Take the dot product of the orthogonal basis expansion with ${\mathbf{u}}_j$. All terms except the $j$th vanish by orthogonality, leavingThereforeThe basis is not assumed normalized, so its squared lengths must remain in the denominators. Applying $S$ and using $S{\mathbf{u}}_i={\lambda }_i{\mathbf{u}}_i$ gives

Chooseor add any constant. Then $f^{\prime }(z)=1$, so the vector field is exactly $\mathbf{r}$ and necessarily has the same flux as in part (a). The formula from part (b)(i) confirms this:

Use the two pieces of the surface integral directly; this avoids introducing an unnecessary second derivative of $f$. On the unit sphere, $\mathbf{n}=\mathbf{r}$ and $\mathbf{F}\cdot \mathbf{n}=f^{\prime }(z)$. A spherical band between heights $z$ and $z+dz$ has area $2\pi dz$, so the curved flux isby the fundamental theorem of calculus. On the horizontal top disk, $\mathbf{F}\cdot \mathbf{n}=z_0{f}^{\prime }(z_0)$ and the area is $\pi (1-z_0^2)$. ThusAs usual for these surface integrals, the first derivative is assumed sufficiently integrable for the displayed fundamental theorem of calculus evaluation; a continuous first derivative suffices. Mere pointwise differentiability without such an integrability assumption need not make an ordinary flux integral exist.

Differentiate the flux:On $-1<z_0<1$, $\Phi$ is strictly increasing. Its limiting endpoint values are $\Phi (-1)=0$ and $\Phi (1)=4\pi$, and $\Phi (0)=2\pi$. It has horizontal limiting tangents at the endpoints, curves upward for $z_0<0$, curves downward for $z_0>0$, and has an inflection point at zero.

The derivative graph is a downward-opening parabola, approaching zero at both endpoints and attaining its maximum $3\pi$ at $z_0=0$.

The divergence of the position vector field is $\nabla \cdot \mathbf{r}=3$. At height $z$, the horizontal cross-section is a disk of area $\pi (1-z^2)$. By the divergence theorem, the outward flux integral is three times the enclosed volume:ThereforeA direct check separates the two pieces of the surface integral. On the unit sphere, the outward unit normal is $\mathbf{r}$, so $\mathbf{r}\cdot \mathbf{n}=1$ and the curved contribution is $2\pi (z_0+1)$. On the top disk, $\mathbf{n}={\mathbf{e}}_z$, giving $\pi z_0(1-z_0^2)$. Their sum is the same flux.

The relevant mixed partial derivatives agree:The vector field is smooth on the entire simply connected plane, so this condition establishes that it is a conservative vector field. We can also construct its vector-field potential explicitly. From ${\varphi }_x=y^2-2x$, take an antiderivative in $x$:Then ${\varphi }_y=2xy+h^{\prime }(y)=2xy-y$ requires $h^{\prime }(y)=-y$. HenceThe fundamental theorem for line integrals checks the previous answer: $\varphi (1,1)-\varphi (0,0)=-1/2$.

For the parametric curve, ${\mathbf{r}}^{\prime }(t)=(2t,3t^2{)}^T$ andThe line integral is therefore

Integrating first over $z$ multiplies by $e^y$. Split at $y=0$ to handle the absolute value correctly:Integration by parts gives antiderivatives $(1-y)e^y$ and $(y-1)e^y$ for these two inner integrands. Their contributions arerespectively. ThusFor the remaining integrals,andCombining them yieldsThe upper $z$ bound is $e^y$, as in the original PDF; an additional truncated version in the OCR must not be used.

Here $y\ge 0$, so the absolute value equals $y$. The volume integral, evaluated in Cartesian coordinates, isThe integrand is constant with respect to $z$, giving

Differentiate $u_y=y^{-1/2}(U_{\xi }-U_{\eta })$, retaining the derivative of the prefactor. The product rule and multivariable chain rule giveUsing $y=(\xi -\eta {)}^2/16$, this becomesThe last term is essential because the coordinate change is nonlinear.

Apply ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$ once more. The chain rule and symmetry of second derivatives giveHere and in the next part the second partial derivatives are understood with the usual smoothness required for this partial differential equation.

The coordinate partial derivatives are ${\xi }_y=1/\sqrt{y}$ and ${\eta }_y=-1/\sqrt{y}$. The multivariable chain rule therefore givesSince $\xi -\eta =4\sqrt{y}>0$, the expression entirely in the new coordinates isThe inverse coordinate formulas are $x=(\xi +\eta )/2$ and $y=(\xi -\eta {)}^2/16$, with domain $\xi >\eta$.

To distinguish the old and new coordinate descriptions, write $u(x,y)=U(\xi (x,y),\eta (x,y))$. Since ${\xi }_x={\eta }_x=1$, the multivariable chain rule givesEquivalently, the differential operator is ${\partial }_x={\partial }_{\xi }+{\partial }_{\eta }$.

The characteristic equation of the reduced homogeneous linear differential equation is $m^2-\beta =0$. Its polynomial roots change type with the sign of $\beta$, givingFor $\beta =0$, the zero polynomial root is repeated, so the second independent solution is $t$. For $\beta <0$, the polynomial roots are the complex conjugate pair $\pm i\sqrt{-\beta }$, and the displayed sine and cosine give a real basis of solutions. The constants are arbitrary real numbers. In every case $x(t)=y^{\prime }(t)-\cos t$ reconstructs the corresponding general solution of the coupled system.

Differentiate the second ordinary differential equation and use the first to eliminate $x^{\prime }$:Thus the reduced second-order linear differential equation isNo solutions are lost: given a solution $y$, define $x=y^{\prime }-\cos t$; then both original ordinary differential equations hold.

There are two equally likely balls bearing label eight, one of each colour. The conditional probability that the removed ball was green is therefore $1/2$. By the law of total probability,Both possible first colours must be included because the observed label alone does not determine the colour.

The specified colour tells us that one green ball has been removed. Thus sampling without replacement again leaves nine green balls among nineteen:The numerical label supplies no further change once this first colour is known.

Only a green ball carries label four. After it is removed, nine of the remaining nineteen balls are green. The conditional probability is

Suppose the first draw is green. The excess of green over red must remain positive after draws one through five and return to zero at draw six. After deleting the first green and final red, the remaining four steps must form a Dyck path: they have equal colour counts and never go below zero relative to the new starting level. There are $C_2=2$ such orders, $GGRR$ and $GRGR$, giving $GGGRRR$ and $GGRGRR$ in the original game.

Interchanging green and red gives $RRRGGG$ and $RRGRGG$. Thus there areorders. More generally, the Catalan number $C_{m-1}$ counts each sign of excursion with first return time $2m$.

The two possible colour orders are $GGRR$ and $RRGG$. Thus the answer isThey correspond to the positive and negative four-step excursions with first return time four.