On the curved path, $\mathbf{r}(t)=(t,t,t^2{)}^T$ and ${\mathbf{r}}^{\prime }=(1,1,2t{)}^T$. Direct substitution givesThusBoth values agree with part (i), for every $a,b,c$. Agreement on these two selected paths alone does not establish a conservative vector field.

On the first segment, parameterize ${\mathbf{r}}_1(t)=(t,t,0{)}^T$, $0\le t\le 1$. The dot products with ${\mathbf{r}}_1^{\prime }=(1,1,0{)}^T$ areTheir line integrals are one and $\sin 1$, respectively.

On the second segment take ${\mathbf{r}}_2(t)=(1,1,t{)}^T$, with tangent vector $(0,0,1{)}^T$. The corresponding integrands are $a-2+2t$ and $c$, whose integrals are $a-1$ and $c$. Adding the segments yields

Now $h=e^x-e^y-x+y$ and the origin has stationary value zero. Near it,The level curves therefore resemble hyperbolas near the saddle point of a scalar function. Along the $x$-axis $h=e^x-x-1>0$ for $x\ne 0$, while along the $y$-axis $h=1-e^y+y<0$ for $y\ne 0$. The zero level set includes the exact diagonal $x=y$ and another branch crossing it at the origin; this second branch is tangent to $y=-x$ but is curved away from the origin. The right panel of the contour figure displays these features.

Here the stationary point is the origin, of value two. The Taylor expansion isThus nearby level curves are approximately concentric circles. Globally the level curves above two are closed around the unique minimum, symmetric under interchanging $x$ and $y$, but not under changing either coordinate's sign. The faster exponential growth for positive coordinates compresses the curves on their positive sides.

The three displacements from $Q_1$ areHere $\mathbf{u}\times \mathbf{v}=(-1,-2,3{)}^T$ and its dot product with $\mathbf{w}$ is zero. Because the cross product itself is nonzero, the first three points determine a plane and the fourth lies on it. Explicitly all four satisfyThus the four points are coplanar.

Use displacements from $P_1$, since coplanarity of points concerns an affine plane that need not contain the origin. The three displacement vectors areTheir cross product is $\mathbf{u}\times \mathbf{v}=(-1,1,-1{)}^T$ and its dot product with $\mathbf{w}$ is $-1$. Since the scalar triple product is nonzero,

The acute angles follow from part (ii) and the Snell's law relation:The path lengths below and above the interface are respectivelyHence the total distance and minimum travel time areFor a sketch to scale, choose the interface as $y=0$, with $A=(0,-\sqrt{3})$ and $B=(2,1)$. The crossing is $C=(1,0)$ because $d_1\tan \theta =d_2\tan \varphi =1$. The figure uses equal axis scales and shows the two normal-relative angles.

Put $s=\sin \theta$. By part (i), $\sin \varphi =\sqrt{2}s$. The acute-angle geometry requires $0<s<1/\sqrt{2}$. Expressing the tangents in terms of their sines in the geometric constraint gives the required implicit equationThe hint suggests $s=1/m$ for an integer $m\ge 2$. Testing $m=2$ givesThus $\boxed{\sin \theta =1/2}$. Each term on the left is strictly increasing in $s$ over the allowed interval, so this solution is unique.

The angles in the PDF are measured from the vertical normal to the interface. The horizontal displacements of the two path segments are $d_1\tan \theta$ and $d_2\tan \varphi$, while their lengths are $d_1\sec \theta$ and $d_2\sec \varphi$. Thus the geometric constraint and travel time areThe Lagrange multiplier equations $\nabla T=\lambda \nabla g$ giveCancelling the positive geometric factors gives $\lambda =2\sin \theta /c=\sqrt{2}\sin \varphi /c$, henceThis is Snell's law, $\sin \theta /(c/2)=\sin \varphi /(c/\sqrt{2})$. The relation depends only on the ratio of the two speeds, so changing $D,d_1,d_2$ leaves the relation unchanged; it changes which angles satisfy it together with the geometric constraint.

For a minimum check, let $u$ be the horizontal distance to the crossing point. ThenandThus the travel time is a strictly convex function of $u$, and its stationary point is the unique global minimum. For positive $D,d_1,d_2$, $T^{\prime }(0)<0<T^{\prime }(D)$, placing the crossing between the two endpoint projections.

