla U=\lambda\nabla g+\mu\nabla h$: $(1,2,1)=\lambda(2x,2y,8z)+\mu(0,0,1)$. Component eqns: $1=2\lambda x,\;2=2\lambda y,\; 1=8\lambda z+\mu$. With $z=0$: third gives $\mu=1$ (free, determined) (3). From first two: $x=\frac{1}{2\lambda}, y=\frac1\lambda$. Constraint $x^2+y^2=12$: $\frac{1}{4\lambda^2}+\frac1{\lambda^2}=12\Rightarrow\frac{5}{4\lambda^2}=12\Rightarrow\lambda^2=\frac{5}{48}\Rightarrow\lambda=\pm\sqrt{\frac{5}{48}}$ (4). Points: $\left(\frac{1}{2\lambda},\frac1\lambda,0\right)$. Numerically $\lambda=\pm0.3227$, giving $(\pm1.549,\pm3.098,0)$ (same sign). $U=x+2y=\frac{1}{2\lambda}+\frac2\lambda=\frac{5}{2\lambda}$. Max: $\lambda>0$, $U=\frac{5}{2\lambda}=\frac52\cdot\frac{1}{\sqrt{5/48}}=\frac52\sqrt{48/5}=\sqrt{60}=2\sqrt{15}\approx7.746$; min $=-2\sqrt{15}$ (3).

(d) [8 marks] Algorithm (Newton on the KKT/Lagrange system) — award for correctness:

# unknowns (x,y,z,lam)
F(x,y,z,lam) = [ grad U - lam*grad g ;  g(x,y,z) ]   # 4 eqns
solve F=0 by Newton: 
  guess w0
  repeat: w <- w - J(w)^{-1} F(w)  until ||F||<tol
J = Jacobian of the 4 residuals (includes Hessian of Lagrangian block)

(4) for a coherent scheme (projected-gradient alternative equally valid: project ∇U onto tangent plane of S, step, renormalize onto S). Classification via bordered Hessian: form

For a single constraint in $n=3$ vars, examine the sign of the last two leading principal minors of $\bar H$: an alternating sign pattern signals a constrained max, uniform sign a constrained min (4). Concretely $\lambda>0$ makes $H_L$ negative definite on the tangent space → local max ($U=3\sqrt7$); $\lambda<0$ → min.

Question 3

(a) [8 marks] Region: $0\le y\le1$, $\sqrt y\le x\le1$, i.e. $y\le x^2$. Reverse: $0\le x\le1$, $0\le y\le x^2$ (3).