Use the root limit of a finite sum of powers. Put $M=\max \{1,a,a^2/2\}\ge 1$. For every $x>0$, at least one of the three terms is $M^x$ and none is larger, soTaking the positive $x$th root givesSince $3^{1/x}\to 1$, the squeeze theorem givesIndeed, for $a\le 1$ both remaining bases are at most $1$; for $1\le a\le 2$, $a\ge a^2/2$; and for $a\ge 2$, $a^2/2\ge a$. The endpoint cases $a=0,1,2$ are included in the same bound.

The original PDF specifies $x\to a$. The quotient is the difference quotient for the function $f(x)=x^x=e^{x\ln x}$ at the positive point $a$. The chain rule gives $f^{\prime }(x)=x^x(\ln x+1)$, so

Write the product as a quotient with indeterminate form $(-\infty )/(+\infty )$. By L'Hôpital's rule,

For $u_k=a^{2k+1}/(2k+1)>0$, the consecutive ratio isThus the ratio test proves convergence when $0<a<1$ and divergence when $a>1$. At $a=1$, the series is ${\sum }_{k\ge 0}1/(2k+1)$, which diverges by comparison with $\frac{1}{2}{\sum }_{k\ge 0}1/(k+1)$. ThereforeAll terms are nonnegative, so convergence here is also absolute convergence.

The term at $k=0$ is finite. For every $k\ge 1$,The comparison P-series $\sum k^{-2}$ converges, so the comparison test for series proves that the given series converges absolutely.

Again $C=\sin (2\pi /3)=\sqrt{3}/2$, but the principal inverse sine would give $\pi /3$ at $x=1$. The required branch is $v=\pi -\arcsin (Cx)$, corresponding to $k=1$. ThereforeAt $x=1$ the bracket is $2\pi /3$, as required. Both the initial condition and the original differential equation are thus satisfied on the stated branch.

The initial condition gives $C=\sin (\pi /3)=\sqrt{3}/2$. Since $\pi /3$ belongs to the principal inverse sine range, choose $k=0$ in part (i). The particular solution isThe upper endpoint is excluded because $y/x\to \pi /2$ there and the original tangent ceases to be defined.

Use the substitution $v=y/x$ for a homogeneous first-order differential equation. Since $y^{\prime }=v+x{v}^{\prime }$, it givesWhere $\sin v\ne 0$, this separable differential equation integrates to $\ln |\sin v|=\ln |x|+C$, hence $\sin v=C_{1}x$. One can obtain the same relation without division by $\sin v$: the equation gives $(\sin v{)}^{\prime }=(\sin v)/x$, so $\sin v=C_{1}x$, including $\sin v\equiv 0$.

All real branches of the inverse sine are captured byEach branch is used on an interval with $x\ne 0$ and $|Cx|<1$, with fixed $k$. The strict inequality excludes zeros of $\cos (y/x)$, where the original tangent is undefined. When $C=0$, the formula includes all straight-line solutions $y=k\pi x$. Keeping both families of inverse sine branches is essential for the two different initial conditions below.

The horizontal cross-section is a triangle, so its area is half its width times its depth:Thus the volume integral, with $t=z/H$, isFor the substitution rule, put $u=1-t/2$. The endpoint values are $1$ and $1/2$, and $1-t=2u-1$, givingTherefore $\boxed{V=\frac{(4+2\sqrt{2})LDH}{15}}$.

At height $z$, the horizontal triangle has half-width $L\sqrt{1-z/(2H)}$ along $x$ and depth $D(1-z/H)$ along $y$. These shrink with $z$, so its orthogonal projection onto the $(x,y)$ plane is the base triangle with vertices $(-L,0)$, $(L,0)$ and $(0,D)$:The orthogonal projection onto the $(y,z)$ plane is the triangle with vertices $(0,0)$, $(D,0)$ and $(0,H)$:The orthogonal projection onto the $(x,z)$ plane is bounded by $z=0$, $z=H$ and the two curvesEquivalently, these curves lie on the parabola $z=2H(1-x^2/L^2)$, truncated at height $H$. In particular the top width is $\sqrt{2}L$, so this projection has a horizontal upper edge from $-L/\sqrt{2}$ to $L/\sqrt{2}$; it is not triangular. The top horizontal cross-section has zero depth, consistent with this edge.

At each height, the removed circular sector has the same angle and $4/5$ of the original radius. Its area and hence its volume are therefore multiplied by $(4/5{)}^2=16/25$. Equating the removed volume to that of the reformed solid sphere givesConsequently